Re: [R] Importing Data for a two sample t-test
?subset --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. nilsonern nilson...@gmail.com wrote: I am trying to do a two sample t-test with data that i received in a text document. one list has the slab weights and the second has the company it is associated with. here is an example. weights company 1 A 2 A 2 B 3 B I was able to import the data but i cannot figure out how separate the data. I want to put them in two separate sets, one being A and one being B. Any help would be appreciated. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Importing-Data-for-a-two-sample-t-test-tp4649565.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.dll Reference Guide?
On 14/11/2012 23:10, BayesForever wrote: Is there a reference guide to the complete set of functions and their arguments that are available in the R.dll library for Windows, i.e. the set of functions that includes Rf_initEmbeddedR and Rf_endEmbeddedR listed in section 8.2.2 of the Writing R Extensions manual? What does 'available' mean? From that manual: We can classify the entry points as API Entry points which are documented in this manual and declared in an installed header file. These can be used in distributed packages and will only be changed after deprecation. public Entry points declared in an installed header file that are exported on all R platforms but are not documented and subject to change without notice. private Entry points that are used when building R and exported on all R platforms but are not declared in the installed header files. Do not use these in distributed code. hidden Entry points that are where possible (Windows and some modern Unix-alike compilers/loaders when using R as a shared library) not exported. So everything which we intend to be documented is documented in that manual, and it is the 'reference guide'. This may be more of an indictment of my lexical searching skills than anything else, but I was unable to turn up anything comprehensive after an hour with various search engines. -- View this message in context: http://r.789695.n4.nabble.com/R-dll-Reference-Guide-tp4649543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replacing string in matrix with zero
Dear Rolf, You are of course quite right, however I wish the diagonal to be ignored by setting the value to zero. The inf was just recognising the flow value from self to self. A very fair point though. Best, Nick On 15 November 2012 04:10, Rolf Turner rolf.tur...@xtra.co.nz wrote: On 15/11/12 11:13, Nick Duncan wrote: Dear All, I have a matrix in which the diagonal has the string Inf in it. In order to be able to do cluster analysis this needs to be replaced with a Zero. I am very sceptical about this assertion. Inf means infinity which is of course *NOT* the same as 0. By anyone's standards. I know nothing about cluster analysis, so I may be toadally out to lunch in this, but I suspect that you need to re-think. cheers, Rolf Turner SNIP -- An Englishman, even if he is alone, forms an orderly queue of one. --George Mikes __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Function for extracting Lambda Sets
Dear All, I would like to extract Lambda Sets from a binary matrix that represents a social network. The calculation of Lambda Sets is set out in Borgatti 1990 (which can be downloaded freely). The package SNA goes part of the way with the function MaxFlow but does not have the Lambda partitioning - but perhaps it is called differently in R. Does anyone know if it exists as a function or if there is a sequence functions to replicate it? Many thanks for any guidance on this. Best, Nick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Extracting list names within a looped function
Hello all,
I have the following problem:
1) A list was defined as 'a'
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned to x
using the deparse(substitute()) functionality. This name will then be used
within the function to generate specific output files, etc.
foo - function(x) {
print( deparse(substitute(x)) )
}
However, I am currently interested in looping through all list variables and
extract the list variable name from within the function. The current loop (see
below) will result in
for(i in 1:length(a)) {
foo(a[[i]])
}
[1] a[[i]]
which actually does what I expected of deparse(substitute(x)), but is not what
I wanted. I would like to end up with something like
[1] var1
[1] var2
[1] var3
or
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
Keep in mind that x has to be a matrix, and not a list. This to keep the
function as general as possible.
Does anyone have any idea on how to tackle this? Is deparse(substitute(x)) here
the best way to go? Are there alternatives?
Any help would be greatly appreciated!
Many thanks,
-- Stan
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Re: [R] Extracting list names within a looped function
On 15-11-2012, at 11:14, Gaj Stan (BIGCAT) wrote:
Hello all,
I have the following problem:
1) A list was defined as 'a'
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned to x
using the deparse(substitute()) functionality. This name will then be used
within the function to generate specific output files, etc.
foo - function(x) {
print( deparse(substitute(x)) )
}
However, I am currently interested in looping through all list variables and
extract the list variable name from within the function. The current loop
(see below) will result in
for(i in 1:length(a)) {
foo(a[[i]])
}
[1] a[[i]]
which actually does what I expected of deparse(substitute(x)), but is not
what I wanted. I would like to end up with something like
[1] var1
[1] var2
[1] var3
or
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
Keep in mind that x has to be a matrix, and not a list. This to keep the
function as general as possible.
Does anyone have any idea on how to tackle this? Is deparse(substitute(x))
here the best way to go? Are there alternatives?
I'm not sure if I understand what you want but will this give you what you seem
to want:
names(a)
Berend
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting list names within a looped function
Hello Berend,
Thanks for your quick response.
I am aware of the names() function, but this is not the answer to my question,
since it will not work within the foo function.
Let me try to explain it again! (:
I want to analyze several datasets in a similar fashion and stored all data as
seperate list components in a (here: $var1, $var2, $var3). Each component of
that list has a unique name and will be sent through the foo()-function using a
for-loop. This foo() function would (in my case) generate several plots and
output files where I would like to use the name of the component ($var1, $var2
or $var3) in the labels and output file names. I currently am unable to do so.
As such, my end-goal would be that the foo function returns the name of the
list components and not [1] a[[i]], as deparse(substitute(x)) does.
The output of foo() needs to be:
[1] var1
[1] var2
[1] var3
But I'm also happy with
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
(or something similar)
This leads to the question: how can I adjust my foo() so it gives me the output
mentioned above?
I hope that this clears things up.
-- Stan
-Original Message-
From: Berend Hasselman [mailto:b...@xs4all.nl]
Sent: 15 November 2012 11:27
To: Gaj Stan (BIGCAT)
Cc: r-help@r-project.org
Subject: Re: [R] Extracting list names within a looped function
On 15-11-2012, at 11:14, Gaj Stan (BIGCAT) wrote:
Hello all,
I have the following problem:
1) A list was defined as 'a'
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned to x
using the deparse(substitute()) functionality. This name will then be used
within the function to generate specific output files, etc.
foo - function(x) {
print( deparse(substitute(x)) )
}
However, I am currently interested in looping through all list variables and
extract the list variable name from within the function. The current loop
(see below) will result in
for(i in 1:length(a)) {
foo(a[[i]])
}
[1] a[[i]]
which actually does what I expected of deparse(substitute(x)), but is not
what I wanted. I would like to end up with something like
[1] var1
[1] var2
[1] var3
or
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
Keep in mind that x has to be a matrix, and not a list. This to keep the
function as general as possible.
Does anyone have any idea on how to tackle this? Is deparse(substitute(x))
here the best way to go? Are there alternatives?
I'm not sure if I understand what you want but will this give you what you seem
to want:
names(a)
Berend
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Getting information encoded in a SAS, SPSS or Stata command file into R.
What I do not understand is how SAS knows where the variables begin and end. I managed to break off a little hunk of the beginning of my file and look at it in an editor, and it is numbers without any obvious delimiters. Is the delimiter a particular numeric string? I thought the SAS command file would contain the starting location for each of the fixed-length fields, but I do not see anything in the file that could be interpreted that way just a little wraparound code and then a long list of variable names followed by triplets of a code, an equals sign, and a text string, terminating with a semicolon. Search around for the word INPUT. If that doesn't exist, you're probably looking at a formatting-only script -- and the delimiting happens elsewhere. Here are some examples of what the R SAScii package looks for when it uses a SAS importation script to read in ASCII data: library(SAScii) ?parse.SAScii note that there are additional examples under ?read.SAScii that include IPUMS data ;) also: David's idea is totally awesome [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to modify a S4 function (in the package NADA)
I want to get access to the code of an S4 method in order to possibly modify a function to accomplish my particular needs: in my case the function in is cenfit() from the package NADA Now, given my reproducible example: my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072), LR = c(0L, 0L, 1L, 0L, 1L, 0L)), .Names = c(TEC, LR), class = data.frame, row.names = c(NA, -6L)) and the following few lines of code: library(“NADA) attach(my.ex) cenfit(TEC,LR) giving this output: n n.cen median mean sd 6.00 2.00 0.033 0.058 0.083 I would like to add one more result to the above, namely “sum”, very simply computed as the product of “n” times “mean” Do you think that is possible or convenient? I’ve been reading that editing a S4 methods needs particular attention as by using showMethods(cenfit) and then getMethod(cenfit); in fact, I’ve been trying that but without much success (i.e. understanding of the cumbersome output) can anyone give me some help on how is probably better to proceed in order to get my new result? different alternative solutions or hints or advices are also more than welcomed (I’m pretty new on R and specially on the field of function handling and customizing) all the best max -- View this message in context: http://r.789695.n4.nabble.com/How-to-modify-a-S4-function-in-the-package-NADA-tp4649586.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R exponential regression
Hi
your model
f - function(x,a,b) {a * I(x^b)}
can be expressed as
log(a)+b*log(x)
and for that it shall result in straight line and you can use lm for estimate
of b and log(a)
It is also better to use 1:33 instead of 1980:2012
Based on values you get from linear realation you can set sensible starting
values.
fm - nls(y~a*z^b, data=test, start=c(a=0.2,b=2))
summary(fm)
Formula: y ~ a * z^b
Parameters:
Estimate Std. Error t value Pr(|t|)
a 0.133910.01848 7.248 3.74e-08 ***
b 2.225630.04140 53.763 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 6.239 on 31 degrees of freedom
Number of iterations to convergence: 6
Achieved convergence tolerance: 7.673e-07
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of muzammil786
Sent: Wednesday, November 14, 2012 12:12 PM
To: r-help@r-project.org
Subject: Re: [R] R exponential regression
I am having a similar problem on this data (given below). I have tried
several starting values for a and b, but it is consistently giving me:
fm - nls(y~f(x,a,b), data.frame(x,y), start=c(a=1,b=1)) Error in
numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
The function f(x,a,b) is defined here:
f - function(x,a,b) {a * I(x^b)}
The data is given here:
x y
1 1980 1
2 1981 1
3 1982 1
4 1983 3
5 1984 3
6 1985 5
7 1986 8
8 1987 8
9 1988 9
10 1989 12
11 1990 15
12 1991 24
13 1992 33
14 1993 44
15 1994 62
16 1995 68
17 1996 81
18 1997 87
19 1998 102
20 1999 114
21 2000 123
22 2001 135
23 2002 144
24 2003 158
25 2004 172
26 2005 188
27 2006 197
28 2007 224
29 2008 234
30 2009 254
31 2010 278
32 2011 312
33 2012 317
I shall be grateful if you could guide me what's wrong here.
Cheers.
Muzammil
--
View this message in context: http://r.789695.n4.nabble.com/R-
exponential-regression-tp1009449p4649474.html
Sent from the R help mailing list archive at Nabble.com.
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guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] heatmap with symbols
Use the col= argument to the heatmap() function to specify the three colors you want. For example: x - matrix(rnorm(50), ncol=5) heatmap(x, col=1:3) Jean furor furor1...@gmail.com wrote on 11/14/2012 03:31:42 AM: Hi all, I've made a heatmap using '-', '=' and '+' as possible values. However, the heatmap itself shows more than three colors. How to avoid this? Thanks in advance. regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to modify a S4 function (in the package NADA)
Hi,
Your reproducible example isn't reproducible for me, as I get an error message.
But regardless, the easiest thing to do is write your own wrapper,
something like:
mycenfit - function(x, y) {
result - cenfit(x, y)
c(result, sum = result$n * result$mean)
}
No need to change anything in the package itself. Incidentally, using
attach() is generally a really bad idea. with() is much safer.
Sarah
On Thu, Nov 15, 2012 at 7:00 AM, maxbre mbres...@arpa.veneto.it wrote:
I want to get access to the code of an S4 method in order to possibly modify
a function to accomplish my particular needs: in my case the function in is
cenfit() from the package NADA
Now, given my reproducible example:
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072), LR =
c(0L, 0L, 1L, 0L, 1L, 0L)), .Names = c(TEC, LR), class = data.frame,
row.names = c(NA, -6L))
and the following few lines of code:
library(“NADA)
attach(my.ex)
cenfit(TEC,LR)
giving this output:
n n.cen median mean sd
6.00 2.00 0.033 0.058 0.083
I would like to add one more result to the above, namely “sum”, very simply
computed as the product of “n” times “mean”
Do you think that is possible or convenient?
I’ve been reading that editing a S4 methods needs particular attention as by
using showMethods(cenfit) and then getMethod(cenfit); in fact, I’ve been
trying that but without much success (i.e. understanding of the cumbersome
output)
can anyone give me some help on how is probably better to proceed in order
to get my new result?
different alternative solutions or hints or advices are also more than
welcomed
(I’m pretty new on R and specially on the field of function handling and
customizing)
all the best
max
--
Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] Send Email from R
THE ERROR WAS : ERROR: compilation failed for package 'base64' Problem is with version. Any idea, how can we do this without any packages ? - Thanks Antony. -- View this message in context: http://r.789695.n4.nabble.com/Send-Email-from-R-tp4645565p4649585.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] configure: error: --with-x=yes (default) and X11 headers/libs are not a
Hi I'm trying to install R-2.15.2 in Unix environment and when i run ./configure i'm getting the following error configure: error: --with-x=yes (default) and X11 headers/libs are not available. i just unzip .tar file and ran ./configure. I'm not able to get x11() anywhere. Please help me - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/configure-error-with-x-yes-default-and-X11-headers-libs-are-not-a-tp4649592.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] correct use with fligner test
Hi R users, Iwant to run a model, but want to know previously if my data present homoscedascity: the model is something like this: glm(ax~a*b*c*d) I´ve found four different ways to test for homoscedascity, but i want to know which is correct: 1.-fligner.test(glm(x~a*b*c*d)) 2.-fligner.test(x~a*b*c*d) 3.-fligner.test(x~a,b,c,d) 4.-interc-interaction(a,b,c,d) fligner.test(x,interc) Thanks in advance - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/correct-use-with-fligner-test-tp4649583.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] combine similar variables in chart
I want to make a chart on variables which are taken in 2 different years on 2 the same locations, at different depths, where for every location the species and the amount of species are determined. The first chart I want to make is to see how much of every species is collected in total in a bar or piechart. Then I want to make a few charts with the species per location and per depth Here is a subset of my dataset Station Latitude(S) Longitude(E)Depth(m)Sa.Code Sp.Code N UPG02 -5.122355036119.339401 4 UPG02_4 Amphis_lesson 398 UPG02 -5.122355036119.339401 4 UPG02_4 Amphis_radiat 58 UPG02 -5.122355036119.339401 4 UPG02_4 Penero_planat 3 UPG02 -5.122355036119.339401 4.8 UPG02_4.8 Sphero_sp 5 UPG02 -5.122355036119.339401 4.8 UPG02_4.8 Amphis_lesson 204 UPG02 -5.122355036119.339401 4.8 UPG02_4.8 Amphis_radiat 18 UPG02 -5.122355036119.339401 9 UPG02_9 Sphero_sp 1 UPG02 -5.122355036119.339401 9 UPG02_9 Eponid_repand 2 UPG02 -5.122355036119.339401 9 UPG02_9 Planor_sp 7 UPG02 -5.122355036119.339401 15 UPG02_15Opercu_ammono 1 UPG02 -5.122355036119.339401 15 UPG02_15Hetero_depres 94 UPG02 -5.122355036119.339401 15 UPG02_15Amphis_lesson 101 UPG02 -5.122355036119.339401 18 UPG02_18Sphero_sp 3 UPG02 -5.122355036119.339401 18 UPG02_18Millio_sp 1 UPG02 -5.122355036119.339401 18 UPG02_18Calcar_mayori 56 UPG03 -5.136863021119.387289 3 UPG03_3 Millio_sp 2 UPG03 -5.136863021119.387289 3 UPG03_3 Sahuli_sp1 1 UPG03 -5.136863021119.387289 3 UPG03_3 Elphid_cratic 20 UPG03 -5.136863021119.387289 4 UPG03_4 Planor_sp 1 UPG03 -5.136863021119.387289 4 UPG03_4 Opercu_elegan 1 UPG03 -5.136863021119.387289 4 UPG03_4 Septot_sp 1 UPG03 -5.136863021119.387289 5 UPG03_5 Sphero_sp 1 UPG03 -5.136863021119.387289 5 UPG03_5 Millio_sp 1 UPG03 -5.136863021119.387289 5 UPG03_5 Ammoni_sp1 1 UPG03 -5.136863021119.387289 6 UPG03_6 Calcar_sp1 2 UPG03 -5.136863021119.387289 6 UPG03_6 Calcar_mayori 68 UPG03 -5.136863021119.387289 6 UPG03_6 Elphid_cratic 17 UPG03 -5.136863021119.387289 9 UPG03_9 Sphero_sp 2 UPG03 -5.136863021119.387289 9 UPG03_9 Calcar_mayori 323 UPG03 -5.136863021119.387289 9 UPG03_9 Calcar_spengl 2 UPG28 -5.122355036119.339401 5 UPG28_5 Penero_planat 8 UPG28 -5.122355036119.339401 5 UPG28_5 Amphis_latera 1 UPG28 -5.122355036119.339401 5 UPG28_5 Sorite_sp 1 UPG28 -5.122355036119.339401 8 UPG28_8 Bilocu_sp3 1 UPG28 -5.122355036119.339401 8 UPG28_8 Verteb_sp 1 UPG28 -5.122355036119.339401 8 UPG28_8 Penero_planat 7 UPG28 -5.122355036119.339401 11 UPG28_11Placop_c.f. 1 UPG28 -5.122355036119.339401 11 UPG28_11Sphero_sp 2 UPG28 -5.122355036119.339401 11 UPG28_11Siphon_siphon 1 UPG28 -5.122355036119.339401 14 UPG28_14Sphero_sp 1 UPG28 -5.122355036119.339401 14 UPG28_14Sorite_sp 1 UPG28 -5.122355036119.339401 14 UPG28_14Spirol_hadaii 1 UPG28 -5.122355036119.339401 17 UPG28_17Sphero_sp 3 UPG28 -5.122355036119.339401 17 UPG28_17Siphon_siphon 1 UPG28 -5.122355036119.339401 17 UPG28_17Sorite_sp 1 UPG32 -5.136863021119.387289 4 UPG32_4 Pyrgo_striol1 UPG32 -5.136863021119.387289 4 UPG32_4 Ammoma_alveol 1 UPG32 -5.136863021119.387289 4 UPG32_4 Quinqu_netstr 1 UPG32 -5.136863021119.387289 6 UPG32_6 Neorot_calcar 1 UPG32 -5.136863021119.387289 6 UPG32_6 Quinqu_c.f.ag 1 UPG32 -5.136863021119.387289 6 UPG32_6 Amphis_lesson 50 UPG32 -5.136863021119.387289 8 UPG32_8 Spirol_sp3 1 UPG32 -5.136863021119.387289 8 UPG32_8 Triloc_tricar 1 UPG32 -5.136863021119.387289 8 UPG32_8 Amphis_lesson 34 -- View this message in context: http://r.789695.n4.nabble.com/combine-similar-variables-in-chart-tp4649581.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting
[R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
Hi everyone, I have a data frame one of whose columns is a character vector and the rest are numeric, and in debugging a script, I noticed that an ifelse call seems to be coercing the character column to a numeric column, and producing unintended values as a result. Roughly, here's what I tried to do: df: a data frame with, say, the first column as a character column and the second and third columns numeric. also: NA's occur only in the numeric columns, and if they occur in one, they occur in the other as well. I wanted to replace the NA's in column 2 with 0's and the ones in column 3 with 1's, so first I did this: na.replacements -ifelse(col(df)==2,0,1). Then I used a second ifelse call to try to remove the NA's as I wanted, first by doing this: clean.df - ifelse(is.na(df), na.replacements, df), which produced a list of lists vaguely resembling df, with the NA's mostly intact, and so then I tried this: clean.df - ifelse(is.na(df), na.replacements, unlist(df)), which seems to work if all the columns are numeric, but otherwise changes strings to numbers. I can't make sense of the help documentation enough to clear this up, but my guess is that the yes and no values passed to ifelse need to be vectors, in which case it seems I'll have to use another approach entirely, but even if is not the case and lists are acceptable, I'm not sure how to convert a mixed-mode data frame into a vector-like list of elements (which I would hope would work). I'd be grateful for any suggestions! Thanks, David Romano [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting list names within a looped function
No this doesn't seem like what I want.
