Re: [R] Namibia becoming NA
Hi Suresh, I think you will need to use read.table() rather than the read.csv() wrapper for it. Try: input - read.table(file = padded.csv, sep = ,, header = TRUE, na.strings = NULL) HTH, Josh On Sat, Jul 17, 2010 at 10:47 PM, Suresh Singh singh@osu.edu wrote: I have a data file in which one of the columns is country code and NA is the code for Namibia. When I read the data file using read.csv, NA for Namibia is being treated as null or NA How can I prevent this from happening? I tried the following but it didn't work input - read.csv(padded.csv,header = TRUE,as.is = c(code2)) thanks, Suresh [[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. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.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 with a problem
Hi all,
zoo::rollmean() is a nice idea. But if I understand Mike correctly, he
wants 5 out of any 7 consecutive logicals to be TRUE, where these 5 do
not necessarily need to be consecutive themselves. (remaining open
question: could, e.g., the condition on c1 be TRUE for rows 1,2,3,4,5
and on c2 for rows 3,4,5,6,7, or would it need to be TRUE for the same
rows?). Then something like this would make sense:
any(rollmean(dat$c1=100,7)=5/7-.01 rollmean(dat$c2=8,7)=5/7-.01))
or
any(rollmean(dat$c1=100,7)=5/7-.01 dat$c2=8,7)=5/7-.01))
depending on the open question above.
The -.01 above may be necessary in light of FAQ 7.31.
HTH,
Stephan
Joshua Wiley schrieb:
Hi Michael,
The days in your example do not look continuous (at least from my
thinking), so you may have extra requirements in mind, but take a look
at this code. My general thought was first to turn each column into a
logical vector (c1 = 100 and c2 = 8). Taking advantage of the fact
that R treats TRUE as 1 and FALSE as 0, compute a rolling mean. If
(and only if) 5 consecutive values are TRUE, the mean will be 1. Next
I added the rolling means for each column, and then tested whether any
were 2 (i.e., 1 + 1).
Cheers,
Josh
###
#Load required package
library(zoo)
#Your data with ds converted to Date
#from dput()
dat -
structure(list(ds = structure(c(14702, 14729, 14730, 14731, 14732,
14733, 14734, 14735, 14736, 14737, 14738, 14739, 14740, 14741,
14742, 14743, 14744), class = Date), c1 = c(100L, 11141L, 3L,
7615L, 6910L, 5035L, 3007L, 4L, 8335L, 2897L, 6377L, 3177L, 7946L,
8705L, 9030L, 8682L, 8440L), c2 = c(0L, 15L, 16L, 14L, 17L, 3L,
15L, 14L, 17L, 13L, 17L, 17L, 15L, 0L, 16L, 16L, 1L)), .Names = c(ds,
c1, c2), row.names = c(NA, -17L), class = data.frame)
#Order by ds
dat - dat[order(dat$ds), ]
yourvar - 0
#Test that 5 consecutive values from c1 AND c2 meet requirements
if(any(
c(rollmean(dat$c1 = 100, 5) + rollmean(dat$c2 = 8, 5)) == 2)
) {yourvar - 1}
###
On Sat, Jul 17, 2010 at 2:38 PM, Michael Hess mlh...@med.umich.edu wrote:
Sorry for not being clear.
In the dataset there are around 100 or so days of data (in the case also rows
of data)
I need to make sure that the person meets that c1 is at least 100 AND c2 is at
least 8 for 5 of 7 continuous days.
I will play with what I have and see if I can find out how to do this.
Thanks for the help!
Michael
Stephan Kolassa 07/17/10 4:50 PM
Mike,
I am slightly unclear on what you want to do. Do you want to check rows
1 and 7 or 1 *to* 7? Should c1 be at least 100 for *any one* or *all*
rows you are looking at, and same for c2?
You can sort your data like this:
data - data[order(data$ds),]
Type ?order for help. But also do this for added enlightenment...:
library(fortunes)
fortune(dog)
Next, your analysis on the sorted data frame. As I said, I am not
entirely clear on what you are looking at, but the following may solve
your problem with choices 1 to 7 and any one above.
foo - 0
for ( ii in 1:(nrow(data)-8) ) {
if (any(data$c1[ii+seq(0,6)]=100) any(data$c2[ii+seq(0,6)]=8)) {
foo - 1
break
}
}
The variable foo should contain what you want it to. Look at ?any
(and, if this does not do what you want it to, at ?all) for further info.
No doubt this could be vectorized, but I think the loop is clear enough.
Good luck!
Stephan
Michael Hess schrieb:
Hello R users,
I am a researcher at the University of Michigan looking for a solution to an R
problem. I have loaded my data in from a mysql database and it looks like this
data
ds c1 c2
1 2010-04-03100 0
2 2010-04-30 11141 15
3 2010-05-01 3 16
4 2010-05-02 7615 14
5 2010-05-03 6910 17
6 2010-05-04 5035 3
7 2010-05-05 3007 15
8 2010-05-06 4 14
9 2010-05-07 8335 17
10 2010-05-08 2897 13
11 2010-05-09 6377 17
12 2010-05-10 3177 17
13 2010-05-11 7946 15
14 2010-05-12 8705 0
15 2010-05-13 9030 16
16 2010-05-14 8682 16
17 2010-05-15 8440 15
What I am trying to do is sort by ds, and take rows 1,7, see if c1 is at least
100 AND c2 is at least 8. If it is not, start with check rows 2,8 and if not
there 3,9until it loops over the entire file. If it finds a set that
matches, set a new variable equal to 1, if never finds a match, set it equal to
0.
I have done this in stata but on this project we are trying to use R. Is this
something that can be done in R, if so, could someone point me in the correct
direction.
Thanks,
Michael Hess
University of Michigan
Health System
**
Electronic Mail is not secure, may not be read every day, and should not be
used for urgent or sensitive issues
__
Re: [R] Namibia becoming NA
On 18-Jul-10 05:47:03, Suresh Singh wrote: I have a data file in which one of the columns is country code and NA is the code for Namibia. When I read the data file using read.csv, NA for Namibia is being treated as null or NA How can I prevent this from happening? I tried the following but it didn't work input - read.csv(padded.csv,header = TRUE,as.is = c(code2)) thanks, Suresh I suppose this was bound to happen, and in my view it represent a bit of a mess! With a test file temp.csv: Code,Country DE,Germany IT,Italy NA,Namibia FR,France X - read.csv(temp.csv) X Code Country # 1 DE Germany # 2 IT Italy # 3 NA Namibia # 4 FR France which(is.na(X)) # [1] 3 exactly as Suresh describes. It does not help to surround the NA in temp.csv with quotes: Code,Country DE,Germany IT,Italy NA,Namibia FR,France leads to exactly the same result. And I have tried every variation I can think of of as.is and colClasses, still with exactly the same result! Conclusion: If an entry in a data file is intended to become the character value NA, there seems to be no way of reading it in directly. This should not be so: it should be preventable! As a cure, assuming that no other value in the Country Code is actually missing (and so should be NA), then (with Suresh's naming) I would suggest, subsequent to reading in the file, something like the following. The complication is that the variable code2 is now a factor, and you cannot simply assign a character value NA to its NA value -- you will get an error message. Hence: ix - which(is.na(input$code2)) Y - as.character(input$code2) Y[ix] - NA input$code2) - factor(Y) The corresponding code for my test example is: ix - which(is.na(X$Code)) Y - as.character(X$Code) Y[ix] - NA X$Code - factor(Y) X # Code Country # 1 DE Germany # 2 IT Italy # 3 NA Namibia # 4 FR France which(is.na(X)) # integer(0) So that works. There ought to be an option in read.csv() and friends which suppresses the conversion of a string NA found in input into an NA value. Maybe there is -- but, if so, it is not visible in the documentation! Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 18-Jul-10 Time: 09:25:05 -- XFMail -- __ 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] Namibia becoming NA
G'day Ted, On Sun, 18 Jul 2010 09:25:09 +0100 (BST) (Ted Harding) ted.hard...@manchester.ac.uk wrote: On 18-Jul-10 05:47:03, Suresh Singh wrote: I have a data file in which one of the columns is country code and NA is the code for Namibia. When I read the data file using read.csv, NA for Namibia is being treated as null or NA How can I prevent this from happening? I tried the following but it didn't work input - read.csv(padded.csv,header = TRUE,as.is = c(code2)) thanks, Suresh I suppose this was bound to happen, and in my view it represent a bit of a mess! With a test file temp.csv: Code,Country DE,Germany IT,Italy NA,Namibia FR,France Thanks for providing an example. leads to exactly the same result. And I have tried every variation I can think of of as.is and colClasses, still with exactly the same result! Did you think of trying some variations of na.strings? ;-) IMO, the simplest way of coding missing values in CSV files is to have two consecutive commas; not some code (whether NA, 99, 999, -1, ...) between them. Conclusion: If an entry in a data file is intended to become the character value NA, there seems to be no way of reading it in directly. This should not be so: it should be preventable! It is, through simple use of the na.strings argument: R X - read.csv(temp.csv, na.strings=) R X Code Country 1 DE Germany 2 IT Italy 3 NA Namibia 4 FR France R which(is.na(X)) integer(0) HTH. Cheers, Berwin == Full address Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Maths and Stats (M019)+61 (8) 6488 3383 (self) The University of Western Australia FAX : +61 (8) 6488 1028 35 Stirling Highway Crawley WA 6009e-mail: ber...@maths.uwa.edu.au Australiahttp://www.maths.uwa.edu.au/~berwin __ 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] Continuing on with a loop when there's a failure
Oops, forgot to add the line:
library(Design)
#the lrm function is in the Design library
...but even when I load the Design library, the loop still doesn't work. It
stops after failing on the second run of the loop:
results
y1 y2 y3
[1,] 0.6976063 NA NA
(The third run through the loop should have succeeded and left numeric output
in
the matrix called results, but it did not).
To: Peter Konings peter.l.e.koni...@gmail.com
Cc: R Help r-help@r-project.org
Sent: Sun, July 18, 2010 12:25:42 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hello Peter,
I tried your suggestion, but I was still not able to get it to work. Would you
mind looking at my code again? Here's what I'm trying:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels donuts bagels 2 9
indv.7 bagels donuts bagels 8 5
indv.8 bagels donuts bagels 4 1
indv.9 donuts donuts donuts 3 3
indv.10 bagels donuts bagels 5 9
indv.11 bagels donuts bagels 9 10
indv.12 bagels donuts bagels 3 1
indv.13 donuts donuts donuts 7 10
indv.14 bagels donuts bagels 2 10
indv.15 bagels donuts bagels 9 6), header = TRUE)
results - matrix(nrow = 1, ncol = 3)
colnames(results) - c(y1, y2, y3)
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
if(class(mod.poly3) == 'try-error')
{results[1,i] - NA}
else {mod.poly3 - lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x)
results[1,i] - anova(mod.poly3)[1,3]
}
}
...and here's the output:
results
y1 y2 y3
[1,] NA NA NA
[[elided Yahoo spam]]
From: Peter Konings peter.l.e.koni...@gmail.com
Sent: Tue, July 13, 2010 5:45:17 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hi Josh,
Test the class of the resulting object. If it is 'try-error' fill your result
with NA or do some other error handling.
result - try(somemodel)
if(class(result) == 'try-error')
{
# some error handling
} else {
# whatever happens if the result is ok
}
HTH
Peter.
In my opinion the try and tryCatch commands are written and documented rather
poorly. Thus I am not sure what to program exactly.
For instance, I could query mod.poly3 and use an if/then statement to proceed,
but querying mod.poly3 is weird. For instance, here's the output when it fails:
mod.poly3 - try(lrm(x[,2] ~ pol(x1, 3) + pol(x2, 3), data=x))
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
mod.poly3
[1] Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights
=
weights, : \n NA/NaN/Inf in foreign function call (arg 1)\n
attr(,class)
[1] try-error
...and here's the output when it succeeds:
mod.poly3 - try(lrm(x[,1] ~ pol(x1, 3) + pol(x2, 3), data=x))
mod.poly3
Logistic Regression Model
lrm(formula = x[, 1] ~ pol(x1, 3) + pol(x2, 3), data = x)
Frequencies of Responses
bagels donuts
10 5
Obs Max Deriv Model L.R. d.f. P C
15 4e-04 3.37 6 0.7616 0.76
Dxy Gamma Tau-a R2 Brier g
0.52 0.52 0.248 0.279 0.183 1.411
gr gp
4.1 0.261
Coef S.E.Wald Z P
Intercept -5.68583 5.23295 -1.09 0.2772
x1 1.87020 2.14635 0.87 0.3836
x1^2 -0.42494 0.48286 -0.88 0.3788
x1^3 0.02845 0.03120 0.91 0.3618
x2 3.49560 3.54796 0.99 0.3245
x2^2 -0.94888 0.82067 -1.16 0.2476
x2^3 0.06362 0.05098 1.25 0.2121
...so what exactly would I query to design my if/then statement?
From: David Winsemius dwinsem...@comcast.net
To: David Winsemius dwinsem...@comcast.net
Sent: Tue, July 13, 2010 9:09:04 AM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 13, 2010, at 9:04 AM, David Winsemius wrote:
On Jul 13, 2010, at 8:47 AM, Josh B wrote:
Thanks again, David.
[[elided Yahoo spam]]
(BTW, it did work.)
Here's what I'm trying now:
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
results[1,i] - anova(mod.poly3)[1,3]
}
You need to do some programming.
(Or I suppose you could wrap both the lrm and the anova calls in try.)
You did not get an error from the lrm but rather from the anova call because
you tried to give the results of the try function to anova without first
checking to see if an error had occurred.
--David.
Here's what happens (from the console):
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
Error in UseMethod(anova) :
no applicable method for 'anova' applied to an object of class
Re: [R] Continuing on with a loop when there's a failure
Hello Peter,
I tried your suggestion, but I was still not able to get it to work. Would you
mind looking at my code again? Here's what I'm trying:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels donuts bagels 2 9
indv.7 bagels donuts bagels 8 5
indv.8 bagels donuts bagels 4 1
indv.9 donuts donuts donuts 3 3
indv.10 bagels donuts bagels 5 9
indv.11 bagels donuts bagels 9 10
indv.12 bagels donuts bagels 3 1
indv.13 donuts donuts donuts 7 10
indv.14 bagels donuts bagels 2 10
indv.15 bagels donuts bagels 9 6), header = TRUE)
results - matrix(nrow = 1, ncol = 3)
colnames(results) - c(y1, y2, y3)
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
if(class(mod.poly3) == 'try-error')
{results[1,i] - NA}
else {mod.poly3 - lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x)
results[1,i] - anova(mod.poly3)[1,3]
}
}
...and here's the output:
results
y1 y2 y3
[1,] NA NA NA
The results matrix is empty!
From: Peter Konings peter.l.e.koni...@gmail.com
Sent: Tue, July 13, 2010 5:45:17 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hi Josh,
Test the class of the resulting object. If it is 'try-error' fill your result
with NA or do some other error handling.
result - try(somemodel)
if(class(result) == 'try-error')
{
# some error handling
} else {
# whatever happens if the result is ok
}
HTH
Peter.
In my opinion the try and tryCatch commands are written and documented rather
poorly. Thus I am not sure what to program exactly.
For instance, I could query mod.poly3 and use an if/then statement to proceed,
but querying mod.poly3 is weird. For instance, here's the output when it fails:
mod.poly3 - try(lrm(x[,2] ~ pol(x1, 3) + pol(x2, 3), data=x))
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
mod.poly3
[1] Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights
=
weights, : \n NA/NaN/Inf in foreign function call (arg 1)\n
attr(,class)
[1] try-error
...and here's the output when it succeeds:
mod.poly3 - try(lrm(x[,1] ~ pol(x1, 3) + pol(x2, 3), data=x))
mod.poly3
Logistic Regression Model
lrm(formula = x[, 1] ~ pol(x1, 3) + pol(x2, 3), data = x)
Frequencies of Responses
bagels donuts
10 5
Obs Max Deriv Model L.R. d.f. P C
15 4e-04 3.37 6 0.7616 0.76
Dxy Gamma Tau-a R2 Brier g
0.52 0.52 0.248 0.279 0.183 1.411
gr gp
4.1 0.261
Coef S.E.Wald Z P
Intercept -5.68583 5.23295 -1.09 0.2772
x1 1.87020 2.14635 0.87 0.3836
x1^2 -0.42494 0.48286 -0.88 0.3788
x1^3 0.02845 0.03120 0.91 0.3618
x2 3.49560 3.54796 0.99 0.3245
x2^2 -0.94888 0.82067 -1.16 0.2476
x2^3 0.06362 0.05098 1.25 0.2121
...so what exactly would I query to design my if/then statement?
