Hi,
I would like to know if there is a way to carry out a bivariate (e.g.,
between individuals of two tree species) local pair correlation function in
Spatstat. There is currently a function localpcf for univariate data but
I have not found one for bivariate data and was wondering if there is way
Thanks for the reply.
Goal of my meta analysis is to compare 2 different surgical procedures to fix
the same problem.
Data (after systematic search) are collected from studies with one or the other
procedure applied.
There are a couple of outcome variables that measure the effetivenss of
Achim,
Thank you for your response, and for your work on the zoo package in
general. Also, that implementation of the color scheme looks great.
Frankly, I can't remember exactly what I was originally trying to
accomplish for this particular problem, but I think that your suggestion of
reviewing
Why the result is coming as NULL. Can anyone help. I want to find the outliers
for a reference
setwd(D:/AZ)
library(RODBC)
cdb_cnct - odbcConnectExcel(Book1.xls)
cdb_frame - sqlFetch(cdb_cnct, Sheet1)
odbcClose(cdb_cnct)
rm(cdb_cnct)
x- cdb_frame$Publication =1990 cdb_frame$Publication =2012
Hello
I have dataframe
How to remove dates in hourly time seriesThe example time series likedate
value
2000-01-05 00:00:00 1.0
2000-01-05 01:00:00 1.0
2000-01-05 05:00:00 3.6
2000-01-05 06:00:00 3.6
2000-01-05 07:00:00
Thanks a lot! After thinking about it it definitely makes sense !
Greetings, Matthias
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Not sure, but maybe check the name of the column(s) you are retrieving
you may get a NULL value if the column name is misspelled
R example:
## --
df - data.frame(
Publication = 1991:2000,
Reference = LETTERS[1:10] )
df[ 2, References ]
# NULL
Hello,
You can try the following:
x - read.table(text=2000-01-05 00:00:00 1.0
2000-01-05 01:00:00 1.0
2000-01-05 05:00:00 3.6
2000-01-05 06:00:00 3.6
2000-01-05 07:00:00 2.2
2000-01-05 08:00:00 2.2
2000-01-05 09:00:00 2.2
I assume result means the invalid variable.
works with sample data frame i made, so please provide an example of cdb_frame.
You can use dput(cdb_frame) for that, or dput(head(cdb_frame)) if you have a
lot of data.
For example it could be that $Publication is a string, not numeric.
On
Dear all,
I would like to calculate the optimal cut off (threshold) of a test using
the Epi package. Here I am presenting some data based on the output of two
tests. I am interested in identifying the optimal cut off and its 95% CI.
Running the ROC() function with the Epi package I obtain a
Hi,
I've written a logistic function using nls and I'd like to do cross
validation for this. Is there a package for that? Below is an example of my
data and the function. N terms are presence/absence data and the response is
succesful/failed data.
y1-sample(0:1,100,replace=T)
Hi,
I am working on barplot. I need to plot x-axis but scale on x axis is very
upto 15 while my data is upto 19.
pdf(image.pdf, width=10 , height =13)
par(mar=c(5,22.5,2,2))
barplot(t(data[,2:3]), beside=TRUE, col=c(rgb(.537, .769, .933),rgb(.059,
.412, .659)), width = 1,
Hello everybody,
I need to calculate seasonal means with temperature data for my work.
I have 70 files coming from weather stations, which looks like this for
example:
startdate - as.POSIXct(01/01/2006, format = %d/%m/%Y)
enddate - as.POSIXct(05/01/2006, format = %d/%m/%Y)
date - seq(from =
Dear Dushanthi,
Please keep your e-mails on the R-Help list, where Michael has already given
you some excellent advice. As Michael already explained, metafor can handle
proportions, but does not have any specific functionality for categorical
variables with more than 2 levels (at the moment).
Hi, all:
This is the first time I use mail-list for help and I'm a student
from Chinese. So forgive my poor English skill.
Here is the question: I have an excel file contains some stkcd
information. The stkcd is a string fill with numbers, e.g 02. When i
use sqlFetch function from
On 01/08/2012 10:11, delvin008 delta wrote:
Hi, all:
This is the first time I use mail-list for help and I'm a student
from Chinese. So forgive my poor English skill.
