Re: [R] How to convert c:\a\b to c:/a/b?
If you have 'copied' the path from DOS, then you can use 'scan' to read it
into a variable with the proper characters.
Here is the string that I 'copied'
D:\spencerg\statmtds\R\Rnews
Here is the results after 'scan':
x.1 - scan('clipboard', what='', allowEscapes=FALSE)
Read 1 item
x.1
[1] D:\\spencerg\\statmtds\\R\\Rnews
Jim
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Henrik Bengtsson
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cc
r-help@stat.math.ethz.ch, Dirk
06/27/2005 14:53 Eddelbuettel [EMAIL PROTECTED]
Subject
Re: [R] How to convert c:\a\b to
c:/a/b?
Spencer Graves wrote:
Hi, Henrik:
Several functions, e.g., grep, sub, gsub, and regexpr,
have an argument perl, FALSE by default. Moreover, ?regexp has a
section on Perl Regular Expressions. If you can do it in perl, might
that transfer to gsub(..., perl=TRUE)?
I do not know the details behind the different dialects of regular
expressions, but you can _not_ get the R parser to interpret the two
ASCII characters \n, as the two characters \ and n. The R parser
is used when code is read by source() or when expressions are typed at
the R prompt. The parser will always read it as the newline character
(ASCII 10). The results from the parser is then passed to the R enginee.
Thus, you cannot write your program such that it fools the parser,
because your program is evaluated first after the parser. In other
words, there is no way you can get nchar(\n) to equal 2.
Cheers
Henrik
Thanks,
spencer graves
p.s. I skimmed the discussion of Pearl Regular Expressions, and
experimented with gsub(..., perl=TRUE) without success. However,
there may be a way to do it, and I just don't know perl and regexp well
enough to have figured it out in the time available.
Henrik Bengtsson wrote:
Spencer Graves wrote:
Thanks, Dirk, Gabor, Eric:
You all provided appropriate solutions for the stated problem.
Sadly, I oversimplified the problem I was trying to solve: I copy a
character string giving a DOS path from MS Windows Explorer into an R
script file, and I get something like the following:
D:\spencerg\statmtds\R\Rnews
I want to be able to use this in R with its non-R meaning,
e.g., in readLine, count.fields, read.table, etc., after appending a
file name. Your three solutions all work for my oversimplified toy
example but are inadequate for the problem I really want to solve.
Hmmm. It should work as long as you do not source() the file (see
below). There are two things to watch out for here.
First, you have to be careful with backslashes, that is, a backslash
is a single character ('\') in memory, but to be typed at the R
prompt, you have to escape it (with a backslash), which is why we type
\\, cf. nchar(\\) == 0. Consider the file foo.txt containing the
28 characters (==28 bytes in plain ASCII format)
D:\spencerg\statmtds\R\Rnews
You can create such a file in R by
cat(file=foo.txt, D:\\spencerg\\statmtds\\R\\Rnews)
str(file.info(foo.txt))
`data.frame': 1 obs. of 6 variables:
$ size : num 28
$ isdir: logi FALSE
$ mode :Class 'octmode' int 438
$ mtime:'POSIXct', format: chr 2005-06-27 19:14:20
$ ctime:'POSIXct', format: chr 2005-06-27 19:14:20
$ atime:'POSIXct', format: chr 2005-06-27 19:14:20
Re-read it into R:
bfr - readLines(foo.txt)
Warning message:
incomplete final line found by readLines on 'foo.txt'
bfr
[1] D:\\spencerg\\statmtds\\R\\Rnews
cat(bfr=', bfr, '\n, sep=)
bfr='D:\spencerg\statmtds\R\Rnews'
Now, convert backslashes to forwardslashes:
bfr2 - gsub(, /, bfr)
bfr2
[1] D:/spencerg/statmtds/R
Re: [R] grep negation
?setdiff e.g., txt - c(arm,foot,lefroo, bafoobar) i - grep(foo,txt); i [1] 2 4 setdiff(seq(length(txt)),grep(foo,txt)) [1] 1 3 Jim __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Convergys Labs [EMAIL PROTECTED] +1 (513) 723-2929 Marcus Leinweber [EMAIL PROTECTED]To: 'r-help@stat.math.ethz.ch' r-help@stat.math.ethz.ch Sent by: cc: [EMAIL PROTECTED]Subject: [R] grep negation ath.ethz.ch 06/23/2005 08:59 hi, using the example in the grep help: txt - c(arm,foot,lefroo, bafoobar) i - grep(foo,txt); i [1] 2 4 but how can i get the negation (1,3) when looking for 'foo'? thanks, m. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] (no subject)
'rle' might be your friend. This will find the 'run of a sequence'
Here is some code working off the 'visit' data that you created.