As you can see in the dataset sometimes there are the same species on
different locations. First I want to make a chart on how many of a single
species there are in this dataset and preferrably for all species in one
chart.
Second I want to make such a chart per Location or Station as called in the
data set and third such a chart per Sa.Code
Gr,
Marcella
2012/11/15 Berend Hasselman b...@xs4all.nl
On 15-11-2012, at 11:14, Gaj Stan (BIGCAT) wrote:
Hello all,
I have the following problem:
1) A list was defined as 'a'
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned
to x using the deparse(substitute()) functionality. This name will then be
used within the function to generate specific output files, etc.
foo - function(x) {
print( deparse(substitute(x)) )
}
However, I am currently interested in looping through all list variables
and extract the list variable name from within the function. The current
loop (see below) will result in
for(i in 1:length(a)) {
foo(a[[i]])
}
[1] a[[i]]
which actually does what I expected of deparse(substitute(x)), but is
not what I wanted. I would like to end up with something like
[1] var1
[1] var2
[1] var3
or
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
Keep in mind that x has to be a matrix, and not a list. This to keep the
function as general as possible.
Does anyone have any idea on how to tackle this? Is
deparse(substitute(x)) here the best way to go? Are there alternatives?
I'm not sure if I understand what you want but will this give you what you
seem to want:
names(a)
Berend
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Re: [R] RMySQL install on windows
Uwe, I set the variable with single backslashes, but the output remains the same. The last part reads: gcc -m64 -shared -s -static-libgcc -o RMySQL.dll tmp.def RS-DBI.o RS-MySQL.o C:\Program Files\MySQL\MySQL Server 5.5/bin/libmysql.dll -Ld:/RCompile/CRANpkg/extralibs64/local/lib/x64 -Ld:/Rcompile/CRANpkg/extralibs64/local/lib -LC:/PROGRA~1/R/R-215~1.1/BIN/X64 -lR gcc.exe: error: C:\Program Files\MySQL\MySQL Server 5.5/bin/libmySQL.dll: No such file or directory ERROR: compilation failed for package 'RMySQL' * removing 'C:/Users/Acer/Documents/R/win-library/2.15/RMySQL' -- View this message in context: http://r.789695.n4.nabble.com/RMySQL-install-on-windows-tp4645629p4649590.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] survreg gompertz
Hi all, Sorry if this has been answered already, but I couldn't find it in the archives or general internet. Is it possible to implement the gompertz distribution as survreg.distribution to use with survreg of the survival library? I haven't found anything and recent attempts from my side weren't succefull so far. I know that other packages like 'eha' and 'flexsurv' offer functions similar to survreg with gompertz support. However, due to the run-time environment were this needs to be running in the end, I can't use these packages :( Same questions for the gompertz-makeham distribution. Many thanks! Matthias __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can't see what i did wrong..
Its not scaling.. so.. I guess i'll stay severely frustrated, and yes i know this is probably not enough information for anyone to help. Still, talking helps ;) On 15.11.2012, at 15:15, Jessica Streicher wrote: with pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] dframe-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2))); results[[i]]$classifier-ksvm(class~.,data=dframe,scaled=T,kernel=polydot,type=C-svc, C=C,kpar=list(degree=degree,scale=scale,offset=offset),prob.model=T) and a degree of 5 i get an error of 0 reported by the ksvm object. But when doing pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] pred.svm-kernlab::predict(results[[i]]$classifier,pred.pca,type=probabilities); results[[i]]$trainResults$predicted-pred.svm[,2] the results vary widely from the class vector. Nearly all predictions are somewhat around 0.29. Its just strange. And i have no idea where things go wrong. They're in the same loop with i, so its probably not an indexing issue. Maybe kernlabs predict doesn't scale the data or something? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SVM? Comparison method wanted: 3 Groups, Microarray data
Dear all, i have microarray data of 3 classes of patients. It's not a time course experiment only steady state. I used a rule-based method to classify the groups by the expression of the genes. This works out so far. Nevertheless I want to check my results with an other method. Therefore I look for one and want to ask you, what you suggest. I have 3 different patient groups, only the steady state data, 10/10/13 (total of 33) arrays and the challenge is to classify the groups by the expression of all genes on the hgu133a chip. The named rule-based method found for each group genes, which explain a classification for each sample to one of the groups. Are there packages/method you can suggest? I guess PAMR works not so good for all genes on the chip, right? Would be amazing if you can tell me your opinion. All the best and thanks Peter from the desk of peter kupfer leibnitz institute of natural product research and infection biology - hans-knoell institute - research group: systems biology / bioinformatics beutenbergstrasse 11a D-07745 jena fon: +49 3641 532 1329 fax: +49 3641 532 2329 mail: peter.kup...@hki-jena.de [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] configure: error: --with-x=yes (default) and X11 headers/libs are not a
Then you need to install the x11 headers for your Unix (linux?) system. I can't tell you what or how without knowing more about your particular setup. By far the easiest thing to do if you're not familiar with compiling your own software, headers and such, is to install the binary for your linux distro from the usual repository. Assuming you're running linux: just telling us Unix isn't actually all that informative. Sarah On Thu, Nov 15, 2012 at 8:26 AM, arunkumar akpbond...@gmail.com wrote: Hi I'm trying to install R-2.15.2 in Unix environment and when i run ./configure i'm getting the following error configure: error: --with-x=yes (default) and X11 headers/libs are not available. i just unzip .tar file and ran ./configure. I'm not able to get x11() anywhere. Please help me -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to modify a S4 function (in the package NADA)
sorry sarah
you are right, my fault, this is the correct reproducible example
(a misinterpretation of LR type occurred in previous reproducible example)
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072
), LR = c(FALSE, FALSE, TRUE, FALSE, TRUE, FALSE)), .Names = c(TEC,
LR), row.names = c(NA, -6L), class = data.frame)
but, nevertheless, still some problems exists because...
str(my.ex)
library(NADA)
with(my.ex,
cenfit(TEC,LR)
)
mycenfit - function(x, y) {
result - cenfit(x, y)
c(result, sum = result$n * result$mean)
}
with(my.ex,
mycenfit(TEC,LR)
)
giving me the following error message
Error in result$n : $ operator not defined for this S4 class
again pointing to the peculiar structure of S4 class (at least I think to
understand)
thank you for your reply
max
--
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Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to modify a S4 function (in the package NADA)
Thanks for the clarification, Martin. Since I couldn't reproduce the
example, I didn't get to see what was actually being returned (and
was, admittedly, too lazy to dig through the help).
library(NADA)
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072), LR =
+ c(0L, 0L, 1L, 0L, 1L, 0L)), .Names = c(TEC, LR), class = data.frame,
+ row.names = c(NA, -6L))
with(my.ex, cenfit(TEC,LR))
Error in function (classes, fdef, mtable) :
unable to find an inherited method for function ‘cenfit’ for
signature ‘numeric, integer, missing’
Not that I'm worried about it not working, but it's a good idea to
test reproducible examples on a clean session to ensure the best
possible answers.
Sarah
On Thu, Nov 15, 2012 at 9:32 AM, Martin Morgan mtmor...@fhcrc.org wrote:
On 11/15/2012 06:10 AM, Sarah Goslee wrote:
Hi,
Your reproducible example isn't reproducible for me, as I get an error
message.
But regardless, the easiest thing to do is write your own wrapper,
something like:
mycenfit - function(x, y) {
result - cenfit(x, y)
c(result, sum = result$n * result$mean)
}
No need to change anything in the package itself. Incidentally, using
attach() is generally a really bad idea. with() is much safer.
Sarah
On Thu, Nov 15, 2012 at 7:00 AM, maxbre mbres...@arpa.veneto.it wrote:
I want to get access to the code of an S4 method in order to possibly
modify
a function to accomplish my particular needs: in my case the function in
is
cenfit() from the package NADA
Now, given my reproducible example:
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072),
LR =
c(0L, 0L, 1L, 0L, 1L, 0L)), .Names = c(TEC, LR), class =
data.frame,
row.names = c(NA, -6L))
and the following few lines of code:
library(“NADA)
attach(my.ex)
cenfit(TEC,LR)
giving this output:
n n.cen median mean sd
6.00 2.00 0.033 0.058 0.083
actually, cenfit() returns an object
xx = cenfit(my.ex$TEC, as.logical(my.ex$LR))
and what you're seeing is the result of the object's 'show' method
xx
n n.cen median mean sd
6. 2. 0.0330 0.05836667 0.08350178
You can see the body of the show method with
selectMethod(show, class(xx))
and methods that are available to work on xx with
showMethods(class=class(xx), where=search())
Sarah is right that you'd likely want to write your own function, using a
combination of available methods, e.g.,
censtats - function(x) {
s = summary(x)
c(n = nrow(s), n.cen = nrow(s) - sum(s$n.event), median = median(x),
mean = mean(x)[[mean]], sd = sd(x))
}
Martin
I would like to add one more result to the above, namely “sum”, very
simply
computed as the product of “n” times “mean”
Do you think that is possible or convenient?
I’ve been reading that editing a S4 methods needs particular attention as
by
using showMethods(cenfit) and then getMethod(cenfit); in fact, I’ve
been
trying that but without much success (i.e. understanding of the
cumbersome
output)
can anyone give me some help on how is probably better to proceed in
order
to get my new result?
different alternative solutions or hints or advices are also more than
welcomed
(I’m pretty new on R and specially on the field of function handling and
customizing)
all the best
max
--
Sarah Goslee
http://www.functionaldiversity.org
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[R] Can't see what i did wrong..
with pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] dframe-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2))); results[[i]]$classifier-ksvm(class~.,data=dframe,scaled=T,kernel=polydot,type=C-svc, C=C,kpar=list(degree=degree,scale=scale,offset=offset),prob.model=T) and a degree of 5 i get an error of 0 reported by the ksvm object. But when doing pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] pred.svm-kernlab::predict(results[[i]]$classifier,pred.pca,type=probabilities); results[[i]]$trainResults$predicted-pred.svm[,2] the results vary widely from the class vector. Nearly all predictions are somewhat around 0.29. Its just strange. And i have no idea where things go wrong. They're in the same loop with i, so its probably not an indexing issue. Maybe kernlabs predict doesn't scale the data or something? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can't see what i did wrong..
Now i let it run for one specific set and got the same bad result, then i deactivated the probabilities and got a good result, then i activated the probabilities again and got a good result .. huh??? On 15.11.2012, at 15:32, Jessica Streicher wrote: Its not scaling.. so.. I guess i'll stay severely frustrated, and yes i know this is probably not enough information for anyone to help. Still, talking helps ;) On 15.11.2012, at 15:15, Jessica Streicher wrote: with pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] dframe-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2))); results[[i]]$classifier-ksvm(class~.,data=dframe,scaled=T,kernel=polydot,type=C-svc, C=C,kpar=list(degree=degree,scale=scale,offset=offset),prob.model=T) and a degree of 5 i get an error of 0 reported by the ksvm object. But when doing pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] pred.svm-kernlab::predict(results[[i]]$classifier,pred.pca,type=probabilities); results[[i]]$trainResults$predicted-pred.svm[,2] the results vary widely from the class vector. Nearly all predictions are somewhat around 0.29. Its just strange. And i have no idea where things go wrong. They're in the same loop with i, so its probably not an indexing issue. Maybe kernlabs predict doesn't scale the data or something? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to modify a S4 function (in the package NADA)
On 11/15/2012 06:10 AM, Sarah Goslee wrote:
Hi,
Your reproducible example isn't reproducible for me, as I get an error message.
But regardless, the easiest thing to do is write your own wrapper,
something like:
mycenfit - function(x, y) {
result - cenfit(x, y)
c(result, sum = result$n * result$mean)
}
No need to change anything in the package itself. Incidentally, using
attach() is generally a really bad idea. with() is much safer.
Sarah
On Thu, Nov 15, 2012 at 7:00 AM, maxbre mbres...@arpa.veneto.it wrote:
I want to get access to the code of an S4 method in order to possibly modify
a function to accomplish my particular needs: in my case the function in is
cenfit() from the package NADA
Now, given my reproducible example:
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072), LR =
c(0L, 0L, 1L, 0L, 1L, 0L)), .Names = c(TEC, LR), class = data.frame,
row.names = c(NA, -6L))
and the following few lines of code:
library(“NADA)
attach(my.ex)
cenfit(TEC,LR)
giving this output:
n n.cen median mean sd
6.00 2.00 0.033 0.058 0.083
actually, cenfit() returns an object
xx = cenfit(my.ex$TEC, as.logical(my.ex$LR))
and what you're seeing is the result of the object's 'show' method
xx
n n.cen median mean sd
6. 2. 0.0330 0.05836667 0.08350178
You can see the body of the show method with
selectMethod(show, class(xx))
and methods that are available to work on xx with
showMethods(class=class(xx), where=search())
Sarah is right that you'd likely want to write your own function, using a
combination of available methods, e.g.,
censtats - function(x) {
s = summary(x)
c(n = nrow(s), n.cen = nrow(s) - sum(s$n.event), median = median(x),
mean = mean(x)[[mean]], sd = sd(x))
}
Martin
I would like to add one more result to the above, namely “sum”, very simply
computed as the product of “n” times “mean”
Do you think that is possible or convenient?