From: David Winsemius dwinsem...@comcast.net
To: David Winsemius dwinsem...@comcast.net
Sent: Tue, July 13, 2010 9:09:04 AM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 13, 2010, at 9:04 AM, David Winsemius wrote:
On Jul 13, 2010, at 8:47 AM, Josh B wrote:
Thanks again, David.
[[elided Yahoo spam]]
(BTW, it did work.)
Here's what I'm trying now:
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
results[1,i] - anova(mod.poly3)[1,3]
}
You need to do some programming.
(Or I suppose you could wrap both the lrm and the anova calls in try.)
You did not get an error from the lrm but rather from the anova call because
you tried to give the results of the try function to anova without first
checking to see if an error had occurred.
--David.
Here's what happens (from the console):
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
Error in UseMethod(anova) :
no applicable method for 'anova' applied to an object of class try-error
...so I still can't make my results matrix. Could I ask you for some
specific
code to make this work? I'm not that familiar with the syntax for try or
tryCatch, and the help files for them are pretty bad, in my humble opinion.
I should clarify that I actually don't care about the failed runs per se. I
just want R to keep going in spite of them and give me my results matrix.
From: David Winsemius dwinsem...@comcast.net
Cc: R Help r-help@r-project.org
Sent: Mon, July 12, 2010 8:09:03 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 12, 2010, at 6:18 PM, Josh B wrote:
Re: [R] sort file names in numerical order
Hi Yes it is possible- one way is: fileNames[order(sprintf(%02s, sub([[:upper:]],, fileNames)))] [1] A1 B1 A2 B2 A10 B10 Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England ARMIDALE NSW 2351 Email home: mac...@northnet.com.au At 06:47 18/07/2010, you wrote: I get some file names by list.files(). These names are in alphabetical order. I want to change it to logical numeric order. Example: fileNames - c(A10, A1, A2, B1, B2, B10) sort(fileNames) [1] A1 A10 A2 B1 B10 B2 I want to have: A1 A2 A10 B1 B2 B10 Is this possible? Greetings. I see that you've gotten an elegant solution to your problem. I've appended a poor-man's solution, which I generated more for my own edification than for yours. -- Mike ## modified file names, changed to exhibit sorting on letters fileNames - c(A10, B10, A1, A2, B1, B2); fileNames ## use regular expressions to pick off letters, numbers fnLet - gsub((^[^0-9]+).*$, \\1, fileNames); fnLet fnNum - gsub(^[^0-9]+(.*)$, \\1, fileNames); fnNum ## need to force numeric sorting of the numbers fnNum - as.numeric(fnNum) ## find the order of the numbers, for later use in subsetting fnNumOrd - order(fnNum); fnNumOrd ## pack all the relevant information into a data frame, then select from that fnPieces - data.frame(fnLet, fnNum, fileNames, stringsAsFactors=FALSE); fnPieces ## partial sort by numbers only (gives new df) dfPartialSort - fnPieces[fnNumOrd, ]; dfPartialSort ## find the order of the names, then select from new df with that fnLetOrd - order(dfPartialSort[, 1]); fnLetOrd dfFullSort - dfPartialSort[fnLetOrd, ]; dfFullSort ## the file names have gone along for the ride, so pick them off now sortedFileNames - dfFullSort$fileNames; sortedFileNames __ 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] Namibia becoming NA
Berwin A Turlach wrote: Did you think of trying some variations of na.strings? ;-) IMO, the simplest way of coding missing values in CSV files is to have two consecutive commas; not some code (whether NA, 99, 999, -1, ...) between them. Yes. Arguably, na.strings=NULL should be the default (and na= for write.csv) since delimited formats are (mainly) for communicating with external programs, which are not likely to use the NA code (unless it is for Namibia, North America, Noradrenalin, Niels Andersen, etc.) However, back compatibility (including that with files written with R's own write.csv) probably precludes changing anything at this point. Notice that read.csv and friends do pass ... to read.table, so it is easy to override the default setting. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R graphic help
Dear All, I've done some searching, but to no avail -- I'm plotting x-y data via the plot command, and the log=x command. The graphed x values are in scientific notation (1e-02 1e-01 1e+00 etc). Might you have some idea on how I can get the plot to uses the x values (0.01 0.10 1.0 10.0 etc) instead? A related Q - where might a good place to search for such answers? I've tried using Google searches and reading some posted Manuals but ... Thanks, Tim O'Brien [[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] R graphic help
Hi, There may be a simpler way but try this, plot(10^jitter(seq(-2,4,length=10)), 1:10, log=x, xaxt=n) axis(1, at = axTicks(1),labels = format(axTicks(1),scientific=FALSE)) HTH, baptiste On 18 July 2010 10:58, Timothy O'Brien teobr...@gmail.com wrote: Dear All, I've done some searching, but to no avail -- I'm plotting x-y data via the plot command, and the log=x command. The graphed x values are in scientific notation (1e-02 1e-01 1e+00 etc). Might you have some idea on how I can get the plot to uses the x values (0.01 0.10 1.0 10.0 etc) instead? A related Q - where might a good place to search for such answers? I've tried using Google searches and reading some posted Manuals but ... Thanks, Tim O'Brien [[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] R graphic help
On 07/18/2010 06:58 PM, Timothy O'Brien wrote: Dear All, I've done some searching, but to no avail -- I'm plotting x-y data via the plot command, and the log=x command. The graphed x values are in scientific notation (1e-02 1e-01 1e+00 etc). Might you have some idea on how I can get the plot to uses the x values (0.01 0.10 1.0 10.0 etc) instead? A related Q - where might a good place to search for such answers? I've tried using Google searches and reading some posted Manuals but ... Hi Tim, The scipen option is the one usually recommended for this. R works out how to print a number in a way that is (hopefully) maximally informative and minimally long. By penalizing this method, you can suppress scientific notation most of the time: print(10) [1] 1e+05 options(scipen=4) print(10) [1] 10 On the R home page is an option Search. The first search listed is Jon Baron's searchable archive messages, functions, etc. and is a good place to start. In this case, you would have to take two steps - see the scipen option mentioned and then look at the options help page. Usually you will get the answer straight from the search, _if_ you choose the correct search terms (life is never simple). Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Continuing on with a loop when there's a failure
On Jul 18, 2010, at 4:25 AM, Josh B wrote:
Hello Peter,
I tried your suggestion, but I was still not able to get it to work.
Would you
mind looking at my code again? Here's what I'm trying:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels donuts bagels 2 9
indv.7 bagels donuts bagels 8 5
indv.8 bagels donuts bagels 4 1
indv.9 donuts donuts donuts 3 3
indv.10 bagels donuts bagels 5 9
indv.11 bagels donuts bagels 9 10
indv.12 bagels donuts bagels 3 1
indv.13 donuts donuts donuts 7 10
indv.14 bagels donuts bagels 2 10
indv.15 bagels donuts bagels 9 6), header = TRUE)
closeAllConnections()
results - matrix(nrow = 1, ncol = 3)
colnames(results) - c(y1, y2, y3)
require(rms) # or Design
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x),
silent=TRUE)
if(class(mod.poly3)[1] != 'try-error')
{results[1,i] - anova(mod.poly3)[1,3]}
}
results
y1 y2 y3
[1,] 0.6976063 NA NA
...and here's the output:
results
y1 y2 y3
[1,] NA NA NA
The results matrix is empty!
From: Peter Konings peter.l.e.koni...@gmail.com
Sent: Tue, July 13, 2010 5:45:17 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hi Josh,
Test the class of the resulting object. If it is 'try-error' fill
your result
with NA or do some other error handling.
result - try(somemodel)
if(class(result) == 'try-error')
{
# some error handling
} else {
# whatever happens if the result is ok
}
HTH
Peter.
In my opinion the try and tryCatch commands are written and
documented rather
poorly. Thus I am not sure what to program exactly.
For instance, I could query mod.poly3 and use an if/then statement
to proceed,
but querying mod.poly3 is weird. For instance, here's the output
when it fails:
mod.poly3 - try(lrm(x[,2] ~ pol(x1, 3) + pol(x2, 3), data=x))
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol,
weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
mod.poly3
[1] Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol =
tol, weights
=
weights, : \n NA/NaN/Inf in foreign function call (arg 1)\n
attr(,class)
[1] try-error
...and here's the output when it succeeds:
mod.poly3 - try(lrm(x[,1] ~ pol(x1, 3) + pol(x2, 3), data=x))
mod.poly3
Logistic Regression Model
lrm(formula = x[, 1] ~ pol(x1, 3) + pol(x2, 3), data = x)
Frequencies of Responses
bagels donuts
10 5
Obs Max Deriv Model L.R. d.f. P C
15 4e-04 3.37 6 0.7616 0.76
Dxy Gamma Tau-a R2 Brier g
0.52 0.52 0.248 0.279 0.183 1.411
gr gp
4.1 0.261
Coef S.E.Wald Z P
Intercept -5.68583 5.23295 -1.09 0.2772
x1 1.87020 2.14635 0.87 0.3836
x1^2 -0.42494 0.48286 -0.88 0.3788
x1^3 0.02845 0.03120 0.91 0.3618
x2 3.49560 3.54796 0.99 0.3245
x2^2 -0.94888 0.82067 -1.16 0.2476
x2^3 0.06362 0.05098 1.25 0.2121
...so what exactly would I query to design my if/then statement?
From: David Winsemius dwinsem...@comcast.net
To: David Winsemius dwinsem...@comcast.net
Sent: Tue, July 13, 2010 9:09:04 AM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 13, 2010, at 9:04 AM, David Winsemius wrote:
On Jul 13, 2010, at 8:47 AM, Josh B wrote:
Thanks again, David.
[[elided Yahoo spam]]
(BTW, it did work.)
Here's what I'm trying now:
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
results[1,i] - anova(mod.poly3)[1,3]
}
You need to do some programming.
(Or I suppose you could wrap both the lrm and the anova calls in
try.)
You did not get an error from the lrm but rather from the anova
call because
you tried to give the results of the try function to anova without
first
checking to see if an error had occurred.
--David.
Here's what happens (from the console):
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol,
weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
Error in UseMethod(anova) :
no applicable method for 'anova' applied to an object of class
try-error
...so I still can't make my results matrix. Could I ask you for
some
specific
code to make this work? I'm not that familiar with the syntax for
try or
tryCatch, and the help files for them are pretty bad, in my
humble opinion.
I should clarify that I actually don't care about the failed runs
per se. I
just want R to keep going in spite of them and give me my results
matrix.
From: David Winsemius dwinsem...@comcast.net
Cc: R Help r-help@r-project.org
Sent: Mon, July 12, 2010 8:09:03 PM
Subject: Re: [R] Continuing on
Re: [R] sort file names in numerical order
Another option: require(gtools) ?mixedsort mixedsort(fileNames) [1] A1 A2 A10 B1 B2 B10 -- David On Jul 18, 2010, at 5:16 AM, Duncan Mackay wrote: Hi Yes it is possible- one way is: fileNames[order(sprintf(%02s, sub([[:upper:]],, fileNames)))] [1] A1 B1 A2 B2 A10 B10 Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England ARMIDALE NSW 2351 Email home: mac...@northnet.com.au At 06:47 18/07/2010, you wrote: I get some file names by list.files(). These names are in alphabetical order. I want to change it to logical numeric order. Example: fileNames - c(A10, A1, A2, B1, B2, B10) sort(fileNames) [1] A1 A10 A2 B1 B10 B2 I want to have: A1 A2 A10 B1 B2 B10 Is this possible? Greetings. I see that you've gotten an elegant solution to your problem. I've appended a poor-man's solution, which I generated more for my own edification than for yours. -- Mike ## modified file names, changed to exhibit sorting on letters fileNames - c(A10, B10, A1, A2, B1, B2); fileNames ## use regular expressions to pick off letters, numbers fnLet - gsub((^[^0-9]+).*$, \\1, fileNames); fnLet fnNum - gsub(^[^0-9]+(.*)$, \\1, fileNames); fnNum ## need to force numeric sorting of the numbers fnNum - as.numeric(fnNum) ## find the order of the numbers, for later use in subsetting fnNumOrd - order(fnNum); fnNumOrd ## pack all the relevant information into a data frame, then select from that fnPieces - data.frame(fnLet, fnNum, fileNames, stringsAsFactors=FALSE); fnPieces ## partial sort by numbers only (gives new df) dfPartialSort - fnPieces[fnNumOrd, ]; dfPartialSort ## find the order of the names, then select from new df with that fnLetOrd - order(dfPartialSort[, 1]); fnLetOrd dfFullSort - dfPartialSort[fnLetOrd, ]; dfFullSort ## the file names have gone along for the ride, so pick them off now sortedFileNames - dfFullSort$fileNames; sortedFileNames __ 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 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] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n - c(5, 10, 15)
ybar - c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)-c(Posterior 1, Posterior 2, Posterior 3)
colnames(posterior)-c(Mu, SD)
return(posterior)}
---
Joe King, M.A.
Ph.D. Student
University of Washington - Seattle
206-913-2912
mailto:j...@joepking.com j...@joepking.com
---
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
[[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.
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] loop troubles
I tried that, this is what I tried, but I only get it to do one iteration
and then it wont cycle back. I am not sure how to tell it how to look at the
previous precision to add to the current new sample precision to get the new
precision. This is my first try at a loop so I am not sure if I am doing
anything right, sorry about the elementary question.
n=c(5,10,15)
ybar=c(12,13,14)
sd=c(2,3,4)
mutotals-matrix(nrow=length(n), ncol=1)
mu=1;p0=0
for (i in 1:length(n)){
var = sd[i]^2 #sample variance
p = n[i]/var #sample precision
p0[i]= i
pn = p0[i]+p[i]
mu.out = ((p0[i])/(pn)*mu[i])+((p)/(pn))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu.out[i]
}
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a name
worth remembering. --Theodore Roosevelt
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Sunday, July 18, 2010 5:36 AM
To: Joe P King
Cc: r-help@r-project.org
Subject: Re: [R] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n - c(5, 10, 15)
ybar - c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)-c(Posterior 1, Posterior 2, Posterior 3)
colnames(posterior)-c(Mu, SD)
return(posterior)}
---
Joe King, M.A.
Ph.D. Student
University of Washington - Seattle
206-913-2912
mailto:j...@joepking.com j...@joepking.com
---
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
[[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.
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] slplot issue
It's truly amazing how many people asking for help can't be bothered to report what add-on packages they're using. Shawn's query is a particularly egregious case: slplot is mentioned in pkg:ccmn, which can be found on CRAN, but which depends on pkg:pcaMethods which is on Bioconductor. And pcaMethods has slplot(). Please, folks, do tell us what packages you are using! [Now, I'm having trouble with the pemqzrsTawYY function; can anybody help? :)] -Peter Ehlers On 2010-07-17 16:32, Dennis Murphy wrote: In which package would one find slplot? It's not in any of the 1800+ packages on my system... Dennis __ 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] loop troubles
On Jul 18, 2010, at 9:12 AM, Joe P King wrote:
I tried that, this is what I tried, but I only get it to do one
iteration
and then it wont cycle back. I am not sure how to tell it how to
look at the
previous precision to add to the current new sample precision to get
the new
precision.
And I'm not exactly sure what you are trying to do since I found your
problem statement insufficiently precise, but if you have a
vector, p and an index in a loop is i, then the previous p is
just p[i-1]. Sometimes you can get constructs like the following to
accomplish an indexed assignment:
p[2:length(n)] -a*p[1:(length(n)-1)]
This is my first try at a loop so I am not sure if I am doing
anything right, sorry about the elementary question.
n=c(5,10,15)
ybar=c(12,13,14)
sd=c(2,3,4)
mutotals-matrix(nrow=length(n), ncol=1)
mu=1;p0=0
for (i in 1:length(n)){
var = sd[i]^2 #sample variance
p = n[i]/var #sample precision
At this point you seem to be using p as a constant
p0[i]= i
Comments might help to figure out why you are doing each of htese steps.
pn = p0[i]+p[i]
Wouldn't pn need to be indexed as well? And now you seem to be using
p as a vector but haven't extended it with c() or some other
function, so on the second iteration it should fail.
mu.out = ((p0[i])/(pn)*mu[i])+((p)/(pn))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu.out[i]
}
I get:
Error in sigma[i] = 1/sqrt(pn[i]) : object 'sigma' not found
You are not reporting the error messages this throws (which should
have made it apparent that you needed to do some further setup of the
data structures you are using, not just with p, pn, but also
sigma. )
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Sunday, July 18, 2010 5:36 AM
To: Joe P King
Cc: r-help@r-project.org
Subject: Re: [R] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n - c(5, 10, 15)
ybar - c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)-c(Posterior 1, Posterior 2, Posterior 3)
colnames(posterior)-c(Mu, SD)
return(posterior)}
---
Joe King, M.A.