# $Log$
x.1 - matrix(visit, ncol=4) # your data
x.rle - apply(x.1, 1, rle) # compute 'rle' for each row
Passed - lapply(x.rle, function(x){ # now process each row see if it
meets the criteria
.len - length(x$lengths)
if (x$lengths[.len] 1 x$values[.len] == 1) return(TRUE) # last
two passed
else if (.len == 2){ # two sequences
if (x$lengths[.len] == 1 x$values[.len] == 1) return(TRUE) #
only last passed
}
return(FALSE)
})
cbind(unlist(Passed), x.1) # put results in first column with the data
Jim
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ath.ethz.ch
06/20/2005 11:58
R friends,
I am using R 2.1.0 in a Win XP . I have a problem working with lists,
probably I
do not understand how to use them.
Lets suppose that a set of patients visit a clinic once a year for 4 years
on each visit a test, say 'eib' is performed with results 0 or 1
The patients do not all visit the clinic the 4 times but they missed a lot
of visits.
The test is considered positive if it is positive at the last 2 visits of
that
patient, or a more lenient definition, it is positive in the last visit,
and
never before.
Otherwise it is Negative = always negative or is a YoYo = unstable =
changes
from positive to negative.
So, if I codify the visits with codes 1,2,4,8 if present at year 1,2,3,4
and
similarly the tests positive I get the last2 list codifying the test code
corresponding to the visits patterns possible, similarly the last1 list
20 here means NULL
nobs - 400
# visits 0 1 23 45 6 7 89
last1 - list((20),(1),(2),c(3,2),(4),c(5,4),c(6,4),c(7,6,4),(8),c(9,8),
# visits 10 11 12 13 14 15
c(10,8),c(11,10,8),c(12,8),c(13,12,8),c(14,12,8),c(15,14,12,8))
# visits 0 123 45 67 89
last2 - list((20),(20),(20),(3),(20),(5),(6),c(7,6),(20),(9),
# visits 1011 1213 14 15
(10),c(11,10),(12),c(13,12),c(14,12),c(15,14,12))
#
# simulate the visits
#
visit - rbinom(nobs,1,0.7)
eib - visit
#
# simulate a positive test at a given visit
#
eib - ifelse(runif(nobs) 0.7,visit,0)
#
# create the codes
#
viskode - matrix(visit,ncol=4) %*% c(1,2,4,8)
eibkode - matrix(eib,ncol=4) %*% c(1,2,4,8)
#
#this is the brute force method, slow, of computing the Results
according to
#the 2 definitions above. Add 16 to the test kode to signify YoYos,
Exactly
#16 will be the negatives
#
eibnoyoyo - eibkode+16
eiblst2 - eibkode+16
for(i in 1:nobs){
if(eibkode[i] %in% last1[[viskode[i]+1]])
eibnoyoyo[i] - eibkode[i]
if(eibkode[i] %in% last2[[viskode[i]+1]])
eiblast2[i] - eibkode[i]
}
#
#why is that these statements do not work?
#
eeibnoyoyo - eeiblst2 - rep(0,nobs)
eeibnoyoyo - ifelse(eibkode %in% last1[viskode+1],eibkode,eibkode+16)
eeiblast2 - ifelse(eibkode %in% last2[viskode+1],eibkode,eibkode+16)
#
table(viskode,eibkode)
table(viskode,eibnoyoyo)
table(viskode,eiblast2)
#
# these two tables must be diagonal!!
#
table(eibnoyoyo,eeibnoyoyo)
table(eiblast2,eeiblast2)
#
Thanks for any help
Heberto Ghezzo
McGill University
Canada
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Re: [R] vectorisation suggestion
v3 - numeric()
v3[v1] - table(v2)[v1]
Jim
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Federico Calboli
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suggestion
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ath.ethz.ch
06/20/2005 16:15
Hi All,
I am counting the number of occurrences of the terms listed in one
vector in another vector.
