Re: [R] function for handling time
On 10/16/2011 04:13 AM, Alaios wrote:
Dear all
I have the following time stamps (in the following format)
MeasurementSet$TimeStamps
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 201172 13 43 48.718
[2,] 201172 13 43 54.281
[3,] 201172 13 43 59.843
[4,] 201172 13 44 5.390
[5,] 201172 13 44 10.859
[6,] 201172 13 44 16.375
[7,] 201172 13 44 21.890
[8,] 201172 13 44 27.390
[9,] 201172 13 44 33.015
[10,] 201172 13 44 38.531
[11,] 201172 13 44 44.078
[12,] 201172 13 44 49.546
[13,] 201172 13 44 55.078
[14,] 201172 13 45 0.718
[15,] 201172 13 45 6.281
[16,] 201172 13 45 11.953
[17,] 201172 13 45 17.453
[18,] 201172 13 45 22.984
I would like to write a function that will have inputs like that:
function(data, TimeStamps, timeBegin, timeEnd) {(not fixed though)
and will return the index of start and the end.
I need your help specify how the input arguments should look like (something
simple and compatible with the format I have already should be good).
Then based on that two arguments, how I can search for start and end of
timestamps inside the MeasurementSet$Timestamps and return the indexes of start
and end of the time block?
Hi Alex,
I think what you are trying to do is this:
TimeStamps-matrix(
c(2011,7,2,13,43,48.718,
2011,7,2,13,43,54.281,
2011,7,2,13,43,59.843,
2011,7,2,13,44,5.390,
2011,7,2,13,44,10.859,
2011,7,2,13,44,16.375,
2011,7,2,13,44,21.890,
2011,7,2,13,44,27.390,
2011,7,2,13,44,33.015,
2011,7,2,13,44,38.531,
2011,7,2,13,44,44.078,
2011,7,2,13,44,49.546,
2011,7,2,13,44,55.078,
2011,7,2,13,45,0.718,
2011,7,2,13,45,6.281,
2011,7,2,13,45,11.953,
2011,7,2,13,45,17.453,
2011,7,2,13,45,22.984),
ncol=6,byrow=TRUE)
findBeginEnd-function(x,timeBegin,timeEnd) {
bits2date-function(x) {
the_date-strptime(paste(x,c(-,-, ,:,:,),
sep=,collapse=),format=%Y-%m-%d %H:%M:%S)
return(the_date)
}
dimx-dim(x)
timeBegin-strptime(timeBegin,format=%Y-%m-%d %H:%M:%S)
timeEnd-strptime(timeEnd,format=%Y-%m-%d %H:%M:%S)
start_index-1
nextdate-bits2date(x[1,])
while(nextdate timeBegin start_index dimx[1]) {
start_index-start_index + 1
nextdate-bits2date(x[start_index,])
}
end_index-start_index
while(timeEnd nextdate end_index dimx[1]) {
end_index-end_index + 1
nextdate-bits2date(x[end_index,])
}
return(list(start=start_index,end=end_index))
}
findBeginEnd(TimeStamps,2011-7-2 13:44:20.0,2011-7-2 13:45:12.0)
Jim
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[R] multicore combn
This is a 'rather than re-invent the wheel' post. Has anyone out there re-written combn so that it can be parallelized - with multicore, snow, or otherwise? I have a job that requires large numbers of combinations, and rather than get all of the index values, then crank it through mclapply, I was wondering if there was a way to just do this natively within a function. Just curious. Thanks! -Jarrett -- View this message in context: http://r.789695.n4.nabble.com/multicore-combn-tp3908633p3908633.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] grouped lattice plot with overall regression line
I'd like to draw a lattice plot with groups. The groups (the grouping
condition, fns) are successfully marked with separate symbols, using the
following code:
xyplot(T~A|speaker,groups=fns,pch=1:3,key=list(space=right,points=list(pch=1:3)),type=c(g,p,r))
Here's a hard part. This draws regression lines for each group in each panel
(there are three groups, so three regression lines show up in each panel). But
I want to have only one global regression line for all groups together, still
maintaining separate symbols for each group.
I don't want to have individual regression lines for each group.
I tried many things, with various panel functions, but I've never succeeded.
Could any expert help with this? I'd appreciate it greatly. The closest I could
find was something like the following, which didn't work for me - anyways, it
is supposed to draw by-group regression lines as well as the global one. So
this is not exactly I need.
xyplot(T~A|speaker,data=spk0,layout=c(4,5),type='p',groups=fns),
panel=function(x,y){
panel.superpose(x,y)
panel.abline(lm(y~x))}),
panel.groups=function(x,y,lty){
panel.xyplot(x,y,lty=lty)
panel.abline(lm(y~x),lty=3)})
Thanks so much !!!
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[R] nlrq {quantreg}
Dear all,
I sent an email on Friday asking about nlrq {quantreg}, but I haven't received
any answer.
I need to estimate the quantile regression estimators of a model as: y =
exp(b0+x'b1+u). The model is nonlinear in parameters, although I can linearise
it by using log.When I write:
fitnl - nlrq(y ~ exp(x), tau=0.5)
I have the following error: Error in match.call(func, call = cll) : invalid
'definition' argument
Is there any way to estimate this model, or should I accept the following
change:
fitnl - rq(log(y) ~ x, tau=0.5) ?
Thanks in advance!
Best,
Julia
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Re: [R] write.csv naming file after function argument
Thank you for your help. It works now.
2011/10/13 Jean V Adams jvad...@usgs.gov
Kristian Lind wrote on 10/13/2011 04:52:16 AM:
Dear R-users,
I'm writing a program that constructs a dataset. I wish to save the
dataset
to a file.
Here's a very simple example of what I'm trying to do
function(x=peter){
y - x/2
write.csv(y, file = ...\x)
}
The problem is that I want to name the dataset as whatever the name of
the
input is. In this case peter.
How do I do this?
Thank you in advance.
Kristian
I think you're looking for something like this
foo - function(x){
y - x/2
file.name - paste(...\\, deparse(substitute(x)), .csv,
sep=)
# I include the print() functions just so you can see what happened
print(y)
print(file.name)
write.csv(y, file=file.name)
}
peter - 12
foo(peter)
Jean
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Re: [R] function for handling time
You are too good! Thanks a lot !
Have a nice weekend
B.R
Alexs
From: Jim Lemon j...@bitwrit.com.au
Cc: R-help@r-project.org R-help@r-project.org
Sent: Sunday, October 16, 2011 9:10 AM
Subject: Re: [R] function for handling time
On 10/16/2011 04:13 AM, Alaios wrote:
Dear all
I have the following time stamps (in the following format)
MeasurementSet$TimeStamps
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 2011 7 2 13 43 48.718
[2,] 2011 7 2 13 43 54.281
[3,] 2011 7 2 13 43 59.843
[4,] 2011 7 2 13 44 5.390
[5,] 2011 7 2 13 44 10.859
[6,] 2011 7 2 13 44 16.375
[7,] 2011 7 2 13 44 21.890
[8,] 2011 7 2 13 44 27.390
[9,] 2011 7 2 13 44 33.015
[10,] 2011 7 2 13 44 38.531
[11,] 2011 7 2 13 44 44.078
[12,] 2011 7 2 13 44 49.546
[13,] 2011 7 2 13 44 55.078
[14,] 2011 7 2 13 45 0.718
[15,] 2011 7 2 13 45 6.281
[16,] 2011 7 2 13 45 11.953
[17,] 2011 7 2 13 45 17.453
[18,] 2011 7 2 13 45 22.984
I would like to write a function that will have inputs like that:
function(data, TimeStamps, timeBegin, timeEnd) {(not fixed though)
and will return the index of start and the end.
I need your help specify how the input arguments should look like (something
simple and compatible with the format I have already should be good).
Then based on that two arguments, how I can search for start and end of
timestamps inside the MeasurementSet$Timestamps and return the indexes of
start and end of the time block?
Hi Alex,
I think what you are trying to do is this:
TimeStamps-matrix(
c(2011,7,2,13,43,48.718,
2011,7,2,13,43,54.281,
2011,7,2,13,43,59.843,
2011,7,2,13,44,5.390,
2011,7,2,13,44,10.859,
2011,7,2,13,44,16.375,
2011,7,2,13,44,21.890,
2011,7,2,13,44,27.390,
2011,7,2,13,44,33.015,
2011,7,2,13,44,38.531,
2011,7,2,13,44,44.078,
2011,7,2,13,44,49.546,
2011,7,2,13,44,55.078,
2011,7,2,13,45,0.718,
2011,7,2,13,45,6.281,
2011,7,2,13,45,11.953,
2011,7,2,13,45,17.453,
2011,7,2,13,45,22.984),
ncol=6,byrow=TRUE)
findBeginEnd-function(x,timeBegin,timeEnd) {
bits2date-function(x) {
the_date-strptime(paste(x,c(-,-, ,:,:,),
sep=,collapse=),format=%Y-%m-%d %H:%M:%S)
return(the_date)
}
dimx-dim(x)
timeBegin-strptime(timeBegin,format=%Y-%m-%d %H:%M:%S)
timeEnd-strptime(timeEnd,format=%Y-%m-%d %H:%M:%S)
start_index-1
nextdate-bits2date(x[1,])
while(nextdate timeBegin start_index dimx[1]) {
start_index-start_index + 1
nextdate-bits2date(x[start_index,])
}
end_index-start_index
while(timeEnd nextdate end_index dimx[1]) {
end_index-end_index + 1
nextdate-bits2date(x[end_index,])
}
return(list(start=start_index,end=end_index))
}
findBeginEnd(TimeStamps,2011-7-2 13:44:20.0,2011-7-2 13:45:12.0)
Jim
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and provide commented, minimal, self-contained, reproducible code.