Here is the question: I have an excel file contains some stkcd
information. The stkcd is a string fill with
Hello everyone. Like others on this list, I'm new to R, and really not much
of a programmer, so please excuse any obtuse questions! I'm trying to
repeat a function across all possible combinations of vectors in a data
frame. I'd hugely appreciate any advice!
Here's what I'm doing:
I have some
Hello,
Here's a quick and dirty solution. By setting axes = FALSE in the barplot plot
arguments we suppress the axes. You then use the axis function afterward to
draw the desired axis. Below, i have also used the pretty function.
Have a look at this:
data - rbind(c(3,4,6), c(5,15,19))
Even though the sheet name Publication exists. The error is coming.
Can anyone help?
library(RODBC)
cdb_cnct - odbcConnectExcel(Copy of
AZIF_DC_GVK_NSCLC_MSALL_287papers_02072012_141450_v1_4.xls)
cdb_frame - sqlFetch(cdb_cnct, Publication)
odbcClose(cdb_cnct)
Error in odbcTableExists(channel,
Hi
Something like
aggregate(DF$data, list(quarters(DF$date), format(DF$date, %Y)), mean)
Regards
Petr
Hello everybody,
I need to calculate seasonal means with temperature data for my work.
I have 70 files coming from weather stations, which looks like this for
example:
startdate -
Hi
I need to optimize the below function:
a=function(x){
A=x[1]
B=x[2]
C=B-A
return(C)
}
I need to optimize the above function such that x can be any combination of
these number (0,3,5,8) of vector length 2
(i.e) x can be (3,0), (5,0), (8,0), (3,5), (3,8), (5,8), .. etc
can
Hi
did you find function dist? It seems that it can do directly what you
want.
Regards
Petr
Hello everyone. Like others on this list, I'm new to R, and really not
much
of a programmer, so please excuse any obtuse questions! I'm trying to
repeat a function across all possible
Thank you both for your answers.
I found a best way to delete the first 2 months (Jan + Feb) and the last
month (Dec), which should work everytime:
DF$year - as.numeric(format(DF$Day, format = %Y))
DF$month - as.numeric(format(DF$Day, format = %m))
# delete first 2 months
for(i in DF[1,3]) #
On Wed, Aug 1, 2012 at 12:40 AM, Gene Leynes gley...@gmail.com wrote:
Achim,
Thank you for your response, and for your work on the zoo package in
general. Also, that implementation of the color scheme looks great.
Frankly, I can't remember exactly what I was originally trying to
accomplish
On Wed, Aug 1, 2012 at 7:31 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Aug 1, 2012 at 12:40 AM, Gene Leynes gley...@gmail.com wrote:
Achim,
Thank you for your response, and for your work on the zoo package in
general. Also, that implementation of the color scheme looks
Hello,
Try the following.
dat - structure(list(date = c(2000-01-05 00:00:00, 2000-01-05
01:00:00,
2000-01-05 05:00:00, 2000-01-05 06:00:00, 2000-01-05 07:00:00,
2000-01-05 08:00:00, 2000-01-05 09:00:00, 2000-01-05 10:00:00,
2000-01-05 11:00:00, 2000-02-05 00:00:00, 2000-02-05 01:00:00,
On 31.07.2012 21:14, Duncan Murdoch wrote:
On 12-07-31 2:54 PM, Andras Farkas wrote:
Dear All,
using the example from the package scatterplot3d I created a 3d plot
as follows:
x -rnorm(500,50,2)
y -rnorm(500,5,1)
z -rnorm(500,6,1)
scatterplot3d(x, y, z, highlight.3d=TRUE,
We cannot help you if you don't give us enough information to do so. Please use
the dput function to give us a copy of your cdb_frame as others have already
requested.
You may also find the str function helpful in diagnosing your own problems.
As Petr suggests, the dist() function will do much of the work for you.
For example ...