I’ve been reading that editing a S4 methods needs particular attention as by
using showMethods(cenfit) and then getMethod(cenfit); in fact, I’ve been
trying that but without much success (i.e. understanding of the cumbersome
output)
can anyone give me some help on how is probably better to proceed in order
to get my new result?
different alternative solutions or hints or advices are also more than
welcomed
(I’m pretty new on R and specially on the field of function handling and
customizing)
all the best
max
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Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] Send Email from R
On 15/11/2012 11:49, R_Antony wrote: We had no context. This is R-help, not a Nabble forum. THE ERROR WAS : ERROR: compilation failed for package 'base64' Problem is with version. Any idea, how can we do this without any packages ? Does searching not work for you? ??email returned utils::create.post Ancillary Function for Preparing Emails and Postings Whatever is using the orphaned package base64 (sendmailR?) should be updated to use base64enc. - Thanks Antony. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] configure: error: --with-x=yes (default) and X11 headers/libs are not a
On 15/11/2012 14:28, Sarah Goslee wrote: Then you need to install the x11 headers for your Unix (linux?) system. I can't tell you what or how without knowing more about your particular setup. But note that the manual does cover a lot of the possibilities. From the INSTALL file: 'The main source of information on installation is the `R Installation and Administration Manual', an HTML copy of which is available as file `doc/html/R-admin.html'. Please read that before installing R. But if you are impatient, read on but please refer to the manual to resolve any problems.' By far the easiest thing to do if you're not familiar with compiling your own software, headers and such, is to install the binary for your linux distro from the usual repository. Assuming you're running linux: just telling us Unix isn't actually all that informative. Sarah On Thu, Nov 15, 2012 at 8:26 AM, arunkumar akpbond...@gmail.com wrote: Hi I'm trying to install R-2.15.2 in Unix environment and when i run ./configure i'm getting the following error configure: error: --with-x=yes (default) and X11 headers/libs are not available. i just unzip .tar file and ran ./configure. I'm not able to get x11() anywhere. Please help me -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] hessian fails for box-constrained problems when close to boundary?
Hi I am trying to recover the hessian of a problem optimised with box-constraints. The problem is that in some cases, my estimates are very close to the boundary, which will make optim(..., hessian=TRUE) or optimHessian() fail, as they do not follow the box-constraints, and hence estimate the function in the unfeasible parameter space. As a simple example (my problem is more complex though, simple param transformations do not apply ashere), imagine estimating mu and sigma (restricted to be 0) of a simple normally distributed data, where however the true sigma is very close to zero: LLNorm - function(para, dat=rn) return(-sum(dnorm(dat, para[1], para[2], log=TRUE))) rn2 - c(rep(10.3, 2000), 10.31) optim(c(10,1), fn=LLNorm, method=L-BFGS-B, lower=c(-Inf, 0.001), dat=rn2,hessian=TRUE) Error in optim(c(10, 1), fn = LLNorm, method = L-BFGS-B, lower = c(-Inf, : non-finite finite-difference value [2] The only solution/workaround I found is to do a two steps procedure: use optim() without hessian, then use optimHess, specifying the length of the numerical variations (arg ndeps) as smaller as the distance between the parameter and the bound, i.e.: op-optim(c(10,1), fn=LLNorm, method=L-BFGS-B, lower=c(-Inf, 0.001), dat=rn2,hessian=FALSE) H-optimHess(op$par, LLNorm, dat=rn2, control=list(ndeps=rep((op$par[2]-0)/1000,2))) While this solution works, it is surely not elegant, and I have following questions: 1) is this solution to use smaller steps good at all? 2) I am not sure either I understand the reasons why the hessian in a constrained optim problem is evaluated without the constraints, or at least the problem transformed into a warning? Furthermore, I realised that using the numDeriv package, the function hessian() does not offer either constraints for the parameters, yet in the special case of the example above, it does not fail (although would have problems with parameter even closer to the bound), unlike the optimHessian() function? See: library(numDeriv) hessian(LLNorm, op$par, dat=rn2) This brings me to the final question: is there a technical reason not to allow a constrained hessian, which seems to indicate the fact that the two implementations in R do not do it? I found a very interesting answer by Spencer Grave for a similar question: http://tolstoy.newcastle.edu.au/R/e4/help/08/06/15061.html although this regards more the statistical implications than the numerical issues. Thanks a lot! Matthieu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple groups barplot
Hi Jim, thanks again for your support. Yes, I meant the subject codes; I will add a new variable and set the color to white all over. Thanks, Michele On Nov 15, 2012, at 1:13 AM, Jim Lemon wrote: On 11/15/2012 07:21 AM, michele caseposta wrote: Back again. Is there a quick way to add the sample names in the plot? I was not able to find anything other than creating a new category with the name in it (and the same color all over). Hi Michele, If by sample names you mean the variable names in your data frame: library(plotrix) mcdat-data.frame(smk=c(No,No,Yes),sex=c(M,M,F), age=c(35,24,30)) sizetree(mcdat,toplab=c(Smoker,Sex,Age), col=list(c(gray80,gray20),c(pink,lightblue),rainbow(3))) This displays the variable names at the top of the stacked bars. If you mean the subject codes, then yes, you would have to create a new variable. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting list names within a looped function
Hi Stan,
On Thu, Nov 15, 2012 at 6:02 AM, Gaj Stan (BIGCAT)
stan@maastrichtuniversity.nl wrote:
Hello Berend,
Thanks for your quick response.
I am aware of the names() function, but this is not the answer to my
question, since it will not work within the foo function.
Let me try to explain it again! (:
I want to analyze several datasets in a similar fashion and stored all data
as seperate list components in a (here: $var1, $var2, $var3). Each component
of that list has a unique name and will be sent through the foo()-function
using a for-loop. This foo() function would (in my case) generate several
plots and output files where I would like to use the name of the component
($var1, $var2 or $var3) in the labels and output file names. I currently am
unable to do so.
As such, my end-goal would be that the foo function returns the name of the
list components and not [1] a[[i]], as deparse(substitute(x)) does.
The output of foo() needs to be:
[1] var1
[1] var2
[1] var3
I suggest iterating over the names themselves, like this:
for(i in names(a)) {
print(i)
}
[1] var1
[1] var2
[1] var3
If you define your function with two arguments you can do stuff like this:
foo - function(x, d) {
print(x)
cat(d[[x]], file=paste(x, .txt, sep=))
## other stuff you want to do with d and x
}
for(i in names(a)) {
foo(x=i, d=a)
}
which is (I think) basically what you are after.
Best,
Ista
But I'm also happy with
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
(or something similar)
This leads to the question: how can I adjust my foo() so it gives me the
output mentioned above?
I hope that this clears things up.
-- Stan
-Original Message-
From: Berend Hasselman [mailto:b...@xs4all.nl]
Sent: 15 November 2012 11:27
To: Gaj Stan (BIGCAT)
Cc: r-help@r-project.org
Subject: Re: [R] Extracting list names within a looped function
On 15-11-2012, at 11:14, Gaj Stan (BIGCAT) wrote:
Hello all,
I have the following problem:
1) A list was defined as 'a'
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned to
x using the deparse(substitute()) functionality. This name will then be used
within the function to generate specific output files, etc.
foo - function(x) {
print( deparse(substitute(x)) )
}
However, I am currently interested in looping through all list variables and
extract the list variable name from within the function. The current loop
(see below) will result in
for(i in 1:length(a)) {
foo(a[[i]])
}
[1] a[[i]]
which actually does what I expected of deparse(substitute(x)), but is not
what I wanted. I would like to end up with something like
[1] var1
[1] var2
[1] var3
or
[1] a[[\var1\]]
[1] a[[\var2\]]
[1] a[[\var3\]]
Keep in mind that x has to be a matrix, and not a list. This to keep the
function as general as possible.
Does anyone have any idea on how to tackle this? Is deparse(substitute(x))
here the best way to go? Are there alternatives?
I'm not sure if I understand what you want but will this give you what you
seem to want:
names(a)
Berend
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Re: [R] Can't see what i did wrong..
Hi Jessica, Talking may help you, but it's kind of a waste of time for the thousands of readers of R-help unless you provide a reproducible example and full context. We'd like to help, but can't without adequate information. Sarah On Thu, Nov 15, 2012 at 9:48 AM, Jessica Streicher j.streic...@micromata.de wrote: Now i let it run for one specific set and got the same bad result, then i deactivated the probabilities and got a good result, then i activated the probabilities again and got a good result .. huh??? On 15.11.2012, at 15:32, Jessica Streicher wrote: Its not scaling.. so.. I guess i'll stay severely frustrated, and yes i know this is probably not enough information for anyone to help. Still, talking helps ;) On 15.11.2012, at 15:15, Jessica Streicher wrote: with pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] dframe-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2))); results[[i]]$classifier-ksvm(class~.,data=dframe,scaled=T,kernel=polydot,type=C-svc, C=C,kpar=list(degree=degree,scale=scale,offset=offset),prob.model=T) and a degree of 5 i get an error of 0 reported by the ksvm object. But when doing pred.pca-predict(splits[[i]]$pca,trainingData@samples)[,1:nPCs] pred.svm-kernlab::predict(results[[i]]$classifier,pred.pca,type=probabilities); results[[i]]$trainResults$predicted-pred.svm[,2] the results vary widely from the class vector. Nearly all predictions are somewhat around 0.29. Its just strange. And i have no idea where things go wrong. They're in the same loop with i, so its probably not an indexing issue. Maybe kernlabs predict doesn't scale the data or something? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem in fitting model in NLS function
Bad scaling will waste a lot of everyone's time.
I put the data in a data frame mdat, then
library(nlmrt)
mdat-read.csv(muzzamil.csv, header=T)
fmn - nlxb(y~a * (x^b), data=mdat, start=c(a=1,b=1), trace=T)
fm - nls(y~a * (x^b), data=mdat, start=c(a=1,b=1), trace=T)
fmn2 - nlxb(y~a2 * ((x-1979)^b2), data=mdat, start=c(a2=1,b2=1), trace=T)
fm2 - nls(y~a2 * ((x-1979)^b2), data=mdat, start=c(a2=1,b2=1), trace=T)
found a possible solution quickly with the last 2 forms. Without
adjusting the x data, you get a very ill-conditioned problem.
JN
Message: 21
Date: Wed, 14 Nov 2012 03:04:56 -0800 (PST)
From: muzammil786 m.shah...@sheffield.ac.uk
To: r-help@r-project.org
Subject: Re: [R] problem in fitting model in NLS function
Message-ID: 1352891096449-4649473.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii
Dear David,
I am having a similar problem on this data (given below). I have tried
several starting values for a and b, but it is consistently giving me:
*fm - nls(y~f(x,a,b), data.frame(x,y), start=c(a=1,b=1))
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model*
The function f(x,a,b) is defined here:
*f - function(x,a,b) {a * I(x^b)} *
The data is given here:
* x y
1 1980 1
2 1981 1
3 1982 1
4 1983 3
5 1984 3
6 1985 5
7 1986 8
8 1987 8
9 1988 9
10 1989 12
11 1990 15
12 1991 24
13 1992 33
14 1993 44
15 1994 62
16 1995 68
17 1996 81
18 1997 87
19 1998 102
20 1999 114
21 2000 123
22 2001 135
23 2002 144
24 2003 158
25 2004 172
26 2005 188
27 2006 197
28 2007 224
29 2008 234
30 2009 254
31 2010 278
32 2011 312
33 2012 317*
I shall be grateful if you could guide me what's wrong here.
Cheers.
Muzammil
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Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
Thanks for the suggestion, Bert; I just re-read the introduction with particular attention to the sections you mentioned, but I don't see how any of it bears on my question. Namely -- to rephrase: What constraints are there on the form of the yes and no values required by ifelse? The introduction doesn't really speak to this, and the help documentation seems to suggest that as long the shapes of the test, yes values, and no values agree, that would be sufficient -- I don't see anything that specifies that any of these should be of a particular data type. My example, however, seems to indicate that the yes and no values can't be a mixture of characters and numbers, and I'm trying to figure out what the underlying constraints are on ifelse. Thanks again, David On Thu, Nov 15, 2012 at 6:46 AM, Bert Gunter gunter.ber...@gene.com wrote: David: You seem to be getting lost in basic R tasks. Have you read the Intro to R tutorial? If not, do so, as this should tell you how to do what you need. If so, re-read the sections on indexing ([), replacement, and NA's. Also read about character vectors and factors. -- Bert On Thu, Nov 15, 2012 at 3:19 AM, David Romano drom...@stanford.edu wrote: Hi everyone, I have a data frame one of whose columns is a character vector and the rest are numeric, and in debugging a script, I noticed that an ifelse call seems to be coercing the character column to a numeric column, and producing unintended values as a result. Roughly, here's what I tried to do: df: a data frame with, say, the first column as a character column and the second and third columns numeric. also: NA's occur only in the numeric columns, and if they occur in one, they occur in the other as well. I wanted to replace the NA's in column 2 with 0's and the ones in column 3 with 1's, so first I did this: na.replacements -ifelse(col(df)==2,0,1). Then I used a second ifelse call to try to remove the NA's as I wanted, first by doing this: clean.df - ifelse(is.na(df), na.replacements, df), which produced a list of lists vaguely resembling df, with the NA's mostly intact, and so then I tried this: clean.df - ifelse(is.na(df), na.replacements, unlist(df)), which seems to work if all the columns are numeric, but otherwise changes strings to numbers. I can't make sense of the help documentation enough to clear this up, but my guess is that the yes and no values passed to ifelse need to be vectors, in which case it seems I'll have to use another approach entirely, but even if is not the case and lists are acceptable, I'm not sure how to convert a mixed-mode data frame into a vector-like list of elements (which I would hope would work). I'd be grateful for any suggestions! Thanks, David Romano [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to modify a S4 function (in the package NADA)
Ok, this is my finally (hopefully) clean session
my.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.0072), LR =
c(FALSE, FALSE, TRUE, FALSE, TRUE, FALSE)), .Names = c(TEC,LR),
row.names = c(NA, -6L), class = data.frame)
library(NADA)
xx-with(my.ex,
cenfit(TEC,LR)
)
xx
mycenfit - function(x) {
s = summary(x)
c(n = nrow(s), n.cen = nrow(s) - sum(s$n.event), median = median(x),
mean = mean(x)[[mean]], sd = sd(x), sum=mean(x)[[mean]]*length(x))
}
mycenfit(xx)
thank you so much for your help
(I still have much to fully understand S4 methods, R objects and functions)
max
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Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
David: You seem to be getting lost in basic R tasks. Have you read the Intro to R tutorial? If not, do so, as this should tell you how to do what you need. If so, re-read the sections on indexing ([), replacement, and NA's. Also read about character vectors and factors. -- Bert On Thu, Nov 15, 2012 at 3:19 AM, David Romano drom...@stanford.edu wrote: Hi everyone, I have a data frame one of whose columns is a character vector and the rest are numeric, and in debugging a script, I noticed that an ifelse call seems to be coercing the character column to a numeric column, and producing unintended values as a result. Roughly, here's what I tried to do: df: a data frame with, say, the first column as a character column and the second and third columns numeric. also: NA's occur only in the numeric columns, and if they occur in one, they occur in the other as well. I wanted to replace the NA's in column 2 with 0's and the ones in column 3 with 1's, so first I did this: na.replacements -ifelse(col(df)==2,0,1). Then I used a second ifelse call to try to remove the NA's as I wanted, first by doing this: clean.df - ifelse(is.na(df), na.replacements, df), which produced a list of lists vaguely resembling df, with the NA's mostly intact, and so then I tried this: clean.df - ifelse(is.na(df), na.replacements, unlist(df)), which seems to work if all the columns are numeric, but otherwise changes strings to numbers. I can't make sense of the help documentation enough to clear this up, but my guess is that the yes and no values passed to ifelse need to be vectors, in which case it seems I'll have to use another approach entirely, but even if is not the case and lists are acceptable, I'm not sure how to convert a mixed-mode data frame into a vector-like list of elements (which I would hope would work). I'd be grateful for any suggestions! Thanks, David Romano [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GUI Development reg
Vijayan: Choose one or both from the following. 1. You don't . R is a programming language that you learn. Read the Intro to R Tutorial to start learning. 2. See the R GUI page on CRAN here: http://www.sciviews.org/_rgui/ There are several GUI's available for R and you can choose one that suits. Cheers, Bert On Wed, Nov 14, 2012 at 10:31 PM, Vijayan Padmanabhan v.padmanab...@itc.in wrote: Dear R Group I have a character vector from which I want to select a few elements and create a new character vector. I need a GUI to do this in R Script. Can someone help? a-c(A,B,C,D,E) ## I want to have a GUI in R that will display elements in object a as a drop down list.. from there I want to be able to select a few elements say my requirement is to select A, D and E from the list and want to pass the selection into a new vector b, such that b-c(A, D , E) How do i do this in R? Regards Vijayan Padmanabhan What is expressed without proof can be denied without proof - Euclide. Please visit us at www.itcportal.com ** This Communication is for the exclusive use of the intended recipient (s) and shall not attach any liability on the originator or ITC Ltd./its Subsidiaries/its Group Companies. If you are the addressee, the contents of this email are intended for your use only and it shall not be forwarded to any third party, without first obtaining written authorisation from the originator or ITC Ltd./its Subsidiaries/its Group Companies. It may contain information which is confidential and legally privileged and the same shall not be used or dealt with by any third party in any manner whatsoever without the specific consent of ITC Ltd./its Subsidiaries/its Group Companies. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can't see what i did wrong..