[R] Custom Sort Character and Numeric
Im trying to do a custom sort in this order:
1) Numeric digit furthest right;
2) Alphabetical second furthest to the right;
3) Alphabetical the rest of the string beginning with the first character;
The example code I'm using is an array that follows:
/myArray - c('AFP9','AFR9','TLQP7','AFS9','AFR8','AFP8','AFS7','TLQS8')/
The output I desire is:
/myArray
[1] AFS7 AFP8 AFR8 AFP9 AFR9 AFS9 TLQP7 TLQS8 /
What I'm thinking is writing a function that will order it by analyzing it
from right to left. Ideally there would be a way to look at the individual
strings like the formula in Excel =RIGHT(cell, 1) and drop the furthest
right then do the same thing to the next character. I've been looking into
custom sort for R and haven't found much. Any idea what this function would
look like? Possibly a while loop? Each string would have a length of at
least 3, possibly longer. Thank you in advance.
--
View this message in context:
http://r.789695.n4.nabble.com/Custom-Sort-Character-and-Numeric-tp3909058p3909058.html
Sent from the R help mailing list archive at Nabble.com.
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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.
[R] two-sided p-value?
Dear all, I am a little bit confused because of the returned p-value by summary.lm and also summary.rq I thought if the pvalue is = 0.05 the difference is significant. But the R help says it is a two-sided pvalue. So does that mean the pvalue has to be = 0.025 and = 0.975? Best, Laura -- View this message in context: http://r.789695.n4.nabble.com/two-sided-p-value-tp3909277p3909277.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] Drop ALL Levels of a Data Frame Object
No worries for the brevity. That worked exactly like I wanted. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Drop-ALL-Levels-of-a-Data-Frame-Object-tp3903788p3909062.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] two-sided p-value?
On 16.10.2011 13:08, Laura wrote: Dear all, I am a little bit confused because of the returned p-value by summary.lm and also summary.rq I thought if the pvalue is= 0.05 the difference is significant. But the R help says it is a two-sided pvalue. So does that mean the pvalue has to be = 0.025 and= 0.975? And at the same time. ;-) Actually, the hypothesis is a two sided one, the p-value should still be smaller than your predefined alpha that may or may not be 0.05. If that is not yet clear, you should really ask for local statistical advice or start reading yourself some statistics textbooks. Best, Uwe Ligges Best, Laura -- View this message in context: http://r.789695.n4.nabble.com/two-sided-p-value-tp3909277p3909277.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] Custom Sort Character and Numeric
Try this, but I get a different order especially based on the last digit
myArray - c('AFP9','AFR9','TLQP7','AFS9','AFR8','AFP8','AFS7','TLQS8')
# create a sort key
key - sub(^(.*)(.)(.)$, \\3\\2\\1, myArray)
key
[1] 9PAF 9RAF 7PTLQ 9SAF 8RAF 8PAF 7SAF 8STLQ
# sort, but don't get your output
myArray[order(key)]
[1] TLQP7 AFS7 AFP8 AFR8 TLQS8 AFP9 AFR9 AFS9
On Sun, Oct 16, 2011 at 4:47 AM, swonder03 ramey.ste...@gmail.com wrote:
Im trying to do a custom sort in this order:
1) Numeric digit furthest right;
2) Alphabetical second furthest to the right;
3) Alphabetical the rest of the string beginning with the first character;
The example code I'm using is an array that follows:
/myArray - c('AFP9','AFR9','TLQP7','AFS9','AFR8','AFP8','AFS7','TLQS8')/
The output I desire is:
/myArray
[1] AFS7 AFP8 AFR8 AFP9 AFR9 AFS9 TLQP7 TLQS8 /
What I'm thinking is writing a function that will order it by analyzing it
from right to left. Ideally there would be a way to look at the individual
strings like the formula in Excel =RIGHT(cell, 1) and drop the furthest
right then do the same thing to the next character. I've been looking into
custom sort for R and haven't found much. Any idea what this function would
look like? Possibly a while loop? Each string would have a length of at
least 3, possibly longer. Thank you in advance.
--
View this message in context:
http://r.789695.n4.nabble.com/Custom-Sort-Character-and-Numeric-tp3909058p3909058.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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Re: [R] Custom Sort Character and Numeric
Here is another solution that gets the order you posted:
myArray - c('AFP9','AFR9','TLQP7','AFS9','AFR8','AFP8','AFS7','TLQS8')
# create a sort key
key - sub(^(.*)(.)(.)$, \\3\\2\\1, myArray)
key
[1] 9PAF 9RAF 7PTLQ 9SAF 8RAF 8PAF 7SAF 8STLQ
# sort, but don't get your output
myArray[order(key)]
[1] TLQP7 AFS7 AFP8 AFR8 TLQS8 AFP9 AFR9 AFS9
# to get your output, new key
newKey - sub(^(.*)(.)(.)$, \\1\\3\\2, myArray)
newKey
[1] AF9P AF9R TLQ7P AF9S AF8R AF8P AF7S TLQ8S
myArray[order(newKey)]
[1] AFS7 AFP8 AFR8 AFP9 AFR9 AFS9 TLQP7 TLQS8
On Sun, Oct 16, 2011 at 4:47 AM, swonder03 ramey.ste...@gmail.com wrote:
Im trying to do a custom sort in this order:
1) Numeric digit furthest right;
2) Alphabetical second furthest to the right;
3) Alphabetical the rest of the string beginning with the first character;
The example code I'm using is an array that follows:
/myArray - c('AFP9','AFR9','TLQP7','AFS9','AFR8','AFP8','AFS7','TLQS8')/
The output I desire is:
/myArray
[1] AFS7 AFP8 AFR8 AFP9 AFR9 AFS9 TLQP7 TLQS8 /
What I'm thinking is writing a function that will order it by analyzing it
from right to left. Ideally there would be a way to look at the individual
strings like the formula in Excel =RIGHT(cell, 1) and drop the furthest
right then do the same thing to the next character. I've been looking into
custom sort for R and haven't found much. Any idea what this function would
look like? Possibly a while loop? Each string would have a length of at
least 3, possibly longer. Thank you in advance.
--
View this message in context:
http://r.789695.n4.nabble.com/Custom-Sort-Character-and-Numeric-tp3909058p3909058.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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
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] Use of ICA for sound
On 16.10.2011 07:44, Noah Silverman wrote: Hi, I'm looking at the cocktail party classic problem. I can see how to use ICA to separate the components. But, How do I then create new wav files of the separated sounds so that they can be played? Others suggested tuneR already for a former question. You can make a Wave object from the signal returned by ICA processing using tuneR and also write a wav filke with tuneR: See ?Wave and ?writeWave Best, Uwe Ligges Thanks -- Noah Silverman UCLA Department of Statistics 8208 Math Sciences Building Los Angeles, CA 90095 [[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] multicore combn
Just thought I'd let you know the following: In the gRbase package there is a function called combnPrim which does the same as combn but it is implemented in C - and is quite a bit faster than combn(). Regards Søren Fra: r-help-boun...@r-project.org [r-help-boun...@r-project.org] P#229; vegne af jebyrnes [byr...@msi.ucsb.edu] Sendt: 16. oktober 2011 03:25 Til: r-help@r-project.org Emne: [R] multicore combn This is a 'rather than re-invent the wheel' post. Has anyone out there re-written combn so that it can be parallelized - with multicore, snow, or otherwise? I have a job that requires large numbers of combinations, and rather than get all of the index values, then crank it through mclapply, I was wondering if there was a way to just do this natively within a function. Just curious. Thanks! -Jarrett -- View this message in context: http://r.789695.n4.nabble.com/multicore-combn-tp3908633p3908633.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] Editor for RD file?
Dear all, can somebody please update me on what could be the best editor to write and edit the RD files? I need to something with syntax highlighter, auto-completion etc (like Notepad++ for R etc.). Currently I am using plain Notepad however expect something which could be more professional. Thanks for your help. [[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] Use of ICA for sound
On Oct 16, 2011, at 1:44 AM, Noah Silverman wrote: Hi, I'm looking at the cocktail party classic problem. I can see how to use ICA to separate the components. But, How do I then create new wav files of the separated sounds so that they can be played? FWIW you can play sounds directly without creating any files using play() from the audio package. Cheers, Simon Thanks -- Noah Silverman UCLA Department of Statistics 8208 Math Sciences Building Los Angeles, CA 90095 [[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] grouped lattice plot with overall regression line
If you want to draw the global regression line in each panel, you can try
xyplot(T~A|speaker,data=spk0,layout=c(4,5),type='p',groups=fns),
panel=function(x,y,...){
panel.xyplot(x,y,...)
panel.abline(lm(y~x,data=spk0))})
Weidong Gu
2011/10/16 조혜선 feb...@naver.com:
I'd like to draw a lattice plot with groups. The groups (the grouping
condition, fns) are successfully marked with separate symbols, using the
following code:
xyplot(T~A|speaker,groups=fns,pch=1:3,key=list(space=right,points=list(pch=1:3)),type=c(g,p,r))
Here's a hard part. This draws regression lines for each group in each panel
(there are three groups, so three regression lines show up in each panel).
But I want to have only one global regression line for all groups together,
still maintaining separate symbols for each group.
I don't want to have individual regression lines for each group.
I tried many things, with various panel functions, but I've never succeeded.
Could any expert help with this? I'd appreciate it greatly. The closest I
could find was something like the following, which didn't work for me -
anyways, it is supposed to draw by-group regression lines as well as the
global one. So this is not exactly I need.
xyplot(T~A|speaker,data=spk0,layout=c(4,5),type='p',groups=fns),
panel=function(x,y){
panel.superpose(x,y)
panel.abline(lm(y~x))}),
panel.groups=function(x,y,lty){
panel.xyplot(x,y,lty=lty)
panel.abline(lm(y~x),lty=3)})
Thanks so much !!!