# example matrix of data
nsamples - 40
nreadings - 46
dat - matrix(runif(nsamples*nreadings), nrow=nsamples)
# Euclidean distance between the ROWS of dat
distance - dist(dat)
Jean
JenniferH
Ingmar, many thanks. I get that one from R:
Error in model.frame.default(terms(formula, lhs = lhs, rhs = rhs, data =
data), :
variable lengths differ (found for 'X')
X is the variable I have used.
Any comment would be much appreciated.
Best regards!
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Hello,
I have a big data frame where consecutive time dates and corresponding observed
values for each subject (ID) are on a line. I want to compute the linear slope
for each subject. I would like to use apply but I do
not know how to express the corresponding function. An example using a loop
It's working now!
The problem was not for winter, but with the with you had in your object
DF$season. I got an error: invalid 'envir' argument.
I removed it and now it seems to be OK.
Thank you very much for your help ricardo.
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You could use the xlim argument to barplot. For example,
barplot(t(data[,2:3]), beside=TRUE, xlim=c(0, 20),
col=c(rgb(.537, .769, .933), rgb(.059, .412, .659)),
width=1, horiz=TRUE, cex.names=.9, border =white,
las=1, cex.axis=1, cex.lab=1.2)
Jean
Manish Gupta
Hello,
See if this is it.
fun - function(DF, FUN = mean){
month - as.integer(format(DF$date, format=%m))
year - format(DF$date, format=%Y)
month[month %in% 1:2] - 13
DF$season - NA
DF$season[month %in% 12:13] - paste(year[month %in% 12:13], Winter)
DF$season[month %in%
Hi,
On Wed, Aug 1, 2012 at 9:06 AM, R Heberto Ghezzo, Dr
heberto.ghe...@mcgill.ca wrote:
Hello,
I have a big data frame where consecutive time dates and corresponding
observed values for each subject (ID) are on a line. I want to compute the
linear slope for each subject. I would like to
On 12-08-01 8:42 AM, Uwe Ligges wrote:
On 31.07.2012 21:14, Duncan Murdoch wrote:
On 12-07-31 2:54 PM, Andras Farkas wrote:
Dear All,
using the example from the package scatterplot3d I created a 3d plot
as follows:
x -rnorm(500,50,2)
y -rnorm(500,5,1)
z -rnorm(500,6,1)
scatterplot3d(x, y,
R Heberto Ghezzo, Dr heberto.ghe...@mcgill.ca wrote on 08/01/2012
08:06:30 AM:
Hello,
I have a big data frame where consecutive time dates and
corresponding observed values for each subject (ID) are on a line. I
want to compute the linear slope for each subject. I would like to
use apply
Hello,
Because there's a bug in the way I read the data: data.matrix is not
well used here, it transforms all variables in character vectors and may
cause problems such as
9 10
[1] FALSE
9 10
[1] TRUE
It was David Winsemius that pointed this out.
As a side effect, data.matrix returns a
On 01.08.2012 15:28, Duncan Murdoch wrote:
On 12-08-01 8:42 AM, Uwe Ligges wrote:
On 31.07.2012 21:14, Duncan Murdoch wrote:
On 12-07-31 2:54 PM, Andras Farkas wrote:
Dear All,
using the example from the package scatterplot3d I created a 3d plot
as follows:
x -rnorm(500,50,2)
y
Hi,
maybe working with a data.frame in long format is an option - then you
can use e.g. lmList and so on up to mixed models, depending on your
final goals of analyses (e.g. check for differential slopes).
vmat-matrix(c(X1,X2,X3,X4,Y1,Y2,Y3,Y4),nrow=2,byrow=T)
Hi Everyone,
Thanks so much for all your suggestions! All of these worked but David's
was best suited for my purposes, considering it was something happens
sporadically.
I don't do expenditure analyses often as I mostly do run of the mill survey
analysis, but this will come in handy for the
Hi,
not sure if that is what you are looking for, but have a look at
cmb-t(combn(c(0,3,5,8),2)) #get all pairs of combinations
cbind(cmb,apply(cmb,1,diff)) #for each pair, get the difference
cheers
Am 01.08.2012 12:29, schrieb loyolite270:
Hi
I need to optimize the below function:
combn() gives ordered combinations, while expand.grid() gives all combinations.