I'll try .. lets see dput(dframe) structure(list(PC1 = c(-112.131127281872, -116.596689350263, -128.974817313418, 1196.74257699487, 1216.80975908625, -162.017759589637, -85.3191578683763, -130.528496524968, -6.27985683466795, -110.832877061623, -127.466759201992, -106.169095838249, -135.615623230809, -140.740670692038, -163.111656805033), PC2 = c(-116.519244989851, 127.45362265627, -111.091607020859, -90.5227270764559, -123.103694889442, -92.620202365279, -31.3353634789021, -24.4646703348531, 627.701800243656, -5.87156108239855, -88.8993352468245, -106.166416491185, -95.3261810128253, -112.315850116593, -141.197051597155), PC3 = c(64.9404052536739, 259.296520455491, 81.2617131909924, -40.2969961195199, -49.959279821408, 114.342573713326, -127.625681151677, 293.080433739631, -130.564246301803, 243.838480231849, -188.819145619591, -184.73732844957, 68.0106824426824, -202.00600443294, -3.31973728306762), PC4 = c(56.5233383519436, -188.28891646169, 36.982033091297, 2.96535930848599, 38.9643424847775, 66.1927347587241, -70.2278717299823, 24.2000870994265, 85.41705362572, 81.3070906919206, -49.4076806523943, 44.4291395459566, 44.4189881931216, -47.787706169273, 60.3084338394971), PC5 = c(-24.4330445432631, 119.03248964697, 25.3182533898001, 2.22712423872515, 20.1564988015345, 8.58984063196041, -73.6618858270628, 154.531152822823, 80.0172481588946, -77.3997650093847, 12.5317198856102, 16.7807094602606, 21.9728292529402, 18.0679363256966, 14.8070821345172), PC6 = c(-3.51236160660659, 4.01466215993921, -14.0560600152898, 79.9790057450278, 63.8459665084246, 70.1505245709296, -9.72399636564978, -37.3441332401865, -20.0211154965333, 34.8148111484049, 19.6697124222742, -12.5006014423202, -35.9412566332911, 43.8525592643803, 49.0837309296695), PC7 = c(25.0416790893669, 29.2636829211613, 58.5827246416567, 24.5865832216293, -3.79787182743163, 32.269692748727, 9.13653843864113, 22.2828779024266, -1.7405059119, -81.2339879700331, -8.39734635571791, 4.90076673663989, 26.3079758222854, -16.6096446464957, 27.2347172477201), PC8 = c(-7.21492487063905, 41.4415040543379, -6.8561527559211, -9.6515803005436, 40.5578239936188, 35.6268320766844, 4.33982606720149, -27.7334062748555, 1.58916911406888, 42.2854747927154, -19.0049848668069, 53.1381176037003, -43.7208819443017, -27.4638912504461, 26.0693753249626), PC9 = c(5.07013225478207, 3.80978415808406, -12.9265139792091, 22.3706065788694, 43.9527567691036, 44.7661368799448, -25.6398826082422, 4.37103700086919, -9.0919050870289, 11.9520700564903, -23.250383344361, -30.3220249879799, -46.3887501545308, -14.0817576379036, 16.168581787112), PC10 = c(-17.5630832749372, 52.568055046381, -9.07626573878839, -14.0951768936723, -5.23003343696749, 35.0154213026854, 0.272344896140687, 13.3950601982445, -1.5533667315822, -40.8292771335063, -6.65959388343316, 5.06195872128945, -22.7857465669081, 6.62863071798569, 18.3995069374505), PC11 = c(18.6010453343854, -28.4637751927999, 1.71806569355502, 46.1205174166346, 9.53199967063977, 41.2972754904467, 9.33464956019139, 23.3603727930995, 10.9744437554614, -0.126735591637572, 21.8896623331257, -10.457965922839, -18.8322478793736, 27.7321423838636, 28.4679654920173), PC12 = c(-2.22649901627144, 2.14317119837832, 8.11993688993784, -1.02101828008616, -3.02022884335462, 3.99260060747241, -3.98208300398649, -8.73435109273014, -15.4116191493509, -21.9916390100063, -24.5400095100928, 11.7508407539612, -4.82585090072434, -21.4330004806799, 4.07445502977139), PC13 = c(12.0478528474097, 19.1463355000476, -13.7614159178043, -2.45636501215998, 10.7570602509335, -27.124462677835, 2.50581414841615, -1.43645727084062, -9.92501401886999, 14.541265924675, 4.61925247518254, 31.9444025444124, 15.0123077631076, 1.73726777225125, -21.7438739224478), PC14 = c(-24.3652304503883, 21.1785358765272, 0.221058054284211, -28.7788948242716, -0.840491052309543, 40.1941623968201, -17.1958234033114, 7.5770624168308, 7.45003285540585, 15.5283651855595, -8.65376069406027, 9.56538784228923, 4.23345479961019, -1.50016336971708, 19.755969304), PC15 = c(9.92727520847547, -16.0078543011331, 3.37100252722285, 26.8689400893287, 16.2203912046009, -6.07396584265131, -7.32119707525572, 0.194098009190007, -10.3216068605436, -23.0845947221256, 0.696161095884098, -5.29057246485069, -1.79824229741562, -3.26507441975894, 1.08088312906883), class = c(0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0)), .Names = c(PC1, PC2, PC3, PC4, PC5, PC6, PC7, PC8, PC9, PC10, PC11, PC12, PC13, PC14, PC15, class), row.names = c(2396, 2754, 3645, 7122, 7106, 7172, 6761, 1599, 2107, 1865, 6795, 6913, 3329, 6836, 7229), class = data.frame) These are 15 of my samples, and it seems to work with just that. classifier-ksvm(class~.,data=dframe,scaled=T,kernel=polydot,type=C-svc, C=1,kpar=list(degree=5,scale=1,offset=0),prob.model=T) pred.svm-kernlab::predict(classifier,dframe[,1:15],type=probabilities);
Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
Hi, df1-read.table(text= col1 col2 col3 A 15.5 8.5 A 8.5 7.5 A NA NA B 8.0 6.0 B NA NA B 9.0 10.0 ,sep=,header=TRUE,stringsAsFactors=FALSE) str(df1) #'data.frame': 6 obs. of 3 variables: # $ col1: chr A A A B ... # $ col2: num 15.5 8.5 NA 8 NA 9 # $ col3: num 8.5 7.5 NA 6 NA 10 df1$col2[is.na(df1$col2)]-0 df1$col3[is.na(df1$col3)]-1 df1 # col1 col2 col3 #1 A 15.5 8.5 #2 A 8.5 7.5 #3 A 0.0 1.0 #4 B 8.0 6.0 #5 B 0.0 1.0 #6 B 9.0 10.0 #or if you want to use ifelse() from the original df1 ifelse(is.na(df1$col2),0,df1$col2) #[1] 15.5 8.5 0.0 8.0 0.0 9.0 ifelse(is.na(df1$col3),1,df1$col2) #[1] 15.5 8.5 1.0 8.0 1.0 9.0 A.K. - Original Message - From: David Romano drom...@stanford.edu To: r-help@r-project.org Cc: Sent: Thursday, November 15, 2012 6:19 AM Subject: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers Hi everyone, I have a data frame one of whose columns is a character vector and the rest are numeric, and in debugging a script, I noticed that an ifelse call seems to be coercing the character column to a numeric column, and producing unintended values as a result. Roughly, here's what I tried to do: df: a data frame with, say, the first column as a character column and the second and third columns numeric. also: NA's occur only in the numeric columns, and if they occur in one, they occur in the other as well. I wanted to replace the NA's in column 2 with 0's and the ones in column 3 with 1's, so first I did this: na.replacements -ifelse(col(df)==2,0,1). Then I used a second ifelse call to try to remove the NA's as I wanted, first by doing this: clean.df - ifelse(is.na(df), na.replacements, df), which produced a list of lists vaguely resembling df, with the NA's mostly intact, and so then I tried this: clean.df - ifelse(is.na(df), na.replacements, unlist(df)), which seems to work if all the columns are numeric, but otherwise changes strings to numbers. I can't make sense of the help documentation enough to clear this up, but my guess is that the yes and no values passed to ifelse need to be vectors, in which case it seems I'll have to use another approach entirely, but even if is not the case and lists are acceptable, I'm not sure how to convert a mixed-mode data frame into a vector-like list of elements (which I would hope would work). I'd be grateful for any suggestions! Thanks, David Romano [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Stepwise regression scope: all interacting terms (.^2)
Dear Gurus, Thank you in advance for your assistance. I'm trying to understand scope better when performing stepwise regression using step. I have a model with a binary response variable and 10 predictor variables. When I perform stepwise regression I define scope=.^2 to allow interactions between all terms. But I am missing something. When I perform stepwise regression (both directions) on the main model (y~x1+x2+…+x10) the method returns quickly with an answer; however, when I define all interactions in the main model (y~x1+x2+…+x10+x1:x2+x1:x3+…) and then perform stepwise regression (backward only) it runs so long I have to kill it. So here's my question: what is the difference between scope=.^2 on the additive (proper term?) model and defining all interactions and doing backward regression? My understanding is that .^2 is supposed to allow all interactions! Thank you for your help. Mark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Confidence intervals in Ripley's K function - a little challenge...
Sorry - CSR = Complete Spatial Randomness. -- View this message in context: http://r.789695.n4.nabble.com/Confidence-intervals-in-Ripley-s-K-function-a-little-challenge-tp4649392p4649597.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] finding the lowest which is not an outlier
Hello, If I have a series of numbers, and I want to find the level (the lowest value) on which each series is becam significant in the meaning of of frequency. In other words, how to find the value of lowest point (number) which is not an outlier? Thanks in advance Przemek -- View this message in context: http://r.789695.n4.nabble.com/finding-the-lowest-which-is-not-an-outlier-tp4649595.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] merge dataframes with condition
Hi @ all, I wamnt to combine two dataframes including a condition. I have two dataframes like the following: animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra) val-c(2,42,67,5,12,9,87,1,12) place-c(S,N,N,Z,R,O,E,I,Q) df1-data.frame(animal,val,place) animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra) val-c(21,45,78,6,18,77,89,17,28) place-c(S,N,N,Z,R,G,O,P,Q) df2-data.frame(animal,val,place) I would like to merge them with a condition. If animal and place are eqal in both df's, then add the values. If not add all three parameters (animal,value,place) at the bottom (like rbind). I hope somebody can help me. Thank a lot. geo -- View this message in context: http://r.789695.n4.nabble.com/merge-dataframes-with-condition-tp4649605.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Importing Data for a two sample t-test
HI, If you need to separate into two datasets, dat1[dat1$company==A,] # weights company #1 1 A #2 2 A #or dat1[dat1$company==B,] #or split into a list split(dat1,dat1$company) A.K. - Original Message - From: nilsonern nilson...@gmail.com To: r-help@r-project.org Cc: Sent: Wednesday, November 14, 2012 11:52 PM Subject: [R] Importing Data for a two sample t-test I am trying to do a two sample t-test with data that i received in a text document. one list has the slab weights and the second has the company it is associated with. here is an example. weights company 1 A 2 A 2 B 3 B I was able to import the data but i cannot figure out how separate the data. I want to put them in two separate sets, one being A and one being B. Any help would be appreciated. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Importing-Data-for-a-two-sample-t-test-tp4649565.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
Hi everyone, please put me off that list!!! The unsubscribe function does not function... THANKS!!! BW Sonja -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von David Romano Gesendet: 15 November 2012 12:19 An: r-help@r-project.org Betreff: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers Hi everyone, I have a data frame one of whose columns is a character vector and the rest are numeric, and in debugging a script, I noticed that an ifelse call seems to be coercing the character column to a numeric column, and producing unintended values as a result. Roughly, here's what I tried to do: df: a data frame with, say, the first column as a character column and the second and third columns numeric. also: NA's occur only in the numeric columns, and if they occur in one, they occur in the other as well. I wanted to replace the NA's in column 2 with 0's and the ones in column 3 with 1's, so first I did this: na.replacements -ifelse(col(df)==2,0,1). Then I used a second ifelse call to try to remove the NA's as I wanted, first by doing this: clean.df - ifelse(is.na(df), na.replacements, df), which produced a list of lists vaguely resembling df, with the NA's mostly intact, and so then I tried this: clean.df - ifelse(is.na(df), na.replacements, unlist(df)), which seems to work if all the columns are numeric, but otherwise changes strings to numbers. I can't make sense of the help documentation enough to clear this up, but my guess is that the yes and no values passed to ifelse need to be vectors, in which case it seems I'll have to use another approach entirely, but even if is not the case and lists are acceptable, I'm not sure how to convert a mixed-mode data frame into a vector-like list of elements (which I would hope would work). I'd be grateful for any suggestions! Thanks, David Romano [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Adding two different factors to one observation?
In a data frame, I would like to assign two or more factors to one observation. Is it possible? If so, how? Example: In the variable $ dishes, there are several levels: cup, plate, saucer. In my first observation, I see a saucer only. But in the second one, I see a cup and a saucer. In the third, however, there is only a cup, in the fourth a plate only and so forth. I'm an R beginner and I'll be grateful for any help. Florian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RMySQL install on windows
On Thu, Nov 15, 2012 at 7:49 AM, sbarandiaran sbaran...@gmail.com wrote: Uwe, I set the variable with single backslashes, but the output remains the same. The last part reads: gcc -m64 -shared -s -static-libgcc -o RMySQL.dll tmp.def RS-DBI.o RS-MySQL.o C:\Program Files\MySQL\MySQL Server 5.5/bin/libmysql.dll -Ld:/RCompile/CRANpkg/extralibs64/local/lib/x64 -Ld:/Rcompile/CRANpkg/extralibs64/local/lib -LC:/PROGRA~1/R/R-215~1.1/BIN/X64 -lR gcc.exe: error: C:\Program Files\MySQL\MySQL Server 5.5/bin/libmySQL.dll: No such file or directory ERROR: compilation failed for package 'RMySQL' * removing 'C:/Users/Acer/Documents/R/win-library/2.15/RMySQL' The error message says it can't find a particular file. You may have an incomplete installation or perhaps in your distribution the file is not in the expected place. First look for it: cd %MYSQL_HOME% dir/s/b libmySQL.dll If its missing then there is a problem with your installation of MySQL. If its in a different location than expected copy it to the expected location. Do that as Administrator: copy ...wherever...\libmySQL.dll %MYSQL_HOME%\bin If it happens again with some other file then copy it too. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Installing R on Ubuntu
Hello R-help team, I seek your help (for what is most likely a very simple problem). I'm new to Ubuntu and tried to install R using the Ubuntu Software Center. However, after clicking the install command, I always get prompted with the error Failed to download package files --- check your internet connection. Details: Failed to fetch http://archive.ubuntu.com/ubuntu/pool/main/a/apparmor/dh-apparmor_2.7.102-0ubuntu3.1_all.deb404 Not Found [IP: 91.189.92.200 80] However, my internet connection is perfectly fine and I've already installed a bunch of programs using the Software Center without any problems. I've tried various of the available R packages (also the Rcommander) and always get this same error. Perhaps it has something to do with the server/mirror? Alternatively, I've downloaded a .zip of the package, but so far I've failed to install the program. With me, for now, being an unequivocal Ubuntu dummy I would prefer the straightforward Software Center route and in any case I wonder why it does not work for me. Best regards and thanks in advance! Karel van Duijvenboden (Netherlands) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
Replace your NA's column by column, not all at once.
In your first example, of the form
ifelse(condition, numbers, data.frame)
the second and third arguments are replicated to the length
of the first. A data.frame's length is the number of columns
it has, so ifelse repeats its columns, not what you want.
Also, the 2nd and 3rd arguments to ifelse should be of the same
type, since the output will be a vector that accepts some values
from each. If they don't have the same type the output will be
of some type that can accept values from both types. That type
is often character or list, not what you want
Your second example code used unlist(data.frame). data.frames
contain columns of various classes and unlist(data.frame) creates
a vector with one class, the class is chosen to retain the information,
if not the format, of columns in the data.frame. It is generally not
a useful thing, unless all columns have the same class.