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[R] question: ragged array
Hello,
I have a big problem which Im just not able to solve.
I created the following mean value from the following dataset structure:
Id |value
1 | 2
1 | 3
1 | 4
2 | 2
2 | 1
3 | 5
4 | 3
etc.|etc.
with the command:
mean_rating - tapply(ratok$value, ratok$project_id , mean,simplify = FALSE)
this gives me a ragged array:
mean_rating [1]
$`14` ###==project_id
[1] 3.93 ###==mean value
dim (mean_rating)
[1] 2214
I want to separate now the project_id from the mean value. So that I have
two separate columns (2 dimension).
How can I separate a ragged array into two variables?
*Additional information:*
I need this information to put the mean value into another dataset
(projok) at the correct project_id.
I would do that as follow:
k=projok[,1]
for(i in 1:2442) {
projLineNb = which(mean_rating$project_id==k[i])
projok$mean[i] = mean_rating$value[projLineNb]
}
But with a ragged array I can not refer to project_id. Or is there a
possibility that I can refer to the project_id in a ragged array?
mean_rating [[1]]
[1] 3.93 ==I can refer to the value only
Thank you very much
Best,
Helene
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Re: [R] grouped lattice plot with overall regression line
As usual, try reading the Help files!
?panel.lmline
-- Bert
On Sun, Oct 16, 2011 at 6:50 AM, Weidong Gu anopheles...@gmail.com wrote:
If you want to draw the global regression line in each panel, you can try
xyplot(T~A|speaker,data=spk0,layout=c(4,5),type='p',groups=fns),
panel=function(x,y,...){
panel.xyplot(x,y,...)
panel.abline(lm(y~x,data=spk0))})
Weidong Gu
2011/10/16 ì¡°íì feb...@naver.com:
I'd like to draw a lattice plot with groups. The groups (the grouping
condition, fns) are successfully marked with separate symbols, using the
following code:
xyplot(T~A|speaker,groups=fns,pch=1:3,key=list(space=right,points=list(pch=1:3)),type=c(g,p,r))
Here's a hard part. This draws regression lines for each group in each
panel (there are three groups, so three regression lines show up in each
panel). But I want to have only one global regression line for all groups
together, still maintaining separate symbols for each group.
I don't want to have individual regression lines for each group.
I tried many things, with various panel functions, but I've never
succeeded. Could any expert help with this? I'd appreciate it greatly. The
closest I could find was something like the following, which didn't work for
me - anyways, it is supposed to draw by-group regression lines as well as
the global one. So this is not exactly I need.
xyplot(T~A|speaker,data=spk0,layout=c(4,5),type='p',groups=fns),
panel=function(x,y){
panel.superpose(x,y)
panel.abline(lm(y~x))}),
panel.groups=function(x,y,lty){
panel.xyplot(x,y,lty=lty)
panel.abline(lm(y~x),lty=3)})
Thanks so much !!!
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--
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Internal Contact Info:
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Re: [R] Editor for RD file?
On 11-10-16 8:57 AM, Bogaso Christofer wrote: Dear all, can somebody please update me on what could be the best editor to write and edit the RD files? I need to something with syntax highlighter, auto-completion etc (like Notepad++ for R etc.). Currently I am using plain Notepad however expect something which could be more professional. I believe ESS does syntax highlighting for Rd files if you use Emacs. You could probably modify a LaTeX syntax highlighter for any other editor to work with Rd files: the syntax is quite similar. Duncan Murdoch __ 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] question: ragged array
if you set parameter simplify=TRUE, it returns a vector of the ragged
mean. In your case,
mean_rating - tapply(ratok$value, ratok$project_id , mean,simplify = TRUE)
df-data.frame(ID=dimnames(mean_rating)[[1]], mean=mean_rating)
Weidong Gu
On Sun, Oct 16, 2011 at 9:53 AM, Helene Schreyer
helene.schre...@gmail.com wrote:
Hello,
I have a big problem which I’m just not able to solve.
I created the following mean value from the following dataset structure:
Id |value
1 | 2
1 | 3
1 | 4
2 | 2
2 | 1
3 | 5
4 | 3
etc.|etc.
with the command:
mean_rating - tapply(ratok$value, ratok$project_id , mean,simplify = FALSE)
this gives me a ragged array:
mean_rating [1]
$`14` ###==project_id
[1] 3.93 ###==mean value
dim (mean_rating)
[1] 2214
I want to separate now the project_id from the mean value. So that I have
two separate columns (2 dimension).
How can I separate a ragged array into two variables?
*Additional information:*
I need this information to put the mean value into another dataset
(“projok”) at the correct project_id.
I would do that as follow:
k=projok[,1]
for(i in 1:2442) {
projLineNb = which(mean_rating$project_id==k[i])
projok$mean[i] = mean_rating$value[projLineNb]
}
But with a ragged array I can not refer to project_id. Or is there a
possibility that I can refer to the project_id in a ragged array?
mean_rating [[1]]
[1] 3.93 ==I can refer to the value only
Thank you very much
Best,
Helene
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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[R] background normalization in rma() in the affy package
Hi,
i was looking into the documentation for the rma() function in affy()
package, and was trying to figure out how exactly the background
normalization is done. I read all three papers cited in the rma()
documentation, but the most detailed explanation i could find was in Irizary
et al., 2003, where they state that they compute
B(PM_{ijn}) = E[s_{ijn} | PM_{ijn}]
where s_{ijn} is assumed to be exponential, and bg_{ijn} is normal.
I still don't understand what value is being computed here, neither am i
clear on what the correction looks like. i.e. if s_{ijn} is an
exponentially-distributed random variable, how is bg_{ijn} fit into this?
thanks!
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Re: [R] Help in kmeans
Hi All, For executing kmeans for Iris, we found that there were 2 different ways. dataFrame - read.csv(c:/Iris.csv,header=T) 1. kmeans_model-kmeans(dataFrame[1:5],size=3) *This gave an error as it had Species which is a String column as one of the inputs* 2.attach(dataFrame) kmeans_model-kmeans(cbind(SepalLength,SepalWidth,PetalLength,PetalWidth,Species),3) * But this command worked and gave output.* Does this mean that kmeans can accept String inputs also? Can you please let me know how the second command works? Thanks in advance. Regards, Raji -- View this message in context: http://r.789695.n4.nabble.com/Help-in-kmeans-tp3430433p3909552.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] glmmadmb help
chchjames james.mccarthy at windowslive.com writes: Thanks for the reply Ben. I tried it with verbose=TRUE, and got about 7 pages of a word doc as an output, that ended with the error Error in glmmadmb(stainp ~ beetle.ev + Caged * Section/SegmentT + (1 | : The function maximizer failed. I am not sure how I would best go about posting this, as you put it, voluminous output, but I am happy to post/send that and a similar data-set somewhere. How would I best go about this? Don't bother with the output, please send me the data and any accompanying R code (you can find my e-mail easily). Ben Bolker __ 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] ecdf
Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ 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] ecdf
On Oct 16, 2011, at 11:31 AM, gj wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? ecdf() returns a function rather than a vector. You need to supply arguments to that function to get something that you recognize. Had you passed that function off to plot you would have seen that the information needed to calculate the plot is obviously in there. If you go to the stepfun page you find that the knots function can recover some of htat information for display. plot( ecdf(fs$numstudents) ) knots( ecdf(fs$numstudents) ) [1] 8 10 13 23 25 28 34 140 209 If you count the knots you can deduce the quantile values (the y- values) at which those x-values will start the step dot-line -- David Winsemius, MD West Hartford, CT __ 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] nlrq {quantreg}
The model _is_ linear in parameters, after the log transformation of
the response, so
you don't need nlrq. If you really want something like:
y = exp(a + b x) + u
then you need to make a token effort to look at the documentation.
Here is another
example:
x - exp(rnorm(50))
y - exp(1 + .5*x) + rnorm(50)
nlrq(y ~ exp(a + b * x), start = list(a = 2, b = 1))
Nonlinear quantile regression
model: y ~ exp(a + b * x)
data: parent.frame
tau: 0.5
deviance: 15.39633
a b
1.0348673 0.4962638
Roger Koenker
rkoen...@illinois.edu
On Oct 16, 2011, at 3:59 AM, Julia Lira wrote:
Dear all,
I sent an email on Friday asking about nlrq {quantreg}, but I
haven't received any answer.
I need to estimate the quantile regression estimators of a model as:
y = exp(b0+x'b1+u). The model is nonlinear in parameters, although I
can linearise it by using log.When I write:
fitnl - nlrq(y ~ exp(x), tau=0.5)
I have the following error: Error in match.call(func, call = cll) :
invalid 'definition' argument
Is there any way to estimate this model, or should I accept the
following change:
fitnl - rq(log(y) ~ x, tau=0.5) ?
Thanks in advance!
Best,
Julia
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[R] Which function to use: grep, replace, substr etc.?
Hello, I have a simple question but I don't know which method is best to use for my problem. I have the following strings: str1 - My_name_is_peter str2 - what_is_your_surname_peter I would like to apply predefined abbreviations for peter=p and name=n to both strings so that the new strings look like the followings: str1: My_n_is_p str2: what_is_your_surn_p Which method is the best to use for that particular problem? syrvn -- View this message in context: http://r.789695.n4.nabble.com/Which-function-to-use-grep-replace-substr-etc-tp3909871p3909871.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] Which function to use: grep, replace, substr etc.?