I'd give worked code but this hints at homework to me.
Sarah
On Wed, Aug 1, 2012 at 10:23 AM, Eik Vettorazzi e.vettora...@uke.de wrote:
Hi,
not sure if that is what you are looking for, but have a look at
%/0% - function(x,y) { res - x / y ; res[ is.na(res) ] - 0;
return(res) }
I think this would be more to the point if the output were set to 0
if the numerator were 0, not if the output would be NA.
res[x==0] -0
You may also want to restructure your computations so that
the proportion
Hi,
igraph will give you the mappings via vertex ids. If you want to use
symbolic vertex names, then attach a vertex attribute called 'name'.
Then any vector of numeric vertex ids (v, from graph g) can be
converted to vertex names via
V(g)$name[v]
or the more readable equivalent
I've been playing around with data like the following:
NameDateHrs
Way, S 2-3-12 8
Nun, B 2-3-12 9
Way, S 2-4-12 7.5
Nun, B 2-4-12 9
Gus, T 2-5-12 8
I've been able to take this into a data.frame and even develop a cumsum
for each of the people
d.cum - with(data,by(Hrs,
Hi,
I was trying to run the following, where the labels on the x axis are dates
and need to be converted to characters. It did not work. Any help would be
appreciated. Thanks.
text(axTicks(1),par(usr)[3]-2,srt=45,adj=1,labels=as.character(c(2008-01-08,2008-08-10,2008-08-22,2008-09-03,
On Aug 1, 2012, at 16:34 , Sarah Goslee wrote:
combn() gives ordered combinations, while expand.grid() gives all
combinations.
...and there's one more function that is designed to tabulate a function of two
variables over a grid. And yet another one to find which value is max in an
array
On Aug 1, 2012, at 5:33 AM, Rui Barradas wrote:
Hello,
Try the following.
dat - structure(list(date = c(2000-01-05 00:00:00, 2000-01-05
01:00:00,
2000-01-05 05:00:00, 2000-01-05 06:00:00, 2000-01-05 07:00:00,
2000-01-05 08:00:00, 2000-01-05 09:00:00, 2000-01-05 10:00:00,
2000-01-05
On Aug 1, 2012, at 12:37 AM, Pascal Oettli wrote:
Hello,
You can try the following:
x - read.table(text=2000-01-05 00:00:00 1.0
2000-01-05 01:00:00 1.0
2000-01-05 05:00:00 3.6
2000-01-05 06:00:00 3.6
2000-01-05 07:00:00 2.2
2000-01-05
On Aug 1, 2012, at 9:04 AM, Yolande Tra wrote:
Hi,
I was trying to run the following, where the labels on the x axis
are dates
and need to be converted to characters. It did not work. Any help
would be
appreciated. Thanks.
text(axTicks(1),par(usr)[3]
There is an unmatched quote in
Thanks for the tip for future typos.
Yolande
On Wed, Aug 1, 2012 at 1:38 PM, David Winsemius dwinsem...@comcast.netwrote:
On Aug 1, 2012, at 9:04 AM, Yolande Tra wrote:
Hi,
I was trying to run the following, where the labels on the x axis are
dates
and need to be converted to
Yesterday I changed the headers for a couple of columns in data text files
and removed hyphens from within character strings, too. When I tried to
re-read these data sources using read.table() I encountered an issue I've
not before seen. Both files were read almost instantly until yesterday's
On Wed, 1 Aug 2012, Rich Shepard wrote:
What might cause this?
I restored these two files from last Friday and they are read into R with
no problems. So, I'll make one change at a time and see where things break.
Will post results when I have them.
Rich
An unmatched quote can make read.table run very slowly
when there are lots of lines in the file. E.g.,
z - rep(A B C, 10^6)
z[2] - A \B C # unmatched quote on line 2
tf - tempfile()
cat(file=tf, sep=\n, z)
system.time(z2 - read.table(tf, skip=2)) # skip bad line
user system elapsed
On Wed, 1 Aug 2012, Rich Shepard wrote:
What might cause this?