You showed some code but not data, so I'll make up something like
you described
df - data.frame(stringsAsFactors=FALSE,
Number1 = c(1, 2, 3, NA, 5, 6),
Number2 = c(11, 12, 13, 14, 14, NA),
String = c(one,two,NA,four,five,six),
Factor = factor(c(Group A, NA, Group A, Group B, Group B,
Group B)))
Look at its structure with
str(df)
'data.frame': 6 obs. of 4 variables:
$ Number1: num 1 2 3 NA 5 6
$ Number2: num 11 12 13 14 14 NA
$ String : chr one two NA four ...
$ Factor : Factor w/ 2 levels Group A,Group B: 1 NA 1 2 2 2
To do the sort of conversion you want try something like
f - function(d) {
for(i in seq_along(d)) {
di - d[[i]]
di[is.na(di)] - if (is.numeric(di)) { # could use switch instead
of if-then-else
if (i==2) { 0 } else { 1 }
} else if (is.factor(di)) {
levels(di)[1] # I don't know what
you want here
} else if (is.character(di)) {
Unknown
}
d[[i]] - di
}
d
}
That would give you
str(f(df))
'data.frame': 6 obs. of 4 variables:
$ Number1: num 1 2 3 1 5 6
$ Number2: num 11 12 13 14 14 0
$ String : chr one two Unknown four ...
$ Factor : Factor w/ 2 levels Group A,Group B: 1 1 1 2 2 2
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of David Romano
Sent: Thursday, November 15, 2012 7:58 AM
To: Bert Gunter
Cc: r-help@r-project.org
Subject: Re: [R] using ifelse to remove NA's from specific columns of a data
frame
containing strings and numbers
Thanks for the suggestion, Bert; I just re-read the introduction with
particular attention to the sections you mentioned, but I don't see how any
of it bears on my question. Namely -- to rephrase: What constraints are
there on the form of the yes and no values required by ifelse? The
introduction doesn't really speak to this, and the help documentation seems
to suggest that as long the shapes of the test, yes values, and no
values agree, that would be sufficient -- I don't see anything that
specifies that any of these should be of a particular data type. My
example, however, seems to indicate that the yes and no values can't be
a mixture of characters and numbers, and I'm trying to figure out what the
underlying constraints are on ifelse.
Thanks again,
David
On Thu, Nov 15, 2012 at 6:46 AM, Bert Gunter gunter.ber...@gene.com wrote:
David:
You seem to be getting lost in basic R tasks. Have you read the Intro
to R tutorial? If not, do so, as this should tell you how to do what
you need. If so, re-read the sections on indexing ([), replacement,
and NA's. Also read about character vectors and factors.
-- Bert
On Thu, Nov 15, 2012 at 3:19 AM, David Romano drom...@stanford.edu
wrote:
Hi everyone,
I have a data frame one of whose columns is a character vector and the
rest
are numeric, and in debugging a script, I noticed that an ifelse call
seems
to be coercing the character column to a numeric column, and producing
unintended values as a result. Roughly, here's what I tried to do:
df: a data frame with, say, the first column as a character column and
the
second and third columns numeric.
also: NA's occur only in the numeric columns, and if they occur in one,
they occur in the other as well.
I wanted to replace the NA's in column 2 with 0's and the ones in column
3
with 1's, so first I did this:
na.replacements -ifelse(col(df)==2,0,1).
Then I used a second ifelse call to try to remove the NA's as I wanted,
first by doing this:
clean.df - ifelse(is.na(df), na.replacements, df),
Re: [R] Can't see what i did wrong..
Guess it has something t do with the values in prob.model. The classifiers with bad predicitons have positive values and the ones with good predictions have negative values. On 15.11.2012, at 17:18, Jessica Streicher wrote: I'll try .. lets see dput(dframe) structure(list(PC1 = c(-112.131127281872, -116.596689350263, -128.974817313418, 1196.74257699487, 1216.80975908625, -162.017759589637, -85.3191578683763, -130.528496524968, -6.27985683466795, -110.832877061623, -127.466759201992, -106.169095838249, -135.615623230809, -140.740670692038, -163.111656805033), PC2 = c(-116.519244989851, 127.45362265627, -111.091607020859, -90.5227270764559, -123.103694889442, -92.620202365279, -31.3353634789021, -24.4646703348531, 627.701800243656, -5.87156108239855, -88.8993352468245, -106.166416491185, -95.3261810128253, -112.315850116593, -141.197051597155), PC3 = c(64.9404052536739, 259.296520455491, 81.2617131909924, -40.2969961195199, -49.959279821408, 114.342573713326, -127.625681151677, 293.080433739631, -130.564246301803, 243.838480231849, -188.819145619591, -184.73732844957, 68.0106824426824, -202.00600443294, -3.31973728306762), PC4 = c(56.5233383519436, -188.28891646169, 36.982033091297, 2.96535930848599, 38.9643424847775, 66.1927347587241, -70.2278717299823, 24.2000870994265, 85.41705362572, 81.3070906919206, -49.4076806523943, 44.4291395459566, 44.4189881931216, -47.787706169273, 60.3084338394971), PC5 = c(-24.4330445432631, 119.03248964697, 25.3182533898001, 2.22712423872515, 20.1564988015345, 8.58984063196041, -73.6618858270628, 154.531152822823, 80.0172481588946, -77.3997650093847, 12.5317198856102, 16.7807094602606, 21.9728292529402, 18.0679363256966, 14.8070821345172), PC6 = c(-3.51236160660659, 4.01466215993921, -14.0560600152898, 79.9790057450278, 63.8459665084246, 70.1505245709296, -9.72399636564978, -37.3441332401865, -20.0211154965333, 34.8148111484049, 19.6697124222742, -12.5006014423202, -35.9412566332911, 43.8525592643803, 49.0837309296695), PC7 = c(25.0416790893669, 29.2636829211613, 58.5827246416567, 24.5865832216293, -3.79787182743163, 32.269692748727, 9.13653843864113, 22.2828779024266, -1.7405059119, -81.2339879700331, -8.39734635571791, 4.90076673663989, 26.3079758222854, -16.6096446464957, 27.2347172477201), PC8 = c(-7.21492487063905, 41.4415040543379, -6.8561527559211, -9.6515803005436, 40.5578239936188, 35.6268320766844, 4.33982606720149, -27.7334062748555, 1.58916911406888, 42.2854747927154, -19.0049848668069, 53.1381176037003, -43.7208819443017, -27.4638912504461, 26.0693753249626), PC9 = c(5.07013225478207, 3.80978415808406, -12.9265139792091, 22.3706065788694, 43.9527567691036, 44.7661368799448, -25.6398826082422, 4.37103700086919, -9.0919050870289, 11.9520700564903, -23.250383344361, -30.3220249879799, -46.3887501545308, -14.0817576379036, 16.168581787112), PC10 = c(-17.5630832749372, 52.568055046381, -9.07626573878839, -14.0951768936723, -5.23003343696749, 35.0154213026854, 0.272344896140687, 13.3950601982445, -1.5533667315822, -40.8292771335063, -6.65959388343316, 5.06195872128945, -22.7857465669081, 6.62863071798569, 18.3995069374505), PC11 = c(18.6010453343854, -28.4637751927999, 1.71806569355502, 46.1205174166346, 9.53199967063977, 41.2972754904467, 9.33464956019139, 23.3603727930995, 10.9744437554614, -0.126735591637572, 21.8896623331257, -10.457965922839, -18.8322478793736, 27.7321423838636, 28.4679654920173), PC12 = c(-2.22649901627144, 2.14317119837832, 8.11993688993784, -1.02101828008616, -3.02022884335462, 3.99260060747241, -3.98208300398649, -8.73435109273014, -15.4116191493509, -21.9916390100063, -24.5400095100928, 11.7508407539612, -4.82585090072434, -21.4330004806799, 4.07445502977139), PC13 = c(12.0478528474097, 19.1463355000476, -13.7614159178043, -2.45636501215998, 10.7570602509335, -27.124462677835, 2.50581414841615, -1.43645727084062, -9.92501401886999, 14.541265924675, 4.61925247518254, 31.9444025444124, 15.0123077631076, 1.73726777225125, -21.7438739224478), PC14 = c(-24.3652304503883, 21.1785358765272, 0.221058054284211, -28.7788948242716, -0.840491052309543, 40.1941623968201, -17.1958234033114, 7.5770624168308, 7.45003285540585, 15.5283651855595, -8.65376069406027, 9.56538784228923, 4.23345479961019, -1.50016336971708, 19.755969304), PC15 = c(9.92727520847547, -16.0078543011331, 3.37100252722285, 26.8689400893287, 16.2203912046009, -6.07396584265131, -7.32119707525572, 0.194098009190007, -10.3216068605436, -23.0845947221256, 0.696161095884098, -5.29057246485069, -1.79824229741562, -3.26507441975894, 1.08088312906883), class = c(0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0)), .Names = c(PC1, PC2, PC3, PC4, PC5, PC6, PC7, PC8, PC9, PC10, PC11, PC12, PC13, PC14, PC15, class), row.names = c(2396, 2754, 3645, 7122, 7106, 7172, 6761, 1599, 2107, 1865, 6795, 6913, 3329, 6836, 7229), class = data.frame) These are 15 of my
Re: [R] merge dataframes with condition
Is this what you want:
animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra)
val-c(2,42,67,5,12,9,87,1,12)
place-c(S,N,N,Z,R,O,E,I,Q)
df1-data.frame(animal,val,place)
animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra)
val-c(21,45,78,6,18,77,89,17,28)
place-c(S,N,N,Z,R,G,O,P,Q)
df2-data.frame(animal,val,place)
x - merge(df1, df2, by = c('animal', 'place'), all = TRUE)
# compute the total
x$value - x$val.x + x$val.y
# separate the full/partial matches
x[order(is.na(x$value)), ]
animal place val.x val.y value
2bear N424587
4bear S 22123
5fish R121830
6lion N6778 145
8 monkey Z 5 611
11 zebra Q122840
1bear E87NANA
3bear ONA89NA
7 monkey O 9NANA
9 monkey GNA77NA
10 zebra I 1NANA
12 zebra PNA17NA
On Thu, Nov 15, 2012 at 9:47 AM, Geophagus f...@retposto.net wrote:
Hi @ all,
I wamnt to combine two dataframes including a condition.
I have two dataframes like the following:
animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra)
val-c(2,42,67,5,12,9,87,1,12)
place-c(S,N,N,Z,R,O,E,I,Q)
df1-data.frame(animal,val,place)
animal-c(bear,bear,lion,monkey,fish,monkey,bear,zebra,zebra)
val-c(21,45,78,6,18,77,89,17,28)
place-c(S,N,N,Z,R,G,O,P,Q)
df2-data.frame(animal,val,place)
I would like to merge them with a condition. If animal and place are
eqal in both df's, then add the values.
If not add all three parameters (animal,value,place) at the bottom
(like rbind).
I hope somebody can help me.
Thank a lot.
geo
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[R] Compraring two independent samples
Hi, In my reading of pairing means of two independent samples, I read statements such as the standard error of the meanof X1 minus the mean of X2 is the square root of s1^2/n1+s2^2/n2. Then I read: We could now derive the two independent samples confidence interval and test statistic. However, a problem arises in that the distribution of the test statistic (under the null hypothesis) will not be a t-distribution. I keep seeing this type of thing stated in a variety of readings but I never seem to get an explanation. This is just stated and the author goes on to an alternate approach. So, I am wondering if there is some sort of R simulation that could be used to demonstrate that this distribution is not a t-distribution. David -- View this message in context: http://r.789695.n4.nabble.com/Compraring-two-independent-samples-tp4649636.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Importing Data for a two sample t-test
On 2012-11-14 20:52, nilsonern wrote: I am trying to do a two sample t-test with data that i received in a text document. one list has the slab weights and the second has the company it is associated with. here is an example. weights company 1 A 2 A 2 B 3 B I was able to import the data but i cannot figure out how separate the data. I want to put them in two separate sets, one being A and one being B. Any help would be appreciated. Thanks. Why separate? The 'long' form is generally preferred in statistical analyses. Just use the formula version of t.test(). Peter Ehlers -- View this message in context: http://r.789695.n4.nabble.com/Importing-Data-for-a-two-sample-t-test-tp4649565.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compraring two independent samples
On 2012-11-15 10:17, David Arnold wrote: Hi, In my reading of pairing means of two independent samples, I read statements such as the standard error of the meanof X1 minus the mean of X2 is the square root of s1^2/n1+s2^2/n2. Then I read: We could now derive the two independent samples confidence interval and test statistic. However, a problem arises in that the distribution of the test statistic (under the null hypothesis) will not be a t-distribution. I keep seeing this type of thing stated in a variety of readings but I never seem to get an explanation. This is just stated and the author goes on to an alternate approach. So, I am wondering if there is some sort of R simulation that could be used to demonstrate that this distribution is not a t-distribution. David If the populations are Normal, see Wikipedia (or elsewhere) for the Behrens–Fisher problem. If the populations are not Normal, I don't see why a t-distribution would be expected. I seem to recall that Welch included some simulation results in his Biometrika paper (1947? 1953?; I'm getting senile). Shouldn't be difficult to generate in R. Maybe Greg Snow's TeachingDemos package has something. Peter Ehlers -- View this message in context: http://r.789695.n4.nabble.com/Compraring-two-independent-samples-tp4649636.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to view source code of a function inside a package?
Dear list,
I am trying to look at the function inside a package. I know that
methods() would do the trick, but what if the function is hidden? I have a
problem displaying the hidden function.
Say, for example the MCMC package. How do you view the code of that
function?
something like this:
which
function (x, arr.ind = FALSE, useNames = TRUE)
{
wh - .Internal(which(x))
if (arr.ind !is.null(d - dim(x)))
arrayInd(wh, d, dimnames(x), useNames = useNames)
else wh
}
bytecode: 0x1021eef50
environment: namespace:base
[[alternative HTML version deleted]]
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Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
#Data
df-data.frame(id=letters[1:10],var1=rnorm(10,10,5),var2=rnorm(10,5,2),var3=rnorm(10,1,1))
#Missing
df$var1[2]-df$var2[c(2,6)]-df$var3[c(2,5)]-NA
na.replace-seq(1:ncol(df))-1
df[,names(df)]-sapply(1:dim(df)[2], function(ii)
{ifelse(is.na(df[,ii]),na.replace[ii],df[,ii])} )
David Romano-2 wrote
Hi everyone,
I have a data frame one of whose columns is a character vector and the
rest
are numeric, and in debugging a script, I noticed that an ifelse call
seems
to be coercing the character column to a numeric column, and producing
unintended values as a result. Roughly, here's what I tried to do:
df: a data frame with, say, the first column as a character column and the
second and third columns numeric.
also: NA's occur only in the numeric columns, and if they occur in one,
they occur in the other as well.
I wanted to replace the NA's in column 2 with 0's and the ones in column 3
with 1's, so first I did this:
na.replacements -ifelse(col(df)==2,0,1).
Then I used a second ifelse call to try to remove the NA's as I wanted,
first by doing this:
clean.df - ifelse(is.na(df), na.replacements, df),
which produced a list of lists vaguely resembling df, with the NA's mostly
intact, and so then I tried this:
clean.df - ifelse(is.na(df), na.replacements, unlist(df)),
which seems to work if all the columns are numeric, but otherwise changes
strings to numbers.
I can't make sense of the help documentation enough to clear this up, but
my guess is that the yes and no values passed to ifelse need to be
vectors, in which case it seems I'll have to use another approach
entirely,
but even if is not the case and lists are acceptable, I'm not sure how to
convert a mixed-mode data frame into a vector-like list of elements (which
I would hope would work).
I'd be grateful for any suggestions!
Thanks,
David Romano
[[alternative HTML version deleted]]
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Re: [R] how to view source code of a function inside a package?
On 15/11/2012 2:21 PM, C W wrote:
Dear list,
I am trying to look at the function inside a package. I know that
methods() would do the trick, but what if the function is hidden? I have a
problem displaying the hidden function.
Say, for example the MCMC package. How do you view the code of that
function?
You can use the prefix MCMC:::foo to display the foo function from MCMC,
whether or not it is exported.
Duncan Murdoch
something like this:
which
function (x, arr.ind = FALSE, useNames = TRUE)
{
wh - .Internal(which(x))
if (arr.ind !is.null(d - dim(x)))
arrayInd(wh, d, dimnames(x), useNames = useNames)
else wh
}
bytecode: 0x1021eef50
environment: namespace:base
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to view source code of a function inside a package?
This is R FAQ 7.40, and that document gives some good pointers:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-access-the-source-code-for-a-function_003f
I prefer to simply download the source code version of the package
from CRAN, so that comments aren't stripped, but there are several
other options.
Sarah
On Thu, Nov 15, 2012 at 2:21 PM, C W tmrs...@gmail.com wrote:
Dear list,
I am trying to look at the function inside a package. I know that
methods() would do the trick, but what if the function is hidden? I have a
problem displaying the hidden function.