Ph.D. Student
University of Washington - Seattle
206-913-2912
mailto:j...@joepking.com j...@joepking.com
---
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
[[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.
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.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
Re: [R] Identify points (was Plot error)
This is exactly what I want as I will have several thousand data points on the final graph i make, so the scroll over option is ideal. I've read the TeachingDemos pdf, I'm working on a Mac so Cannot use HWidentify. I have installed and loaded the TeachingDemos package but im having trouble loading the tkrplot package after installing it won't load. library(tcltk) Loading Tcl/Tk interface ... Error : .onLoad failed in loadNamespace() for 'tcltk', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared library '/Library/Frameworks/ R.framework/Resources/library/tcltk/libs/i386/tcltk.so': dlopen(/Library/Frameworks/R.framework/Resources/library/tcltk/libs/ i386/tcltk.so, 10): Library not loaded: /usr/local/lib/libtcl8.5.dylib Referenced from: /Library/Frameworks/R.framework/Resources/library/ tcltk/libs/i386/tcltk.so Reason: image not found Error: package/namespace load failed for 'tcltk' Cheers, James On 17 Jul 2010, at 18:12, Peter Ehlers wrote: On 2010-07-17 9:50, James Platt wrote: The other question I have: Is there any way to link the data point on the graph to the name of a row i.e in my table: name value_1 value_2 bill 14 ben 2 2 jane 3 1 I click on the data point at 2,2 and it would read out ben Check out HWidentify or HTKidentify in pkg:TeachingDemos. -Peter Ehlers __ 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 with a problem
You can also use 'embed' to create a list of indices you can use to do the test:
dat
dsc1 c2
1 2010-04-03 100 0
2 2010-04-30 11141 15
3 2010-05-01 3 16
4 2010-05-02 7615 14
5 2010-05-03 6910 17
6 2010-05-04 5035 3
7 2010-05-05 3007 15
8 2010-05-06 4 14
9 2010-05-07 8335 17
10 2010-05-08 2897 13
11 2010-05-09 6377 17
12 2010-05-10 3177 17
13 2010-05-11 7946 15
14 2010-05-12 8705 0
15 2010-05-13 9030 16
16 2010-05-14 8682 16
17 2010-05-15 8440 1
# create index to check against
indx - embed(seq(nrow(dat)), 7)
result - apply(indx, 1, function(x){
+ # whatever condition you want
+ sum(dat$c1[x] = 5000 dat$c2[x] 0)
+ })
result
[1] 4 4 4 4 4 3 3 3 4 4 5
On Sun, Jul 18, 2010 at 3:06 AM, Stephan Kolassa stephan.kola...@gmx.de wrote:
Hi all,
zoo::rollmean() is a nice idea. But if I understand Mike correctly, he wants
5 out of any 7 consecutive logicals to be TRUE, where these 5 do not
necessarily need to be consecutive themselves. (remaining open question:
could, e.g., the condition on c1 be TRUE for rows 1,2,3,4,5 and on c2 for
rows 3,4,5,6,7, or would it need to be TRUE for the same rows?). Then
something like this would make sense:
any(rollmean(dat$c1=100,7)=5/7-.01 rollmean(dat$c2=8,7)=5/7-.01))
or
any(rollmean(dat$c1=100,7)=5/7-.01 dat$c2=8,7)=5/7-.01))
depending on the open question above.
The -.01 above may be necessary in light of FAQ 7.31.
HTH,
Stephan
Joshua Wiley schrieb:
Hi Michael,
The days in your example do not look continuous (at least from my
thinking), so you may have extra requirements in mind, but take a look
at this code. My general thought was first to turn each column into a
logical vector (c1 = 100 and c2 = 8). Taking advantage of the fact
that R treats TRUE as 1 and FALSE as 0, compute a rolling mean. If
(and only if) 5 consecutive values are TRUE, the mean will be 1. Next
I added the rolling means for each column, and then tested whether any
were 2 (i.e., 1 + 1).
Cheers,
Josh
###
#Load required package
library(zoo)
#Your data with ds converted to Date
#from dput()
dat -
structure(list(ds = structure(c(14702, 14729, 14730, 14731, 14732,
14733, 14734, 14735, 14736, 14737, 14738, 14739, 14740, 14741,
14742, 14743, 14744), class = Date), c1 = c(100L, 11141L, 3L,
7615L, 6910L, 5035L, 3007L, 4L, 8335L, 2897L, 6377L, 3177L, 7946L,
8705L, 9030L, 8682L, 8440L), c2 = c(0L, 15L, 16L, 14L, 17L, 3L,
15L, 14L, 17L, 13L, 17L, 17L, 15L, 0L, 16L, 16L, 1L)), .Names = c(ds,
c1, c2), row.names = c(NA, -17L), class = data.frame)
#Order by ds
dat - dat[order(dat$ds), ]
yourvar - 0
#Test that 5 consecutive values from c1 AND c2 meet requirements
if(any(
c(rollmean(dat$c1 = 100, 5) + rollmean(dat$c2 = 8, 5)) == 2)
) {yourvar - 1}
###
On Sat, Jul 17, 2010 at 2:38 PM, Michael Hess mlh...@med.umich.edu
wrote:
Sorry for not being clear.
In the dataset there are around 100 or so days of data (in the case also
rows of data)
I need to make sure that the person meets that c1 is at least 100 AND c2
is at least 8 for 5 of 7 continuous days.
I will play with what I have and see if I can find out how to do this.
Thanks for the help!
Michael
Stephan Kolassa 07/17/10 4:50 PM
Mike,
I am slightly unclear on what you want to do. Do you want to check rows
1 and 7 or 1 *to* 7? Should c1 be at least 100 for *any one* or *all*
rows you are looking at, and same for c2?
You can sort your data like this:
data - data[order(data$ds),]
Type ?order for help. But also do this for added enlightenment...:
library(fortunes)
fortune(dog)
Next, your analysis on the sorted data frame. As I said, I am not
entirely clear on what you are looking at, but the following may solve
your problem with choices 1 to 7 and any one above.
foo - 0
for ( ii in 1:(nrow(data)-8) ) {
if (any(data$c1[ii+seq(0,6)]=100) any(data$c2[ii+seq(0,6)]=8)) {
foo - 1
break
}
}
The variable foo should contain what you want it to. Look at ?any
(and, if this does not do what you want it to, at ?all) for further info.
No doubt this could be vectorized, but I think the loop is clear enough.
Good luck!
Stephan
Michael Hess schrieb:
Hello R users,
I am a researcher at the University of Michigan looking for a solution
to an R problem. I have loaded my data in from a mysql database and it
looks like this
data
ds c1 c2
1 2010-04-03 100 0
2 2010-04-30 11141 15
3 2010-05-01 3 16
4 2010-05-02 7615 14
5 2010-05-03 6910 17
6 2010-05-04 5035 3
7 2010-05-05 3007 15
8 2010-05-06 4 14
9 2010-05-07 8335 17
10 2010-05-08 2897 13
11 2010-05-09 6377 17
12 2010-05-10 3177 17
13 2010-05-11 7946 15
14 2010-05-12 8705
[R] Giovanna Jonalasinio è fuori ufficio, Out of Office
Risposta automatica dal 18/7/10 fino al 27/7/10 I'm not in Rome and I won't read my email regularly untill the 27th of July Non sono a Roma e leggerò la posta sporadicamente fino al 27 luglio [[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] Identify points (was Plot error)
On Jul 18, 2010, at 9:43 AM, James Platt wrote: This is exactly what I want as I will have several thousand data points on the final graph i make, so the scroll over option is ideal. I've read the TeachingDemos pdf, I'm working on a Mac so Cannot use HWidentify. Not true. I have installed and loaded the TeachingDemos package but im having trouble loading the tkrplot package after installing it won't load. library(tcltk) In the Mac GUI Package Installer, I cannot even find a package by that name. However, then running the HWidentify example after require(Teaching Demos), tcltk does get loaded as well as X11, making me think you need to provide further information such as that requested in the Posting Guide. You may also want to review the material in the R for Mac OS X FAQ. -- David. sessionInfo() R version 2.11.1 Patched (2010-06-14 r52281) x86_64-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] tcltk splines stats graphics grDevices utils [7] datasets methods base other attached packages: [1] tkrplot_0.0-18TeachingDemos_2.6 gtools_2.6.2 [4] Design_2.3-0 Hmisc_3.8-1 survival_2.35-8 [7] lattice_0.18-8 loaded via a namespace (and not attached): [1] cluster_1.12.3 grid_2.11.1tools_2.11.1 Loading Tcl/Tk interface ... Error : .onLoad failed in loadNamespace() for 'tcltk', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared library '/Library/Frameworks/ R.framework/Resources/library/tcltk/libs/i386/tcltk.so': dlopen(/Library/Frameworks/R.framework/Resources/library/tcltk/libs/ i386/tcltk.so, 10): Library not loaded: /usr/local/lib/libtcl8.5.dylib Referenced from: /Library/Frameworks/R.framework/Resources/library/ tcltk/libs/i386/tcltk.so Reason: image not found Error: package/namespace load failed for 'tcltk' On 17 Jul 2010, at 18:12, Peter Ehlers wrote: On 2010-07-17 9:50, James Platt wrote: The other question I have: Is there any way to link the data point on the graph to the name of a row i.e in my table: name value_1 value_2 bill 14 ben 2 2 jane 3 1 I click on the data point at 2,2 and it would read out ben Check out HWidentify or HTKidentify in pkg:TeachingDemos. -Peter Ehlers 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] producing biplot of canonical correlation analysis
Dear List, I would like to obtain biplot containing arrows and variable labels. However, after checking the previous message and package CCA and function anacor, my attempt remains an open question. Please kindly advise code or package I might miss to achieve it. Thank you. Elaine [[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] Identify points (was Plot error)
You can always use the identify() function which will leave the id of the point on the plot; see ?identify. But you do eventually have to sort out your Mac problem. Sorry, I can't help with that. -Peter Ehlers On 2010-07-18 7:43, James Platt wrote: This is exactly what I want as I will have several thousand data points on the final graph i make, so the scroll over option is ideal. I've read the TeachingDemos pdf, I'm working on a Mac so Cannot use HWidentify. I have installed and loaded the TeachingDemos package but im having trouble loading the tkrplot package after installing it won't load. library(tcltk) Loading Tcl/Tk interface ... Error : .onLoad failed in loadNamespace() for 'tcltk', details: call: dyn.load(file, DLLpath = DLLpath, ...) error: unable to load shared library '/Library/Frameworks/R.framework/Resources/library/tcltk/libs/i386/tcltk.so': dlopen(/Library/Frameworks/R.framework/Resources/library/tcltk/libs/i386/tcltk.so, 10): Library not loaded: /usr/local/lib/libtcl8.5.dylib Referenced from: /Library/Frameworks/R.framework/Resources/library/tcltk/libs/i386/tcltk.so Reason: image not found Error: package/namespace load failed for 'tcltk' Cheers, James __ 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] re. Mathematica and R
The wiki on rcom.univie.ac.at in section R(D)COM, rcom, and other software systems has an example how to use R from within Mathematica on Windows. You will need to install statconnDCOM and rcom available from the same site. On Jul 17, 2010, at 1:06 PM, Alan Kelly wrote: David - information on calling R from within Mathematica can be found at the following link: http://www.mofeel.net/1164-comp-soft-sys-math-mathematica/13022.aspx HTH, Alan Kelly __ 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. -- Erich Neuwirth Didactic Center for Computer Science and Institute for Scientific Computing University of Vienna [[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] heatmap.2 - change column row locations; angle / rotate
Esteemed R user's, I'm struggling to achieve some details of a heatmap using heatmap.2(): 1. Change label locations, for both rows columns from the default right bottom, to left and top. Can this be done within heatmap.2()? Or do i need to suppress this default behavior (how) and call a new function to relabel (what) specifying locations? 2. Change the angle of the labels. By default column labels are 90deg anti-clock-wise from horizontal. How to bring them back to horizontal? Or better, rotate 45deg clock-wise from horizontal (ie., rotate 135deg a.clock.wise from default)? Any suggestions or pointers to helpful resources greatly appreciated, Karl -- Karl Brand Department of Genetics Erasmus MC Dr Molewaterplein 50 3015 GE Rotterdam T +31 (0)10 704 3457 |F +31 (0)10 704 4743 |M +31 (0)642 777 268 __ 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] package plotrix
I installed package plotrix because reading its vignette it looks like it can help me solve a legend problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function lgendg. library(plotrix) plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type=n, + main=Test of grouped legend function) legendg(0.5,0.8,c(one,two,three),pch=list(1,2:3,4:6), + col=list(2,3:4,5:7)) Error: could not find function legendg My problem with the regular R legend function is that I cannot indicate in the legend the line plotted by the command abline Here is the code. Attached is the plot. Any suggestion is welcome. Thank you. Maura x11(width=10,heigh=8) plot(NN,LB,xlim=c(1,300),ylim=c(0,3), main=Intrinsic Dimensionality of 1D Helix,font.main=2,cex.main=2,col.main=red, xlab=Number of Nearest-Neighbors,ylab=Estimated Dimension,col.lab=red,cex.lab=1,font.lab=2, xaxt=n,yaxt=n,pch=19,col=red4) axis(1,at=NN[1:40],labels=NN[1:40],cex.lab=1,cex.axis=0.8,col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) axis(2,at=c(0,0.5,1,1.5,2,2.5,3,3.5,4),labels=c(0,0.5,1,1.5,2,2.5,3,3.5,4),cex.lab=1,cex.axis=0.8, col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) lines(NN,McG,pch=18,col=green) lines(NN,PBJD,pch=20,col=magenta1) points(NN,GRPR,pch=18,col=cyan) abline(h=1,pch= 1,lty=1,col=black) legend(topright,legend = c(Levina-Bickel,MacKay-Ghahramani,Pettis-Bailey-Jain-Dubes,Grassberger-Procaccia,Helix Dimension), text.width = strwidth(Pettis-Bailey-Jain-Dubes),pch = c(19,18,20,18,1), col = c(red4,green,magenta1,cyan,black),text.col = black,lty=c(-1,-1,-1,1), bg =snow) e tutti i telefonini TIM! Vai su __ 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] NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
I am confused by the behavior of the below piece of code. The NAs are
making it past the logical call ==0. I am sure that I am missing
something. I just don't understand this behavior. Thanks for your
help in advance.