My code runs:
for( i in 1:length(vector3)){
vector3[i] = sum(1*is.element(vector2, vector1[i]))
}
where
vector1 = vector containing the terms whose occurrences I want to count
vector2 = made up of a number of repetitions of all the elements of
vector1
vector3 = a vector of NAs that is meant to get the result of the
counting
My problem is that vector1 is about 6 terms, and vector2 is
62... can anyone suggest a faster code than the one I wrote?
Cheers,
Federico Calboli
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax +44 (0)20 75943193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
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Re: [R] vectorization
try this:
x.1 - data.frame(income=runif(100)*1,
educ=sample(c('hs','col','none'),100,T))
x.1
income educ
1 5930.30882 col
2 5528.83222 hs
3 5967.04041 hs
4 3926.30682 hs
5 2603.75924 none
...
x.2 - tapply(x.1$income, x.1$educ, mean)
x.2
col hs none
5575.310 4994.921 5481.962
x.1$median - x.2[x.1$educ]
x.1
income educ median
1 5930.30882 col 5575.310
2 5528.83222 hs 4994.921
3 5967.04041 hs 4994.921
4 3926.30682 hs 4994.921
5 2603.75924 none 5481.962
6 7398.83325 col 5575.310
7265.06895 hs 4994.921
.
Jim
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Dimitri Joe
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ath.ethz.ch
06/17/2005 14:00
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for
each education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end
of this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
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Re: [R] Manipulating dates
Use POSIX. To convert: my.dates - strptime(your.characters, format='%d/%m/%Y') once you have that, you can use 'min' to find the minimum. 'difftime' will give you the differences. Jim __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Convergys Labs [EMAIL PROTECTED] +1 (513) 723-2929 Richard Hillary [EMAIL PROTECTED]To: r-help@stat.math.ethz.ch c.ukcc: Sent by: Subject: [R] Manipulating dates [EMAIL PROTECTED] ath.ethz.ch 06/14/2005 10:07 Please respond to r.hillary Hello, Given a vector of characters, or factors, denoting the date in the following way: 28/03/2000, is there a method of 1) Computing the earliest of these dates; 2) Using this as a base, then converting all the other dates into merely the number of days after this minimum date Many thanks Richard Hillary __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Dateticks
try this example: x.1 - strptime(6/17/03,'%m/%d/%y') x.1 [1] 2003-06-17 plot(0:250, xaxt='n') dates - x.1 + c(0,50,100,150,200,250) * 86400 # 'dates' is in seconds, so add the appropriate number of days dates [1] 2003-06-17 00:00:00 EDT 2003-08-06 00:00:00 EDT 2003-09-25 00:00:00 EDT [4] 2003-11-13 23:00:00 EST 2004-01-02 23:00:00 EST 2004-02-21 23:00:00 EST axis(1, at=c(0,50,100,150,200,250), labels=format(dates,%m/%d/%y)) # format the output Jim __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Convergys Labs [EMAIL PROTECTED] +1 (513) 723-2929 Bernard L. Dillard [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent by: cc: [EMAIL PROTECTED]Subject: [R] Dateticks ath.ethz.ch 06/14/2005 12:27 Hello. I am having the worst time converting x-axis date ticks to real dates. I have tried several suggestions in online help tips and books to no avail. For example, the x-axis has 0, 50, 100, etc, and I want it to have 6/17/03, 8/6/03 etc. See attached (sample). Can anybody help me with this. Here's my code: ts.plot(date.attackmode.table[,1], type=l, col=blue, lty=2,ylab=IED Attacks, lwd=2,xlab=Attack Dates,main=Daily Summary of Attack Mode) grid() Thanks for your help if possible. (See attached file: sample.pdf) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html sample.pdf Description: Adobe PDF document __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] transform large matrix into list
x.1 [,1] [,2] [1,]14 [2,]25 [3,] NA6 cbind(x.1[!is.na(x.1)], which(!is.na(x.1), arr.ind=TRUE)) row col [1,] 1 1 1 [2,] 2 2 1 [3,] 4 1 2 [4,] 5 2 2 [5,] 6 3 2 Jim __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Convergys Labs [EMAIL PROTECTED] +1 (513) 723-2929 Stefan Mischke [EMAIL PROTECTED]To: r-help@stat.math.ethz.ch .ch cc: Sent by: Subject: [R] transform large matrix into list [EMAIL PROTECTED] ath.ethz.ch 06/07/2005 08:55 Dear List I need to transform a large matrix M with many NAs into a list L with one row for each non missing cell. Every row should contain the cell value in the first column, and its coordinates of the matrix in column 2 and 3. M: x1 x2 y1 1 2 y2 4 5 y3 7 8 L: vx y 11 1 41 2 71 2 22 1 52 2 82 3 I'm trying to do this with a loop, but since my matrix is quite large (around 10k^2) this just takes a very long time. There must be a more efficient and elegant way to do this. Any hints? Thanks, Stefan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] weighted.mean and tapply (again)
x.1 - read.table('clipboard',header=T)
x.1
GROUP VALUE FREQUENCY
1 2 278
2 2 340
3 2 416
4 2 5 3
5 2 6 1
6 2 8 1
7 3 319
8 3 410
9 3 519
10 3 6 4
by(x.1, x.1$GROUP, function(x) weighted.mean(x$VALUE, x$FREQUENCY))
x.1$GROUP: 2
[1] 2.654676
---
x.1$GROUP: 3
[1] 4.153846
Jim
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Dan Bolser
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and tapply (again)
[EMAIL PROTECTED]
ath.ethz.ch
05/25/2005 11:33
I read answers to questions including the words tapply and
weighted.mean, but I didn't understand either the problem (data) or the
solution provided.