Say, for example the MCMC package. How do you view the code of that
function?
something like this:
which
function (x, arr.ind = FALSE, useNames = TRUE)
{
wh - .Internal(which(x))
if (arr.ind !is.null(d - dim(x)))
arrayInd(wh, d, dimnames(x), useNames = useNames)
else wh
}
bytecode: 0x1021eef50
environment: namespace:base
--
Sarah Goslee
http://www.functionaldiversity.org
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Optimizing
Hey, It's actually not homework, what gave you that impression? I graduated in May and studied Math, economics, and international relations, so I don't have much of a programming background. This is a project that I'm working on out of personal interest. Obviously, I've tried doing some homework, but after 45 minutes of digging around without even really having any leads I figured I would post here. The optimization problems I see more generally seem tailored towards maximizing a function subject to some constraint on that function. For example, maximizing U(x,y)=2x^2+4y s.t. x+y=3. I don't really see a way to frame my current maximization like that at all. We aren't choosing one observation in our dataset, we are choosing a group. And, the constraints aren't something like the types sum to a number, but rather that our choice of observations meet some specific condition. Sorry if I'm being clueless here, but when I look at http://cran.r-project.org/web/views/Optimization.html I just see a giant list of packages, most of which I believe are optimizing a function subject to constraints. Maybe my problem actually is a common one, there is some package out there that would do it, and I just need to find it. If so, my googling and looking on the help pages has been so far unsuccessful. I apologize if the answer really is sitting there on some help page, but I really haven't found it. If this were some homework problem, then I would know that it's probably a common, normal thing, and feel more confident looking on the Cran page. But, since it isn't, I wasn't even sure if the answer was out there and my searching had so far been fruitless. That's why I wanted to know if there would be some function that would do this sort of thing or if it's the sort of thing that I just need to manually work through. The help pages are over my head in so far as the help pages are highly technical and focused on methods. They assume a knowledge about the basic type of optimization problem that I don't have, and I think that's because there is a standard type of optimization that is most common. Anyways, I am thinking I might just try to manually do the calculation because I can't follow the optimize commands. The general strategy I think is to somehow make a dataset that has all combinations of 6 people as rows. Then, for each row, make a vector that is just all the types of the people in that row, a vector for the sum of the values, and a vector for the sum of the wages. Then, keep only those rows which have the correct types and wages under the max. Then, sort by the sum of the values. Let me know if you have any better ideas! Sam On Wed, Nov 14, 2012 at 7:35 PM, Bert Gunter gunter.ber...@gene.com wrote: Sam: 1. Homework? R has a no homework policy. 2. But in any case, check out the Optimization task view on CRAN. You should be able to find something there that meets your needs. Of course, if something is a little over your head, that's not an excuse, but rather an admission that you have to do some homework on your own. Cheers, Bert On Wed, Nov 14, 2012 at 5:23 PM, Sam Asin asin@gmail.com wrote: Hello, I am fairly new with R and am having trouble finding an optimal group. I checked the help functions for the various optimize commands and it was a little over my head. I have a dataset with 4 columns, name, type, value, and cost. The set consists of a list of people, which have 3 types. I want to choose 6 people, two of each type, and maximize the sum of their values. However, I'm subject to the constraint that the wage of the three people has to sum to less than 20 dollars. Here is some sample data. people - c(A, B, C, D, E, F, G, H, I) type- c(1, 1, 1, 1, 2, 2, 3, 3, 3) value-c(25.20, 24, 38, 20, 14, 20, 31, 11, 8) wage- c(4, 3.8, 5.1, 3.5, 2.4, 3, 6, 2.4, 2) data- data.frame(people, type, value, wage) With this small dataset the question isn't very interesting, but the answer would be something like person C, D, E, F, G, and I (I didn't check to see that those prices sum to less than $20). How can I write a program that will do this? Can I just use the optimize command? Do I have to transform my dataset into something that is easier to use the optimize command on? Or should I write my own code that does the process? Thanks, Sam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm [[alternative HTML version
Re: [R] Adding two different factors to one observation?
Florian, There are a number of different ways to handle data like this. Two that come to my mind are shown below. You could allow each observation to be represented by multiple rows in the data frame: obs dishes 11 saucer 22cup 32 plate 42 saucer 53cup 63 plate 74cup 84 saucer 96 plate 10 6 saucer 11 7 plate 12 8cup Or you could let each level be its own variable: obs cup plate saucer 1 1 0 0 1 2 2 1 1 1 3 3 1 1 0 4 4 1 0 1 5 5 0 0 0 6 6 0 1 1 7 7 0 1 0 8 8 1 0 0 The arrangement that you choose will likely depend on what you want to do with the data. Jean Florian Ahrweiler florian.ahrwei...@uni-wh.de wrote on 11/15/2012 10:10:53 AM: In a data frame, I would like to assign two or more factors to one observation. Is it possible? If so, how? Example: In the variable $ dishes, there are several levels: cup, plate, saucer. In my first observation, I see a saucer only. But in the second one, I see a cup and a saucer. In the third, however, there is only a cup, in the fourth a plate only and so forth. I'm an R beginner and I'll be grateful for any help. Florian [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can you have a by variable in Lag function as in SAS
Hello, I want to use lag on a time variable but I have to take date into consideration ie I don't want days to overlap ie: I don't want my first time of today to match my last time of yeterday. In SAS I would use : data x; set y; by date tim; previous=lag(tim); if first.date then do; previous=.; end; run; How can I do something similar in R? I can't find any examples anywhere. Thank you all for your help. -- View this message in context: http://r.789695.n4.nabble.com/Can-you-have-a-by-variable-in-Lag-function-as-in-SAS-tp4649647.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to view source code of a function inside a package?
thanks all. The mcmc:::foo is exactly what I am looking for.
On Thu, Nov 15, 2012 at 2:35 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 15/11/2012 2:21 PM, C W wrote:
Dear list,
I am trying to look at the function inside a package. I know that
methods() would do the trick, but what if the function is hidden? I have
a
problem displaying the hidden function.
Say, for example the MCMC package. How do you view the code of that
function?
You can use the prefix MCMC:::foo to display the foo function from MCMC,
whether or not it is exported.
Duncan Murdoch
something like this:
which
function (x, arr.ind = FALSE, useNames = TRUE)
{
wh - .Internal(which(x))
if (arr.ind !is.null(d - dim(x)))
arrayInd(wh, d, dimnames(x), useNames = useNames)
else wh
}
bytecode: 0x1021eef50
environment: namespace:base
[[alternative HTML version deleted]]
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posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] combine similar variables in chart
You've included three I wants in your message, but no question. Are you looking for functions? Try searching ... in R, on google, on http://rseek.org/, ... Have you tried some code, but can't get it to work? Send your code with error messages, and tell us precisely what you trying to get your code to do. Did you want us to do your homework? Good luck with that. Jean Marcella m.e.j.meerk...@students.uu.nl wrote on 11/15/2012 04:36:56 AM: I want to make a chart on variables which are taken in 2 different years on 2 the same locations, at different depths, where for every location the species and the amount of species are determined. The first chart I want to make is to see how much of every species is collected in total in a bar or piechart. Then I want to make a few charts with the species per location and per depth Here is a subset of my dataset Station Latitude(S) Longitude(E) Depth(m) Sa.Code Sp.Code N UPG02 -5.122355036 119.339401 4 UPG02_4 Amphis_lesson 398 UPG02 -5.122355036 119.339401 4 UPG02_4 Amphis_radiat 58 UPG02 -5.122355036 119.339401 4 UPG02_4 Penero_planat 3 UPG02 -5.122355036 119.339401 4.8 UPG02_4.8 Sphero_sp 5 UPG02 -5.122355036 119.339401 4.8 UPG02_4.8 Amphis_lesson 204 UPG02 -5.122355036 119.339401 4.8 UPG02_4.8 Amphis_radiat 18 UPG02 -5.122355036 119.339401 9 UPG02_9 Sphero_sp 1 UPG02 -5.122355036 119.339401 9 UPG02_9 Eponid_repand 2 UPG02 -5.122355036 119.339401 9 UPG02_9 Planor_sp 7 UPG02 -5.122355036 119.339401 15 UPG02_15 Opercu_ammono 1 UPG02 -5.122355036 119.339401 15 UPG02_15 Hetero_depres 94 UPG02 -5.122355036 119.339401 15 UPG02_15 Amphis_lesson 101 UPG02 -5.122355036 119.339401 18 UPG02_18 Sphero_sp 3 UPG02 -5.122355036 119.339401 18 UPG02_18 Millio_sp 1 UPG02 -5.122355036 119.339401 18 UPG02_18 Calcar_mayori 56 UPG03 -5.136863021 119.387289 3 UPG03_3 Millio_sp 2 UPG03 -5.136863021 119.387289 3 UPG03_3 Sahuli_sp1 1 UPG03 -5.136863021 119.387289 3 UPG03_3 Elphid_cratic 20 UPG03 -5.136863021 119.387289 4 UPG03_4 Planor_sp 1 UPG03 -5.136863021 119.387289 4 UPG03_4 Opercu_elegan 1 UPG03 -5.136863021 119.387289 4 UPG03_4 Septot_sp 1 UPG03 -5.136863021 119.387289 5 UPG03_5 Sphero_sp 1 UPG03 -5.136863021 119.387289 5 UPG03_5 Millio_sp 1 UPG03 -5.136863021 119.387289 5 UPG03_5 Ammoni_sp1 1 UPG03 -5.136863021 119.387289 6 UPG03_6 Calcar_sp1 2 UPG03 -5.136863021 119.387289 6 UPG03_6 Calcar_mayori 68 UPG03 -5.136863021 119.387289 6 UPG03_6 Elphid_cratic 17 UPG03 -5.136863021 119.387289 9 UPG03_9 Sphero_sp 2 UPG03 -5.136863021 119.387289 9 UPG03_9 Calcar_mayori 323 UPG03 -5.136863021 119.387289 9 UPG03_9 Calcar_spengl 2 UPG28 -5.122355036 119.339401 5 UPG28_5 Penero_planat 8 UPG28 -5.122355036 119.339401 5 UPG28_5 Amphis_latera 1 UPG28 -5.122355036 119.339401 5 UPG28_5 Sorite_sp 1 UPG28 -5.122355036 119.339401 8 UPG28_8 Bilocu_sp3 1 UPG28 -5.122355036 119.339401 8 UPG28_8 Verteb_sp 1 UPG28 -5.122355036 119.339401 8 UPG28_8 Penero_planat 7 UPG28 -5.122355036 119.339401 11 UPG28_11 Placop_c.f. 1 UPG28 -5.122355036 119.339401 11 UPG28_11 Sphero_sp 2 UPG28 -5.122355036 119.339401 11 UPG28_11 Siphon_siphon 1 UPG28 -5.122355036 119.339401 14 UPG28_14 Sphero_sp 1 UPG28 -5.122355036 119.339401 14 UPG28_14 Sorite_sp 1 UPG28 -5.122355036 119.339401 14 UPG28_14 Spirol_hadaii 1 UPG28 -5.122355036 119.339401 17 UPG28_17 Sphero_sp 3 UPG28 -5.122355036 119.339401 17 UPG28_17 Siphon_siphon 1 UPG28 -5.122355036 119.339401 17 UPG28_17 Sorite_sp 1 UPG32 -5.136863021 119.387289 4 UPG32_4 Pyrgo_striol 1 UPG32 -5.136863021 119.387289 4 UPG32_4 Ammoma_alveol 1 UPG32 -5.136863021 119.387289 4 UPG32_4 Quinqu_netstr 1 UPG32 -5.136863021 119.387289 6 UPG32_6 Neorot_calcar 1 UPG32 -5.136863021 119.387289 6 UPG32_6 Quinqu_c.f.ag 1 UPG32 -5.136863021 119.387289 6 UPG32_6 Amphis_lesson 50 UPG32 -5.136863021 119.387289 8 UPG32_8 Spirol_sp3 1 UPG32 -5.136863021 119.387289 8 UPG32_8 Triloc_tricar 1 UPG32 -5.136863021 119.387289 8 UPG32_8 Amphis_lesson 34 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] confidence intervals with glmmPQL
Hi - I am using R version 2.13.0. I have run several GLMMs using the glmmPQL function to model the proportion of fish caught in one net to the total caught in both nets by length. I started with a polynomial regression full model with three length terms: l, l^2, and l^3 (l=length). The length terms and intercept were the fixed effects and the random effect was a paired haul (n=18). m1-glmmPQL(fixed=Proportion~1+Length+second+third,random=~1|Pair,family=binomial,data=species,verbose=T,niter=2,weight=(Experimental+Control)) For the majority of the models, I ended up with a constant model with no length effect. The issue I am having is with the confidence intervals that were calculated. For two models the CIs are not symmetrical around the mean proportion from the model. The CIs for the other constant models are symmetrical around the mean. I was wondering if anyone has an idea why this would be or if anyone has any suggestions. Thanks Sally -- View this message in context: http://r.789695.n4.nabble.com/confidence-intervals-with-glmmPQL-tp4649637.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R KNN Regression Help
Hi Experts,I'm writing up my thesis on open source data mining at the moment and I have to do some simple R experiments; however, I have very little experience with R.. And I am hoping one of you can please help me out:)I need need to run a regression task that uses the KNN algorithm with k = 5 and cross validation = 10 on the computer hardware datset from the uci repo and report the RMSE. All other parameters should be set to their default values (if possible). I am hoping one of you experts can please help me out with the code?http://archive.ics.uci.edu/ml/datasets/Computer+Hardwarethanks,Bonjovi -- View this message in context: http://r.789695.n4.nabble.com/R-KNN-Regression-Help-tp4649638.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Changing start and end in a time serie
Hello everyone, I have a time serie created with the function ts(), starting in 2006/01 and ending in 2011/01. If I want to use the same data of the firsts two years of that serie in a new one that starts and ends in years later, say 2012/01 and 2013/12, how can I do it? Thanks in advance! -- Laura Catalina Echeverri Guzmán [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Switch Groups in ehplot (x-axis)
Hello Peter, I checked for the levels() and reorder(), but I can't make it work. R classifies the Groups (factor) by alphabetical order, but I would just like to make him understand that it is ordered (is.ordered()) in my data frame. Here is my code : abs1-spleen*p1*p2*cd8 length(abs1) # = 13 abs-c(0.499,0.73,0.5,0.75,abs1) symb-rep(c(17,15,17,15,19),times=c(2,2,5,5,3)) noms-rep(c(WT (NI),Bim-/-Bid-/- (NI),WT,Bim-/-Bid-/-,Bim-/-Noxa-/-),times=c(2,2,5,5,3)) df-data.frame(abs,symb,noms) ehplot(df$abs,df$noms,offset=0.1,intervals=10,median=FALSE,pch=df$symb, xlab=Recipient mice strain (CD45.2),ylab=# CD8alpha DC x 10^6, main=Absolute number of CD8alpha DCs in spleen,ylim=c(0,3)) Thank you for your help Dusan -- View this message in context: http://r.789695.n4.nabble.com/Switch-Groups-in-ehplot-x-axis-tp4649485p4649632.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Importing Data for a two sample t-test
Hi, Not sure why you wanted to separate A and B for a two sample t-test. dat1-read.table(text= weights company 1 A 2 A 2 B 3 B ,sep=,header=TRUE,stringsAsFactors=FALSE) t.test(weights~company,data=dat1) # Welch Two Sample t-test # #data: weights by company #t = -1.4142, df = 2, p-value = 0.2929 #alternative hypothesis: true difference in means is not equal to 0 #95 percent confidence interval: #-4.042435 2.042435 #sample estimates: #mean in group A mean in group B # 1.5 2.5 In case, you wanted to separate A and B: library(reshape2) dcast(dat1,weights~company,value.var=weights) # weights A B #1 1 1 NA #2 2 2 2 #3 3 NA 3 Hope it helps. A.K. - Original Message - From: nilsonern nilson...@gmail.com To: r-help@r-project.org Cc: Sent: Wednesday, November 14, 2012 11:52 PM Subject: [R] Importing Data for a two sample t-test I am trying to do a two sample t-test with data that i received in a text document. one list has the slab weights and the second has the company it is associated with. here is an example. weights company 1 A 2 A 2 B 3 B I was able to import the data but i cannot figure out how separate the data. I want to put them in two separate sets, one being A and one being B. Any help would be appreciated. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Importing-Data-for-a-two-sample-t-test-tp4649565.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'scan' in a script?