code###
left - (structure(list(measurment_num = c(2, 2.2, 2.4, 2.6, 2.8, 2.82,
3, NA, NA, NA), bankfull_depths_m = c(1.29, 1.28, 1.23, 0.18,
-0.05, 0, -0.09, NA, NA, NA)), .Names = c(measurment_num, bankfull_depths_m
), row.names = c(10L, 11L, 12L, 13L, 14L, 20L, 15L, 16L, 17L,
18L), class = data.frame))
if(sum(left[,bankfull_depths_m]==0, na.rm=TRUE) == 1){
left_min - left[left[,bankfull_depths_m]==0, measurment_num]
}
left
##
--
Stephen Sefick
| Auburn University |
| Department of Biological Sciences |
| 331 Funchess Hall |
| Auburn, Alabama |
| 36849 |
|___|
| sas0...@auburn.edu |
| http://www.auburn.edu/~sas0025 |
|___|
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] log transformation
Hi, I am reading an article in which, first, some given p-results are obtained, and, second, these are afterwards transformed and expressed as the NEGATIVE base-10 logarithm of the p-value. My question is if anyone could indicate how is such transformation achieved with R and how are they expressed exponentially, as in for example 1.8E-18. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/log-transformation-tp2293162p2293162.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] Reading files with varying number of columns
Dear R list, I am trying to read files with a varying number of columns and can't figure out how to get them in the proper format. Some rows have five value and some have seven. Is there a way to read them in so that two empty columns are added to every row that has only five values? An example of the data follows. Thanks! Tim 46,B,00301,2004-02-03,11:59:16 46,B,00301,2004-02-03,12:03:43 46,B,00301,2004-02-03,12:39:17 46,B,00301,2004-02-03,23:07:28 43,C,00232,2004-02-04,07:39:57,-1.5,meters 44,A,00230,2004-02-04,07:44:59 44,A,00230,2004-02-04,07:46:19 44,A,00230,2004-02-04,07:48:24 44,A,00230,2004-02-04,07:49:12 43,C,00232,2004-02-04,07:53:08,2.5,meters 44,A,00230,2004-02-04,07:54:34 44,A,00230,2004-02-04,07:55:08 43,C,00232,2004-02-04,07:56:18,2.9,meters 43,C,00232,2004-02-04,07:56:39,2.5,meters 46,B,00301,2004-02-04,08:00:49 43,C,00232,2004-02-04,08:02:12,-1.1,meters 43,C,00232,2004-02-04,08:04:01,2.2,meters 43,C,00232,2004-02-04,08:04:26,3.3,meters 43,C,00232,2004-02-04,08:40:26,2.9,meters 43,C,00232,2004-02-04,08:41:04,-2.2,meters 43,C,00232,2004-02-04,08:42:44,2.2,meters 43,C,00232,2004-02-04,08:44:31,2.9,meters 44,A,00231,2004-02-04,09:04:43 44,A,00231,2004-02-04,09:10:30 44,A,00231,2004-02-04,09:12:08 44,A,00231,2004-02-04,09:13:32 44,A,00231,2004-02-04,09:14:06 43,C,00232,2004-02-04,09:18:24,0.4,meters 43,C,00232,2004-02-04,09:20:13,2.5,meters 44,A,00231,2004-02-04,09:21:04 44,A,00231,2004-02-04,09:22:53 44,A,00231,2004-02-04,09:25:32 44,A,00231,2004-02-04,09:29:31 44,A,00231,2004-02-04,09:30:05 43,C,00232,2004-02-04,09:53:25,2.5,meters 43,C,00232,2004-02-04,09:53:53,1.5,meters 46,B,00301,2004-02-04,22:16:24 46,B,00301,2004-02-04,22:19:32 46,B,00258,2004-02-04,23:44:01 46,B,00258,2004-02-04,23:44:28 Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, American Samoa 96799 __ 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] package plotrix
On 18.07.2010 17:59, mau...@alice.it wrote: I installed package plotrix because reading its vignette it looks like it can help me solve a legend problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function lgendg. You mean legendg. library(plotrix) plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type=n, + main=Test of grouped legend function) legendg(0.5,0.8,c(one,two,three),pch=list(1,2:3,4:6), + col=list(2,3:4,5:7)) Error: could not find function legendg That works for me. Try to upgrade bopth R and plotrix if you have not already. My problem with the regular R legend function is that I cannot indicate in the legend the line plotted by the command abline Here is the code. Attached is the plot. Any suggestion is welcome. Thank you. Maura x11(width=10,heigh=8) plot(NN,LB,xlim=c(1,300),ylim=c(0,3), main=Intrinsic Dimensionality of 1D Helix,font.main=2,cex.main=2,col.main=red, xlab=Number of Nearest-Neighbors,ylab=Estimated Dimension,col.lab=red,cex.lab=1,font.lab=2, xaxt=n,yaxt=n,pch=19,col=red4) axis(1,at=NN[1:40],labels=NN[1:40],cex.lab=1,cex.axis=0.8,col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) axis(2,at=c(0,0.5,1,1.5,2,2.5,3,3.5,4),labels=c(0,0.5,1,1.5,2,2.5,3,3.5,4),cex.lab=1,cex.axis=0.8, col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) lines(NN,McG,pch=18,col=green) lines(NN,PBJD,pch=20,col=magenta1) points(NN,GRPR,pch=18,col=cyan) abline(h=1,pch= 1,lty=1,col=black) legend(topright,legend = c(Levina-Bickel,MacKay-Ghahramani,Pettis-Bailey-Jain-Dubes,Grassberger-Procaccia,Helix Dimension), text.width = strwidth(Pettis-Bailey-Jain-Dubes),pch = c(19,18,20,18,1), col = c(red4,green,magenta1,cyan,black),text.col = black,lty=c(-1,-1,-1,1), bg =snow) We do not have the data nor do we see the figure, hence this is not reproducible for us. Uwe Ligges e tutti i telefonini TIM! Vai su __ 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] Eliminating case numbers in a dendrogram
Hi folks, I conducted a hierarchical cluster analysis. As I wanted to illustrate the result, I created a dendrogram with the code plot(as.dendrogram(fit),sub=,xlab=,ylab=Heterogeneity,nodePar = list(lab.cex=.5,pch=NA,xlab=)) However, the dendrogram contains the case numbers and, as I have N = 300, looks not very nice. Can anybody tell me how to prevent the case numbers? Thanks in advance Holger http://r.789695.n4.nabble.com/file/n2293207/Dendrogram.jpeg -- View this message in context: http://r.789695.n4.nabble.com/Eliminating-case-numbers-in-a-dendrogram-tp2293207p2293207.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] NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
On 18.07.2010 18:02, stephen sefick wrote:
I am confused by the behavior of the below piece of code. The NAs are
making it past the logical call ==0. I am sure that I am missing
something. I just don't understand this behavior. Thanks for your
help in advance.
code###
left- (structure(list(measurment_num = c(2, 2.2, 2.4, 2.6, 2.8, 2.82,
3, NA, NA, NA), bankfull_depths_m = c(1.29, 1.28, 1.23, 0.18,
-0.05, 0, -0.09, NA, NA, NA)), .Names = c(measurment_num, bankfull_depths_m
), row.names = c(10L, 11L, 12L, 13L, 14L, 20L, 15L, 16L, 17L,
18L), class = data.frame))
if(sum(left[,bankfull_depths_m]==0, na.rm=TRUE) == 1){
left_min- left[left[,bankfull_depths_m]==0, measurment_num]
}
NA == 0 is ??? -- NA!
Use is.na() to check for missing values.
Uwe Ligges
left
##
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading files with varying number of columns
See argument fill in ?read.table Uwe Ligges On 18.07.2010 12:14, Tim Clark wrote: Dear R list, I am trying to read files with a varying number of columns and can't figure out how to get them in the proper format. Some rows have five value and some have seven. Is there a way to read them in so that two empty columns are added to every row that has only five values? An example of the data follows. Thanks! Tim 46,B,00301,2004-02-03,11:59:16 46,B,00301,2004-02-03,12:03:43 46,B,00301,2004-02-03,12:39:17 46,B,00301,2004-02-03,23:07:28 43,C,00232,2004-02-04,07:39:57,-1.5,meters 44,A,00230,2004-02-04,07:44:59 44,A,00230,2004-02-04,07:46:19 44,A,00230,2004-02-04,07:48:24 44,A,00230,2004-02-04,07:49:12 43,C,00232,2004-02-04,07:53:08,2.5,meters 44,A,00230,2004-02-04,07:54:34 44,A,00230,2004-02-04,07:55:08 43,C,00232,2004-02-04,07:56:18,2.9,meters 43,C,00232,2004-02-04,07:56:39,2.5,meters 46,B,00301,2004-02-04,08:00:49 43,C,00232,2004-02-04,08:02:12,-1.1,meters 43,C,00232,2004-02-04,08:04:01,2.2,meters 43,C,00232,2004-02-04,08:04:26,3.3,meters 43,C,00232,2004-02-04,08:40:26,2.9,meters 43,C,00232,2004-02-04,08:41:04,-2.2,meters 43,C,00232,2004-02-04,08:42:44,2.2,meters 43,C,00232,2004-02-04,08:44:31,2.9,meters 44,A,00231,2004-02-04,09:04:43 44,A,00231,2004-02-04,09:10:30 44,A,00231,2004-02-04,09:12:08 44,A,00231,2004-02-04,09:13:32 44,A,00231,2004-02-04,09:14:06 43,C,00232,2004-02-04,09:18:24,0.4,meters 43,C,00232,2004-02-04,09:20:13,2.5,meters 44,A,00231,2004-02-04,09:21:04 44,A,00231,2004-02-04,09:22:53 44,A,00231,2004-02-04,09:25:32 44,A,00231,2004-02-04,09:29:31 44,A,00231,2004-02-04,09:30:05 43,C,00232,2004-02-04,09:53:25,2.5,meters 43,C,00232,2004-02-04,09:53:53,1.5,meters 46,B,00301,2004-02-04,22:16:24 46,B,00301,2004-02-04,22:19:32 46,B,00258,2004-02-04,23:44:01 46,B,00258,2004-02-04,23:44:28 Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, American Samoa 96799 __ 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] Eliminating case numbers in a dendrogram
On 18.07.2010 19:02, Holger Steinmetz wrote: Hi folks, I conducted a hierarchical cluster analysis. As I wanted to illustrate the result, I created a dendrogram with the code plot(as.dendrogram(fit),sub=,xlab=,ylab=Heterogeneity,nodePar = list(lab.cex=.5,pch=NA,xlab=)) However, the dendrogram contains the case numbers and, as I have N = 300, looks not very nice. Can anybody tell me how to prevent the case numbers? You do not want any labels??? In that case, see argument leaflab in ?plot.dendrogram Uwe Ligges Thanks in advance Holger http://r.789695.n4.nabble.com/file/n2293207/Dendrogram.jpeg __ 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] NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
On Jul 18, 2010, at 12:02 PM, stephen sefick wrote:
I am confused by the behavior of the below piece of code. The NAs are
making it past the logical call ==0. I am sure that I am missing
something. I just don't understand this behavior. Thanks for your
help in advance.
code###
left - (structure(list(measurment_num = c(2, 2.2, 2.4, 2.6, 2.8,
2.82,
3, NA, NA, NA), bankfull_depths_m = c(1.29, 1.28, 1.23, 0.18,
-0.05, 0, -0.09, NA, NA, NA)), .Names = c(measurment_num,
bankfull_depths_m
), row.names = c(10L, 11L, 12L, 13L, 14L, 20L, 15L, 16L, 17L,
18L), class = data.frame))
if(sum(left[,bankfull_depths_m]==0, na.rm=TRUE) == 1){
left_min - left[left[,bankfull_depths_m]==0, measurment_num]
}
left
Why aren't you looking at left_min? That is what the code is supposed
to be changing. I do not get an error and left_min[1] (now) =='s 2.82.
Regarding why there are the 3 NA's, ... you need to look at the
documentation for Extract in the section very appropriately entitled:
NAs in indexing
I have been bitten by that behavior more times than I care to admit
and I am not the only person that thinks that aspect of the R language
is badly designed. See Aniko Szabo's comments:
http://radfordneal.wordpress.com/2008/09/21/design-flaws-in-r-3-%E2%80%94-zero-subscripts/
The problem is in data.frame[ and any NA in a logical vector will
return a row of NA's. This can be avoid by wrapping which() around the
logical vector which seems entirely wasteful or using subset().
David Winsemius, MD
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Continuing on with a loop when there's a failure
On Jul 18, 2010, at 12:28 PM, Josh B wrote:
Thanks very much again David for your helpful answers. However, the
code STILL does not appear to be working properly!
Even though the third time through the loop *should* work, it
appears that R has given up after the second time through the loop.
What I mean is: although y2 causes the lrm function to fail, y3 is a
kosher variable. If the loop continues on, it should give data for
the model with y3.
Right, but your loop index only went to 2!!! Your design, not mine.
--
David.
But if you look at the matrix called results, it returns NA for
the third spot corresponding to the model of y3:
results
y1 y2 y3
[1,] 0.6976063 NA NA
If you run y3 in isolation, rather than through the loop, you can
see that it should work and contribute data to the matrix called
results:
mod.poly3 - lrm(x[,3] ~ pol(x1, 3) + pol(x2, 3), data=x)
anova(mod.poly3)[1,3]
[1] 0.6976063
Any ideas?
From: David Winsemius dwinsem...@comcast.net
To: Josh B josh...@yahoo.com
Cc: Peter Konings peter.l.e.koni...@gmail.com; R Help r-help@r-project.org
Sent: Sun, July 18, 2010 3:33:07 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 18, 2010, at 4:25 AM, Josh B wrote:
Hello Peter,
I tried your suggestion, but I was still not able to get it to
work. Would you
mind looking at my code again? Here's what I'm trying:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels donuts bagels 2 9
indv.7 bagels donuts bagels 8 5
indv.8 bagels donuts bagels 4 1
indv.9 donuts donuts donuts 3 3
indv.10 bagels donuts bagels 5 9
indv.11 bagels donuts bagels 9 10
indv.12 bagels donuts bagels 3 1
indv.13 donuts donuts donuts 7 10
indv.14 bagels donuts bagels 2 10
indv.15 bagels donuts bagels 9 6), header = TRUE)
closeAllConnections()
results - matrix(nrow = 1, ncol = 3)
colnames(results) - c(y1, y2, y3)
require(rms) # or Design
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x),
silent=TRUE)
if(class(mod.poly3)[1] != 'try-error')
{results[1,i] - anova(mod.poly3)[1,3]}
}
results
y1 y2 y3
[1,] 0.6976063 NA NA
...and here's the output:
results
y1 y2 y3
[1,] NA NA NA
The results matrix is empty!
From: Peter Konings peter.l.e.koni...@gmail.com
Sent: Tue, July 13, 2010 5:45:17 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hi Josh,
Test the class of the resulting object. If it is 'try-error' fill
your result
with NA or do some other error handling.
result - try(somemodel)
if(class(result) == 'try-error')
{
# some error handling
} else {
# whatever happens if the result is ok
}
HTH
Peter.
In my opinion the try and tryCatch commands are written and
documented rather
poorly. Thus I am not sure what to program exactly.
For instance, I could query mod.poly3 and use an if/then
statement to proceed,
but querying mod.poly3 is weird. For instance, here's the output
when it fails:
mod.poly3 - try(lrm(x[,2] ~ pol(x1, 3) + pol(x2, 3), data=x))
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol,
weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
mod.poly3
[1] Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol =
tol, weights
=
weights, : \n NA/NaN/Inf in foreign function call (arg 1)\n
attr(,class)
[1] try-error
...and here's the output when it succeeds:
mod.poly3 - try(lrm(x[,1] ~ pol(x1, 3) + pol(x2, 3), data=x))
mod.poly3
Logistic Regression Model
lrm(formula = x[, 1] ~ pol(x1, 3) + pol(x2, 3), data = x)
Frequencies of Responses
bagels donuts
10 5
Obs Max Deriv Model L.R. d.f. P C
15 4e-04 3.37 60.7616 0.76
Dxy Gamma Tau-aR2 Brier g
0.52 0.52 0.248 0.279 0.183 1.411
grgp
4.1 0.261
CoefS.E.Wald Z P
Intercept -5.68583 5.23295 -1.09 0.2772
x11.87020 2.14635 0.87 0.3836
x1^2 -0.42494 0.48286 -0.88 0.3788
x1^3 0.02845 0.03120 0.91 0.3618
x23.49560 3.54796 0.99 0.3245
x2^2 -0.94888 0.82067 -1.16 0.2476
x2^3 0.06362 0.05098 1.25 0.2121
...so what exactly would I query to design my if/then statement?
From: David Winsemius dwinsem...@comcast.net
To: David Winsemius dwinsem...@comcast.net
Sent: Tue, July 13, 2010 9:09:04 AM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 13, 2010, at 9:04 AM, David Winsemius wrote:
On Jul 13, 2010, at 8:47 AM, Josh B wrote:
Thanks again, David.
[[elided Yahoo spam]]
(BTW, it did work.)