Here is my question ...
dat[1:10,]
GROUP VALUE FREQUENCY
1 2 278
2 2 340
3 2 416
4 2 5 3
5 2 6 1
6 2 8 1
7 3 319
8 3 410
9 3 519
1 3 6 4
For each GROUP, I would like to calculate the weighted.mean of VALUE using
the FREQUENCY as the weight, so for the snippet of data shown that would
be...
group.2 - weighted.mean(c(2,3,4,5,6,8),c(78,40,16,3,1,1))
group.3 - weighted.mean(c(3,4,5,6),c(19,10,19,4))
cbind(rbind(2,3),rbind(group.2,group.3))
[,1] [,2]
group.22 2.654676
group.33 4.153846
I would like to use tapply to automatically do this across the whole
dataset (dat) - which includes lots of other distinct grouping factors,
however, like I said, I couldn't understand (and therefore apply to my
data) any of the other solutions I found, so any help here would be
greatly appreciated!
All the best,
Dan.
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Re: [R] plot question
tt - data.frame(c(0.5, 1, 0.5))
names(tt) - a
plot(tt$a, type = 'o',xlim=c(0,4))
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Christoph Lehmann
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ath.ethz.ch
03/03/2005 11:29
I have the following simple situation:
tt - data.frame(c(0.5, 1, 0.5))
names(tt) - a
plot(tt$a, type = 'o')
gives the following plot ('I' and '.' represent the axis):
I
I
I X
I
I
I X X
I...
1 2 3
what do I have to change to get the following:
I
I
I X
I
I
I X X
I.
1 2 3
i.e. the plot-region should be widened at the left and right side
thanks for a hint
christoph
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Re: [R] (no subject)
use 'gsub'
x - c('1,200.44', '23,345.66')
gsub(',','',x)
[1] 1200.44 23345.66
as.numeric(gsub(',','',x))
[1] 1200.44 23345.66
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Jim Gustafsson
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ath.ethz.ch
02/16/2005 09:08
R-people
I wonder if one could change a list of table with number of the form
1,200.44 , to 1200.44
Regards
JG
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Re: [R] Programming/scripting with expressions - variables
Here is one way. It is the custom to return a value that will be assigned
to the variable, so I changed your 'macro' to a function that returns the
value and then assigns it to your variable:
test - function(name, value){
+ .result - NULL # initialize to NULL
+ .result[name] - value
+ .result[paste('other_', name, sep='')] - paste(other_, value,
sep='')
+ .result
+ }
Gregor - test('Gorjanc', '25')
Gregor# print out the vector
Gorjanc other_Gorjanc
25other_25
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Programming/scripting with expressions - variables
[EMAIL PROTECTED]
ath.ethz.ch
02/07/2005 09:52
Hello to Rusers!
I am puzzled with R and I really do not know where to look
in for my problem. I am moving from SAS and I have difficulties
in translating SAS to R world. I hope I will get some hints
or pointers so I can study from there on.
I would like to do something like this. In SAS I can write
a macro as example bellow, which is afcourse a silly one but
shows what I don't know how to do in R.
%macro test(data, colname, colvalue);
data data;
...
colname=colvalue;
other_colname=other_colvalue;
run;
%mend;
And if I run it with this call:
%test(Gregor, Gorjanc, 25);
I get a table with name 'Gregor' and columns 'Gorjanc',
and 'other_Gorjanc' with values:
Gorjanc other_Gorjanc
25other_25
So can one show me the way to do the same thing in R?