On Oct 16, 2011, at 12:35 PM, syrvn wrote: Hello, I have a simple question but I don't know which method is best to use for my problem. I have the following strings: str1 - My_name_is_peter str2 - what_is_your_surname_peter I would like to apply predefined abbreviations for peter=p and name=n to both strings so that the new strings look like the followings: str1: My_n_is_p str2: what_is_your_surn_p Which method is the best to use for that particular problem? ?sub # on same page as grep sub((p)eter, \\1, vec) [1] My_name_is_p what_is_your_surname_p -- David Winsemius, MD West Hartford, CT __ 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] need help in pausing a script
Hey, I'm new to R. I wrote a script for doing several statistic tests and plot. is there any way to add a kind of pause function which halts script execution until a key is pressed. Please help fast if you can -- View this message in context: http://r.789695.n4.nabble.com/need-help-in-pausing-a-script-tp3909815p3909815.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] Which function to use: grep, replace, substr etc.?
Hi,
thanks for the tip! I do it as follows now but I still have a problem I do
not understand:
abbrvs - data.frame(c(peter, name, male, female),
c(P, N, m, f))
colnames(abbrvs) - c(pattern, replacement)
str - My name is peter and I am male
for(m in 1:nrow(abbrvs)) {
str - sub(abbrvs$pattern[m], abbrvs$replacement[m], str,
fixed=TRUE)
print(str)
}
This works perfectly fine as I get: My N is P and I am m
However, when I replace male by female then I get the following: My N is P
and I am fem
but I want to have My N is P and I am f.
Even with the parameter fixed=true I get the same result. Why is that?
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Re: [R] Which function to use: grep, replace, substr etc.?
Note that male comes before female in your data frame.
---
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syrvn ment...@gmx.net wrote:
Hi,
thanks for the tip! I do it as follows now but I still have a problem I do
not understand:
abbrvs - data.frame(c(peter, name, male, female),
c(P, N, m, f))
colnames(abbrvs) - c(pattern, replacement)
str - My name is peter and I am male
for(m in 1:nrow(abbrvs)) {
str - sub(abbrvs$pattern[m], abbrvs$replacement[m], str,
fixed=TRUE)
print(str)
}
This works perfectly fine as I get: My N is P and I am m
However, when I replace male by female then I get the following: My N is P
and I am fem
but I want to have My N is P and I am f.
Even with the parameter fixed=true I get the same result. Why is that?
--
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Re: [R] right justify right-axis tick values in lattice
Perfect. Thank you David. Since I almost always want right-axis numeric ticks to be right justified, I will include this function as part of the next version of the HH package (any day now), listing you as author. Would you consider sending this as a proposed patch to lattice? As a patch it might be necessary to make the justification direction an argument, rather than a hard-wired change. Rich On Sun, Oct 16, 2011 at 9:50 AM, David Winsemius dwinsem...@comcast.netwrote: On Oct 16, 2011, at 1:17 AM, Richard M. Heiberger wrote: [[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] need help in pausing a script
Perhaps something like this (stolen from the demo() code): readline(\nType Return\t to start : ) # If you don't want the auto-print and are running it interactively, a call to invisible() might help. Michael Weylandt On Sun, Oct 16, 2011 at 12:08 PM, M3Mph15 d.niem...@onlinehome.de wrote: Hey, I'm new to R. I wrote a script for doing several statistic tests and plot. is there any way to add a kind of pause function which halts script execution until a key is pressed. Please help fast if you can -- View this message in context: http://r.789695.n4.nabble.com/need-help-in-pausing-a-script-tp3909815p3909815.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] need help in pausing a script
In addition to Michael's suggestion, if what you want to pause is the creation of graphs, set: par(ask = TRUE) see ?par for details. It makes it so that user input is required between each graph plotting. Cheers, Josh On Sun, Oct 16, 2011 at 9:08 AM, M3Mph15 d.niem...@onlinehome.de wrote: Hey, I'm new to R. I wrote a script for doing several statistic tests and plot. is there any way to add a kind of pause function which halts script execution until a key is pressed. Please help fast if you can Generally speaking, telling volunteers to respond fast is rather rude (also may hurt your chances of a response because it makes academics suspicious that it is for an assignment or homework due soon). At least some justification (e.g., I work at an animal rescue and we just saved 3000 baby whales, but we do not have facilities for all 3000 of them, have little funds and need to find the most cost effective way to get them to special whale vets all around the worldalso 35 need medical attention quickly, so I need to get this script working in the next 6 hours so they have time to make it to a treatment facility and receive life-saving care.) might make some of us feel better about prioritizing a response to you. -- View this message in context: http://r.789695.n4.nabble.com/need-help-in-pausing-a-script-tp3909815p3909815.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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] need help in pausing a script
On Sun, 16-Oct-2011 at 09:08AM -0700, M3Mph15 wrote:
| Hey, I'm new to R. I wrote a script for doing several statistic tests and
| plot. is there any way to add a kind of pause function which halts script
| execution until a key is pressed. Please help fast if you can
?browser
|
| --
| View this message in context:
http://r.789695.n4.nabble.com/need-help-in-pausing-a-script-tp3909815p3909815.html
| Sent from the R help mailing list archive at Nabble.com.
|
| __
| 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.
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{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in kmeans
Hi, I suspect your column Species is of class factor (as it is in R's built in iris dataset). This means that in your case Species is an integer vector with the additional information of the levels names. kmeans is internally calling as.matrix(), which creates a character matrix of your dataframe, because one column is factor and you get an error. After binding the columns with cbind, the result is an integer matrix with the Species columns as the internal levels (1,2 and 3 instead of setosa versicolor virginica ) and kmeans is not throwing a error any more. Furthermore kmeans wouldn't work in the first case, because there is no size= - argument in kmeans. You probably meant centers=3. For additional information try ?kmeans Christoph 2011/10/16 Raji raji.sanka...@gmail.com Hi All, For executing kmeans for Iris, we found that there were 2 different ways. dataFrame - read.csv(c:/Iris.csv,header=T) 1. kmeans_model-kmeans(dataFrame[1:5],size=3) *This gave an error as it had Species which is a String column as one of the inputs* 2.attach(dataFrame) kmeans_model-kmeans(cbind(SepalLength,SepalWidth,PetalLength,PetalWidth,Species),3) * But this command worked and gave output.* Does this mean that kmeans can accept String inputs also? Can you please let me know how the second command works? Thanks in advance. Regards, Raji -- View this message in context: http://r.789695.n4.nabble.com/Help-in-kmeans-tp3430433p3909552.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] Help in kmeans
Hi, Thank you .. The information was very helpful. Yes.It was meant to be centers=3.Even with that , kmeans gives error if we give the index of Species columns. So, *is it ok to use kmeans for String data by using cbind*.But, kmeans*works even if we give a column which contains distinct String values *. For example,a column which contains names like country names.How does this work in such cases? Is it expected behavior? Country --- England Germany China Thanks, Raji On Mon, Oct 17, 2011 at 1:02 AM, Christoph Molnar christoph.mol...@googlemail.com wrote: Hi, I suspect your column Species is of class factor (as it is in R's built in iris dataset). This means that in your case Species is an integer vector with the additional information of the levels names. kmeans is internally calling as.matrix(), which creates a character matrix of your dataframe, because one column is factor and you get an error. After binding the columns with cbind, the result is an integer matrix with the Species columns as the internal levels (1,2 and 3 instead of setosa versicolor virginica ) and kmeans is not throwing a error any more. Furthermore kmeans wouldn't work in the first case, because there is no size= - argument in kmeans. You probably meant centers=3. For additional information try ?kmeans Christoph 2011/10/16 Raji raji.sanka...@gmail.com Hi All, For executing kmeans for Iris, we found that there were 2 different ways. dataFrame - read.csv(c:/Iris.csv,header=T) 1. kmeans_model-kmeans(dataFrame[1:5],size=3) *This gave an error as it had Species which is a String column as one of the inputs* 2.attach(dataFrame) kmeans_model-kmeans(cbind(SepalLength,SepalWidth,PetalLength,PetalWidth,Species),3) * But this command worked and gave output.* Does this mean that kmeans can accept String inputs also? Can you please let me know how the second command works? Thanks in advance. Regards, Raji -- View this message in context: http://r.789695.n4.nabble.com/Help-in-kmeans-tp3430433p3909552.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] question: ragged array
Hi:
Try this:
ratok - data.frame(Id = rep(1:3, 3:1), value = c(2, 3, 4, 2, 1, 5))
aggregate(value ~ Id, data = ratok, FUN = mean)
Id value
1 1 3.0
2 2 1.5
3 3 5.0
aggregate() returns a data frame with the Id variable and mean(value).
HTH,
Dennis
On Sun, Oct 16, 2011 at 6:53 AM, Helene Schreyer
helene.schre...@gmail.com wrote:
Hello,
I have a big problem which I’m just not able to solve.
I created the following mean value from the following dataset structure:
Id |value
1 | 2
1 | 3
1 | 4
2 | 2
2 | 1
3 | 5
4 | 3
etc.|etc.
with the command:
mean_rating - tapply(ratok$value, ratok$project_id , mean,simplify = FALSE)
this gives me a ragged array:
mean_rating [1]
$`14` ###==project_id
[1] 3.93 ###==mean value
dim (mean_rating)
[1] 2214
I want to separate now the project_id from the mean value. So that I have
two separate columns (2 dimension).
How can I separate a ragged array into two variables?
*Additional information:*
I need this information to put the mean value into another dataset
(“projok”) at the correct project_id.
I would do that as follow:
k=projok[,1]
for(i in 1:2442) {
projLineNb = which(mean_rating$project_id==k[i])
projok$mean[i] = mean_rating$value[projLineNb]
}
But with a ragged array I can not refer to project_id. Or is there a
possibility that I can refer to the project_id in a ragged array?
mean_rating [[1]]
[1] 3.93 ==I can refer to the value only
Thank you very much
Best,
Helene
[[alternative HTML version deleted]]
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Re: [R] ecdf
Hi: I don't understand what you're attempting to do. Wouldn't courseid be a categorical variable with a numeric label? If that is so, why are you trying to compute an EDF? An EDF computes cumulative relative frequency of a random variable, which by definition is numeric. If we were talking about EDFs for a distribution of student course grades on a numeric point system by course, that would make some sense, but I don't see how the course IDs themselves qualify as being on an interval scale of measurement. Could you clarify your intent? Dennis On Sun, Oct 16, 2011 at 8:31 AM, gj gaw...@gmail.com wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ 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] Editor for RD file?
