On May 3, 2013, at 21:36 , David Winsemius wrote:
>
> On May 3, 2013, at 10:46 AM, peter dalgaard wrote:
>>
>>
>> Because comparison with an unknown value yields an unknown result.
>
> Anything else would violate the Second Law of Thermodynamics. We cannot have
> comparisons reducing entrop
On 05/03/2013 11:49 PM, Vahe nr wrote:
Hi all,
I have a big .csv file (21Mb with 100 rows) it has this shape:
x
1 NaN
2 NaN
3 0.23
and so on.
So the first column has x as a header then row number, the second column
contains values between -1,1 and NaN for empty values.
What should I n
On May 3, 2013, at 3:32 PM, Iuri Gavronski wrote:
> Hi,
> I would like to know the criteria by which R removes a factor in linear
> models. For example, I have a four level factor, and R creates 3 dummies to
> estimate coefficients. Which level is chosen? Can I chance it?
The default order is al
Sorry, Jeff, I misspoke: the 'matrix' data.m is really a data frame -- I
was just thinking about it as a matrix since it's the numeric part of df1,
and didn't realize the thought made it's way in the message. So the
memory issues are unrelated to converting between data frames and
matrices. -Da
Hi,
I would like to know the criteria by which R removes a factor in linear
models. For example, I have a four level factor, and R creates 3 dummies to
estimate coefficients. Which level is chosen? Can I chance it?
Thanks,
Iuri
[[alternative HTML version deleted]]
Hello,
I built R 2.15.2 on Solaris X64,
I have an issue when trying to execute the "check" target to test if everything
goes ok.
Do you have any idea what could be causing this issue?
Error message:
Examples/tools-Ex.Rout.fail
> cat("Time elapsed: ", proc.time() - get("ptime", pos = 'CheckExEn
I have a version that uses bigmemory on my blog, but looks at distance on
a sphere for a 36k * 36K matrix
not hundreds of Gb so I dont know if the approach will work for you
http://stevemosher.wordpress.com/2012/04/12/nick-stokes-distance-code-now-with-big-memory/
Steve
However, I never
I am not seeing any good justification in your description for converting to
matrix if you are planning to convert it back to data frame. Memory is going to
be inefficiently-used if you do this.
---
Jeff Newmiller
HI,
Not sure I understand it correctly.
dat1<- read.table(text="
site Year doy fish Feed swim agr_1 agr_2 agr_3 rest hide
3 2012 203 1 1 0 0 0 0 0 0
3 2012 203 1 0 1 0 0 0 0 0
3 2012 203 1 0 1 0 0 0 0 0
3 2012 203 2 0 0 0 0 0 1 0
3 2012 203 2 1 0 0 0 0 0 0
3 2012 203 2 1 0 0 0 0 0 0
4 2012 197 1 0
Hi everyone,
I have large data frame, say df1, with 165K columns, and all but the first
four columns of df1 are numeric. I transformed the numeric data and
obtained a matrix, call it data.m, with 165K - 4 columns, and then tried to
create a second data frame by replacing the numeric columns of
Hi everyone,
I'm trying to use apply (with a call to zoo's rollapply within) on the
columns of a 1.5Kx165K matrix, and I'd like to make use of the other cores
on my machine to speed it up. (And hopefully also leave more memory free: I
find that after I create a big object like this, I have to save
On May 3, 2013, at 3:21 PM, Sarah Goslee wrote:
> Hi Aldi,
>
> You might want
> ?layout
> instead.
>
Indeed. In particular a matrix argument might be:
matrix(c(1,2,3, 4,4,4)
> Sarah
>
> On Fri, May 3, 2013 at 5:54 PM, Aldi Kraja wrote:
>> Hmm,
>> I had a typo paste by mistake in my x vect
Hello ,
I want to compare two quadratic regression models with non-parametric
bootstrap.
However, I do not know which R package can serve the purpose,
such as boot, rms, or bootstrap, DeltaR.
Please kindly advise and thank you.