Here's what I'm
Re: [R] Toggle between the various pages for multi-page figures
On 16.07.2010 18:38, mahesh samtani wrote: Hello, I am a new R user having transitioned over from S-plus recently. I have a question that is probably very trivial but I am having trouble finding a solution. In S-plus, graphic pages are created as tabs when multi-page figures are created. I have shown the R code for xpose.VPC (a function within library xpose4 for R) where I want the figure from each Strata (STRT) to displayed on a different page. When I run the code, the pages just zoom by and I was wondering if I could toggle between the various pages. I can only export the last of the 8 pages that are created with the code below (and I wish to export all 8 pages not just the last one): In order to export: Better start the appropriate graphics device (such as pdf()) before you start plotting and when finished plotting ask R to dev.off(). All your graphics are collected in the file(s) in the end which is a bit device dependent. See you favorite device's help page for details. Uwe Ligges xpose.VPC (vpc.info=vpc1/vpc_results.csv, vpctab=vpc1/vpctab1, PI=NULL, by=STRT, PI.ci=area, PI.real=TRUE,type=n, PI.limits=c(0.05,0.95),*layout=c(1,1,8)*,xlb=Time (week),ylb=Concentration (ng/mL), scales = list(x = list(at=seq(0,3360,by=336) , labels=seq(0,20,by=2 )), y = list(at=log(c(.3,1,3,7.5,20,50,150)), labels=c(.3,1,3,7.5,20,50,150)) )) Please help, MNS [[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] Error using the mi package
Since we do not have the data and cannot reproduce: What about sending a reproducible example to the mi maintainer in the first step? Best, Uwe Ligges On 15.07.2010 16:48, Andrew Miles wrote: I'm trying to impute data using the mi package, but after running through almost the entire first round of imputations (which takes quite a while), it throws this error (I'll include the whole output prior to the error for context). Does anyone know what is causing it, or how I can fix it? More specifically, how can I tell what is throwing the error so I know what to fix? Is it a problem with a variable? With the formula I am using for imputations? Beginning Multiple Imputation ( Thu Jul 15 09:10:33 2010 ): Iteration 1 Imputation 1 : min.func* Tenure* Salary* Housing* ord_stat* rural_now* a3* a7* a8* a9* a10* a12* a13* a14* a15* a16* a17a* a17b* a17c* a17d* a17e* a17f* a18* a21* b1a* b1b* b1c* b1d* b1e* b1f* b1g* b1h* b1i* b3a* b3b* b3c* b3d* b3e* b3f* b3g* b4a* b4b* b4c* b4d* b4e* b4f* b4g* b4h* b4i* b4j* b4k* b4l* b4m* b4n* b4o* b4p* b5a* b5b* b5c* b5d* b5e* b5f* b5g* b5h* b5i* b5j* b5k* b5l* b5m* b5n* b5o* b6a* b6b* b6c* b6d* b6e* b6f* b6g* b6h* b6i* b6j* b6k* b6l* b7a* b7b* b7c* b7d* b7e* b7f* b7g* b7h* b7i* b7j* b7k* b9b* b10a* b10b* b13a* b13b* b13c* b14* c3* c4* c5* c12a* c12b* c12c* c12d* c12e* c12f* c19* d1* d2* d3* d6* d7a* d7b* d7c* d7d* d7e* d7f* d7g* d7h* d7i* d8a* d8b* d8c* d8d* d8e* d8f* d8g* d8h* d8i* d9a* d9b* d9c* d9d* d9e* d10a* d10b* d10c* d10d* d10e* d10f* d11* d13* d14* e1* e2* e3* e5* e6* f2* f3a* f3b* f3c* f3d* f3e* f4* f6* f7* f8* f9* f12* f14* f15* f16* f20* f21* f22* f23* f25* g2* g3* g6* g8* g9* g12* g13* g15* g17* g18* g22* g25* g26* g27* g28* g31* g34* g43* g55* g58* g59* g60* g61* g63* g65* g68a* g68b* g69* g70* g71* g91* g92* g93* g94* g95* Error in AveVar[s, i, ] - c(avevar.mean, avevar.sd) : number of items to replace is not a multiple of replacement length And here is what traceback() gives: traceback() 3: .local(object, ...) 2: mi(imp.data, info = info2, n.iter = 6, preprocess = FALSE) 1: mi(imp.data, info = info2, n.iter = 6, preprocess = FALSE) Andrew Miles __ 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] latex table question
Why do you think R-help is a mailing list about LaTeX?
Best,
Uwe Ligges
On 13.07.2010 23:05, Felipe Carrillo wrote:
Hi:
My head is spinning with this latex doc so hopefully after I align my tables to
the left of the page
my headache are going to be over. I always use:
\hspace*{-0.1in}
to move my figures horizontally to the left margin of the page but the table
below doesn't move at all,
but instead it gets sideways. Not sure if \begin{landscape} has something to do
with it or is just me.
I hope someone could lend some help on this matter.
\documentclass[11pt]{article}
\usepackage{longtable,verbatim}
\usepackage{longtable,pdflscape,graphicx}
\usepackage{fmtcount,hyperref} % displaying latex counters
%\usepackage[top=0.2inch, bottom=0.2inch, left=2cm, right=2cm]{geometry}
\usepackage{fullpage}
\usepackage{ctable}
\title{United States Department of the Interior}
\begin{document}
%\setlength{\topmargin}{-1inch} % Just an example
\setkeys{Gin}{width=1\textwidth} % makes all the graphics scales
\maketitle
echo=F,results=hide=
reportDF- structure(list(IDDate = c(3/12/2010, 3/13/2010, 3/14/2010,
3/15/2010), FirstRunoftheYear = c(33 (119 ? 119), n (0 ? 0), 893 (110 ?
146),
140 (111 ? 150)), SecondRunoftheYear = c(33 (71 ? 71), n (0 ? 0),
337 (67 ? 74), 140 (68 ? 84)), ThirdRunoftheYear = c(890 (32 ? 47),
n (0 ? 0), 10,602 (32 ? 52), 2,635 (34 ? 66)), FourthRunoftheYear = c(0 (
? ),
n (0 ? 0), 0 ( ? ), 0 ( ? )), LastRunoftheYear = c(0 ( ? ), n (0 ? 0),
0 ( ? ), 0 ( ? ))), .Names = c(IDDate, First Run of the Year, Second
Run of the Year,
Third Run of the Year, Fourth Run of the Year, Last Run of the Year),
row.names = c(NA, 4L), class = data.frame)
@
\begin{landscape}
\begin{table}[!tbp]
\begin{center}
\begin{tabular}{ll}\hline\hline
\multicolumn{1}{c}{IDDate}
\multicolumn{1}{c}{First Run of the Year}
\multicolumn{1}{c}{Second Run of the Year}
\multicolumn{1}{c}{Third Run of the Year}
\multicolumn{1}{c}{Fourth Run of the Year}
\multicolumn{1}{c}{Last Run of the Year}\tabularnewline
\hline
13/12/201033 (119 ? 119)33 (71 ? 71)890 (32 ? 47)0 ( ? )0 ( ?
)\tabularnewline
23/13/2010n (0 ? 0)n (0 ? 0)n (0 ? 0)n (0 ? 0)n (0 ? 0)\tabularnewline
33/14/2010893 (110 ? 146)337 (67 ? 74)10,602 (32 ? 52)0 ( ? )0 ( ?
)\tabularnewline
43/15/2010140 (111 ? 150)140 (68 ? 84)2,635 (34 ? 66)0 ( ? )0 ( ?
)\tabularnewline
\hline
\end{tabular}
\end{center}
\end{table}\end{landscape}
\end{document}
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Correct function name for text display of S4 object
On 13.07.2010 03:35, Andrew Liu wrote: Hello, I am working on an R package for storing order book data. I currently have a display method that has the following output (ob is an S4 object): display(ob) Current time is 09:35:02 Price Ask Size -- 11.42 900 11.41 1,400 11.40 1,205 11.39 1,600 11.38 400 -- 2,700 11.36 1,100 11.35 1,100 11.34 1,600 11.33 700 11.32 -- Bid Size Price The package already has show, summary, and plot methods. Is there a more conventional name than display for the above output, or is display as good as any other name? Well, if show/print and summary are already dong different things, then go with a new name Uwe Ligges Thanks, Andrew __ 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] latex table question
I' ve seen latex questions being solved here that's why I sent my question
to this list but someone already told me about the latex forum so the same
question has already been solved there, Sorry about that.
- Original Message
From: Uwe Ligges lig...@statistik.tu-dortmund.de
To: Felipe Carrillo mazatlanmex...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Sent: Sun, July 18, 2010 10:24:39 AM
Subject: Re: [R] latex table question
Why do you think R-help is a mailing list about LaTeX?
Best,
Uwe Ligges
On 13.07.2010 23:05, Felipe Carrillo wrote:
Hi:
My head is spinning with this latex doc so hopefully after I align my
tables
to
the left of the page
my headache are going to be over. I always use:
\hspace*{-0.1in}
to move my figures horizontally to the left margin of the page but the table
below doesn't move at all,
but instead it gets sideways. Not sure if \begin{landscape} has something
to
do
with it or is just me.
I hope someone could lend some help on this matter.
\documentclass[11pt]{article}
\usepackage{longtable,verbatim}
\usepackage{longtable,pdflscape,graphicx}
\usepackage{fmtcount,hyperref} % displaying latex counters
%\usepackage[top=0.2inch, bottom=0.2inch, left=2cm, right=2cm]{geometry}
\usepackage{fullpage}
\usepackage{ctable}
\title{United States Department of the Interior}
\begin{document}
%\setlength{\topmargin}{-1inch} % Just an example
\setkeys{Gin}{width=1\textwidth} % makes all the graphics scales
\maketitle
echo=F,results=hide=
reportDF- structure(list(IDDate = c(3/12/2010, 3/13/2010, 3/14/2010,
3/15/2010), FirstRunoftheYear = c(33 (119 ? 119), n (0 ? 0), 893
(110
?
146),
140 (111 ? 150)), SecondRunoftheYear = c(33 (71 ? 71), n (0 ? 0),
337 (67 ? 74), 140 (68 ? 84)), ThirdRunoftheYear = c(890 (32 ? 47),
n (0 ? 0), 10,602 (32 ? 52), 2,635 (34 ? 66)), FourthRunoftheYear =
c(0 (
? ),
n (0 ? 0), 0 ( ? ), 0 ( ? )), LastRunoftheYear = c(0 ( ? ), n (0 ?
0),
0 ( ? ), 0 ( ? ))), .Names = c(IDDate, First Run of the Year,
Second
Run of the Year,
Third Run of the Year, Fourth Run of the Year, Last Run of the Year),
row.names = c(NA, 4L), class = data.frame)
@
\begin{landscape}
\begin{table}[!tbp]
\begin{center}
\begin{tabular}{ll}\hline\hline
\multicolumn{1}{c}{IDDate}
\multicolumn{1}{c}{First Run of the Year}
\multicolumn{1}{c}{Second Run of the Year}
\multicolumn{1}{c}{Third Run of the Year}
\multicolumn{1}{c}{Fourth Run of the Year}
\multicolumn{1}{c}{Last Run of the Year}\tabularnewline
\hline
13/12/201033 (119 ? 119)33 (71 ? 71)890 (32 ? 47)0 ( ? )0 ( ?
)\tabularnewline
23/13/2010n (0 ? 0)n (0 ? 0)n (0 ? 0)n (0 ? 0)n (0 ? 0)\tabularnewline
33/14/2010893 (110 ? 146)337 (67 ? 74)10,602 (32 ? 52)0 ( ? )0 ( ?
)\tabularnewline
43/15/2010140 (111 ? 150)140 (68 ? 84)2,635 (34 ? 66)0 ( ? )0 ( ?
)\tabularnewline
\hline
\end{tabular}
\end{center}
\end{table}\end{landscape}
\end{document}
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
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Re: [R] log transformation
jarry jarryjarry58 at hotmail.com writes: Hi, I am reading an article in which, first, some given p-results are obtained, and, second, these are afterwards transformed and expressed as the NEGATIVE base-10 logarithm of the p-value. My question is if anyone could indicate how is such transformation achieved with R and how are they expressed exponentially, as in for example 1.8E-18. ?? presumably -log10(pvalue) ?? so a p-value of 1.8e-18 (i.e., 1.8 times 10^-18) would correspond to a value of 17.74 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: package plotrix
-Messaggio originale- Da: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Inviato: dom 18/07/2010 18.58 A: mau...@alice.it Cc: j...@bitwrit.com.au; r-h...@stat.math.ethz.ch Oggetto: Re: [R] package plotrix On 18.07.2010 17:59, mau...@alice.it wrote: I installed package plotrix because reading its vignette it looks like it can help me solve a legend problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function lgendg. You mean legendg. Yes. Sorry for my typo. library(plotrix) plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type=n, + main=Test of grouped legend function) legendg(0.5,0.8,c(one,two,three),pch=list(1,2:3,4:6), + col=list(2,3:4,5:7)) Error: could not find function legendg That works for me. Try to upgrade bopth R and plotrix if you have not already. I am running R 2.9.2 on my Mac OS/X 10.5.8 I always feel uncomfortable with upgrades. Shall I uninstall R 2.9.2 in advance of instaling the ew bversion for Mac ? I ave no reason for keeping two versions. My problem with the regular R legend function is that I cannot indicate in the legend the line plotted by the command abline Here is the code. Attached is the plot. Any suggestion is welcome. Thank you. Maura x11(width=10,heigh=8) plot(NN,LB,xlim=c(1,300),ylim=c(0,3), main=Intrinsic Dimensionality of 1D Helix,font.main=2,cex.main=2,col.main=red, xlab=Number of Nearest-Neighbors,ylab=Estimated Dimension,col.lab=red,cex.lab=1,font.lab=2, xaxt=n,yaxt=n,pch=19,col=red4) axis(1,at=NN[1:40],labels=NN[1:40],cex.lab=1,cex.axis=0.8,col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) axis(2,at=c(0,0.5,1,1.5,2,2.5,3,3.5,4),labels=c(0,0.5,1,1.5,2,2.5,3,3.5,4),cex.lab=1,cex.axis=0.8, col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) lines(NN,McG,pch=18,col=green) lines(NN,PBJD,pch=20,col=magenta1) points(NN,GRPR,pch=18,col=cyan) abline(h=1,pch= 1,lty=1,col=black) legend(topright,legend = c(Levina-Bickel,MacKay-Ghahramani,Pettis-Bailey-Jain-Dubes,Grassberger-Procaccia,Helix Dimension), text.width = strwidth(Pettis-Bailey-Jain-Dubes),pch = c(19,18,20,18,1), col = c(red4,green,magenta1,cyan,black),text.col = black,lty=c(-1,-1,-1,1), bg =snow) We do not have the data nor do we see the figure, hence this is not reproducible for us. Actually I attached the TIFF plot that I am attaching again. Thank you . Maura Uwe Ligges e tutti i telefonini TIM! Vai su __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: package plotrix
On 18.07.2010 19:57, David Winsemius wrote: On Jul 18, 2010, at 1:32 PM, mau...@alice.it wrote: -Messaggio originale- Da: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Inviato: dom 18/07/2010 18.58 A: mau...@alice.it Cc: j...@bitwrit.com.au; r-h...@stat.math.ethz.ch Oggetto: Re: [R] package plotrix On 18.07.2010 17:59, mau...@alice.it wrote: I installed package plotrix because reading its vignette it looks like it can help me solve a legend problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function lgendg. You mean legendg. Yes. Sorry for my typo. library(plotrix) plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type=n, + main=Test of grouped legend function) legendg(0.5,0.8,c(one,two,three),pch=list(1,2:3,4:6), + col=list(2,3:4,5:7)) Error: could not find function legendg That works for me. Try to upgrade bopth R and plotrix if you have not already. I am running R 2.9.2 on my Mac OS/X 10.5.8 I always feel uncomfortable with upgrades. Shall I uninstall R 2.9.2 in advance of instaling the ew bversion for Mac ? No need to do so. The appropriate 2.11.1 folders will be created in the R.framework folder and the links in the Current folder will be updated. This is ancient, hence please upgrade anyway - and you have still not reported your plotrix version nor if you updated the package. I ave no reason for keeping two versions. My problem with the regular R legend function is that I cannot indicate in the legend the line plotted by the command abline Here is the code. Attached is the plot. Any suggestion is welcome. Thank you. Maura x11(width=10,heigh=8) plot(NN,LB,xlim=c(1,300),ylim=c(0,3), main=Intrinsic Dimensionality of 1D Helix,font.main=2,cex.main=2,col.main=red, xlab=Number of Nearest-Neighbors,ylab=Estimated Dimension,col.lab=red,cex.lab=1,font.lab=2, xaxt=n,yaxt=n,pch=19,col=red4) axis(1,at=NN[1:40],labels=NN[1:40],cex.lab=1,cex.axis=0.8,col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) axis(2,at=c(0,0.5,1,1.5,2,2.5,3,3.5,4),labels=c(0,0.5,1,1.5,2,2.5,3,3.5,4),cex.lab=1,cex.axis=0.8, col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) lines(NN,McG,pch=18,col=green) lines(NN,PBJD,pch=20,col=magenta1) points(NN,GRPR,pch=18,col=cyan) abline(h=1,pch= 1,lty=1,col=black) legend(topright,legend = c(Levina-Bickel,MacKay-Ghahramani,Pettis-Bailey-Jain-Dubes,Grassberger-Procaccia,Helix Dimension), text.width = strwidth(Pettis-Bailey-Jain-Dubes),pch = c(19,18,20,18,1), col = c(red4,green,magenta1,cyan,black),text.col = black,lty=c(-1,-1,-1,1), bg =snow) We do not have the data nor do we see the figure, hence this is not reproducible for us. Actually I attached the TIFF plot that I am attaching again. Re-read the Posting Guide. It specifies what can be attached and TIFF files are NOT in the list of acceptable types. Uwe might have gotten a copy but the rest of us did not. No, the tiff probably did not pass our e-mail filters at all. Best, Uwe Ligges __ 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] Import of specific column of many space-delimited text files
Hi, I have about 300 space-delimited text files and from each file I want to import one specific column into R to create a data frame where all imported columns are included. Is there a smart way to do so? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Import-of-specific-column-of-many-space-delimited-text-files-tp2293273p2293273.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] NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
The problem is in data.frame[ and any NA in a logical vector will return a row of NA's. This can be avoid by wrapping which() around the logical vector which seems entirely wasteful or using subset(). The basic philosophy that causes this behaviour is sensible in my opinion: missing values must always be handled explicitly. Sure it's a bit of a hassle, but it being explicit prevents you from making major errors in an analysis. In my opinion, the flaw is that R doesn't apply this philosophy entirely consistently: I think factor and table should default to exclude = NULL. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ 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] Import of specific column of many space-delimited text files
Hi,
There might be better ways, but here's something that might help you
get started. Obviously you'll need to tweak a lot of things. If your
files do not have a uniform number of columns and the column you want
is not always in the same position, it may be easier to just read it
all in and then extract the column by name ocne they are in R.