On Sun, Oct 16, 2011 at 7:20 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 11-10-16 8:57 AM, Bogaso Christofer wrote: Dear all, can somebody please update me on what could be the best editor to write and edit the RD files? I need to something with syntax highlighter, auto-completion etc (like Notepad++ for R etc.). Currently I am using plain Notepad however expect something which could be more professional. I believe ESS does syntax highlighting for Rd files if you use Emacs. Yes, it does. It is also nice because it runs R, so if I am writing examples in the Rd files, I can run them directly. Here is the little blurb regarding the major mode for Rd files: (Rd-mode) Major mode for editing R documentation source files. This mode makes it easier to write R documentation by helping with indentation, doing some of the typing for you (with Abbrev mode) and by showing keywords, strings, etc. in different faces (with Font Lock mode on terminals that support it). Cheers, Josh You could probably modify a LaTeX syntax highlighter for any other editor to work with Rd files: the syntax is quite similar. Duncan Murdoch __ 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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] Help in kmeans
Hi, no, don't use kmeans with factors. The kmeans algorithm does, besides other things, calculate the mean of the k clusters. But you don't get a useful mean from factors, because the internally used integers are arbitrary. In this case its 1,2 and 3. But it could be 42, 7 and 10 as well, which would change any calculation of a mean. Thats why the kmeans() function wants numeric matrices. Maybe you should think about how kmeans works: http://en.wikipedia.org/wiki/K-means_clustering Christoph 2011/10/16 raji sankaran raji.sanka...@gmail.com Hi, Thank you .. The information was very helpful. Yes.It was meant to be centers=3.Even with that , kmeans gives error if we give the index of Species columns. So, *is it ok to use kmeans for String data by using cbind*.But, kmeans*works even if we give a column which contains distinct String values *. For example,a column which contains names like country names.How does this work in such cases? Is it expected behavior? Country --- England Germany China Thanks, Raji On Mon, Oct 17, 2011 at 1:02 AM, Christoph Molnar christoph.mol...@googlemail.com wrote: Hi, I suspect your column Species is of class factor (as it is in R's built in iris dataset). This means that in your case Species is an integer vector with the additional information of the levels names. kmeans is internally calling as.matrix(), which creates a character matrix of your dataframe, because one column is factor and you get an error. After binding the columns with cbind, the result is an integer matrix with the Species columns as the internal levels (1,2 and 3 instead of setosa versicolor virginica ) and kmeans is not throwing a error any more. Furthermore kmeans wouldn't work in the first case, because there is no size= - argument in kmeans. You probably meant centers=3. For additional information try ?kmeans Christoph 2011/10/16 Raji raji.sanka...@gmail.com Hi All, For executing kmeans for Iris, we found that there were 2 different ways. dataFrame - read.csv(c:/Iris.csv,header=T) 1. kmeans_model-kmeans(dataFrame[1:5],size=3) *This gave an error as it had Species which is a String column as one of the inputs* 2.attach(dataFrame) kmeans_model-kmeans(cbind(SepalLength,SepalWidth,PetalLength,PetalWidth,Species),3) * But this command worked and gave output.* Does this mean that kmeans can accept String inputs also? Can you please let me know how the second command works? Thanks in advance. Regards, Raji -- View this message in context: http://r.789695.n4.nabble.com/Help-in-kmeans-tp3430433p3909552.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.
[R] Suppressing the Intercept in lm() when using a dataframe for the model
It's easy to run a linear regression on a simple model without an intercept just by doing this: lm(y ~ x1 + x2 -1) Is there a similar trick to suppress the intercept when your model is in a large dataframe and you don't want to write out the names of individual columns? -- View this message in context: http://r.789695.n4.nabble.com/Suppressing-the-Intercept-in-lm-when-using-a-dataframe-for-the-model-tp3910327p3910327.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] position of first and last axis tick
Hi, how can i set the position of the first and last tick to the borderline of a plot? The plot should look like this one made in Gnuplot [1]. Gnu-R adds some space between the ticks and the end of plot. [1] http://commons.wikimedia.org/wiki/File:Atmospheric_radiocarbon_1954-1993.svg kind regards, -- Jonas Stein n...@jonasstein.de __ 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] Suppressing the Intercept in lm() when using a dataframe for the model
On Sun, Oct 16, 2011 at 12:55 PM, Cliff Clive cliffcl...@gmail.com wrote: It's easy to run a linear regression on a simple model without an intercept just by doing this: lm(y ~ x1 + x2 -1) Is there a similar trick to suppress the intercept when your model is in a large dataframe and you don't want to write out the names of individual columns? Yep...same trick. -- View this message in context: http://r.789695.n4.nabble.com/Suppressing-the-Intercept-in-lm-when-using-a-dataframe-for-the-model-tp3910327p3910327.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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] Which function to use: grep, replace, substr etc.?
On Oct 16, 2011, at 1:32 PM, Jeff Newmiller wrote:
Note that male comes before female in your data frame.
---
Jeff Newmiller The . . Go Live...
syrvn ment...@gmx.net wrote:
Hi,
thanks for the tip! I do it as follows now but I still have a
problem I do
not understand:
abbrvs - data.frame(c(peter, name, male, female),
c(P, N, m, f))
colnames(abbrvs) - c(pattern, replacement)
str - My name is peter and I am male
for(m in 1:nrow(abbrvs)) {
str - sub(abbrvs$pattern[m], abbrvs$replacement[m], str,
fixed=TRUE)
print(str)
}
This works perfectly fine as I get: My N is P and I am m
However, when I replace male by female then I get the following: My
N is P
and I am fem
but I want to have My N is P and I am f.
Even with the parameter fixed=true I get the same result. Why is that?
Because male is in female? This reminds me of a comment on a
posting I made this morning on SO.
http://stackoverflow.com/questions/7782113/counting-keyword-occurrences-in-r
The problem was slightly different, but the greppish principle was
that in order to match only complete words, you need to specific ^,
$ or at each end of the word:
dataset - c(corn, cornmeal, corn on the cob, meal)
grep(^corn$|^corn | corn$, dataset)
[1] 1 3
In such cases you may want to look at the gsubfn package. It offers
higher level matching functions and I think strapply might be more
efficient and expressive here. I can imagine construction in a loop
such as yours, but you would probably want to build a pattern outside
the sub() call.
After struggling to fix your loop (and your data.frame which
definitely should not be using factor variables), I am even more
convinced you should be learning gubfn facilities. (Tate out the
debugging print statements.)
abbrvs - data.frame(c(peter, name, male, female),
+c( P , N , m , f ), stringsAsFactors=FALSE)
colnames(abbrvs) - c(pattern, replacement)
for(m in 1:nrow(abbrvs)) { patt - paste(^,abbrvs$pattern[m], $|
,
+ abbrvs$pattern[m], | ,
+ abbrvs$pattern[m], $, sep=)
+ print(c( patt, abbrvs$replacement[m]))
+ str - sub(patt, abbrvs$replacement[m], str)
+ print(str)
+ }
[1] ^peter$| peter | peter$ P
[1] My name is P and I am female
[1] ^name$| name | name$ N
[1] My N is P and I am female
[1] ^male$| male | male$ m
[1] My N is P and I am female
[1] ^female$| female | female$ f
[1] My N is P and I am f
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] ecdf
On Oct 16, 2011, at 3:53 PM, Dennis Murphy wrote: Hi: I don't understand what you're attempting to do. Wouldn't courseid be a categorical variable with a numeric label? If that is so, why are you trying to compute an EDF? An EDF computes cumulative relative frequency of a random variable, which by definition is numeric. If we were talking about EDFs for a distribution of student course grades on a numeric point system by course, that would make some sense, but I don't see how the course IDs themselves qualify as being on an interval scale of measurement. Could you clarify your intent? Huh? gawesh asked for ecdf on numstrudents (not courseid) ... pretty clearly a numeric value for which an ECDF should make sense. -- David. -- Dennis On Sun, Oct 16, 2011 at 8:31 AM, gj gaw...@gmail.com wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] position of first and last axis tick
On Oct 16, 2011, at 3:08 PM, Jonas Stein wrote: Hi, how can i set the position of the first and last tick to the borderline of a plot? The plot should look like this one made in Gnuplot [1]. Gnu-R adds some space between the ticks and the end of plot. do you mean like this? plot(rnorm(25),rnorm(25), xaxs =i, yaxs=i, xlim=c(-2,2), ylim=c(-2,2)) [1] http://commons.wikimedia.org/wiki/File:Atmospheric_radiocarbon_1954-1993.svg kind regards, -- Jonas Stein n...@jonasstein.de __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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] Suppressing the Intercept in lm() when using a dataframe for the model
On Oct 16, 2011, at 3:55 PM, Cliff Clive wrote: It's easy to run a linear regression on a simple model without an intercept just by doing this: lm(y ~ x1 + x2 -1) Is there a similar trick to suppress the intercept when your model is in a large dataframe and you don't want to write out the names of individual columns? Something along the lines of: lm(y ~ . -1, data=dfrm) But you do need to offer specific examples to get specific answers. (Hence the usual advice to read the Posting Guide.) -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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 plot CI's (llim ulim) on ecodist mgram