Must be computers acting like computers. Restored files from backup, made
changes one at a time, and there are no problems reading them into R data
frames. My apologies for taking up space here.
Rich
Dear All,
I am writing code of Gauss Hermite Approximation for 2-d case.
Suppose I want to calculate the integral \int g(b)*exp(-b' W b) db, where b
is a 2 by 1 vector, W is a 2 by 2 positive definite matrix,
In order to get the basic form, I need decompose W = L' L, and define
x=L*b, i.e. b=
On Wed, 1 Aug 2012, William Dunlap wrote:
An unmatched quote can make read.table run very slowly when there are lots
of lines in the file. E.g.,
Bill,
Yes. Turns out that there was no closing quote on a changed header. I
found this by an error message on one data file; the other data file
There are still some seats available in the workshop on R that John
Fox will be teaching next week in Berkeley, CA. The workshop title is
The R Statistical Computing Environment: The Basics and Beyond. It
will run from Monday, August 6 through Thursday, August 9 (9:00 a.m.
through 5:00 p.m.
Your code almost gave me a headache. :-/
There are a lot of unnecessary tests and conditions. I tried to break
down the code and write the tests that have been done when assigning a
variable. I simplified your the first part but cannot guarantee that all
criteria are met.
#COMMENTS#
Hello,
Em 01-08-2012 20:02, Mercier Eloi escreveu:
Your code almost gave me a headache. :-/
Agree. There's a good way of avoiding headaches, package formatR,
function tidy.source.
My simplification is different in many places.
A common one is to treat 'observable' as a logical variable,
On Aug 1, 2012, at 12:02 PM, Mercier Eloi wrote:
Your code almost gave me a headache. :-/
I had a similar reaction. However, my approach might have been to
request a more complete verbal description of the data structures
being operated on and the methods and assumptions being used.
Domi wrote on 08/01/2012 Jul 29, 2012; 8:59am:
Hello erverybody,
I have a problem with my second for-loop.
1. First i read the tables.
datos.mx1 - read.table('PETmx1.csv',head=TRUE,sep=';')
datos.min - read.table('PETmin.csv',head=TRUE,sep=';')
You are most likely to get good
I defined a data.frame by a two-dimensional array.
aa = data.frame(rbind( 11:20, 1:10))
aa
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 11 12 13 14 15 16 17 18 19 20
2 1 2 3 45 6 78 9 10
Now I want to use the data in the second line of aa which from 1 to 10 to
output.But I find
It is not clear to me from your description what exactly you are trying to
do, but to get the numbers in the second line of aa you could use
as.numeric(aa[2,])
HTH,
Jorge.-
On Wed, Aug 1, 2012 at 5:13 PM, Jie Tang wrote:
I defined a data.frame by a two-dimensional array.
aa =
Hello,
I come from using different programming languages (C++, Mathematica, Perl) but
have been using R extensively for several months. I see the data frame as a
key piece of the language and wanted to inquire people's experience regarding
its use.
Say you have a data frame D
D -
Please read An Introduction to R, which ships with every copy of R,
where you will learn how to properly subscript data frames and lists,
among other things. There is a reason for such tutorials -- they
enable you to avoid elementary mistakes like this and wasting time and
effort with such posts
I have no answer to your question, but note:
1. You do not need to return a data frame at all, of course. Most
functions do not -- e.g., say, lm() .
2. See ?with and ?within for perhaps relevant functionality.
-- Bert
On Wed, Aug 1, 2012 at 2:19 PM, Ramiro Barrantes
ram...@precisionbioassay.com
Thanks Michael. Now I switched my approach after doing some google.
Following are my new codes:
###
library(foreign)
readin - read.dta(ordfile.dta, convert.factors=FALSE)
myvars - c(depvar, x1, x2, x3)
mydta - readin[myvars]
# remove all
Disclaimer: I have not followed this thread at all, but only wish to note:
1) Indicator variables are (almost?) never needed in R -- that you are
fooling with them suggests that there is probably a better approach.