Elaine
The two quadratic regression models are
y1=a1x^2+b1x+c1
y1=
Hi Aldi,
You might want
?layout
instead.
Sarah
On Fri, May 3, 2013 at 5:54 PM, Aldi Kraja wrote:
> Hmm,
> I had a typo paste by mistake in my x vector
> It has to be:
>
> x<-rnorm(1000,mean=0,sd=1)
> wheat1<-rnorm(100,mean=0,sd=1)
> wheat2<-rnorm(150,mean=0,sd=2)
> tomatos3<-rnorm(200,mean=0,sd
Hmm,
I had a typo paste by mistake in my x vector
It has to be:
x<-rnorm(1000,mean=0,sd=1)
wheat1<-rnorm(100,mean=0,sd=1)
wheat2<-rnorm(150,mean=0,sd=2)
tomatos3<-rnorm(200,mean=0,sd=3)
tomatos4<-rnorm(250,mean=0,sd=4)
cucumbers5<-rnorm(300,mean=0,sd=5)
cucumbers6<-rnorm(400,mean=0,sd=6)
par(mfro
Hi,
Based on par function, I can split the screen into two parts left and
right.
I wish x occupies the half left screen, and all plants occupy half right
screen, which happens right now.
But I wish the right screen, to be split in 3 or more vertical parts
where each pair of the same type of
On May 3, 2013, at 1:57 PM, Ye Lin wrote:
> I want to plot the values of "Var1" and "Var2" on the same plot, with x-axis
> labeling as the list of IDs. Sth like this:
>
>
> But I want to color the points based on the category in "Group", I dont know
> how to do it with ggplot.
You didn't sa
On 02.05.2013 05:10, ren_az wrote:
Hello every one:
I get following warning when building my R package with R-3.0.0.
building 'SPEEDY.tar.gz' Warning in utils::tar(filepath, pkgname,
compression = "gzip", compression_level = 9L, : number of items to replace
is not a multiple of replacement
On 03.05.2013 15:59, Manta wrote:
Dear all,
I have a dataset with one column being of class Date. When I write the
output, I would like that column being written as number of days from
1970-01-01. I could not find anywhere a way to do it.
as.numeric(x)
where x is the Date object.
Uwe Ligg
On 02.05.2013 14:37, Ramon Hofer wrote:
Hi all
I'm trying to analyse the network speed and used iperf to create a csv
file containing the link test data. It's only about 6 MB big but
contains about 40'000 samples.
I can do boxplots (apart from printing the number of samples but I ask
separate
Thanks A.K
I also add "shape=variable" so that it is much easier to tell two variables
by color +shape
On Fri, May 3, 2013 at 2:14 PM, arun wrote:
> HI,
> May be this helps:
>
> dat1<- read.table(text="
> IDVar1 Var2Group
> A111BB
> A21
HI,
May be this helps:
dat1<- read.table(text="
ID Var1 Var2 Group
A1 1 1 BB
A2 1 2 AA
B1 2 1 CC
B2 1 3 DD
C1 1 2 EE
",sep="",header=TRUE)
lib
I want to plot the values of "Var1" and "Var2" on the same plot, with
x-axis labeling as the list of IDs. But I want to color the points by
their category in "Group". Is it possible to do in ggplot, or do i have to
plot from scratch using basic plot?
On Fri, May 3, 2013 at 1:49 PM, David Winsemi
On May 3, 2013, at 1:37 PM, Ye Lin wrote:
> Hey,
>
> I have a dataset like this:
>
> ID Var1 Var2 Group
> A1 11BB
> A2 1 2AA
> B1 2 1 CC
> B2 13DD
> C1
Hi,
You can use ?split()
lst1<-split(DF,DF$ID)
lst1[1:2]
#$`1`
# ID drugs month
#1 1 drug x 1
#4 1 drug x 1
#5 1 drug y 2
#6 1 drug z 3
#
#$`2`
# ID drugs month
#2 2 drug y 2
#7 2 drug x 1
mean(sapply(lst1,nrow))
#[1] 2.4
#or
library(plyr)
mean(ddply(DF,.(ID),
Hey,
I have a dataset like this:
ID Var1 Var2 Group
A1 11BB
A2 1 2AA
B1 2 1 CC
B2 13DD
C1 12EE
I would like to plot the point
On May 3, 2013, at 9:44 AM, Tien trung Dinh wrote:
> Hi.