I would like to put confidence intervals on a mantel corellogram they are already calculated in the pmgram object but I am unsure how I get the x value in order to plot them? package(ecodist) X-1:100 Y-rnorm(1:100) Z-rnorm(1:100) XY-dist(data.frame(X,Y)) YX-dist(data.frame(Y,X)) my.mgram-mgram(XY,XZ) plot(my.mgram) print(my.mgram) print(my.mgram) $mgram lag ngroup mantelr pvalllim ulim [1,] 3.770055672 0.500012737 0.001 0.49689923 0.504301550 [2,] 11.310165691 0.383960457 0.001 0.38000201 0.387324434 [3,] 18.850274584 0.232086251 0.001 0.22670074 0.237501735 [4,] 26.390384587 0.114097397 0.001 0.10243901 0.122973735 [5,] 33.930494463 -0.003113351 0.835 -0.01928101 0.008839295 [6,] 41.470603468 -0.106354446 0.001 -0.12682280 -0.089539628 [7,] 49.010713357 -0.181250278 0.001 -0.20154017 -0.164863572 [8,] 56.550823348 -0.266397615 0.001 -0.28498271 -0.251134864 [9,] 64.090933252 -0.298705798 0.001 -0.31421396 -0.284154643 [10,] 71.631042228 -0.353134525 0.001 -0.36468910 -0.341422330 [11,] 79.171152147 -0.337181781 0.001 -0.34854961 -0.322161075 [12,] 86.711262108 -0.334465576 0.001 -0.35500933 -0.309763543 [13,] 94.251371 43 -0.238965642 0.001 -0.26437371 -0.196038662 $resids [1] NA attr(,class) [1] mgram Thanks Nevil Amos __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot CI's (llim ulim) on ecodist mgram
Hi, The x value you want is lag from my.mgram$mgram You can use lines() to add them to the plot. Sarah On Sun, Oct 16, 2011 at 7:10 PM, Nevil Amos nevil.a...@gmail.com wrote: I would like to put confidence intervals on a mantel corellogram they are already calculated in the pmgram object but I am unsure how I get the x value in order to plot them? package(ecodist) X-1:100 Y-rnorm(1:100) Z-rnorm(1:100) XY-dist(data.frame(X,Y)) YX-dist(data.frame(Y,X)) my.mgram-mgram(XY,XZ) plot(my.mgram) print(my.mgram) print(my.mgram) $mgram lag ngroup mantelr pval llim ulim [1,] 3.770055 672 0.500012737 0.001 0.49689923 0.504301550 [2,] 11.310165 691 0.383960457 0.001 0.38000201 0.387324434 [3,] 18.850274 584 0.232086251 0.001 0.22670074 0.237501735 [4,] 26.390384 587 0.114097397 0.001 0.10243901 0.122973735 [5,] 33.930494 463 -0.003113351 0.835 -0.01928101 0.008839295 [6,] 41.470603 468 -0.106354446 0.001 -0.12682280 -0.089539628 [7,] 49.010713 357 -0.181250278 0.001 -0.20154017 -0.164863572 [8,] 56.550823 348 -0.266397615 0.001 -0.28498271 -0.251134864 [9,] 64.090933 252 -0.298705798 0.001 -0.31421396 -0.284154643 [10,] 71.631042 228 -0.353134525 0.001 -0.36468910 -0.341422330 [11,] 79.171152 147 -0.337181781 0.001 -0.34854961 -0.322161075 [12,] 86.711262 108 -0.334465576 0.001 -0.35500933 -0.309763543 [13,] 94.251371 43 -0.238965642 0.001 -0.26437371 -0.196038662 $resids [1] NA attr(,class) [1] mgram Thanks Nevil Amos -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] right justify right-axis tick values in lattice
Hi
On 16/10/2011 6:17 p.m., Richard M. Heiberger wrote:
How can I right justify the right-axis tick values? They appear in the
example below as left-justified.
I have tried several different ways and all fail in different ways.
The example below creates the right axis tick value with no attempt at
adjustment.
alternates I have tried are
1. formatting the values. This doesn't work because they are in a
proportional font and the blanks
are too narrow.
2. gsub all leading characters into twocharacters. This
overcompenates because a blank
is slightly wider than half a digit width.
I prefer to keep the default font. I am willing to go to a fixed width font
(courier for example), but I haven't
figured out the incantation to make that work in graphics.
here is my example:
panel.right- function(x, y, ...) {
panel.barchart(x, y, ...)
print(x);print(y)
panel.axis(side=right, at=pretty(y), outside=TRUE)
}
mybar- function(...) {
args- list(...)
args$par.settings=list(clip=list(panel=off))
args$par.settings$layout.widths$axis.key.padding- 4
do.call(barchart, args)
}
mybar(c(1,10,100,10,1) ~ abcd, panel=panel.right, ylab.right=right)
You could do this ...
library(grid)
oldx - grid.get(plot_01.ticklabels.right.panel.1.1)$x
grid.edit(plot_01.ticklabels.right.panel.1.1,
just=right,
x=oldx + stringWidth(000))
Paul
thanks
Rich
[[alternative HTML version deleted]]
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--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
p...@stat.auckland.ac.nz
http://www.stat.auckland.ac.nz/~paul/
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Re: [R] glmmadmb help
[cc'ed back to r-help] I've started to take a look, and there's nothing immediately obvious about the problem with the fit (the warnings and errors are about a non-positive-definite Hessian, which usually means an overfitted/poorly identified model) -- still working on whether there's a way to get more useful information. As it turns out, glmmADMB's default behavior in this case is to just stop with an error. I could re-write things to allow it to tell you something (but it wouldn't be able to compute standard errors on the parameters). It's conceivable that you might get slightly different results (different enough that it would work in one case and fail in another) on different operating systems, different machines, etc. ... because the failure is really a case of something that's numerically on the edge, where negative values in a vector are illegal but the minimum might be -1e-7 in one case and 1e-7 in another -- i.e., different but just by numeric fuzz. My first response to this sort of problem would be to see if you can simplify the model slightly -- e.g. are both random effects really necessary, etc.. Several things do suggest themselves from looking at the data: * your response variable ranges from 0 to 1, which doesn't make much sense for a negative binomial response distribution. What does your response variable represent? If it is overdispersed *binomial* data, then (1) you need to use something like a beta-binomial distribution instead (unfortunately not yet implemented in glmmADMB, but if you were desperate I might give it a whack) (2) you need to specify the denominators -- just the proportions won't do it. If you have just proportions, then you might be looking for a beta distributed response (although that has its own trickiness in dealing with response values of exactly 0 or 1). * your Site variable has only two levels. That makes dealing with it as a random variable extremely questionable (from a *philosophical* or experimental-design point of view it may be sensible to treat it as a random variable, but not practical: see http://glmm.wikidot.com/faq for discussion of this point). When I made it a fixed effect instead, I got (apparently) sensible results. * The plot of the data (see attached, I hope the attachment gets through) makes it clear that you have some potential balance problems. In section/segment combinations (C-D) x (3-5) you have only a few (often only one) observation in the (beetle.ev=1) case. Mild lack of balance is no problem in mixed models, but such severe lack of balance can be. (Note this doesn't guarantee a problem, but it is one of the factors that makes estimation less stable.) On 11-10-16 05:41 PM, James McCarthy wrote: Hello Ben, Many thanks for offering to help with this. I have attached the data set “stain.csv”. I have also attached a script file with all of the code that should (hopefully) work for you. What I would like to get working is the function: nbin1-glmmadmb(stainp~beetle.ev+Caged*Section/Segment+(1|Site)+(1|Log.code),data=dat1,family=nbinom) The issue is with the “beetle.ev” column (binary, presence/absence of insects). Below is some code that works, all that has changed is that “beetle.ev” is substituted with “Test”. This one works, and has some data in pretty much the same format as “beetle.ev” (as you can see in the .csv file). nbin2-glmmadmb(stainp~Test+Caged*Section/Segment+(1|Site)+(1|Log.code),data=dat1,family=nbinom) For the life of me, I can’t figure out what’s wrong with this column. Thanks again for talking a look at it. Cheers, James James McCarthy MSc (Ecology) candidate, University of Canterbury c/o Scion (New Zealand Forest Research Institute) Forestry Rd, P.O. Box 29-237, Christchurch, New Zealand Ph: (03) 364 2987 Ext. 7820 Cell: 027 268 5506 Fax: (03) 364 2812 Email: james.mccar...@scionresearch.com or james.mccar...@pg.canterbury.ac.nz www.scionresearch.com This email may be confidential and subject to legal privilege, it may not reflect the views of the University of Canterbury, and it is not guaranteed to be virus free. If you are not an intended recipient, please notify the sender immediately and erase all copies of the message and any attachments. Please refer to http://www.canterbury.ac.nz/emaildisclaimer for more information. ## install.packages(glmmADMB,repos=http://glmmadmb.r-forge.r-project.org/repos,type=source;) library(glmmADMB) dat1-read.csv(stain.csv) ## attach(dat1) ## dangerous names(dat1) dat1-transform(dat1, Site=factor(Site), ## BMB: fixed typo Log.code=factor(Log.code), Caged=factor(Caged), beetle.ev=factor(beetle.ev), Test=factor(Test)) ## BMB: this could also be done via colClasses= argument to read.csv nbin1-glmmadmb(stainp~beetle.ev+Caged*Section/Segment+(1|Site)+(1|Log.code),data=dat1,family=nbinom,
[R] What does \Sexpr[results=rd]{} exactly mean in Rd?
Hi,
I have spent a few hours on the R-exts manual and the documentation of
parse_Rd() (as well as the PDF document in the references), but I
still have not figured out what results=rd means. I thought I could
use an R code fragment to create an Rd fragment dynamically. Here is
an example, in which I was expected the output to be a describe list
DL in HTML, but it turns out not to be true.