Thanks!
--
Lep pozdrav / With regards,
Gregor GORJANC
---
University of Ljubljana
Biotechnical Faculty URI: http://www.bfro.uni-lj.si
Zootechnical Departmentemail: gregor.gorjanc at bfro.uni-lj.si
Groblje 3 tel: +386 (0)1 72 17 861
SI-1230 Domzalefax: +386 (0)1 72 17 888
Slovenia
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Re: [R] Frequency of Data
Try this using strsplit and table:
dates - c('29.02.1997','15.02.2001','15.02.2001','23.12.2002')
x.1 - do.call('rbind',strsplit(dates,'\\.'))
x.1
[,1] [,2] [,3]
[1,] 29 02 1997
[2,] 15 02 2001
[3,] 15 02 2001
[4,] 23 12 2002
class(x.1) - 'integer'
x.1
[,1] [,2] [,3]
[1,] 292 1997
[2,] 152 2001
[3,] 152 2001
[4,] 23 12 2002
table(list(x.1[,2], x.1[,3]))
.2
.1 1997 2001 2002
2 120
12 001
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Carsten Steinhoff
[EMAIL PROTECTED]To:
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ttingen.decc:
Sent by: Subject: [R]
Frequency of Data
[EMAIL PROTECTED]
h
02/02/2005 14:44
Hello,
just another problem in R, maybe it's simple to solve for you. I didn't
find
a solution up to now, but I'm convinced that I'm not the only one who
has/had a similar problem. Maybe there's a ready-made function in R?
The prob:
I've imported a CSV-file into R with 1000 dates of an observed event
(there's only information of the date. When there happend no event the date
is not recorded, when there have been two events it's recordet twice). Now
I
want to COUNT the frequency of events in every month or year.
The CSV-data is structured as:
date
25.02.2003
29.07.1997
...
My desired output would be a matrix with n rows for the years and m columns
for the month.
How could a solution look like ? If the format is no matrix it doesn't
matter. Importend is the extraction of frequency from my data.
Thanks for all reply,
Carsten
[[alternative HTML version deleted]]
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Re: [R] Finding runs of TRUE in binary vector
use 'rle'; a - rnorm(20) b - a .5 b [1] FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE TRUE TRUE [13] FALSE FALSE TRUE TRUE TRUE FALSE TRUE FALSE rle(b) Run Length Encoding lengths: int [1:9] 1 7 2 2 2 3 1 1 1 values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 Sean Davis [EMAIL PROTECTED]To: r-help r-help@stat.math.ethz.ch cc: Sent by: Subject: [R] Finding runs of TRUE in binary vector [EMAIL PROTECTED] ath.ethz.ch 01/27/2005 17:13 I have a binary vector and I want to find all regions of that vector that are runs of TRUE (or FALSE). a - rnorm(10) b - a0.5 b [1] TRUE TRUE TRUE FALSE TRUE FALSE TRUE TRUE TRUE TRUE My function would return something like a list: region[[1]] 1,3 region[[2]] 5,5 region[[3]] 7,10 Any ideas besides looping and setting start and ends directly? Thanks, Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Avoiding a Loop?
Does this do what you want?
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(myconstant, nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
t(apply(mymatrix, 1, function(x) cumprod(x)))
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
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Rau, Roland
[EMAIL PROTECTED] To:
r-help@stat.math.ethz.ch
Sent by: cc:
[EMAIL PROTECTED]Subject: [R] Avoiding a Loop?
ath.ethz.ch
01/21/2005 07:31
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
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Re: [R] recoding large number of categories (select in SAS)
Here is a way of doing it by setting up a matrix of values to test against.