>
> After I installed R 3.0.0.pkg for mac version , when click the icon R to
> startup . I receive the annoucement in red color to inform that something
> wrongs , but I do not know how to fix them .
> R version 3.0.0 (2013-04-03) -- "Ma
On May 3, 2013, at 10:46 AM, peter dalgaard wrote:
>
> On May 3, 2013, at 17:24 , jpm miao wrote:
>
>> I have a program, when I write
>>
>> if(num!=NA)
>>
>> it yields an error message.
>>
>> However, if I write
>>
>> if(is.na(num)==FALSE)
>>
>> it works.
>>
>> Why doesn't the first state
Hi.
After I installed R 3.0.0.pkg for mac version , when click the icon R to
startup . I receive the annoucement in red color to inform that something
wrongs , but I do not know how to fix them .
R version 3.0.0 (2013-04-03) -- "Masked Marvel"
Copyright (C) 2013 The R Foundation for Statistical
On Fri, May 3, 2013 at 3:36 PM, David Winsemius wrote:
>>
>> On May 3, 2013, at 17:24 , jpm miao wrote:
>>
>>> I have a program, when I write
>>>
>>> if(num!=NA)
>>>
>> snipped
>
> On May 3, 2013, at 10:46 AM, peter dalgaard wrote:
>
>> Because comparison with an unknown value yields an unknown re
>
> On May 3, 2013, at 17:24 , jpm miao wrote:
>
>> I have a program, when I write
>>
>> if(num!=NA)
>>
> snipped
On May 3, 2013, at 10:46 AM, peter dalgaard wrote:
> Because comparison with an unknown value yields an unknown result.
Anything else would violate the Second Law of Thermody
Just got it right please ignore the previous posting...
It worked!
Prof Ripley made my day!! :) THANK YOU!
On Fri, May 3, 2013 at 11:23 AM, Santosh wrote:
> Thanks for your suggestion... I upgraded to R.3.0.0 in 64-bit Windows 7
> environment..
>
> This time when I use file.link..
> I get
Thanks for your suggestion... I upgraded to R.3.0.0 in 64-bit Windows 7
environment..
This time when I use file.link..
I get the following error message: 'Cannot create a file when that file
already exists"
And I don't see the link.
The other function, file.copy, correctly copies to the target lo
Thi is great! Thank you so much for taking the time to give is this hint.
Â
mike
>
> From: Jeff Newmiller
>To:Mihai Nica ; Mihai Nica ; Jorge I
>Velez ; Katarzyna Kulma
>Cc: R mailing list
>Sent: Friday, May 3, 2013 8:16 AM
>Subject: Re: [R] R does not subse
Dear users
I am reposting this and hope it will be accepted this time.
I am using copula package to fit my bivariate data and simulation. As
explained in package documentation we can use our own data distribution to
feed on copula as long as we have d, p and q (pdf, cdf and quantile)
functions are
On 03.05.2013 15:36, David Carlson wrote:
Here's the result on R 3.0.0 64 bit under Windows 8:
A<-matrix(1:365000*144,nrow=365000,ncol=144)
dim(A)
[1] 365000144
d <- dist(mydata_nor, method = "euclidean")
Error in as.matrix(x) : object 'mydata_nor' not found
d <- dist(A, method = "euc
On May 3, 2013, at 17:24 , jpm miao wrote:
> I have a program, when I write
>
> if(num!=NA)
>
> it yields an error message.
>
> However, if I write
>
> if(is.na(num)==FALSE)
>
> it works.
>
> Why doesn't the first statement work?
Because comparison with an unknown value yields an unknown
Hi,
I'm sorry that it takes me so much time to respond, finally yesterday I got
time to try your suggestions. Thank you for them!