(I was actually building a package with Rd's containing \Sexpr{}
instead of really using Rd2HTML(); the content was not rendered after
I run R CMD build.)
des - \\describe{\\item{def}{ghi}}
con - textConnection(c(\\title{abc}\\name{abc},
\\details{\\Sexpr[results=rd,stage=build]{des}}))
z - parse_Rd(con)
Rd2HTML(z, stages = build)
close(con)
!DOCTYPE html PUBLIC -//W3C//DTD HTML 4.01 Transitional//EN
htmlheadtitleR: abc/title
meta http-equiv=Content-Type content=text/html; charset=utf-8
link rel=stylesheet type=text/css href=R.css
/headbody
table width=100% summary=page for abctrtdabc/tdtd
align=rightR Documentation/td/tr/table
h2abc/h2
h3Details/h3
pdefghi/p
/body/html
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] tools stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] devtools_0.4
loaded via a namespace (and not attached):
[1] RCurl_1.6-10
Thanks!
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
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Re: [R] ecdf
David is right. I am looking for the ecfd for fs$numstudents. The other column is just an id. I guess I don't know how to read the R documentation when it comes to functions. looking at the documentation, i now notice that it says Compute an empirical cummulative distribution function and not a vector. But still I would had assumed that in ecdf(x) ... the x is the argument. So ecdf(fs$numstudents)(unique(fs$numstudents)) === == function arguments Yes? But I can't read that from the documentation? I suspect it has something to those dots in the arguments which I don't understand. Why it says usage ecdf(x) when it's clearly not the case? I don't get it. Gawesh On Sun, Oct 16, 2011 at 11:02 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 16, 2011, at 3:53 PM, Dennis Murphy wrote: Hi: I don't understand what you're attempting to do. Wouldn't courseid be a categorical variable with a numeric label? If that is so, why are you trying to compute an EDF? An EDF computes cumulative relative frequency of a random variable, which by definition is numeric. If we were talking about EDFs for a distribution of student course grades on a numeric point system by course, that would make some sense, but I don't see how the course IDs themselves qualify as being on an interval scale of measurement. Could you clarify your intent? Huh? gawesh asked for ecdf on numstrudents (not courseid) ... pretty clearly a numeric value for which an ECDF should make sense. -- David. -- Dennis On Sun, Oct 16, 2011 at 8:31 AM, gj gaw...@gmail.com wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ecdf
Hi, On Sun, Oct 16, 2011 at 8:48 PM, gj gaw...@gmail.com wrote: David is right. I am looking for the ecfd for fs$numstudents. The other column is just an id. I guess I don't know how to read the R documentation when it comes to functions. looking at the documentation, i now notice that it says Compute an empirical cumulative distribution function and not a vector. But still I would had assumed that in ecdf(x) ... the x is the argument. ecdf() is the function you're calling. x is your vector, for which you want the ECDF. num.ecdf - ecdf(fs$numstudents) There. That's the ECDF. But the ECDF is a *function* - that's what the F stands for, after all. If you're looking for the percentiles for your data, you might try: num.ecdf(fs$numstudents) You might also try working the examples given in ?ecdf yourself, so that you can see exactly what's going on before you try it with your own data. So ecdf(fs$numstudents)(unique(fs$numstudents)) === == function arguments Yes? But I can't read that from the documentation? I suspect it has something to those dots in the arguments which I don't understand. Yes. That's the condensed version of what I just proposed, done in one step, instead of two. The two-step version is definitely in the help. It doesn't have anything to do with the ..., which simply allow for other arguments to be passed. Why it says usage ecdf(x) when it's clearly not the case? I don't get it. Clearly that is the case. ecdf(x) returns the empirical cumulative distribution *function* of the vector of data x. I'm not entirely sure what you think you should be getting. Perhaps if you explained your expectations, the list would be able to help you achieve them. Sarah Gawesh On Sun, Oct 16, 2011 at 11:02 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 16, 2011, at 3:53 PM, Dennis Murphy wrote: Hi: I don't understand what you're attempting to do. Wouldn't courseid be a categorical variable with a numeric label? If that is so, why are you trying to compute an EDF? An EDF computes cumulative relative frequency of a random variable, which by definition is numeric. If we were talking about EDFs for a distribution of student course grades on a numeric point system by course, that would make some sense, but I don't see how the course IDs themselves qualify as being on an interval scale of measurement. Could you clarify your intent? Huh? gawesh asked for ecdf on numstrudents (not courseid) ... pretty clearly a numeric value for which an ECDF should make sense. -- David. -- Dennis On Sun, Oct 16, 2011 at 8:31 AM, gj gaw...@gmail.com wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ 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. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com 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,
Re: [R] ecdf
Thanks for the clarification. I stand corrected. Dennis On Sun, Oct 16, 2011 at 5:48 PM, gj gaw...@gmail.com wrote: David is right. I am looking for the ecfd for fs$numstudents. The other column is just an id. I guess I don't know how to read the R documentation when it comes to functions. looking at the documentation, i now notice that it says Compute an empirical cummulative distribution function and not a vector. But still I would had assumed that in ecdf(x) ... the x is the argument. So ecdf(fs$numstudents)(unique(fs$numstudents)) === == function arguments Yes? But I can't read that from the documentation? I suspect it has something to those dots in the arguments which I don't understand. Why it says usage ecdf(x) when it's clearly not the case? I don't get it. Gawesh On Sun, Oct 16, 2011 at 11:02 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 16, 2011, at 3:53 PM, Dennis Murphy wrote: Hi: I don't understand what you're attempting to do. Wouldn't courseid be a categorical variable with a numeric label? If that is so, why are you trying to compute an EDF? An EDF computes cumulative relative frequency of a random variable, which by definition is numeric. If we were talking about EDFs for a distribution of student course grades on a numeric point system by course, that would make some sense, but I don't see how the course IDs themselves qualify as being on an interval scale of measurement. Could you clarify your intent? Huh? gawesh asked for ecdf on numstrudents (not courseid) ... pretty clearly a numeric value for which an ECDF should make sense. -- David. -- Dennis On Sun, Oct 16, 2011 at 8:31 AM, gj gaw...@gmail.com wrote: Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367 34 I load my data using: fs-read.csv(file=C:\\num_students_inallmodules.csv,header=T, sep=',') I want to get the ecdf. So, I looked at the ?ecdf which says usage:ecdf(x) So I expected ecdf(fs$numstudents) to work Instead it just returned: Call: ecdf(fs$numstudents) x[1:210] = 1, 2, 3, ..., 3717, 4538 After Googling, got this to work: ecdf(fs$numstudents)(unique(fs$numstudents)) But I don't understand why if the ?ecdf says usage is ecdf(x) ... I need to use ecdf(fs$numstudents)(unique(fs$numstudents)) to get this to work? Can somebody explain this to me? Regards Gawesh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Which function to use: grep, replace, substr etc.?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Winsemius
Sent: Sunday, October 16, 2011 1:59 PM
To: Jeff Newmiller
Cc: r-help@r-project.org; syrvn
Subject: Re: [R] Which function to use: grep, replace, substr etc.?
On Oct 16, 2011, at 1:32 PM, Jeff Newmiller wrote:
Note that male comes before female in your data frame.
---
Jeff Newmiller The . . Go Live...
syrvn ment...@gmx.net wrote:
Hi,
thanks for the tip! I do it as follows now but I still have a
problem I do
not understand:
abbrvs - data.frame(c(peter, name, male, female),
c(P, N, m, f))
colnames(abbrvs) - c(pattern, replacement)
str - My name is peter and I am male
for(m in 1:nrow(abbrvs)) {
str - sub(abbrvs$pattern[m], abbrvs$replacement[m], str,
fixed=TRUE)
print(str)
}
This works perfectly fine as I get: My N is P and I am m
However, when I replace male by female then I get the following: My
N is P
and I am fem
but I want to have My N is P and I am f.
Even with the parameter fixed=true I get the same result. Why is that?
Because male is in female? This reminds me of a comment on a
posting I made this morning on SO.
http://stackoverflow.com/questions/7782113/counting-keyword-occurrences-in-r
The problem was slightly different, but the greppish principle was
that in order to match only complete words, you need to specific ^,
$ or at each end of the word:
dataset - c(corn, cornmeal, corn on the cob, meal)
grep(^corn$|^corn | corn$, dataset)
[1] 1 3
You can use the 2 character sequences \\ and \\ to match
the beginning and end of a word (where the match takes up zero
characters):
dataset - c(corn, cornmeal, corn on the cob, popcorn, this corn
is sweet)
grep(^corn$|^corn | corn$, dataset)
[1] 1 3
grep(\\corn\\, dataset)
[1] 1 3 5
gsub(\\corn\\, CORN, dataset)
[1] CORN
[2] cornmeal
[3] CORN on the cob
[4] popcorn
[5] this CORN is sweet
If your definition of a word is more expansive it gets complicated.
E.g., if words might include letters, numbers, and periods but not
underscores or anything else, you could use:
gsub((^|[^.[:alpha:][:digit:]])?corn($|[^.[:alpha:][:digit:]])?,
\\1CORN.BY.ITSELF\\2,
c(corn.1, corn_2, corn, 4corn, 1.corn))
[1] corn.1
[2] CORN.BY.ITSELF_2
[3] CORN.BY.ITSELF
[4] 4corn
[5] 1.corn
Moving to perl regular expressions would probably make this simpler.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
In such cases you may want to look at the gsubfn package. It offers
higher level matching functions and I think strapply might be more
efficient and expressive here. I can imagine construction in a loop
such as yours, but you would probably want to build a pattern outside
the sub() call.