Easier than writing all the 'select' statements.
x.trans - matrix(c( # translation matrix; first column is min, second
is max,
+ 149, 150, 150, # and third is the value to be returned
+ 186, 187, 187,
+ 438, 438, 438,
+ 430, 430, 430,
+ 808, 826, 808,
+ 830, 832, 808,
+ 997, 998, 792,
+ 792, 796, 792), ncol=3, byrow=T)
colnames(x.trans) - c('min', 'max', 'value')
x.default - # default/nomatch value
x.test - c(150, 149, 148, 438, 997, 791, 795, 810, 820, 834) # test
data
#
# this function will test each value and if between the min/max, return 3
column
#
newValues - sapply(x.test, function(x){
+ .value - x.trans[(x = x.trans[,'min']) (x =
x.trans[,'max']),'value']
+ if (length(.value) == 0) .value - x.default# on no match, take
default
+ .value[1] # return first value if multiple matches
+ })
newValues
[1] 150 150 438 792 792 808 808
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Denis Chabot
[EMAIL PROTECTED]To:
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.netcc:
Sent by: Subject: [R] recoding large
number of categories (select in SAS)
[EMAIL PROTECTED]
ath.ethz.ch
01/19/2005 08:56 AM
Hi,
I have data on stomach contents. Possible prey species are in the
hundreds, so a list of prey codes has been in used in many labs doing
this kind of work.
When comes time to do analyses on these data one often wants to regroup
prey in broader categories, especially for rare prey.
In SAS you can nest a large number of if-else, or do this more
cleanly with select like this:
select;
when (149 = prey =150) preyGr= 150;
when (186 = prey = 187) preyGr= 187;
when (prey= 438) preyGr= 438;
when (prey= 430) preyGr= 430;
when (prey= 436) preyGr= 436;
when (prey= 431) preyGr= 431;
when (prey= 451) preyGr= 451;
when (prey= 461) preyGr= 461;
when (prey= 478) preyGr= 478;
when (prey= 572) preyGr= 572;
when (692 = prey = 695 )
preyGr= 692;
when (808 = prey = 826, 830 = prey = 832 ) preyGr= 808;
when (997 = prey = 998, 792 = prey = 796) preyGr= 792;
when (882 = prey = 909)
preyGr= 882;
when (prey in (999, 125, 994))
preyGr= 9994;
otherwise preyGr= 1;
end; *select;
The number of transformations is usually much larger than this short
example.
What is the best way of doing this in R?
Sincerely,
Denis Chabot
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Re: [R] plotting percent of incidents within different 'bins'
You can use 'cut' to create the breaks.. Actually there are 8 in the 3-4 range: Outcome predictor 10 1 21 2 31 2 40 3 50 3 60 2 71 3 81 4 91 4 10 0 4 11 0 4 12 0 4 cut(x.1$p, breaks=c(0,2,4)) [1] (0,2] (0,2] (0,2] (2,4] (2,4] (0,2] (2,4] (2,4] (2,4] (2,4] (2,4] (2,4] Levels: (0,2] (2,4] x.c - cut(x.1$p, breaks=c(0,2,4)) tapply(x.1$O, x.c, function(x)sum(x==1)/length(x)) (0,2] (2,4] 0.500 0.375 __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 Stephen Choularton [EMAIL PROTECTED] To: R Help r-help@stat.math.ethz.ch Sent by: cc: [EMAIL PROTECTED]Subject: [R] plotting percent of incidents within different 'bins' ath.ethz.ch 01/05/2005 14:34 Hi Say I have some data, two columns in a table being a binary outcome plus a predictor and I want to plot a graph that shows the percentage positives of the binary outcome within bands of the predictor, e.g. Outcome predictor 0 1 1 2 1 2 0 3 0 3 0 2 1 3 1 4 1 4 0 4 0 4 0 4 etc In this case there are 4 cases in the band 1 - 2 of the predictor, 2 of them are true so the percent is 50% and there are 7 cases in the band 3 - 4, 3 of which are true making the percentage 43% . Is there some function in R that will sum these outcomes by bands of predictor and produce a one by two data set with the percentages in one column and the ordered bands in the other, or alternately is there some sort of special plot. that does it all for you? Thanks Stephen [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lists within a list / data-structure problem
construct you list in the loop:
x.all - list() # initialize
for (i in 1:limit){
...
x.all[[i]] - result.list
}
now you want to name them, e.g., run1
names(x.all) - paste('run', seq(length(x.all)), sep='')
To access, you can dox.all$run1$Dom
To extract all the 'Dom's
lapply(x.all, function(x) x$Dom)
HTH
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Jan Wantia
[EMAIL PROTECTED]To:
[EMAIL PROTECTED]
Sent by: cc:
[EMAIL PROTECTED]Subject: [R] lists within a
list / data-structure problem
ath.ethz.ch
12/13/2004 10:59
Dear all,
this is a rather basic question; i am not sure how to structure my data
well:
I want to extraxt various measures from my raw-data. These measures are
of different sizes, so I decided to store them in a list, like:
run1 - list(Dom = (my_vector), mean = (my_single_number))
I can do that in a for loop for 40 runs, ending up with 40 lists: run1,
run2, ..., run40.