It looks like you are trying to use the scan function to pause some script being called with the source function until you provide user input. To do this, you need to specify the argument file=stdin when scan runs inside a process. EG scan(file='stdin', nmax=1). In my Linux terminal, at least, (using 2.15) I lose the prompt (which formerly defaulted to '1:') and have to enter some arbitrary value like '1' (whereas hitting enter with no value entered counted as input when running from an interactive session). -- View this message in context: http://r.789695.n4.nabble.com/scan-in-a-script-tp898837p4649634.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cluster analysis in R
I have two issues. 1-I am trying to use morphology to identify gender. I have 9 variables, both continuous and categorical. I was using two-step cluster analysis in SPSS because two-step could deal with different types of variables. But the output tells me that an animal is in cluster 1 or 2, it does not give me a probability (ex. 0.70 cluster 2). I also did not want to specify that I want two clusters, I wanted to see if analysis would naturally give me two clusters. These were all advantages to using SPSS but now I'm having trouble. Does cluster analysis in R give probabilities? Which type of cluster analysis in R is best to use? I did not think hierarchical analysis was a great choice, but maybe I'm wrong. I don't want to create the average variable, I want the analysis to do it on its own. I'm also new to R so would have to figure out the right codes to enter, etc. 2-I was also told to analyze each variable on its own before including it in cluster analysis. I had first included them all then teased out which ones were not important, but now have been asked to do the reverse. I cannot do cluster analysis on one variable -for example, one variable is either present or absent on an individual so of course cluster analysis gives me two clusters, one representing present and one representing absent. I was told to use regression, but how can regression also not give the same result? I feel like it would give me a line connecting a bunch of 0s to 1s. I don't know what to use, or if I can analyze each variable like this before putting them into cluster analysis. I ultimately want to only use the smallest number of variables necessary to identify gender. I have tried reading manuals etc and talking to people at my school, but nothing has helped. If anyone has any insight, that would be much appreciated Thank you! -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Selecting the non-attribute part of an object
I have two matrices, generated by R functions that I don't understand. I want to confirm that they're the same, but I know that they have different attributes. If I want to compare the dimnames, I can say identical(attr(tm, dimnames), attr(tmm, dimnames)) [1] FALSE or even: identical(dimnames(tm), dimnames(tmm)) [1] FALSE But I can't find any good way to compare the main part of objects. What I'm doing now is: tm_new - tm tmm_new - tmm attributes(tm_new) - attributes(tmm_new) - NULL identical(tm_new, tmm_new) [1] TRUE But that seems very inaesthetic, besides requiring that I create two pointless objects. I have read ?attributes, ?attr and some web introductions to how R objects work, but have not found an answer. Thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to create a 95 percent confidence interval using the diference of the mean using Bootstrap
Hello all, could you please tell me how to create a 95 percent confidence interval using R, if I have the next data: blue [1] 4 69 87 35 39 79 31 79 65 95 68 62 70 80 84 79 66 75 59 77 36 86 39 85 74 [26] 72 69 85 85 72 red [1] 62 80 82 83 0 81 28 69 48 90 63 77 0 55 83 85 54 72 58 68 88 83 78 30 58 [26] 45 78 64 87 65 Build a confidence interval of 95 % for the difference of the medias using BOOTSTRAP. Thank you, Fjaril [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using ifelse to remove NA's from specific columns of a data frame containing strings and numbers
HI,
But, this replace second column NAs to 1. May be, the na.replace() should be
applied to df1[,-1]
df1-read.table(text=
col1 col2 col3
A 15.5 8.5
A 8.5 7.5
A NA NA
B 8.0 6.0
B NA NA
B 9.0 10.0
,sep=,header=TRUE,stringsAsFactors=FALSE)
df2-df1[,-1]
na.replace-seq(1:ncol(df2))-1
df2[,names(df2)]-sapply(1:dim(df2)[2],function(ii){ifelse(is.na(df2[,ii]),na.replace[ii],df2[,ii])})
df2$col1-df1$col1
df2[order(names(df2))]
# col1 col2 col3
#1 A 15.5 8.5
#2 A 8.5 7.5
#3 A 0.0 1.0
#4 B 8.0 6.0
#5 B 0.0 1.0
#6 B 9.0 10.0
A.K.
- Original Message -
From: soon yi soon...@ymail.com
To: r-help@r-project.org
Cc:
Sent: Thursday, November 15, 2012 2:29 PM
Subject: Re: [R] using ifelse to remove NA's from specific columns of a data
frame containing strings and numbers
#Data
df-data.frame(id=letters[1:10],var1=rnorm(10,10,5),var2=rnorm(10,5,2),var3=rnorm(10,1,1))
#Missing
df$var1[2]-df$var2[c(2,6)]-df$var3[c(2,5)]-NA
na.replace-seq(1:ncol(df))-1
df[,names(df)]-sapply(1:dim(df)[2], function(ii)
{ifelse(is.na(df[,ii]),na.replace[ii],df[,ii])} )
David Romano-2 wrote
Hi everyone,
I have a data frame one of whose columns is a character vector and the
rest
are numeric, and in debugging a script, I noticed that an ifelse call
seems
to be coercing the character column to a numeric column, and producing
unintended values as a result. Roughly, here's what I tried to do:
df: a data frame with, say, the first column as a character column and the
second and third columns numeric.
also: NA's occur only in the numeric columns, and if they occur in one,
they occur in the other as well.
I wanted to replace the NA's in column 2 with 0's and the ones in column 3
with 1's, so first I did this:
na.replacements -ifelse(col(df)==2,0,1).
Then I used a second ifelse call to try to remove the NA's as I wanted,
first by doing this:
clean.df - ifelse(is.na(df), na.replacements, df),
which produced a list of lists vaguely resembling df, with the NA's mostly
intact, and so then I tried this:
clean.df - ifelse(is.na(df), na.replacements, unlist(df)),
which seems to work if all the columns are numeric, but otherwise changes
strings to numbers.
I can't make sense of the help documentation enough to clear this up, but
my guess is that the yes and no values passed to ifelse need to be
vectors, in which case it seems I'll have to use another approach
entirely,
but even if is not the case and lists are acceptable, I'm not sure how to
convert a mixed-mode data frame into a vector-like list of elements (which
I would hope would work).
I'd be grateful for any suggestions!
Thanks,
David Romano
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Re: [R] Selecting the non-attribute part of an object
max(abs(x-y)) numerical tolerance of your choice -- Bert On Thu, Nov 15, 2012 at 11:52 AM, Jonathan Dushoff dush...@mcmaster.ca wrote: I have two matrices, generated by R functions that I don't understand. I want to confirm that they're the same, but I know that they have different attributes. If I want to compare the dimnames, I can say identical(attr(tm, dimnames), attr(tmm, dimnames)) [1] FALSE or even: identical(dimnames(tm), dimnames(tmm)) [1] FALSE But I can't find any good way to compare the main part of objects. What I'm doing now is: tm_new - tm tmm_new - tmm attributes(tm_new) - attributes(tmm_new) - NULL identical(tm_new, tmm_new) [1] TRUE But that seems very inaesthetic, besides requiring that I create two pointless objects. I have read ?attributes, ?attr and some web introductions to how R objects work, but have not found an answer. Thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] strip.custom() with more than one conditioning variable
Suppose I wanted to plot the barley data like this: dotplot(variety ~ yield | year+ site, barley, strip = strip.custom(style = 4)) The factor levels are far too long for that to be useful. I can overcome that problem if I shorten the levels like this: dotplot(variety ~ yield | year + site, barley, strip = strip.custom(style = 4, factor.levels = substring(levels(barley$site), 1, 1))) But that messes up the year levels. So I try to have it influence only the site levels: dotplot(variety ~ yield | year+ site, barley, strip = strip.custom(which.given = 2, style = 4, factor.levels = substring(levels(barley$site), 1, 1))) But then the year strip is omitted altogether. How do I specify a different list for different which.given numbers? In this case, I know I could simply redefine the levels in the factor but that's not an option where I wish to apply this method. TIA Patrick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] assigning NULL to a list element without changing the length of the list
Hi, I have a list: tmp0 - list(a=1, b=2, c=3) And I realize that I can append a NULL element to the end of this list, just by writing: length(tmp0) - 4 Now, the question is, how can I assign NULL to one of the existing list elements without changing the length of the list? Please note I am working from inside a for loop that is working on one list element at a time and in some circumstances I want to set one element to NULL. So, specifying the whole list again as in: tmp0 - list(a=1,b=NULL,c=3) is not an option. But writing, say: tmp0 [[2]] - NULL is not an option either, because it leaves me with a list of length 2. Is there a solution for this? Thank you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] assigning NULL to a list element without changing the length of the list
tmp0 - list(a=1, b=2, c=3) tmp0[b] - list(NULL) # single [, list(NULL), not double [[ and bare NULL str(tmp0) # List of 3 # $ a: num 1 # $ b: NULL # $ c: num 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Gonçalo Ferraz Sent: Thursday, November 15, 2012 1:01 PM To: r-help@r-project.org Subject: [R] assigning NULL to a list element without changing the length of the list Hi, I have a list: tmp0 - list(a=1, b=2, c=3) And I realize that I can append a NULL element to the end of this list, just by writing: length(tmp0) - 4 Now, the question is, how can I assign NULL to one of the existing list elements without changing the length of the list? Please note I am working from inside a for loop that is working on one list element at a time and in some circumstances I want to set one element to NULL. So, specifying the whole list again as in: tmp0 - list(a=1,b=NULL,c=3) is not an option. But writing, say: tmp0 [[2]] - NULL is not an option either, because it leaves me with a list of length 2. Is there a solution for this? Thank you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cluster analysis in R
Dear KitKat, After installing R and reading some introductory material on getting started with R you may want to check the CRAN task view on cluster analysis: http://cran.r-project.org/web/views/Cluster.html which has many useful references to all kinds and flavors of clustering techniques, hierarchical or not, selecting the nr of clusters based on some model selection statistic, et cetera. hth, Ingmar On Thu, Nov 15, 2012 at 7:14 PM, KitKat katherinewri...@trentu.ca wrote: I have two issues. 1-I am trying to use morphology to identify gender. I have 9 variables, both continuous and categorical. I was using two-step cluster analysis in SPSS because two-step could deal with different types of variables. But the output tells me that an animal is in cluster 1 or 2, it does not give me a probability (ex. 0.70 cluster 2). I also did not want to specify that I want two clusters, I wanted to see if analysis would naturally give me two clusters. These were all advantages to using SPSS but now I'm having trouble. Does cluster analysis in R give probabilities? Which type of cluster analysis in R is best to use? I did not think hierarchical analysis was a great choice, but maybe I'm wrong. I don't want to create the average variable, I want the analysis to do it on its own. I'm also new to R so would have to figure out the right codes to enter, etc. 2-I was also told to analyze each variable on its own before including it in cluster analysis. I had first included them all then teased out which ones were not important, but now have been asked to do the reverse. I cannot do cluster analysis on one variable -for example, one variable is either present or absent on an individual so of course cluster analysis gives me two clusters, one representing present and one representing absent. I was told to use regression, but how can regression also not give the same result? I feel like it would give me a line connecting a bunch of 0s to 1s. I don't know what to use, or if I can analyze each variable like this before putting them into cluster analysis. I ultimately want to only use the smallest number of variables necessary to identify gender. I have tried reading manuals etc and talking to people at my school, but nothing has helped. If anyone has any insight, that would be much appreciated Thank you! -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] depmixS4 prediction
Dear EJ, The depmixS4 package has no forecasting or predict functions, but as you note, the forecast distribution is a relatively straightforward function of the parameters. The posterior function provides you with the probability distribution over the states at the end of your sequence and the transition matrix can be used to compute the state distributions ahead. Let me know if you have more questions, best, Ingmar PS: you may have gotten a quicker answer had the message also been sent to me directly. On Thu, Nov 15, 2012 at 5:48 AM, Epic John epicj...@gmail.com wrote: I am getting started with using the depmixS4 package. First, I would like to see I am very impressed with its speed and flexibility. The question I have is regarding predicting on new data. I want to fit the model on some sequences with observed responses, and then make predictions on the right end of the sequences where the responses are not observed. I see no prediction functionality anywhere, and am not sure what the best way to formulate something like is with the package without reinventing the wheel. I once i have a fitted model, i would like to apply it to sequences where the response variables on the right end of the sequence are unobserved, and get the prediction for those (conditioned on observed covariates for the response) using the filtering or smoothing distributions. I could ultimately pull out the relevant parameters of the conditional distribution of the response in each hidden state, the transition probabilities, rightmost posterior probability on the fully observed sequence , and write my own code to make predictions, but am wondering if there is a more direct way of doing it in the package. Thanks in advance for any suggestions, EJ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] survreg gompertz
On Nov 15, 2012, at 5:38 AM, Matthias Ziehm wrote: Hi all, Sorry if this has been answered already, but I couldn't find it in the archives or general internet. A Markmail/Rhelp search on: gompertz survreg ...brings this link to a reply by Terry Therneau. Seems to address everything you asked and a bit more http://markmail.org/search/?q=list%3Aorg.r-project.r-help+gompertz+survreg#query:list%3Aorg.r-project.r-help%20gompertz%20survreg+page:1+mid:6xdlsmo272oa7zkw+state:results (Depending on how your mailer breaks URLs you may need to paste it back together.) Is it possible to implement the gompertz distribution as survreg.distribution to use with survreg of the survival library? I haven't found anything and recent attempts from my side weren't succefull so far. I know that other packages like 'eha' and 'flexsurv' offer functions similar to survreg with gompertz support. However, due to the run-time environment were this needs to be running in the end, I can't use these packages :( Same questions for the gompertz-makeham distribution. Many thanks! Matthias -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting the non-attribute part of an object
all.equal() will give some details on the differences between your objects. If you don't care that some attribute will differ, either ignore all.equal's output concerning it or remove it before giving the object to all.equal. E.g., if you don't care about dimnames but do care about dimensions do canonicalize - function(x) structure(x, dimnames=NULL) all.equal(canonicalize(tm), canonicalize(tmm)) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jonathan Dushoff Sent: Thursday, November 15, 2012 11:53 AM To: r-help@r-project.org Subject: [R] Selecting the non-attribute part of an object I have two matrices, generated by R functions that I don't understand. I want to confirm that they're the same, but I know that they have different attributes. If I want to compare the dimnames, I can say identical(attr(tm, dimnames), attr(tmm, dimnames)) [1] FALSE or even: identical(dimnames(tm), dimnames(tmm)) [1] FALSE But I can't find any good way to compare the main part of objects. What I'm doing now is: tm_new - tm tmm_new - tmm attributes(tm_new) - attributes(tmm_new) - NULL identical(tm_new, tmm_new) [1] TRUE But that seems very inaesthetic, besides requiring that I create two pointless objects. I have read ?attributes, ?attr and some web introductions to how R objects work, but have not found an answer. Thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to view source code of a function inside a package?
It might be a C code wrapped around.
-m
On Thu, Nov 15, 2012 at 8:21 PM, C W tmrs...@gmail.com wrote:
Dear list,
I am trying to look at the function inside a package. I know that
methods() would do the trick, but what if the function is hidden? I have a
problem displaying the hidden function.