I tried both, they give the same results, but in both there are some things
I still need to solve. I would appreciate your help.
I include a little bigger dataframe (t
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Katarzyna Kulma
> Sent: Friday, May 03, 2013 4:21 AM
> To: David Kulp
> Cc: r-help@r-project.org
> Subject: Re: [R] R does not subset
>
> Jorge, thanks for your suggestions, but t
At a minimum, the first statement needs "==".
Also, is.na() gives TRUE/FALSE. While a logical comparison to NA gives NA
as a value.
Kevin
On Fri, May 3, 2013 at 10:24 AM, jpm miao wrote:
> I have a program, when I write
>
> if(num!=NA)
>
> it yields an error message.
>
> However, if I write
> if(num!=NA)
> Why doesn't the first statement work?
An NA value means that the value is unknown. E.g.,
age <- NA
means the you do not know the age of your subject.
(The subject has an age, NA means you did not collect
that data.) Thus you do not know the value of
age == 6
either, the subj
you can also try:
temp[] <- lapply(temp, as.numeric)
On Fri, May 3, 2013 at 11:54 AM, David Winsemius wrote:
>
> On May 2, 2013, at 11:00 PM, jpm miao wrote:
>
> > Hi Anthony,
> >
> > Thank you very much. It works very well. However, after this line
> >
> >> temp <- sapply( temp , as.numeric
Hi ST,
Try this:
set.seed(51)
df1<- as.data.frame(matrix(sample(1:40,60,replace=TRUE),ncol=10))
df2<- df1
check<- c("V3","V7","V9")
df1[,match(check,colnames(df1))]<-lapply(df1[,match(check,colnames(df1))],as.factor)
str(df1)
#'data.frame': 6 obs. of 10 variables:
# $ V1 : int 32 9 12 40 9
num1<- c(0,NA,1,3)
num1==NA
#[1] NA NA NA NA
num1!=NA
#[1] NA NA NA NA
is.na(num1)
#[1] FALSE TRUE FALSE FALSE
A.K.
- Original Message -
From: jpm miao
To: r-help
Cc:
Sent: Friday, May 3, 2013 11:24 AM
Subject: [R] Why can't R understand if(num!=NA)?
I have a program, when I wri
Hi,
I am not sure about what you meant.
lapply(1:5,function(i) data.frame())
[[1]]
data frame with 0 columns and 0 rows
[[2]]
data frame with 0 columns and 0 rows
[[3]]
data frame with 0 columns and 0 rows
[[4]]
data frame with 0 columns and 0 rows
[[5]]
data frame with 0 columns and 0 rows
Just read in and plot the data. The NaN will not be plotted:
> input <- read.table(text = "x
+ 1 NaN
+ 2 NaN
+ 3 0.23
+ 4 .34
+ 5 .55
+ 6 .66
+ 7 NaN
+ 8 .88", header = TRUE)
> plot(input$x)
>
On Fri, May 3, 2013 at 9:49 AM, Vahe nr wrote:
> Hi all,
>
> I have a big .csv file (21Mb with 1000
On 03-05-2013, at 17:24, jpm miao wrote:
> I have a program, when I write
>
> if(num!=NA)
>
> it yields an error message.
>
it?
What is unclear about the error message?
> However, if I write
>
> if(is.na(num)==FALSE)
>
> it works.
>
> Why doesn't the first statement work?
>
Read sectio
On May 3, 2013, at 14:59 , Ozgul Inceoglu wrote:
> Dear All, I am trying to perform MANOVA. I have table with 504
> columns(species) and 36 rows) with two grouping (season and location)
>
> Zx <- Z[c(4:504)]
> Zxm <- as.matrix(Z)
> m<- manova(Zxm~Season*location, data=Z)
>
> when I do summary.
On May 3, 2013, at 8:24 AM, jpm miao wrote:
> I have a program, when I write
>
> if(num!=NA)
>
> it yields an error message.
>
> However, if I write
>
> if(is.na(num)==FALSE)
>
> it works.
>
> Why doesn't the first statement work?