2) Your bols is just least regression, no? -- If so, there are far
better ways to
Hi Petr and Jean,
thanks very much, problem solved! Really appreciate your help.
Jennifer
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Sent from the R help mailing list archive at Nabble.com.
Hi
I've to find the correlation of a variable and lag of other variable.
My code currently looks like this
y= 32 1 45 26 33 55 15 24 31 78
x=51 69 96 67 71 89 61 42 78 24
lagTimes=2
i want to find the correlation between y and Lag(x,1) and Y Lag(x,2)
i'm using loop to find the lag and then
Here is my approximation:
# Creation of the temporal variables
DF$year - as.numeric(format(DF$date, format = %Y))
DF$month - as.numeric(format(DF$date, format = %m))
# For years with data from 2006 to 2008
DF_type1 - DF [ - which (year == 2006 month ==1 | year == 2006 month ==
2 | year ==
Hi,
Try this:
dat1- read.table(text=2000-01-05 00:00:00 1.0
2000-01-05 01:00:00 1.0
2000-01-05 05:00:00 3.6
2000-01-05 06:00:00 3.6
2000-01-05 07:00:00 2.2
2000-01-05 08:00:00 2.2
2000-01-05 09:00:00 2.2
2000-01-05 10:00:00
Hi Petr,
It been sometime since I wrote. But here are the latest developments.
When I give the binary linear opt problem to lpSolve optimizer than for small
instance it gives correct answer but when size of nodes increase let's say 16
then there are about 2000 binary variables and lpSolve
HI,
Try this:
#This might get you started:
startdate - as.POSIXct(01/01/2006, format = %d/%m/%Y)
enddate - as.POSIXct(01/12/2006, format = %d/%m/%Y)
date - seq(from = startdate, to = enddate, by = months,format = %d/%m/%Y)
DF -
Hi Friends,
I'm new to R ,I have a data frame Z16 which is genarated from another data
frame, and I want to add “%” “$” in row 4 and 5 respectively. when I’m
trying using below logic, I’m getting warning message. I'm using R 2.14.2
Version
Can anyone help me out on this.
Note: Initially
Hi everyone
I'm trying to estimate both the rho parameter and the degree of freedom of
t-ev copula from the data using fitCopula function in the copula
package.
However, there is always an error and it says:
Error: length(copula@parameters) == length(param) is not TRUE
I also tried the same
HI,
The code was working perfectly fine yesterday and today, until half an hour
ago. Couldn't find any problems in the code. Still, I am getting error message.
myMatrix - data.matrix(read.table(text=
Name Age
ANTONY 27
IMRAN 30
RAJ 22
NAHAS
Hi Petr,
The following is different line of thought which is posted in different form,
maybe you have some wise input on it.
I need to find Efficient(tracktable) deterministic algorithm for Matching
Weighted Graphs with bounded degree. Now we all know Graph matching is
non-tractable but when
Dear Peter,
I have another question about WGCNA. I am using the package for
meta-analysis to find modules preserved in several datasets. However, I am
unsure how to handle the softpower, because each dataset has its own ideal
scale indepence value. When combining several datasets what should I
Hello,
I have a vector with positive integer numbers, e.g.
numbers - c(1,2,1,2,3,4,5)
and want to split the vector whenever an element in the vector is smaller or
equal to its predecessor.
Hence I want to obtain two vectors: c(1,2) and c(1,2,3,4,5).
I tried with which(), but it is not so
Good morning everyone,
I have attached an Excel file that contains a macro from which I call and
use R's auto.arima function to generate forecasts. The program runs
perfectly and it gets me the results; however, those results are pretty
unusual. I also tried using the auto.arima function directly
Hello. Please forgive me if this problem has already been posted (and solved)
by someone else ... I can't find it anywhere though it seems so very basic.