Say, for example the MCMC package. How do you view the code of that
function?
something like this:
which
function (x, arr.ind = FALSE, useNames = TRUE)
{
wh - .Internal(which(x))
if (arr.ind !is.null(d - dim(x)))
arrayInd(wh, d, dimnames(x), useNames = useNames)
else wh
}
bytecode: 0x1021eef50
environment: namespace:base
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su...@acm.org
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Re: [R] Installing R on Ubuntu
Try Prof. Wilkinson's recent blog entry: http://darrenjw.wordpress.com/2012/11/10/keeping-r-up-to-date-on-ubuntu-linux/ On Thu, Nov 15, 2012 at 5:14 PM, Karel van Duijvenboden viera...@gmail.com wrote: Hello R-help team, I seek your help (for what is most likely a very simple problem). I'm new to Ubuntu and tried to install R using the Ubuntu Software Center. However, after clicking the install command, I always get prompted with the error Failed to download package files --- check your internet connection. Details: Failed to fetch http://archive.ubuntu.com/ubuntu/pool/main/a/apparmor/dh-apparmor_2.7.102-0ubuntu3.1_all.deb404 Not Found [IP: 91.189.92.200 80] However, my internet connection is perfectly fine and I've already installed a bunch of programs using the Software Center without any problems. I've tried various of the available R packages (also the Rcommander) and always get this same error. Perhaps it has something to do with the server/mirror? Alternatively, I've downloaded a .zip of the package, but so far I've failed to install the program. With me, for now, being an unequivocal Ubuntu dummy I would prefer the straightforward Software Center route and in any case I wonder why it does not work for me. Best regards and thanks in advance! Karel van Duijvenboden (Netherlands) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Mehmet Süzen, Ph.D. ( Dr.phil.nat. ) su...@acm.org | PRIVILEGED AND CONFIDENTIAL COMMUNICATION This e-mail transmission, and any documents, files or previous e-mail messages attached to it, may contain confidential information that is legally privileged. If you are not the intended recipient or a person responsible for delivering it to the intended recipient, you are hereby notified that any disclosure, copying, distribution or use of any of the information contained in or attached to this transmission is STRICTLY PROHIBITED within the applicable law. If you have received this transmission in error, please: (1) immediately notify me by reply e-mail to su...@acm.org, and (2) destroy the original transmission and its attachments without reading or saving in any manner. | __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to create a 95 percent confidence interval using the diference of the mean using Bootstrap
Hello, Try the following. ?boot::boot x - scan(text= 4 69 87 35 39 79 31 79 65 95 68 62 70 80 84 79 66 75 59 77 36 86 39 85 74 72 69 85 85 72) y - scan(text= 62 80 82 83 0 81 28 69 48 90 63 77 0 55 83 85 54 72 58 68 88 83 78 30 58 45 78 64 87 65) diffmeans - function(x, i) mean(x[i, 1] - x[i, 2]) res - boot::boot(cbind(x, y), diffmeans, R = 1000) quantile(res$t, probs = c(0.025, 0.975)) # CI res$t0 # estimate Hope this helps, Rui Barradas Em 15-11-2012 19:52, Tania Patiño escreveu: Hello all, could you please tell me how to create a 95 percent confidence interval using R, if I have the next data: blue [1] 4 69 87 35 39 79 31 79 65 95 68 62 70 80 84 79 66 75 59 77 36 86 39 85 74 [26] 72 69 85 85 72 red [1] 62 80 82 83 0 81 28 69 48 90 63 77 0 55 83 85 54 72 58 68 88 83 78 30 58 [26] 45 78 64 87 65 Build a confidence interval of 95 % for the difference of the medias using BOOTSTRAP. Thank you, Fjaril [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can you have a by variable in Lag function as in SAS
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of ramoss Sent: Thursday, November 15, 2012 11:56 AM To: r-help@r-project.org Subject: [R] Can you have a by variable in Lag function as in SAS Hello, I want to use lag on a time variable but I have to take date into consideration ie I don't want days to overlap ie: I don't want my first time of today to match my last time of yeterday. In SAS I would use : data x; set y; by date tim; previous=lag(tim); if first.date then do; previous=.; end; run; How can I do something similar in R? I can't find any examples anywhere. Thank you all for your help. I haven't seen a response to this question yet, so here is an approach that may work for you. Let's say you have a data frame called dat that contains your variables date and tim. Then using the Lag() function from the Hmisc package you could do something like this library(Hmisc) dat$previous - ave(dat$tim,dat$date,FUN=Lag) Hope this is helpful, Dan Daniel J. Nordlund Washington State Department of Social and Health Services Planning, Performance, and Accountability Research and Data Analysis Division Olympia, WA 98504-5204 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strip.custom() with more than one conditioning variable
Hi Patrick Not sure what you finally want to achieve but will this do? I have reduced the size of the text to make them readible dotplot(variety ~ yield | year+ site, barley, strip = strip.custom(which.given = 2, style = 4, factor.levels = paste(levels(barley$year), substring(levels(barley$site), 1, 1)), par.strip.text = list(cex = 0.75)) ) Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au At 06:53 16/11/2012, you wrote: Suppose I wanted to plot the barley data like this: dotplot(variety ~ yield | year+ site, barley, strip = strip.custom(style = 4)) The factor levels are far too long for that to be useful. I can overcome that problem if I shorten the levels like this: dotplot(variety ~ yield | year + site, barley, strip = strip.custom(style = 4, factor.levels = substring(levels(barley$site), 1, 1))) But that messes up the year levels. So I try to have it influence only the site levels: dotplot(variety ~ yield | year+ site, barley, strip = strip.custom(which.given = 2, style = 4, factor.levels = substring(levels(barley$site), 1, 1))) But then the year strip is omitted altogether. How do I specify a different list for different which.given numbers? In this case, I know I could simply redefine the levels in the factor but that's not an option where I wish to apply this method. TIA Patrick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create function to solve derivative
On 16/11/12 02:40, e-letter wrote: Readers, A data set comprises A B C 10 6 .2 20 7 .4 30 8 .16 40 9 .0256 My requirement is to obtain the derivative for values of A with respect to B, create a function in R and plot this derivative against another variable (e.g. values in column C). Have searched the mailing list and found reference to a function 'D', but the help accessed via '?D' and '?deriv' does not explain how to write a derivative function. Could someone please direct me to the relevant parts of the manual to begin? Your question makes little sense. Functions have derivatives --- at least some of them do. Data sets do not have derivatives. The functions D(), deriv() etc. work on specified analytic expressions for functions --- data sets do not come into the picture. What you might possibly want to do is to fit a spline to your data and take the derivative of the spline. The function splinefun() could be what you need. cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cluster analysis in R
Have a look at the package mclust. Jose From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Ingmar Visser [i.vis...@uva.nl] Sent: 15 November 2012 21:10 To: KitKat Cc: r-help@r-project.org Subject: Re: [R] cluster analysis in R Dear KitKat, After installing R and reading some introductory material on getting started with R you may want to check the CRAN task view on cluster analysis: http://cran.r-project.org/web/views/Cluster.html which has many useful references to all kinds and flavors of clustering techniques, hierarchical or not, selecting the nr of clusters based on some model selection statistic, et cetera. hth, Ingmar On Thu, Nov 15, 2012 at 7:14 PM, KitKat katherinewri...@trentu.ca wrote: I have two issues. 1-I am trying to use morphology to identify gender. I have 9 variables, both continuous and categorical. I was using two-step cluster analysis in SPSS because two-step could deal with different types of variables. But the output tells me that an animal is in cluster 1 or 2, it does not give me a probability (ex. 0.70 cluster 2). I also did not want to specify that I want two clusters, I wanted to see if analysis would naturally give me two clusters. These were all advantages to using SPSS but now I'm having trouble. Does cluster analysis in R give probabilities? Which type of cluster analysis in R is best to use? I did not think hierarchical analysis was a great choice, but maybe I'm wrong. I don't want to create the average variable, I want the analysis to do it on its own. I'm also new to R so would have to figure out the right codes to enter, etc. 2-I was also told to analyze each variable on its own before including it in cluster analysis. I had first included them all then teased out which ones were not important, but now have been asked to do the reverse. I cannot do cluster analysis on one variable -for example, one variable is either present or absent on an individual so of course cluster analysis gives me two clusters, one representing present and one representing absent. I was told to use regression, but how can regression also not give the same result? I feel like it would give me a line connecting a bunch of 0s to 1s. I don't know what to use, or if I can analyze each variable like this before putting them into cluster analysis. I ultimately want to only use the smallest number of variables necessary to identify gender. I have tried reading manuals etc and talking to people at my school, but nothing has helped. If anyone has any insight, that would be much appreciated Thank you! -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Wrap Up Run 10k next March to raise vital funds for Age UK Six exciting new 10k races are taking place throughout the country and we want you to join in the fun! Whether you're a runner or not, these are events are for everyone ~ from walking groups to serious athletes. The Age UK Events Team will provide you with a training plan to suit your level and lots of tips to make this your first successful challenge of 2012. Beat the January blues and raise some vital funds to help us prevent avoidable deaths amongst older people this winter. Sign up now! www.ageuk.org.uk/10k Coming to; London Crystal Palace, Southport, Tatton Park, Cheshire Harewood House, Leeds,Coventry, Exeter Age UK Improving later life www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended
[R] Step-wise method for large dimension
Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0-lm(Y~1, data= mydata) fit.final- lm(Y~X1+X2+X3+.+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction=forward) step(fit.final, scope=list(lower=fit.final, upper=fit0), data=mydata, direction=backward) Best,Farnoosh Sheikhi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with XML package
Hi List, I have used XML in R version 2.9.2. The code is working fine using Rv2.9.2 and its related XML package. Now I am using Rv2.13.1 and its related XML package, but I get the following error: Error in UseMethod(xmlName, node) : no applicable method for 'xmlName' applied to an object of class NULL Any idea? Thanks Arvin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error when launching rJava
Hi all I'd like to launch rJava but get that error: library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared object '/home/cloudera/R/x86_64-redhat-linux-gnu-library/2.15/rJava/libs/rJava.so': libjvm.so: cannot open shared object file: No such file or directory Error: package/namespace load failed for rJava Here are some more information if needed: [cloudera@localhost ~]$ java -version java version 1.6.0_31 Java(TM) SE Runtime Environment (build 1.6.0_31-b04) Java HotSpot(TM) 64-Bit Server VM (build 20.6-b01, mixed mode) I need rJava running to be able to install rhdfs for my cloudera demo vm cdh3u4.Is ther something wrong with LD_LIBRARY_PATH settings? If yes, how can I fix that? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Removing default loess line from scatterplot ({car})
scatterplot() is autogenerating a regression and loess line. I want to remove the loess line, but can only find samples for how to add it. I definitely do not have code that adds it (see below), so I am puzzling... scatterplot(Aroused ~ anxious | group, data=data, xlab=Aroused, ylab=Anxious, main=Arousal and anxiety ratings, xlim=c(0.5, 9.5), ylim=c(0.5, 9.5), legend.plot=F, ) legend(6,2,c(Controls, Patients),group) -- View this message in context: http://r.789695.n4.nabble.com/Removing-default-loess-line-from-scatterplot-car-tp4649668.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stepwise
Hi, May be this helps: set.seed(232) mat1-matrix(sample(1:100,80,replace=TRUE),ncol=8) #with 8 columns dat1-data.frame(mat1) names(dat1)[1]-Y form1-as.formula(paste(Y~,paste(names(dat1)[2:8],collapse=+))) #should change 8 to 400 fit.final-lm(form1,data=dat1) fit.final #Call: #lm(formula = form1, data = dat1) # #Coefficients: #(Intercept) X2 X3 X4 X5 X6 # 143.763081 -0.009030 -1.844810 1.436647 -0.005407 -0.050101 # X7 X8 # 0.563285 -1.901918 A.K. From: farnoosh sheikhi farnoosh...@yahoo.com To: arun smartpink...@yahoo.com Sent: Thursday, November 15, 2012 6:14 PM Subject: Stepwise Hi there, I want to apply the following code fo my data with 400 predictors. I was wondering if there is an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0-lm(Y~1, data= mydata) fit.final- lm(Y~X1+X2+X3+.+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction=forward) step(fit.final, scope=list(lower=fit.final, upper=fit0), data=mydata, direction=backward) Best,Farnoosh Sheikhi __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting the non-attribute part of an object
I think that what you are looking for is: all.equal(tm,tmm,check.attributes=FALSE) But BEWARE: m - matrix(1:36,4,9) mm - matrix(1:36,12,3) all.equal(m,mm,check.attributes=FALSE) gives TRUE!!! I.e. sometimes attributes really are vital characteristics. cheers, Rolf Turner On 16/11/12 08:52, Jonathan Dushoff wrote: I have two matrices, generated by R functions that I don't understand. I want to confirm that they're the same, but I know that they have different attributes. If I want to compare the dimnames, I can say identical(attr(tm, dimnames), attr(tmm, dimnames)) [1] FALSE or even: identical(dimnames(tm), dimnames(tmm)) [1] FALSE But I can't find any good way to compare the main part of objects. What I'm doing now is: tm_new - tm tmm_new - tmm attributes(tm_new) - attributes(tmm_new) - NULL identical(tm_new, tmm_new) [1] TRUE But that seems very inaesthetic, besides requiring that I create two pointless objects. I have read ?attributes, ?attr and some web introductions to how R objects work, but have not found an answer. Thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Step-wise method for large dimension
Probably you just want Y ~ . where . means everything which doesn't appear elsewhere in the formula. Michael On Thu, Nov 15, 2012 at 8:45 PM, farnoosh sheikhi farnoosh...@yahoo.com wrote: Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0-lm(Y~1, data= mydata) fit.final- lm(Y~X1+X2+X3+.+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction=forward) step(fit.final, scope=list(lower=fit.final, upper=fit0), data=mydata, direction=backward) Best,Farnoosh Sheikhi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] assigning NULL to a list element without changing the length of the list
HI,
You could try these:
tmp1-list(a=1,b=NA,c=3,d=NA)
lapply(tmp1,function(x) if(is.na(x)) NULL else x) #changing NA to NULL
#$a
#[1] 1
#
#$b
#NULL
#
#$c
#[1] 3
#
#$d
#NULL
#other case;
tmp2-list(a=1,b=3,c=4,d=5) #want to change list elements a, c to NULL
tmp3-lapply(tmp2,function(x) {if(names(tmp2)[match.call()[[2]][[3]]]%in%
c(a,c)) NULL else x})
tmp3
#$a
#NULL
#
#$b
#[1] 3
#
#$c
#NULL
#
#$d
#[1] 5
A.K.
- Original Message -
From: Gonçalo Ferraz gferra...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Thursday, November 15, 2012 4:01 PM
Subject: [R] assigning NULL to a list element without changing the length of
the list
Hi,
I have a list:
tmp0 - list(a=1, b=2, c=3)
And I realize that I can append a NULL element to the end of this list, just by
writing:
length(tmp0) - 4
Now, the question is, how can I assign NULL to one of the existing list
elements without changing the length of the list?
Please note I am working from inside a for loop that is working on one list
element at a time and in some circumstances I want to
set one element to NULL. So, specifying the whole list again as in:
tmp0 - list(a=1,b=NULL,c=3)
is not an option.
But writing, say:
tmp0 [[2]] - NULL
is not an option either, because it leaves me with a list of length 2.
Is there a solution for this?
Thank you
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing default loess line from scatterplot ({car})
Always look at the help page. If you type, in this case, ?scatterplot, you'll
see that all you need to add is the argument smooth=FALSE to omit the loess
line.
Jose
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
nprause [nicole.pra...@gmail.com]
Sent: 15 November 2012 23:01
To: r-help@r-project.org
Subject: [R] Removing default loess line from scatterplot ({car})
scatterplot() is autogenerating a regression and loess line. I want to remove
the loess line, but can only find samples for how to add it. I definitely do
not have code that adds it (see below), so I am puzzling...
scatterplot(Aroused ~ anxious | group,
data=data,
xlab=Aroused,
ylab=Anxious,
main=Arousal and anxiety ratings,
xlim=c(0.5, 9.5), ylim=c(0.5, 9.5),
legend.plot=F,
)
legend(6,2,c(Controls, Patients),group)
--
View this message in context:
http://r.789695.n4.nabble.com/Removing-default-loess-line-from-scatterplot-car-tp4649668.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Wrap Up Run 10k next March to raise vital funds for Age UK
Six exciting new 10k races are taking place throughout the country and we want
you to join in the fun! Whether you're a runner or not, these are
events are for everyone ~ from walking groups to serious athletes. The Age UK
Events Team will provide you with a training plan to suit your
level and lots of tips to make this your first successful challenge of 2012.
Beat the January blues and raise some vital funds to help us
prevent avoidable deaths amongst older people this winter.
Sign up now! www.ageuk.org.uk/10k
Coming to; London Crystal Palace, Southport, Tatton Park, Cheshire Harewood
House, Leeds,Coventry, Exeter
Age UK Improving later life
www.ageuk.org.uk
---
Age UK is a registered charity and company limited by guarantee, (registered
charity number 1128267, registered company number 6825798).
Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA.
For the purposes of promoting Age UK Insurance, Age UK is an Appointed
Representative of Age UK Enterprises Limited, Age UK is an Introducer
Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth
Access for the purposes of introducing potential annuity and health
cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit
Solutions Limited and Simplyhealth Access are all authorised and
regulated by the Financial Services Authority.
--
This email and any files transmitted with it are confidential and intended
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Except where this email is sent in the usual course of our business, any
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Age Concern England (charity number 261794) and Help the Aged (charity number
272786) and their trading and other associated companies merged
on 1st April 2009. Together they have formed the Age UK Group, dedicated to
improving the lives of people in later life. The three national
Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help
the Aged in these nations to form three registered charities:
Age Scotland, Age NI, Age Cymru.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Selecting the non-attribute part of an object
Thanks for all of these useful answers. Thanks also to Ben Bolker, who told me offline that c() is a general way to access the main part of an object (not tested). I also tried: identical(matrix(tm), matrix(tmm)) [1] TRUE which also works, but does _not_ solve the problem Rolf warns about below (to my disappointment). JD On Thu, Nov 15, 2012 at 6:47 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote: I think that what you are looking for is: all.equal(tm,tmm,check.attributes=FALSE) But BEWARE: m - matrix(1:36,4,9) mm - matrix(1:36,12,3) all.equal(m,mm,check.attributes=FALSE) gives TRUE!!! I.e. sometimes attributes really are vital characteristics. cheers, Rolf Turner On 16/11/12 08:52, Jonathan Dushoff wrote: I have two matrices, generated by R functions that I don't understand. I want to confirm that they're the same, but I know that they have different attributes. If I want to compare the dimnames, I can say identical(attr(tm, dimnames), attr(tmm, dimnames)) [1] FALSE or even: identical(dimnames(tm), dimnames(tmm)) [1] FALSE But I can't find any good way to compare the main part of objects. What I'm doing now is: tm_new - tm tmm_new - tmm attributes(tm_new) - attributes(tmm_new) - NULL identical(tm_new, tmm_new) [1] TRUE But that seems very inaesthetic, besides requiring that I create two pointless objects. I have read ?attributes, ?attr and some web introductions to how R objects work, but have not found an answer. Thanks for any help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