Read the "manual":
?"NA"
--
David Winsemius
Alameda,
On May 3, 2013, at 10:24 AM, jpm miao wrote:
> I have a program, when I write
>
> if(num!=NA)
>
> it yields an error message.
>
> However, if I write
>
> if(is.na(num)==FALSE)
>
> it works.
>
> Why doesn't the first statement work?
>
> Thanks,
>
> Miao
NA is undefined:
> NA == NA
[1]
On May 2, 2013, at 11:00 PM, jpm miao wrote:
> Hi Anthony,
>
> Thank you very much. It works very well. However, after this line
>
>> temp <- sapply( temp , as.numeric )
>
> the data becomes a series of numbers instead of a matrix. Is there any
> way to keep it a matrix?
Perhaps (assuming
A logical operation involving NA returns NA, never TRUE or FALSE:
See the 8th Circle of the R Inferno (8.1.4):
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
> num <- 1
> num==NA
[1] NA
> is.na(num)
[1] FALSE
-
David L Carlson
Associate Professor of Anth
Hello,
I can't say I understand the question, but if you want a list of empty
dfs and a list of empty matrices, the following will do.
replicate(10, data.frame())
replicate(10, matrix(NA, nrow = 0, ncol = 0))
Hope this helps,
Rui Barradas
Em 03-05-2013 16:20, jpm miao escreveu:
Hi,
I
On May 2, 2013, at 4:15 PM, T P Kharel wrote:
> I have posted a R copula question yesterday but it is not accepted yet. How
> long does it take?
Generally moderated postings are accepted within 4-6 hours, usually sooner.
> I am waiting if some one can help me on my Copula
> package related ques
You can use only
if(!is.na(num))
[[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 com
> -Original Message-
> if(num!=NA)
> it yields an error message.
> Why doesn't the first statement work?
Because you just compared something with NA (usually interpreted as 'missing')
and because of that the comparison result is also NA.
'if' then tells you that you have a missing va
Well, I have uploaded the data in the public folder of my dropbox. Due
to data confidentiality, I haved to change the labels. To load the data:
con <- url( "http://dl.dropboxusercontent.com/u/101865137/datEx.rda"; )
print(load(con))
# The replicate weights were created according to the jackknife
Dear R-Experts,
I seem to be dealing with a so called "headless" problem in R.
I wrote a quite extensive program that generates a Bayesian network
from a query protein's Phylogenetic Tree and subsequently uses a
message passing algorithm to infer the most likely annotation for the
query leaf in t
Dear All, I am trying to perform MANOVA. I have table with 504 columns(species)
and 36 rows) with two grouping (season and location)
Zx <- Z[c(4:504)]
Zxm <- as.matrix(Z)
m<- manova(Zxm~Season*location, data=Z)
when I do summary.aov, I get respond for each species but summary.manova
summary.man
On Fri, May 3, 2013 at 11:31 AM, jawad hussain wrote:
> Dear Sir
> I tried to find cURL on web but I do not find reliable file; there are some
> files on http://curl.haxx.se/. But I do not know which is suitable for R and
> how to install?
> Kind Regards
As usual, the OS is relevant here. What
I have a program, when I write
if(num!=NA)
it yields an error message.
However, if I write
if(is.na(num)==FALSE)
it works.
Why doesn't the first statement work?
Thanks,
Miao
[[alternative HTML version deleted]]
__
R-help@r-project.org ma
Hi,
I would like to read several datasets and would like to create a set
(list? sequence?) of many empty dataframes. How could this be done? How
could I declare a set (list? sequence?) of many empty matrices?
Thanks,
Miao
[[alternative HTML version deleted]]
Something like this?
library(gridExtra)
grid.arrange(one,two)
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
http://www.fws.gov/redbluff/rbdd_jsmp.aspx
>
>From: Christophe Bouffioux
>To: "r-
sorry, i had assumed readWorksheetFromFile would give you back a data
frame. all of the operations i recommended work on data.frame objects
at different points in the code, check if it's a data.frame or a matrix..
class( temp )
..you can check its current class at any point.
and if it's a mat
Dear all,
I have a dataset with one column being of class Date. When I write the
output, I would like that column being written as number of days from
1970-01-01. I could not find anywhere a way to do it.