# a little sample data
var1 var2 var3
54 6326 45
66 3452 44
52 5436 43
64 7865 48
54 6534 44
#how you can read in just one column
read.table(clipboard, header = TRUE, sep = ,
colClasses = c(NULL, integer, NULL))
#Grab a files in your WD that have .txt in them
file.names - grep(.txt, list.files(), value = TRUE)
#Initialize a list to store all the file you read in
mydata - vector(list, length = length(file.names))
#Read in data files and assign them to your list
for(i in seq_along(file.names)) {
mydata[[i]] - read.table(file = file.names[i], sep = ,
colClasses = c(NULL, integer, NULL))
}
Cheers,
Josh
On Sun, Jul 18, 2010 at 11:22 AM, LogLord nils.sch...@web.de wrote:
Hi,
I have about 300 space-delimited text files and from each file I want to
import one specific column into R to create a data frame where all imported
columns are included.
Is there a smart way to do so?
Thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/Import-of-specific-column-of-many-space-delimited-text-files-tp2293273p2293273.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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
On Jul 18, 2010, at 2:52 PM, Hadley Wickham wrote: The problem is in data.frame[ and any NA in a logical vector will return a row of NA's. This can be avoid by wrapping which() around the logical vector which seems entirely wasteful or using subset(). The basic philosophy that causes this behaviour is sensible in my opinion: missing values must always be handled explicitly. Sure it's a bit of a hassle, but it being explicit prevents you from making major errors in an analysis. How does this apply to the treatment of logical indexing in the function data.frame[ ? The returned value is not expected to have the same number of rows as the source object and the profusion of duplicate NA rows serves what purpose? Wouldn't it be more consistent to have the constructs: dfrm[which(logical_vector), ] dfrm[logical_vector , ] return the same object? In my opinion, the flaw is that R doesn't apply this philosophy entirely consistently: I think factor and table should default to exclude = NULL. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ 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] Import of specific column of many space-delimited text files
That worked perfectly well. Thanks a lot! -- View this message in context: http://r.789695.n4.nabble.com/Import-of-specific-column-of-many-space-delimited-text-files-tp2293273p2293342.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] loop troubles
This is the latest code I have been trying, but it when I try to turn the
vectors of SD into a vector of variances, it turns into a scalar.
n=c(NULL,13,89,50)
ybar=c(NULL,1.58,1.26,0.80)
sd=c(NULL,2.19,4.18,5.47)
mu=1;sigma=1;p0=0;var1=NULL;p=NULL;pn=NULL
for (i in 2:length(n)){
var1[i] = sd[i]^2 #sample variance
p[i] = n[i]/var1[i] #sample precision
p0=p[i-1]
pn[i] = p0[i]+p[i]
mu[i] = ((p0[i])/(pn[i])*mu[i])+((p[i])/(pn[i]))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu[i]
}
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a name
worth remembering. --Theodore Roosevelt
-Original Message-
From: Joe P King [mailto:j...@joepking.com]
Sent: Sunday, July 18, 2010 6:13 AM
To: 'r-help@r-project.org'
Subject: RE: [R] loop troubles
I tried that, this is what I tried, but I only get it to do one iteration
and then it wont cycle back. I am not sure how to tell it how to look at the
previous precision to add to the current new sample precision to get the new
precision. This is my first try at a loop so I am not sure if I am doing
anything right, sorry about the elementary question.
n=c(5,10,15)
ybar=c(12,13,14)
sd=c(2,3,4)
mutotals-matrix(nrow=length(n), ncol=1)
mu=1;p0=0
for (i in 1:length(n)){
var = sd[i]^2 #sample variance
p = n[i]/var #sample precision
p0[i]= i
pn = p0[i]+p[i]
mu.out = ((p0[i])/(pn)*mu[i])+((p)/(pn))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu.out[i]
}
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a name
worth remembering. --Theodore Roosevelt
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Sunday, July 18, 2010 5:36 AM
To: Joe P King
Cc: r-help@r-project.org
Subject: Re: [R] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n - c(5, 10, 15)
ybar - c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)-c(Posterior 1, Posterior 2, Posterior 3)
colnames(posterior)-c(Mu, SD)
return(posterior)}
---
Joe King, M.A.
Ph.D. Student
University of Washington - Seattle
206-913-2912
mailto:j...@joepking.com j...@joepking.com
---
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
[[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.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] loop troubles
On Jul 18, 2010, at 4:21 PM, Joe P King wrote:
This is the latest code I have been trying, but it when I try to
turn the
vectors of SD into a vector of variances, it turns into a scalar.
n=c(NULL,13,89,50)
ybar=c(NULL,1.58,1.26,0.80)
sd=c(NULL,2.19,4.18,5.47)
mu=1;sigma=1;p0=0;var1=NULL;p=NULL;pn=NULL
If you are going to passing values to the above using indices then
they should be defined as vectors of the appropriate length. It would
not be absolutely necessary to to so , but if you don't, then you need
to extend them using the c() function, which you do not seem to have
caught onto yet.
for (i in 2:length(n)){
var1[i] = sd[i]^2 #sample variance
p[i] = n[i]/var1[i] #sample precision
p0=p[i-1]
pn[i] = p0[i]+p[i]
mu[i] = ((p0[i])/(pn[i])*mu[i])+((p[i])/(pn[i]))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu[i]
}
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
-Original Message-
From: Joe P King [mailto:j...@joepking.com]
Sent: Sunday, July 18, 2010 6:13 AM
To: 'r-help@r-project.org'
Subject: RE: [R] loop troubles
I tried that, this is what I tried, but I only get it to do one
iteration
and then it wont cycle back. I am not sure how to tell it how to
look at the
previous precision to add to the current new sample precision to get
the new
precision. This is my first try at a loop so I am not sure if I am
doing
anything right, sorry about the elementary question.
n=c(5,10,15)
ybar=c(12,13,14)
sd=c(2,3,4)
mutotals-matrix(nrow=length(n), ncol=1)
mu=1;p0=0
for (i in 1:length(n)){
var = sd[i]^2 #sample variance
p = n[i]/var #sample precision
p0[i]= i
pn = p0[i]+p[i]
mu.out = ((p0[i])/(pn)*mu[i])+((p)/(pn))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]-mu.out[i]
}
Joe King
206-913-2912
j...@joepking.com
Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Sunday, July 18, 2010 5:36 AM
To: Joe P King
Cc: r-help@r-project.org
Subject: Re: [R] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n - c(5, 10, 15)
ybar - c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)-c(Posterior 1, Posterior 2, Posterior 3)
colnames(posterior)-c(Mu, SD)
return(posterior)}
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] Continuing on with a loop when there's a failure
Thanks very much again David for your helpful answers. However, the code STILL
does not appear to be working properly!
Even though the third time through the loop *should* work, it appears that R
has
given up after the second time through the loop. What I mean is: although y2
causes the lrm function to fail, y3 is a kosher variable. If the loop continues
on, it should give data for the model with y3. But if you look at the matrix
called results, it returns NA for the third spot corresponding to the model
of
y3:
results
y1 y2 y3
[1,] 0.6976063 NA NA
If you run y3 in isolation, rather than through the loop, you can see that it
should work and contribute data to the matrix called results:
mod.poly3 - lrm(x[,3] ~ pol(x1, 3) + pol(x2, 3), data=x)
anova(mod.poly3)[1,3]
[1] 0.6976063
Any ideas?
From: David Winsemius dwinsem...@comcast.net
Cc: Peter Konings peter.l.e.koni...@gmail.com; R Help r-help@r-project.org
Sent: Sun, July 18, 2010 3:33:07 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 18, 2010, at 4:25 AM, Josh B wrote:
Hello Peter,
I tried your suggestion, but I was still not able to get it to work. Would you
mind looking at my code again? Here's what I'm trying:
x - read.table(textConnection(y1 y2 y3 x1 x2
indv.1 bagels donuts bagels 4 6
indv.2 donuts donuts donuts 5 1
indv.3 donuts donuts donuts 1 10
indv.4 donuts donuts donuts 10 9
indv.5 bagels donuts bagels 0 2
indv.6 bagels donuts bagels 2 9
indv.7 bagels donuts bagels 8 5
indv.8 bagels donuts bagels 4 1
indv.9 donuts donuts donuts 3 3
indv.10 bagels donuts bagels 5 9
indv.11 bagels donuts bagels 9 10
indv.12 bagels donuts bagels 3 1
indv.13 donuts donuts donuts 7 10
indv.14 bagels donuts bagels 2 10
indv.15 bagels donuts bagels 9 6), header = TRUE)
closeAllConnections()
results - matrix(nrow = 1, ncol = 3)
colnames(results) - c(y1, y2, y3)
require(rms) # or Design
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x), silent=TRUE)
if(class(mod.poly3)[1] != 'try-error')
{results[1,i] - anova(mod.poly3)[1,3]}
}
results
y1 y2 y3
[1,] 0.6976063 NA NA
...and here's the output:
results
y1 y2 y3
[1,] NA NA NA
[[elided Yahoo spam]]
From: Peter Konings peter.l.e.koni...@gmail.com
Sent: Tue, July 13, 2010 5:45:17 PM
Subject: Re: [R] Continuing on with a loop when there's a failure
Hi Josh,
Test the class of the resulting object. If it is 'try-error' fill your result
with NA or do some other error handling.
result - try(somemodel)
if(class(result) == 'try-error')
{
# some error handling
} else {
# whatever happens if the result is ok
}
HTH
Peter.
In my opinion the try and tryCatch commands are written and documented rather
poorly. Thus I am not sure what to program exactly.
For instance, I could query mod.poly3 and use an if/then statement to
proceed,
but querying mod.poly3 is weird. For instance, here's the output when it
fails:
mod.poly3 - try(lrm(x[,2] ~ pol(x1, 3) + pol(x2, 3), data=x))
Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol, weights =
weights, :
NA/NaN/Inf in foreign function call (arg 1)
mod.poly3
[1] Error in fitter(X, Y, penalty.matrix = penalty.matrix, tol = tol,
weights
=
weights, : \n NA/NaN/Inf in foreign function call (arg 1)\n
attr(,class)
[1] try-error
...and here's the output when it succeeds:
mod.poly3 - try(lrm(x[,1] ~ pol(x1, 3) + pol(x2, 3), data=x))
mod.poly3
Logistic Regression Model
lrm(formula = x[, 1] ~ pol(x1, 3) + pol(x2, 3), data = x)
Frequencies of Responses
bagels donuts
10 5
Obs Max Deriv Model L.R. d.f. P C
15 4e-04 3.37 6 0.7616 0.76
Dxy Gamma Tau-a R2 Brier g
0.52 0.52 0.248 0.279 0.183 1.411
gr gp
4.1 0.261
Coef S.E.Wald Z P
Intercept -5.68583 5.23295 -1.09 0.2772
x1 1.87020 2.14635 0.87 0.3836
x1^2 -0.42494 0.48286 -0.88 0.3788
x1^3 0.02845 0.03120 0.91 0.3618
x2 3.49560 3.54796 0.99 0.3245
x2^2 -0.94888 0.82067 -1.16 0.2476
x2^3 0.06362 0.05098 1.25 0.2121
...so what exactly would I query to design my if/then statement?
From: David Winsemius dwinsem...@comcast.net
To: David Winsemius dwinsem...@comcast.net
Sent: Tue, July 13, 2010 9:09:04 AM
Subject: Re: [R] Continuing on with a loop when there's a failure
On Jul 13, 2010, at 9:04 AM, David Winsemius wrote:
On Jul 13, 2010, at 8:47 AM, Josh B wrote:
Thanks again, David.
[[elided Yahoo spam]]
(BTW, it did work.)
Here's what I'm trying now:
for (i in 1:2) {
mod.poly3 - try(lrm(x[,i] ~ pol(x1, 3) + pol(x2, 3), data=x))
[R] any one knowing how to install library (equate)?