After struggling to fix your loop (and your data.frame which
definitely should not be using factor variables), I am even more
convinced you should be learning gubfn facilities. (Tate out the
debugging print statements.)
abbrvs - data.frame(c(peter, name, male, female),
+ c( P , N , m , f ), stringsAsFactors=FALSE)
colnames(abbrvs) - c(pattern, replacement)
for(m in 1:nrow(abbrvs)) { patt - paste(^,abbrvs$pattern[m], $|
,
+ abbrvs$pattern[m], | ,
+ abbrvs$pattern[m], $, sep=)
+ print(c( patt, abbrvs$replacement[m]))
+ str - sub(patt, abbrvs$replacement[m], str)
+ print(str)
+ }
[1] ^peter$| peter | peter$ P
[1] My name is P and I am female
[1] ^name$| name | name$ N
[1] My N is P and I am female
[1] ^male$| male | male$ m
[1] My N is P and I am female
[1] ^female$| female | female$ f
[1] My N is P and I am f
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] Beginner's question about ptrend. What package to use in R and a general explanation of the statistics.
Hello I'm wanting to understand more about ptrend (a statistical explanation via an internet website if possible) and also to know what package in R would produce a ptrend. Appreciate any help I can get on this as I'm trying to read and understand an epidemiological paper and data and reproduce similar results for my own understanding. Thanks Meredith [[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] Suppressing the Intercept in lm() when using a dataframe for the model
Well don't I feel silly now. Thanks for the help! -- View this message in context: http://r.789695.n4.nabble.com/Suppressing-the-Intercept-in-lm-when-using-a-dataframe-for-the-model-tp3910327p3910443.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] mgp and axis title positions
Hi all, I consider myself a somewhat experienced user of R, but have struggled with this for a while now. to the point where I just end up pulling the entire graph together in powerpoint and fixing it up from there. How does one adjust the horizontal/vertical positions of axis titles? I've tried using mgp in the par function, but that never produces anything satisfactory as x and y-axes are modified simultaneously. It would appear that you can specify mgp in the axis function, but that doesn't seem to work at all. It is particularly frustrating when you need to adjust axis title positions separately for x and y axes - i.e. specify mgp twice - once for each axis within the axis function. But this doesn't seem to work - is this a bug? Cheers, Ben [[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] simultaneously maximizing two independent log likelihood functions using mle2
Hello,
I have a log likelihood function that I was able to optimize using
mle2. I have two years of the data used to fit the function and I would
like to fit both years simultaneously to test if the model parameter
estimates differ between years, using likelihood ratio tests and AIC.
Can anyone give advice on how to do this?
My likelihood functions are long so I'll use the tadpole predation
example from Ben Bolker's book, Ecological Data and Models in R (p.
268-270).
library(emdbook)
data(ReedfrogFuncresp)
attach(ReedfrogFuncresp)
# Holling Type II Equation
holling2.pred = function(N0, a, h, P, T) {
a * N0 * P * T/(1 + a * h * N0)
}
# Negative log likelihood function
NLL.holling2 = function(a, h, P = 1, T = 1) {
-sum(dbinom(Killed, prob = a * T * P/(1 + a * h * Initial),
size = Initial, log = TRUE))
}
# MLE statement
FFR.holling2 = mle2(NLL.holling2, start = list(a = 0.012,
h = 0.84), data = list(T = 14, P = 3))
I have my negative log likelihood function setup similarly to the above
example. Again, my goal is to simultaneously estimate parameters from
the same function for two years, such that I can test if the parameters
from the two years are different. Perhaps an important difference from
the above example is that I am using a multinomial distribution (dmnom)
because my data are trinomially distributed.
Any help would be greatly appreciated.
Adam Zeilinger
--
Adam Zeilinger
Ph. D Candidate
Conservation Biology Program
University of Minnesota
Saint Paul, MN
www.linkedin.com/in/adamzeilinger
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Re: [R] Opening Screen
On 17/10/11 12:12, Matt Curcio wrote:
Greetings All,
What is the procedure to make the open screen for R silent. I would
to have my opening screen in Ubuntu 10.04 linux open to an empty
terminal. Instead of the list of licenses and version of R that is
being run. By the way, I am using RStudio as well. I have entered
the following lines into my '/home/user/.Rprofile' but this is more of
a 'cheat' ;) to me.
.First- function(){
cat(rep(\n,10))
}
If you are starting from the command line, type
R -q # q for quiet.
On my (Ubuntu) system ``man R'' tells me this.
cheers,
Rolf Turner
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Re: [R] How to plot CI's (llim ulim) on ecodist mgram
It is good to provide the code but please make sure it is reproducible? e.g. XZ is not defined in mgram(XY,XZ)? Weidong Gu On Sun, Oct 16, 2011 at 7:10 PM, Nevil Amos nevil.a...@gmail.com wrote: I would like to put confidence intervals on a mantel corellogram they are already calculated in the pmgram object but I am unsure how I get the x value in order to plot them? package(ecodist) X-1:100 Y-rnorm(1:100) Z-rnorm(1:100) XY-dist(data.frame(X,Y)) YX-dist(data.frame(Y,X)) my.mgram-mgram(XY,XZ) plot(my.mgram) print(my.mgram) print(my.mgram) $mgram lag ngroup mantelr pval llim ulim [1,] 3.770055 672 0.500012737 0.001 0.49689923 0.504301550 [2,] 11.310165 691 0.383960457 0.001 0.38000201 0.387324434 [3,] 18.850274 584 0.232086251 0.001 0.22670074 0.237501735 [4,] 26.390384 587 0.114097397 0.001 0.10243901 0.122973735 [5,] 33.930494 463 -0.003113351 0.835 -0.01928101 0.008839295 [6,] 41.470603 468 -0.106354446 0.001 -0.12682280 -0.089539628 [7,] 49.010713 357 -0.181250278 0.001 -0.20154017 -0.164863572 [8,] 56.550823 348 -0.266397615 0.001 -0.28498271 -0.251134864 [9,] 64.090933 252 -0.298705798 0.001 -0.31421396 -0.284154643 [10,] 71.631042 228 -0.353134525 0.001 -0.36468910 -0.341422330 [11,] 79.171152 147 -0.337181781 0.001 -0.34854961 -0.322161075 [12,] 86.711262 108 -0.334465576 0.001 -0.35500933 -0.309763543 [13,] 94.251371 43 -0.238965642 0.001 -0.26437371 -0.196038662 $resids [1] NA attr(,class) [1] mgram Thanks Nevil Amos __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] right justify right-axis tick values in lattice
Hi,
You could also pad the text labels with phantom 0s,
ghostrighter - function(x, ...){
n - sapply(x, nchar)
nmax - max(n)
padaone - function(ii){
si - paste(rep(0, length= nmax - n[ii]), collapse=)
as.expression(bquote(phantom(.(si)) * .(x[ii]) ))
}
sapply(seq_along(x), padaone)
}
## ghostrighter(c(1, 23, 145))
panel.right- function(x, y, ...) {
panel.barchart(x, y, ...)
print(x);print(y)
panel.axis(side=right, at=pretty(y), lab=ghostrighter(pretty(y)),
outside=TRUE)
}
mybar- function(...) {
args- list(...)
args$par.settings=list(clip=list(panel=off))
args$par.settings$layout.widths$axis.key.padding- 4
do.call(barchart, args)
}
mybar(c(1,10,100,10,1) ~ abcd, panel=panel.right, ylab.right=right)
HTH,
baptiste
On 17 October 2011 13:12, Paul Murrell p.murr...@auckland.ac.nz wrote:
Hi
On 16/10/2011 6:17 p.m., Richard M. Heiberger wrote:
How can I right justify the right-axis tick values? They appear in the
example below as left-justified.
I have tried several different ways and all fail in different ways.
The example below creates the right axis tick value with no attempt at
adjustment.
alternates I have tried are
1. formatting the values. This doesn't work because they are in a
proportional font and the blanks
are too narrow.
2. gsub all leading characters into two characters. This
overcompenates because a blank
is slightly wider than half a digit width.
I prefer to keep the default font. I am willing to go to a fixed width
font
(courier for example), but I haven't
figured out the incantation to make that work in graphics.
here is my example:
panel.right- function(x, y, ...) {
panel.barchart(x, y, ...)
print(x);print(y)
panel.axis(side=right, at=pretty(y), outside=TRUE)
}
mybar- function(...) {
args- list(...)
args$par.settings=list(clip=list(panel=off))
args$par.settings$layout.widths$axis.key.padding- 4
do.call(barchart, args)
}
mybar(c(1,10,100,10,1) ~ abcd, panel=panel.right, ylab.right=right)
You could do this ...
library(grid)
oldx - grid.get(plot_01.ticklabels.right.panel.1.1)$x
grid.edit(plot_01.ticklabels.right.panel.1.1,
just=right,
x=oldx + stringWidth(000))
Paul
thanks
Rich
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--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
p...@stat.auckland.ac.nz
http://www.stat.auckland.ac.nz/~paul/
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[R] position of the end of a text file
Hi! I need to find the position of the last character (could be empty space) of a vector of text files (my.text.vector) that I have read into R. I tried Google, but all I have found needs a pattern. I tried the following as.integer(regexpr(??, my.text.vector) but it returns 1 or even the correct number, but not consistently. I understand that ?? is an operator, thus should not be used but itself. Any ideas? Thanks, Henri-Paul -- Henri-Paul Indiogine Curriculum Instruction Texas AM University TutorFind Learning Centre Email: hindiog...@gmail.com Skype: hindiogine Website: http://people.cehd.tamu.edu/~sindiogine __ 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] position of the end of a text file
I think I have it and my apologies for spamming the list. I should not work this late on Sunday :-) I think that it should be as.integer(regexpr($$, my.text.vector) 2011/10/16 Henri-Paul Indiogine hindiog...@gmail.com: I tried the following as.integer(regexpr(??, my.text.vector) -- Henri-Paul Indiogine Curriculum Instruction Texas AM University TutorFind Learning Centre Email: hindiog...@gmail.com Skype: hindiogine Website: http://people.cehd.tamu.edu/~sindiogine __ 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.