Thanks,
Marco
--
View this message in context:
http://r.789695.n4.nabble.com/Write-date-
Hi all,
I have a big .csv file (21Mb with 100 rows) it has this shape:
x
1 NaN
2 NaN
3 0.23
and so on.
So the first column has x as a header then row number, the second column
contains values between -1,1 and NaN for empty values.
What should I need to do is: create a new .csv file from
Hi,
May be this helps:
set.seed(24)
dat1<-
data.frame(date1=sample(seq(as.Date("2012-09-14",format="%Y-%m-%d"),length.out=40,by="day"),20,replace=FALSE),
value=sample(1:60,20,replace=TRUE))
dat1$days1<- as.numeric(difftime(dat1$date1,as.Date("1970-01-01")))
#or
library(lubridate)
dat1$days2<- da
Interesting conclusion. Alternatively, that representation of your object model
may not be computationally effective. This discrepancy may be less exaggerated
in C++, but you may still find that large numbers of objects are less efficient
in their use of memory or cpu time than vector processing
untar != gunzip
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research
I recommend you read the Introduction to R document that comes with R. Look for
making vectors with the c() function, and using the mean() and sd() functions.
Note that this is not a homework help forum (read the Posting Guide mentioned
at the bottom of every message). If this is not homework,
Here's the result on R 3.0.0 64 bit under Windows 8:
> A<-matrix(1:365000*144,nrow=365000,ncol=144)
> dim(A)
[1] 365000144
> d <- dist(mydata_nor, method = "euclidean")
Error in as.matrix(x) : object 'mydata_nor' not found
> d <- dist(A, method = "euclidean")
Error: cannot allocate vector of s
hello everybody,
I want to print two plots in one png file, I tried several options but i
didn't succeed
the first plot (bwplot) print to the defined position, but the second
(ggplot) doesn't
Any idea?
Thanks a lot
Christophe
# Example:
#-
library(ggplot2)
> - the mean-> mean(x)
> - the uncertainty on-> std.error(x) ? Or sd(x)?
> - the standard deviation of x -> ?
> - the uncertainty on the standard deviation -> ?
>
> Anyone has an idea?
1. Use R's help system to look up 'standard deviation' and 'mean'
e.g.:
??'s
This typically occurs because of sloppy manual data entry outside of R. To
relieve further analysis pain, you can manually clean the data (usually only
effective for one-time analyses) or use R to fix problems right after loading
the data (there are multiple methods for doing this... I prefer us
If you don't know, we certainly don't. This is not a question about R or RCurl
anymore... it is a question about cURL. You need to know what operating system
your computer uses and how to enable SSL for cURL on that operating system...
perhaps you need local technical assistance.
---
Dear list members,
Apologies for cross-posting. Please, find below the information of
two statistical courses with R:
1) Statistical Analysis with R
2) Bayesian Data Analysis with R and WinBUGS
If you have any question don't hesitate to contact me.
Best regards,
Pablo
+++
Dear all,
Very simple question, but apparently uneasy to solve in R:
I have a sampling of a variable x: (3, 4. 5, 2, ...)
I want to know:
- the mean-> mean(x)
- the uncertainty on-> std.error(x) ? Or sd(x)?
- the standard deviation of x -> ?
- the uncertainty on
Hi:
"(note the space after "Infected")"
Since I lost a morning too with this issue, I am just curious, why is there a
space?
I know, it must be a dumb question, a reasonable programming rule, but that's
my level :-)
mike
>
> From: Jorge I Velez
>To:Katarzy
On 03/05/2013 08:31, Hakim Abdi wrote:
Dear List,
I have a list of 600+ *.gz files that I would like to extract and read the
geotiffs contained within them. I tried using the untar() function to
simplify this task but I am stumped by an error. I've combed the Internet
for a solution without luck
Jorge, thanks for your suggestions, but they give the same (empty) result:
> RECinf<-subset(REC2, INFECTION=="Infected")
> head(RECinf)
[1] RINGNOyear ccFLEDGE rec2012 binageINFECTION all.rsLD
<0 rows> (or 0-length row.names)
but David's suggestion worked! :
> RECinf<-REC2[REC2$
Hi Kasia,
You need
subset(REC2, INFECTION=="Infected ")
(note the space after "Infected").