I was told to go to packages on the tool bar and select 'install packages'. Then choose library (equate). My R is version 2.11. I tried multiple times with different locations of R within US. Yet, I didn't see library(equate) being offered in the packages. Thanks. Grace [[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] R: package plotrix
On Jul 18, 2010, at 1:32 PM, mau...@alice.it wrote: -Messaggio originale- Da: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Inviato: dom 18/07/2010 18.58 A: mau...@alice.it Cc: j...@bitwrit.com.au; r-h...@stat.math.ethz.ch Oggetto: Re: [R] package plotrix On 18.07.2010 17:59, mau...@alice.it wrote: I installed package plotrix because reading its vignette it looks like it can help me solve a legend problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function lgendg. You mean legendg. Yes. Sorry for my typo. library(plotrix) plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type=n, + main=Test of grouped legend function) legendg(0.5,0.8,c(one,two,three),pch=list(1,2:3,4:6), + col=list(2,3:4,5:7)) Error: could not find function legendg That works for me. Try to upgrade bopth R and plotrix if you have not already. I am running R 2.9.2 on my Mac OS/X 10.5.8 I always feel uncomfortable with upgrades. Shall I uninstall R 2.9.2 in advance of instaling the ew bversion for Mac ? No need to do so. The appropriate 2.11.1 folders will be created in the R.framework folder and the links in the Current folder will be updated. I ave no reason for keeping two versions. My problem with the regular R legend function is that I cannot indicate in the legend the line plotted by the command abline Here is the code. Attached is the plot. Any suggestion is welcome. Thank you. Maura x11(width=10,heigh=8) plot(NN,LB,xlim=c(1,300),ylim=c(0,3), main=Intrinsic Dimensionality of 1D Helix,font.main=2,cex.main=2,col.main=red, xlab=Number of Nearest-Neighbors,ylab=Estimated Dimension,col.lab=red,cex.lab=1,font.lab=2, xaxt=n,yaxt=n,pch=19,col=red4) axis (1 ,at = NN [1 : 40 ],labels = NN [1 :40],cex.lab=1,cex.axis=0.8,col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) axis (2 ,at = c (0,0.5,1,1.5,2,2.5,3,3.5,4 ),labels=c(0,0.5,1,1.5,2,2.5,3,3.5,4),cex.lab=1,cex.axis=0.8, col.axis=blue,lwd.ticks=0.5,col.ticks =gray3) lines(NN,McG,pch=18,col=green) lines(NN,PBJD,pch=20,col=magenta1) points(NN,GRPR,pch=18,col=cyan) abline(h=1,pch= 1,lty=1,col=black) legend(topright,legend = c(Levina-Bickel,MacKay- Ghahramani,Pettis-Bailey-Jain-Dubes,Grassberger- Procaccia,Helix Dimension), text.width = strwidth(Pettis-Bailey-Jain-Dubes),pch = c(19,18,20,18,1), col = c(red4,green,magenta1,cyan,black),text.col = black,lty=c(-1,-1,-1,1), bg =snow) We do not have the data nor do we see the figure, hence this is not reproducible for us. Actually I attached the TIFF plot that I am attaching again. Re-read the Posting Guide. It specifies what can be attached and TIFF files are NOT in the list of acceptable types. Uwe might have gotten a copy but the rest of us did not. -- 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] Question from day2 beginner
Hello, I just started learning R and have a very basic question: When I try ar model and extract residuals it returns Null. Could someone explain what's going on here? Does that mean the model is null? But it seems residual has a length=# observation of the original series according to the model summary. I am learning it by myself and have no one to ask here so please allow me to ask this very basic question... Thank you very much. summary(ar(logrvar, aic=TRUE)) Length Class Mode order1 -none- numeric ar 40 -none- numeric var.pred 1 -none- numeric x.mean 1 -none- numeric aic 44 -none- numeric n.used 1 -none- numeric order.max1 -none- numeric partialacf 43 -none- numeric resid20731 -none- numeric method 1 -none- character series 1 -none- character frequency1 -none- numeric call 3 -none- call asy.var.coef 1600 -none- numeric resid(ar(logrvar, aic=TRUE)) NULL -- View this message in context: http://r.789695.n4.nabble.com/Question-from-day2-beginner-tp2293221p2293221.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] Reading files with varying number of columns
Thanks for the suggestion. I had read fill but didn't pick up on the details. It only uses the first five columns to determine column dimensions, so it ignored my last two columns. Specifying col.names with fill=TRUE did the trick! Aloha, Tim - Original Message From: Uwe Ligges lig...@statistik.tu-dortmund.de To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Sent: Sun, July 18, 2010 6:03:47 AM Subject: Re: [R] Reading files with varying number of columns See argument fill in ?read.table Uwe Ligges On 18.07.2010 12:14, Tim Clark wrote: Dear R list, I am trying to read files with a varying number of columns and can't figure out how to get them in the proper format. Some rows have five value and some have seven. Is there a way to read them in so that two empty columns are added to every row that has only five values? An example of the data follows. Thanks! Tim 46,B,00301,2004-02-03,11:59:16 46,B,00301,2004-02-03,12:03:43 46,B,00301,2004-02-03,12:39:17 46,B,00301,2004-02-03,23:07:28 43,C,00232,2004-02-04,07:39:57,-1.5,meters 44,A,00230,2004-02-04,07:44:59 44,A,00230,2004-02-04,07:46:19 44,A,00230,2004-02-04,07:48:24 44,A,00230,2004-02-04,07:49:12 43,C,00232,2004-02-04,07:53:08,2.5,meters 44,A,00230,2004-02-04,07:54:34 44,A,00230,2004-02-04,07:55:08 43,C,00232,2004-02-04,07:56:18,2.9,meters 43,C,00232,2004-02-04,07:56:39,2.5,meters 46,B,00301,2004-02-04,08:00:49 43,C,00232,2004-02-04,08:02:12,-1.1,meters 43,C,00232,2004-02-04,08:04:01,2.2,meters 43,C,00232,2004-02-04,08:04:26,3.3,meters 43,C,00232,2004-02-04,08:40:26,2.9,meters 43,C,00232,2004-02-04,08:41:04,-2.2,meters 43,C,00232,2004-02-04,08:42:44,2.2,meters 43,C,00232,2004-02-04,08:44:31,2.9,meters 44,A,00231,2004-02-04,09:04:43 44,A,00231,2004-02-04,09:10:30 44,A,00231,2004-02-04,09:12:08 44,A,00231,2004-02-04,09:13:32 44,A,00231,2004-02-04,09:14:06 43,C,00232,2004-02-04,09:18:24,0.4,meters 43,C,00232,2004-02-04,09:20:13,2.5,meters 44,A,00231,2004-02-04,09:21:04 44,A,00231,2004-02-04,09:22:53 44,A,00231,2004-02-04,09:25:32 44,A,00231,2004-02-04,09:29:31 44,A,00231,2004-02-04,09:30:05 43,C,00232,2004-02-04,09:53:25,2.5,meters 43,C,00232,2004-02-04,09:53:53,1.5,meters 46,B,00301,2004-02-04,22:16:24 46,B,00301,2004-02-04,22:19:32 46,B,00258,2004-02-04,23:44:01 46,B,00258,2004-02-04,23:44:28 Tim Clark Marine Ecologist National Park of American Samoa Pago Pago, American Samoa 96799 __ 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] Neural Network
Hi all, I am working for my master's thesis and I need to do a neural network to forecast stock market price, with also external inputs like technical indicators. I would like to know which function and package of R are more suitable for this study. Thanks a lot for your response, Arnaud TREBAOL. -- Arnaud Trébaol T.I.M.E. Student Ecole Centrale de Lille (09) Politecnico di Milano (10) Mail : arnaud.treb...@centraliens-lille.org Tel1 : +33 (0)6 76 46 42 92 Tel2 : +39 327 280 57 68 [[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] Neural Network
I'd start with the nnet library type: ?nnet CS - Corey Sparks, PhD Assistant Professor Department of Demography and Organization Studies University of Texas at San Antonio 501 West Durango Blvd Monterey Building 2.270C San Antonio, TX 78207 210-458-3166 corey.sparks 'at' utsa.edu https://rowdyspace.utsa.edu/users/ozd504/www/index.htm -- View this message in context: http://r.789695.n4.nabble.com/Neural-Network-tp2293366p2293369.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] simple loop(?) analysing subsets
Hi All,
I have a large data set with many columns of data. One of these columns is a
species identifier and the remainder are variables such as temperature or
mass. Currently I am carrying out a single regression on subsets of the data
set, e.g. separated data sets with only the data from one species at a time.
I have been searching for a thread that will help me to understand how best
to repeat this process for each different species identifier in that
variable column. I can’t seem to find one that is similar to what I am
trying to do. It might be the case that I am not looking for the right thing
or that I do not fully understand the process.
How do I run a simple loop that produces a regression for each species as
identified in the variable species id, which is one column in the large data
set that I am using?
Simple regression that I wish to repeat
data- read.table(…/STUDY.txt,header=T)
names(data)
model- with(data,{lm(o2con~bm)})
summary(model)
sample data set
species id o2con bm
1 0.5 5
1 0.6 2
1 0.4 4
1 0.4 2
1 0.5 3
2 0.3 7
2 0.4 8
2 0.5 3
2 0.7 4
2 0.9 2
3 0.3 6
3 0.7 2
3 0.4 1
3 0.3 7
5 0.3 2
5 0.6 1
5 0.9 7
5 0.2 5
I would be very grateful for some help with this. I really like using R and
I can usually figure out what I want to do but I have been trying to figure
this out for a while now and I am getting nowhere.
Thank you.
--
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http://r.789695.n4.nabble.com/simple-loop-analysing-subsets-tp2293383p2293383.html
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] any one knowing how to install library (equate)?
On 18/07/2010 2:57 PM, ying_chen wang wrote: I was told to go to packages on the tool bar and select 'install packages'. Then choose library (equate). My R is version 2.11. I tried multiple times with different locations of R within US. Yet, I didn't see library(equate) being offered in the packages. You should ask the person who told you that for better instructions. 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] simple loop(?) analysing subsets
Hello,
Take a look at ?by
Something like:
by(data, data[ , species id], function(x) {lm(o2con ~ bm, data = x)})
As an aside, it can be a bit cumbersome to deal with spaces in a
variables name (because it has to be quoted then). Also since data is
a function, something like mydata might make it easier to keep code
straight.
HTH,
Josh
On Sun, Jul 18, 2010 at 2:29 PM, karmakiller roisinmoria...@gmail.com wrote:
Hi All,
I have a large data set with many columns of data. One of these columns is a
species identifier and the remainder are variables such as temperature or
mass. Currently I am carrying out a single regression on subsets of the data
set, e.g. separated data sets with only the data from one species at a time.
I have been searching for a thread that will help me to understand how best
to repeat this process for each different species identifier in that
variable column. I can’t seem to find one that is similar to what I am
trying to do. It might be the case that I am not looking for the right thing
or that I do not fully understand the process.
How do I run a simple loop that produces a regression for each species as
identified in the variable species id, which is one column in the large data
set that I am using?
Simple regression that I wish to repeat
data- read.table(…/STUDY.txt,header=T)
names(data)
model- with(data,{lm(o2con~bm)})
summary(model)
sample data set
species id o2con bm
1 0.5 5
1 0.6 2
1 0.4 4
1 0.4 2
1 0.5 3
2 0.3 7
2 0.4 8
2 0.5 3
2 0.7 4
2 0.9 2
3 0.3 6
3 0.7 2
3 0.4 1
3 0.3 7
5 0.3 2
5 0.6 1
5 0.9 7
5 0.2 5
I would be very grateful for some help with this. I really like using R and
I can usually figure out what I want to do but I have been trying to figure
this out for a while now and I am getting nowhere.
Thank you.
--
View this message in context:
http://r.789695.n4.nabble.com/simple-loop-analysing-subsets-tp2293383p2293383.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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.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.
Re: [R] any one knowing how to install library (equate)?
Hi Grace, I'm using the UCLA mirror and this code installs equate fine for me. #install the package (note that the quoted name) install.packages(equate) #load the package library(equate) It also looks like the packages has passed the CRAN checks for all OSes, so availability is probably not an issue. Best regards, Josh On Sun, Jul 18, 2010 at 11:57 AM, ying_chen wang gracedrop.w...@gmail.com wrote: I was told to go to packages on the tool bar and select 'install packages'. Then choose library (equate). My R is version 2.11. I tried multiple times with different locations of R within US. Yet, I didn't see library(equate) being offered in the packages. Thanks. Grace [[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. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.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] Inserting an image into a PDF file
Hi On 12/07/2010 4:33 a.m., Marc Schwartz wrote: On Jul 11, 2010, at 10:51 AM, David Winsemius wrote: On Jul 11, 2010, at 11:06 AM, Dennis Fisher wrote: Colleagues, I am creating a PDF document from R using pdf(), the document contains text / images created with R. I also have an image in either PDF / TIFF / JPEG format (in this case, a scanned image saved as a file). I would like to insert the image into the PDF file. Any ideas on how to accomplish this? http://www.stat.auckland.ac.nz/~paul/Talks/import.pdf (OS X, R 2.11.1) Just to offer an additional option since Dennis is on OSX, if you are on 10.6.x (Snow Leopard), you can view this YouTube video: http://www.youtube.com/watch?v=zF_WgXAfMP0 which describes how to combine/insert pages from multiple document types that OSX' Preview can open. If you are on Leopard (10.5.x), the drag and drop behavior, as noted in the above video, is somewhat different and you may wish to view this page: http://macintoshhowto.com/leopard/how-to-merge-pdf-files-with-preview-in-leopard.html This presumes that you want the scanned image on a separate page, rather than combined on the same page as your pdf() output. Since PDF is a native format on OSX, you can take advantage of built-in functionality to manipulate these files. Of course using Sweave is yet another option. Also see the new rasterImage() and grid.raster() functions in R 2.11.* Paul HTH, Marc Schwartz __ 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. -- 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/ __ 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] re. Mathematica and R
The wiki on rcom.univie.ac.at in section R(D)COM, rcom, and other software systems has an example how to use R from within Mathematica on Windows. You will need to install statconnDCOM and rcom available from the same site. On Jul 17, 2010, at 1:06 PM, Alan Kelly wrote: David - information on calling R from within Mathematica can be found at the following link: http://www.mofeel.net/1164-comp-soft-sys-math-mathematica/13022.aspx HTH, Alan Kelly __ 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. -- Erich Neuwirth Didactic Center for Computer Science and Institute for Scientific Computing University of Vienna [[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 with Reshaping from Wide to Long
Hi Phil and Jeff, Thanks so much for taking the time to help me solve this issue! Both approaches work perfectly. Each of your approaches helped me learn more about what R can do. I really appreciate your help! Very best regards, John -- View this message in context: http://r.789695.n4.nabble.com/Help-with-Reshaping-from-Wide-to-Long-tp2292462p2293463.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] specifying column names in a vector of characters and the use?
Hi,
What I would like to do is have a data.frame with column names and have
these column names stored as strings in another vector. Then I would like
to be able to access the data.fram columns via referencing the vector of
names. The code below shows the last few executions that failed to retrieve
the values for column named X1. Seth
table.1-cbind(c(1,2,3,2,2),c(0,9,0,7,9),c(7,5,9,8,8))
table.1
[,1] [,2] [,3]
[1,]107
[2,]295
[3,]309
[4,]278
[5,]298
table.1-data.frame(table.1)
table.1
X1 X2 X3
1 1 0 7
2 2 9 5
3 3 0 9
4 2 7 8
5 2 9 8
hold-c(X1,X2,X3)
hold
[1] X1 X2 X3
table.1$X1
[1] 1 2 3 2 2
hold[1]
[1] X1
table.1$hold[1] # FROM HERE DOWN ARE MY ATTEMPTS TO ACCESS X1
NULL
table.1$(hold[1])
Error: unexpected '(' in table.1$(
table.1$get(hold[1])
Error: attempt to apply non-function
table.1$(get(hold[1]))
Error: unexpected '(' in table.1$(
--
View this message in context:
http://r.789695.n4.nabble.com/specifying-column-names-in-a-vector-of-characters-and-the-use-tp2293494p2293494.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] specifying column names in a vector of characters and the use?
Hello,
What I would like to do is have a data.frame with column names and have
these column names stored as strings in another vector. Then I would like
to be able to access the data.fram columns via referencing the vector of
names. The code below shows the last few executions that failed to retrieve
the values for column named X1. Seth
table.1-cbind(c(1,2,3,2,2),c(0,9,0,7,9),c(7,5,9,8,8))
table.1
[,1] [,2] [,3]
[1,]107
[2,]295
[3,]309
[4,]278
[5,]298
See easier way to construct your object below.
snip
[1] X1
table.1$hold[1] # FROM HERE DOWN ARE MY ATTEMPTS TO ACCESS X1
NULL
table.1$(hold[1])
Error: unexpected '(' in table.1$(
table.1$get(hold[1])
Error: attempt to apply non-function
table.1$(get(hold[1]))
Error: unexpected '(' in table.1$(
You need to index with ?[
Here is a complete example:
table.1- data.frame(X1 = c(1,2,3,2,2),
X2 = c(0,9,0,7,9),
X3 = c(7,5,9,8,8))
hold - names(table.1)
## to return a data.frame
table.1[hold[1]]
## to return a vector
table.1[[hold[1]]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] specifying column names in a vector of characters and the use?
Try:
table.1[[hold[1]]]
Cheers,
Simon.
On 19/07/10 12:09, Seth wrote:
Hi,
What I would like to do is have a data.frame with column names and have
these column names stored as strings in another vector. Then I would like
to be able to access the data.fram columns via referencing the vector of
names. The code below shows the last few executions that failed to retrieve
the values for column named X1. Seth
table.1-cbind(c(1,2,3,2,2),c(0,9,0,7,9),c(7,5,9,8,8))
table.1
[,1] [,2] [,3]
[1,]107
[2,]295
[3,]309
[4,]278
[5,]298
table.1-data.frame(table.1)
table.1
X1 X2 X3
1 1 0 7
2 2 9 5
3 3 0 9
4 2 7 8
5 2 9 8
hold-c(X1,X2,X3)
hold
[1] X1 X2 X3
table.1$X1
[1] 1 2 3 2 2
hold[1]
[1] X1
table.1$hold[1] # FROM HERE DOWN ARE MY ATTEMPTS TO ACCESS X1
NULL
table.1$(hold[1])
Error: unexpected '(' in table.1$(
table.1$get(hold[1])
Error: attempt to apply non-function
table.1$(get(hold[1]))
Error: unexpected '(' in table.1$(
--
Simon Blomberg, BSc (Hons), PhD, MAppStat.