Here it is:
I have a list comprised of several matrices, each of which has two columns.
list
[[1]]
[,1] [,2]
[1,]1 3
[2,]
Hi,
just during these vacation days, I'm trying to approach with multicore
package
and I have some troubles with foreach.
ex.
foreach(.combine=c, ii=1:200, jj=1:500) %dopar% makefunction[ii,jj]
Command seem to work but if I see ii and jj value at the and of cycle, they
are both 200
instead than
Hi Rui,
Thanks. It was my fault. I had both of these saved (read.table (),
data.frame(read.table()). I think I accidentally pasted the second one, and
got the error.
A.K.
- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: arun smartpink...@yahoo.com
Cc: r-help
I have an object, which I pull in from a csv file here
http://r.789695.n4.nabble.com/file/n4638691/jan_2011.csv jan_2011.csv
mydata - read.csv(jan_2011.csv, header=TRUE, sep=,)
head(mydata)
Delivery.Date Hour.Ending Repeated.Hour.Flag Settlement.Point
Settlement.Point.Price
101/01/2011
You're totally right Jeff. My mistake! to use with, we write it like this:
DF$season - factor ( with ( *DF*, ifelse (( month == 12 | nonth == 1 |
month == 2 ), Win,
ifelse ((month == 3 | nonth == 4 | month == 5 )
, Spr,
ifelse
For me it owrks when i write it like:
as.Date(paste(mydata$Delivery.Date), %m/%d/%Y)
Hope it works,
Regards,
Ricardo
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Thank you!!! But I realise I've simplified my data to the point that your
solution doesn't actually work -- not your fault, mine! My list is actually
more complicated than what I presented it to be; it's not composed of
numerical matrices but of lists, each being composed of 7 columns, the first
Thank you for the answer, but I did´n find the solution. I'm a bit lost,
cos I am a beginner in R. I hope you can give me a little bit more help.
2012/7/29 jholtman [via R] ml-node+s789695n4638270...@n4.nabble.com
Try this: You had some 'indexing' problems. You were accessing the
columns
Thanks for the reply, but the example i have given above is a sample
function. I would be using a function with input vector of size more than
100 and the function will compute some score based on the 100 vector
combinations.
Since i don't want to do this computation sequentially for all these
Hi
I am using the nls function in R however it is causing the following error:
REAL() can only be applied to a 'numeric', not a 'logical'
Using traceback shows this occurs after the .Call(R_nls_iter, m, ctrl,
trace) call. However I do not know how to debug this further to find out
what is
Hi everyone,
I try to add many vectors (L1,L2,L3) to many list objects (a.list,
b.list) in a workspace. Somethings like below, but it is not working. Any
suggestions will be appreciated. Best, John
lf=ls(pattern=.lst)
for (x in listfiles) {
dat=read.delim(x,header=F)
for
Hi everyone,
Let me have a dataframe named “mydata” and created as below,
* n=c(5,5,5,5) #number of trils
x1=c(2,3,1,3) ) #number of successes
x2=c(5,5,5,5) #number of successes
x3=c(0,0,0,0) #number of successes
x4=c(5,0,5,0) #number of successes
mydata=data.frame(n,x1,x2,x3,x4)
mydata*
Hi,
I´ve got some problems with the labels of the x-axis
I,ve got two factors with two categories each: sex (males and females) and
area (central and peninsulae),
but because of the lenngth in the graphic just appeared two of the four
labels.
I thought the solution could be rotate it 45º
Hello,
Anyone know why the command:
plot(x,y) where y is a 0,1 result
sometimes plots the y values as 1,2 rather than 0,1?
And how to prevent this?
Thank you,
Georgiana May
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R-help@r-project.org mailing
Hi everyone,
I try to add many vectors (L1,L2,L3) to multiple list objects (a.list,
b.list) in a workspace. Somethings like below, but it is not
working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=.lst)
for (x in listfiles) {
dat=read.delim(x,header=F)
Hi -
I'm using the tune.nnet() to identify the optimal tuning parameters for a
dataset in which the dependent variables is binary and has very few 1s. As
a result, tune.nnet() provides parameters that give high accuracy, but by
increasing the specificity and decreasing the sensitivity radically.
Thank you Ricardogg and Arun. I don't know how I missed that. Too many
sleepless nights perhaps. Thank you.
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