HTH,
Jorge.-
On Fri, May 3, 2013 at 7:48 PM, Katarzyna Kulma
wrote:
> Hi everyone,
>
> I know there have been several requests regarding subsetting before, but
> none of them really helps with my prob
Hi Luis,
thanks for the suggestion, but still nothing:
> RECinf2<-subset(REC2, INFECTION==1)
> head(RECinf2)
[1] RINGNOyear ccFLEDGE rec2012 binageINFECTION all.rsLD
<0 rows> (or 0-length row.names)
cheers,
Kasia
Katarzyna Kulma
PhD Student
Department of Ecology and Genetics
$ INFECTION: Factor w/ 2 levels "Infected ","Uninfected ": 2 1 2 1 2 2 1 2
it is a factor variable, so it takes numeric values, for "Infected " it is
assigned value 1.
subset(REC2, INFECTION==1)
2013/5/3 Jorge I Velez
> Hi Kasia,
>
> You need
>
> subset(REC2, INFECTION=="Infected ")
>
> (n
Dear Sir
I tried to find cURL on web but I do not find reliable file; there are some
files on http://curl.haxx.se/. But I do not know which is suitable for R and
how to install?
Kind Regards
Jawad Hussain Ashraf
VPO Aroop, Tehsil and District GujranwalaMobile phone# 03016673275
> Date: Su
Good tip. Thanks Morgan.
I agree that a different structure might (necessarily) be in order. I wanted
to create a tree where nodes in a tree were of different derived sub-classes --
possibly holding more data and behaving polymorphically. OO programming seemed
ideal for this: lots of small th
You have an extra space in the INFECTION factors.
Use REC2[REC2$INFECTION=="Infected ",]
or
subset(REC2, INFECTION=="Infected ")
No need to use which here.
On May 3, 2013, at 5:48 AM, Katarzyna Kulma wrote:
> Hi everyone,
>
> I know there have been several requests regarding subsetting before,
Hi everyone,
I know there have been several requests regarding subsetting before, but
none of them really helps with my problem:
I'm trying to subset only infected individuals from the REC2 data.frame:
> str(REC2)
'data.frame':362 obs. of 7 variables:
$ RINGNO : Factor w/ 370 levels "BL1
> I have run a regression and want to calculate the likelihood
> of obtaining the sample.
> Is there a way in which I can use R to get this likelihood value?
See ?logLik
And see also ?help.search and ??. You would have found the above by typing
??likelihood at the command line in R
S Ellison
On 03.05.2013 07:54, PIKAL Petr wrote:
Hi
Probably others can give you some better insight but copying folder with
package from one machine to another is possible until the installation is
required by a new version of R (about each 3 years).
Reinstallation may be required more often, and w
Hello,
I am work with two quadratic regression models
y=ax^2+bx+c with the function of lm.
y1= observed migration distance of butterflies(y1=a1x^2+b1x+c1)
y2= predicted migration distance of butterflies (based on body mass)
(y2=a2x^2+b2x+c2)
x= body mass of butterflies
Now I would like to c
We would like to announce the following statistics course:
Data exploration, regression, GLM & GAM. With introduction to R
When: 26 - 30 August 2013.
Where: Edmonton, Canada
For details, see: http://www.highstat.com/statscourse.htm
Course flyer: http://www.highstat.com/Courses/Flyer2013_09Cana
Dear List,
I have a list of 600+ *.gz files that I would like to extract and read the
geotiffs contained within them. I tried using the untar() function to
simplify this task but I am stumped by an error. I've combed the Internet
for a solution without luck. The details are below, and any help in
92 matches
Mail list logo