Lecturer and Consultant Statistician
School of Biological Sciences
The University of Queensland
St. Lucia Queensland 4072
Australia
T: +61 7 3365 2506
email: S.Blomberg1_at_uq.edu.au
http://www.uq.edu.au/~uqsblomb/
Policies:
1. I will NOT analyse your data for you.
2. Your deadline is your problem
Statistics is the grammar of science - Karl Pearson.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] specifying column names in a vector of characters and the use?
On Jul 18, 2010, at 10:09 PM, Seth wrote:
Hi,
What I would like to do is have a data.frame with column names and
have
these column names stored as strings in another vector. Then I
would like
to be able to access the data.fram columns via referencing the
vector of
names. The code below shows the last few executions that failed to
retrieve
the values for column named X1. Seth
table.1-cbind(c(1,2,3,2,2),c(0,9,0,7,9),c(7,5,9,8,8))
table.1
[,1] [,2] [,3]
[1,]107
[2,]295
[3,]309
[4,]278
[5,]298
table.1-data.frame(table.1)
table.1
X1 X2 X3
1 1 0 7
2 2 9 5
3 3 0 9
4 2 7 8
5 2 9 8
hold-c(X1,X2,X3)
hold
[1] X1 X2 X3
table.1$X1
[1] 1 2 3 2 2
hold[1]
[1] X1
table.1$hold[1] # FROM HERE DOWN ARE MY ATTEMPTS TO ACCESS X1
NULL
Try instead:
table.1[ , hold[1] ]
The $ formalism does not evaluate its argument, but the [ function
does.
--
David.
table.1$(hold[1])
Error: unexpected '(' in table.1$(
table.1$get(hold[1])
Error: attempt to apply non-function
table.1$(get(hold[1]))
Error: unexpected '(' in table.1$(
--
View this message in context:
http://r.789695.n4.nabble.com/specifying-column-names-in-a-vector-of-characters-and-the-use-tp2293494p2293494.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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about KLdiv and large datasets
Is the 'eps' argument part of KLdiv (was not able to find that in the help pages) or part of a general environment (such as the graphics parameters 'par' ) ? I am asking so that I can read about it what it actually does to resolve the question you already raised about its reliability... Ralf On Fri, Jul 16, 2010 at 10:41 AM, Peter Ehlers ehl...@ucalgary.ca wrote: On 2010-07-16 7:56, Ralf B wrote: Hi all, when running KL on a small data set, everything is fine: require(flexmix) n- 20 a- rnorm(n) b- rnorm(n) mydata- cbind(a,b) KLdiv(mydata) however, when this dataset increases require(flexmix) n- 1000 a- rnorm(n) b- rnorm(n) mydata- cbind(a,b) KLdiv(mydata) KL seems to be not defined. Can somebody explain what is going on? Thanks, Ralf Ralf, You can adjust the 'eps=' argument. But I don't know what this will do to the reliability of the results. KLdiv(mydata, eps = 1e-7) -Peter Ehlers __ 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] Cramer-von Mises test
I have found the Kolmogorov-Smirnov (ks.test) and Anderson-Darling (ad.test in package ADGofTest) goodness of fit tests, but was wondering whether there's a package implementing the Cramer-von Mises test for a general distribution (i.e. not just the one for the normal distribution in the nortest package). Any suggestions? Regards, Sean. -- Sean Carmody Twitter: http://twitter.com/seancarmody Stable: http://mulestable.net/sean The Stubborn Mule Blog: http://www.stubbornmule.net Forum: http://mulestable.net/ __ 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] Connecting to MS Access database
Hi All, Can anyone please suggest me from where should I start to learn about 'how to connect to access db' ? How if someone has some written code and I can go over that to understand and make necessary changes... any help would be highly appreciated, -- Xin Ge. [[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] Connecting to MS Access database
Hey Xin this is a piece I copied from my blog. library(RODBC) mdbConnect-odbcConnectAccess(C:\\temp\\demo.mdb) sqlTables(mdbConnect) demo-sqlFetch(mdbConnect, tblDemo) odbcClose(mdbConnect) rm(demo) On Mon, Jul 19, 2010 at 12:05 AM, Xin Ge xingemaill...@gmail.com wrote: Hi All, Can anyone please suggest me from where should I start to learn about 'how to connect to access db' ? How if someone has some written code and I can go over that to understand and make necessary changes... any help would be highly appreciated, -- Xin Ge. [[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. -- == WenSui Liu wens...@paypal.com statcompute.spaces.live.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] Question about KLdiv and large datasets
I just answered this but realize that I did so off-list. So, for completeness, here's what I said: I think I see the problem. From ?KLdiv, you're getting the modeltools help page. What you need is the flexmix help page for KLdiv. Just get to the flexmix index page (you can do ?flexmix and then click on the 'Index' link at the bottom). Then select KLdiv-method. Here you will find the eps argument described. Its default value is eps=10^-4 and the description says: eps Probabilities below this threshold are replaced by this threshold for numerical stability. It's this comment that makes me doubt the reliability for very small eps values. Cheers, Peter On 2010-07-18 20:50, Ralf B wrote: Is the 'eps' argument part of KLdiv (was not able to find that in the help pages) or part of a general environment (such as the graphics parameters 'par' ) ? I am asking so that I can read about it what it actually does to resolve the question you already raised about its reliability... Ralf On Fri, Jul 16, 2010 at 10:41 AM, Peter Ehlersehl...@ucalgary.ca wrote: On 2010-07-16 7:56, Ralf B wrote: Hi all, when running KL on a small data set, everything is fine: require(flexmix) n- 20 a- rnorm(n) b- rnorm(n) mydata- cbind(a,b) KLdiv(mydata) however, when this dataset increases require(flexmix) n- 1000 a- rnorm(n) b- rnorm(n) mydata- cbind(a,b) KLdiv(mydata) KL seems to be not defined. Can somebody explain what is going on? Thanks, Ralf Ralf, You can adjust the 'eps=' argument. But I don't know what this will do to the reliability of the results. KLdiv(mydata, eps = 1e-7) -Peter Ehlers __ 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] simple loop(?) analysing subsets
Hi:
Time to jack up your level of R knowledge, courtesy of the apply family.
The 'R way' to do what you want is to split the data by species into list
components, run lm() on each component and save the resulting lm objects in
a list. The next trick is to figure out how to extract what you want, which
may require a bit more ingenuity in delving into aRcana :)
-
Aside:
To reinforce Joshua's point, variable names with spaces not explicitly
enclosed in quotes is bad practice, especially when someone who wants to
help tries to copy and paste your data into his/her R session:
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
:
line 1 did not have 4 elements
R expected four columns of data, but you provided three. In the future, it's
a good idea to include your data example with dput(), which outputs
dput(d)
structure(list(species = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 5L, 5L, 5L, 5L), o2con = c(0.5, 0.6, 0.4,
0.4, 0.5, 0.3, 0.4, 0.5, 0.7, 0.9, 0.3, 0.7, 0.4, 0.3, 0.3, 0.6,
0.9, 0.2), bm = c(5L, 2L, 4L, 2L, 3L, 7L, 8L, 3L, 4L, 2L, 6L,
2L, 1L, 7L, 2L, 1L, 7L, 5L)), .Names = c(species, o2con,
bm), class = data.frame, row.names = c(NA, -18L))
This is easily copied and pasted into anyone's R sessionbut I digress.
--
Calling your data frame d, here's how to run the same regression model on
all species:
# Create a function to perform the modeling, taking a data frame df as input
f - function(df) lm(o2con ~ bm, data = df)
# Use lapply() to apply the function to each 'split' of the data, by
species:
v - lapply(split(d, d$species), f)
# v is a list object, where each component of the list is an lm object,
# which itself is a list. In other words, it's a list of lists. do.call() is
a
# very useful function that applies a function to components of a list.
# rbind and cbind are commonly used to slurp together common elements
# from each component of a list.
# Pulling out the coefficients from each model:
do.call(rbind, lapply(v, coef))
(Intercept) bm
1 0.5176471 -0.01176471
2 0.9253731 -0.07611940
3 0.5942308 -0.04230769
5 0.3351648 0.04395604
# Extract the r-squared values from each model:
g - function(m) summary(m)$r.squared
do.call(rbind, lapply(v, g))
[,1]
1 0.03361345
2 0.66932578
3 0.43291592
5 0.14652015
# But you have to be careful...e.g., since you have unequal sample sizes per
species,
do.call(cbind, lapply(v, resid))
1 23 5
1 0.04117647 -0.09253731 -0.040384615 -0.1230769
2 0.10588235 0.08358209 0.190384615 0.2208791
3 -0.07058824 -0.19701493 -0.151923077 0.2571429
4 -0.09411765 0.07910448 0.001923077 -0.3549451
5 0.01764706 0.12686567 -0.040384615 -0.1230769
Warning message:
In function (..., deparse.level = 1) :
number of rows of result is not a multiple of vector length (arg 3)
Notice how the first residual is recycled in each of groups 3 and 5. That's
a potential gotcha.
This gives you a small glimpse into the power that R can deliver in data
analysis.
HTH,
Dennis
On Sun, Jul 18, 2010 at 2:29 PM, karmakiller roisinmoria...@gmail.comwrote:
Hi All,
I have a large data set with many columns of data. One of these columns is
a
species identifier and the remainder are variables such as temperature or
mass. Currently I am carrying out a single regression on subsets of the
data
set, e.g. separated data sets with only the data from one species at a
time.
I have been searching for a thread that will help me to understand how best
to repeat this process for each different species identifier in that
variable column. I cant seem to find one that is similar to what I am
trying to do. It might be the case that I am not looking for the right
thing
or that I do not fully understand the process.
How do I run a simple loop that produces a regression for each species as
identified in the variable species id, which is one column in the large
data
set that I am using?
Simple regression that I wish to repeat
data- read.table(
/STUDY.txt,header=T)
names(data)
model- with(data,{lm(o2con~bm)})
summary(model)
sample data set
species id o2con bm
1 0.5 5
1 0.6 2
1 0.4 4
1 0.4 2
1 0.5 3
2 0.3 7
2 0.4 8
2 0.5 3
2 0.7 4
2 0.9 2
3 0.3 6
3 0.7 2
3 0.4 1
3 0.3 7
5 0.3 2
5 0.6 1
5 0.9 7
5 0.2 5
I would be very grateful for some help with this. I really like using R and
I can usually figure out what I want to do but I have been trying to figure
this out for a while now and I am getting nowhere.
Thank you.
--
View this
Re: [R] Cramer-von Mises test
On 2010-07-18 22:04, Sean Carmody wrote: I have found the Kolmogorov-Smirnov (ks.test) and Anderson-Darling (ad.test in package ADGofTest) goodness of fit tests, but was wondering whether there's a package implementing the Cramer-von Mises test for a general distribution (i.e. not just the one for the normal distribution in the nortest package). Any suggestions? Sean, Here's one way you can investigate this yourself: 1. install and load package 'sos'; 2. issue: findFn(Cramer-von Mises test) or, for a little wider scope: findFn(Cramer-von Mises) (in place of 'findFn', you can use '???') -Peter Ehlers Regards, Sean. __ 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] simple loop(?) analysing subsets
Not to hijack the thread, but for my edification, what are the
advantages/disadvantages of split() + lapply() compared to by()?
Josh
On Sun, Jul 18, 2010 at 9:50 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Time to jack up your level of R knowledge, courtesy of the apply family.
The 'R way' to do what you want is to split the data by species into list
components, run lm() on each component and save the resulting lm objects in
a list. The next trick is to figure out how to extract what you want, which
may require a bit more ingenuity in delving into aRcana :)
-
Aside:
To reinforce Joshua's point, variable names with spaces not explicitly
enclosed in quotes is bad practice, especially when someone who wants to
help tries to copy and paste your data into his/her R session:
Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
:
line 1 did not have 4 elements
R expected four columns of data, but you provided three. In the future, it's
a good idea to include your data example with dput(), which outputs
dput(d)
structure(list(species = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 5L, 5L, 5L, 5L), o2con = c(0.5, 0.6, 0.4,
0.4, 0.5, 0.3, 0.4, 0.5, 0.7, 0.9, 0.3, 0.7, 0.4, 0.3, 0.3, 0.6,
0.9, 0.2), bm = c(5L, 2L, 4L, 2L, 3L, 7L, 8L, 3L, 4L, 2L, 6L,
2L, 1L, 7L, 2L, 1L, 7L, 5L)), .Names = c(species, o2con,
bm), class = data.frame, row.names = c(NA, -18L))
This is easily copied and pasted into anyone's R sessionbut I digress.
--
Calling your data frame d, here's how to run the same regression model on
all species:
# Create a function to perform the modeling, taking a data frame df as input
f - function(df) lm(o2con ~ bm, data = df)
# Use lapply() to apply the function to each 'split' of the data, by
species:
v - lapply(split(d, d$species), f)
# v is a list object, where each component of the list is an lm object,
# which itself is a list. In other words, it's a list of lists. do.call() is
a
# very useful function that applies a function to components of a list.
# rbind and cbind are commonly used to slurp together common elements
# from each component of a list.
# Pulling out the coefficients from each model:
do.call(rbind, lapply(v, coef))
(Intercept) bm
1 0.5176471 -0.01176471
2 0.9253731 -0.07611940
3 0.5942308 -0.04230769
5 0.3351648 0.04395604
# Extract the r-squared values from each model:
g - function(m) summary(m)$r.squared
do.call(rbind, lapply(v, g))
[,1]
1 0.03361345
2 0.66932578
3 0.43291592
5 0.14652015
# But you have to be careful...e.g., since you have unequal sample sizes per
species,
do.call(cbind, lapply(v, resid))
1 2 3 5
1 0.04117647 -0.09253731 -0.040384615 -0.1230769
2 0.10588235 0.08358209 0.190384615 0.2208791
3 -0.07058824 -0.19701493 -0.151923077 0.2571429
4 -0.09411765 0.07910448 0.001923077 -0.3549451
5 0.01764706 0.12686567 -0.040384615 -0.1230769
Warning message:
In function (..., deparse.level = 1) :
number of rows of result is not a multiple of vector length (arg 3)
Notice how the first residual is recycled in each of groups 3 and 5. That's
a potential gotcha.
This gives you a small glimpse into the power that R can deliver in data
analysis.
HTH,
Dennis
On Sun, Jul 18, 2010 at 2:29 PM, karmakiller roisinmoria...@gmail.comwrote:
Hi All,
I have a large data set with many columns of data. One of these columns is
a
species identifier and the remainder are variables such as temperature or
mass. Currently I am carrying out a single regression on subsets of the
data
set, e.g. separated data sets with only the data from one species at a
time.
I have been searching for a thread that will help me to understand how best
to repeat this process for each different species identifier in that
variable column. I can’t seem to find one that is similar to what I am
trying to do. It might be the case that I am not looking for the right
thing
or that I do not fully understand the process.
How do I run a simple loop that produces a regression for each species as
identified in the variable species id, which is one column in the large
data
set that I am using?
Simple regression that I wish to repeat
data- read.table(…/STUDY.txt,header=T)
names(data)
model- with(data,{lm(o2con~bm)})
summary(model)
sample data set
species id o2con bm
1 0.5 5
1 0.6 2
1 0.4 4
1 0.4 2
1 0.5 3
2 0.3 7
2 0.4 8
2 0.5 3
2 0.7 4
2 0.9 2
3 0.3 6
3 0.7 2
3 0.4 1
3 0.3 7
5 0.3 2
5 0.6 1
5 0.9 7
Re: [R] Cramer-von Mises test
Thanks Peter...I hadn't some across the sos package before, but I'm sure I'll be putting it to good use from now on! Sean. On Mon, Jul 19, 2010 at 2:55 PM, Peter Ehlers ehl...@ucalgary.ca wrote: On 2010-07-18 22:04, Sean Carmody wrote: I have found the Kolmogorov-Smirnov (ks.test) and Anderson-Darling (ad.test in package ADGofTest) goodness of fit tests, but was wondering whether there's a package implementing the Cramer-von Mises test for a general distribution (i.e. not just the one for the normal distribution in the nortest package). Any suggestions? Sean, Here's one way you can investigate this yourself: 1. install and load package 'sos'; 2. issue: findFn(Cramer-von Mises test) or, for a little wider scope: findFn(Cramer-von Mises) (in place of 'findFn', you can use '???') -Peter Ehlers Regards, Sean. -- Sean Carmody Twitter: http://twitter.com/seancarmody Stable: http://mulestable.net/sean The Stubborn Mule Blog: http://www.stubbornmule.net Forum: http://mulestable.net/ __ 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.