To have all the measurements neatly together I thought of making another
list, containing 40 sub-lists:
ALL - list(run1, run2,..., run40)
ALL
[[1]]
[[1]]$Dom
[1] my_vector
[[1]]$mean
[1] my_single_number
[[2]]
[[2]]$Dom
[1] my_vector
[[2]]$mean
[1] my_single_number
...
1) This may be a bit clumsy as I have to type all the sub-list's names
in by hand in order to produce my ALL-list: Is there a better way?
2) I have problems of addressing the data now. I can easily access any
single value; for example, for the second component of the second sub-
list:
ALL[[2]][[2]]
[1] my_single_number,
but: how could I get the second component of all sub-lists, to plot, for
example, all the $mean in one plot? For a matrix, mat[,2] would give me
the whole second column, but
ALL[[]][[2]]
does not return all the second components.
I feel that 'lapply' might help me here, but I could not figure out
exactly how to use it, and it always comes down to the problem of how to
correctly address the components in the sublists.
Or is there maybe a smarter way to do that instead of using a list of
lists?
Any hint would be warmly appreciated!
Jan
(R 2.0.1 on windows XP)
--
__
Jan Wantia
Deptartment of Informatics, University of Zurich
Andreasstr. 15
CH 8050 Zurich
Switzerland
Tel.:+41 (0) 1 635 4315
Fax: +41 (0) 1 635 45 07
email: [EMAIL PROTECTED]
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Re: [R] How to duplicate rows in dataframe?
x.1 - data.frame(a=1:5, b=1:5) x.1 a b 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 x.1[c(1,2,2,2,3,3,4,4,5,4,3,2,1),] a b 1 1 1 2 2 2 2.1 2 2 2.2 2 2 3 3 3 3.1 3 3 4 4 4 4.1 4 4 5 5 5 4.2 4 4 3.2 3 3 2.3 2 2 1.1 1 1 __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 cstrato [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED]Subject: [R] How to duplicate rows in dataframe? ath.ethz.ch 12/13/2004 14:02 Dear all: I have the following (simple?) problem: Consider a dataframe where the first column contains integers used as index, e.g. index 24 13 46 32 Now I have the following vector used to sort the dataframe: x - c(13,24,32,46) Sorting the dataframe can be done by using order. However consider the following vector: x - c(13,32,13,24,46,24,24) Now I want to get the dataframe in the order of the rows defined in x, i.e. the dataframe contains duplicate rows. One way to achieve this would be to use rbind in a for-loop. My question is: Is there an easier and - more important - faster way to obtain the dataframe as defined in x? Thank you in advance. Best regards Christian _._._._._._._._._._._._._._._._ C.h.i.s.t.i.a.n S.t.r.a.t.o.w.a V.i.e.n.n.a A.u.s.t.r.i.a _._._._._._._._._._._._._._._._ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] 'object.size' takes a long time to return a value
I was using 'object.size' to see how much memory a list was taking up.
After executing the command, I had thought that my computer had locked up.
After further testing, I determined that it was taking 241 seconds for
object.size to return a value.
I did notice in the release notes that 'object.size' did take longer when
the list contained character vectors. Is the time that it is taking
'object.size' to return a value to be expected for such a list?
Much better results were obtained when the character vectors were converted
to factors.
## Results from the testing ###
str(x.1)
List of 10
$ : chr [1:227299] sadc sar date ksh ...
$ : chr [1:227299] aprperf aprperf aprperf aprperf ...
$ : num [1:227299] 23 23 0 23 23 0 0 0 0 23 ...
$ : num [1:227299] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:227299] 3600 3600 0.01 3600 3600 0.01 0.01 0.01 0.01 3600 ...
$ : num [1:227299] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:227299] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:227299] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:227299] 62608 6796829 10208 13128 ...
$ : num [1:227299] 0 1 0 0 1 0 0 0 0 0 ...
# takes a long time (241 seconds) to report the size
gc();system.time(print(object.size(x.1)))
used (Mb) gc trigger (Mb)
Ncells 711007 19.02235810 59.8
Vcells 5191294 39.7 14409257 110.0
[1] 34154972
[1] 241.07 0.00 241.08 NA NA
# trying list of 1000
x.2 - list.subset(x.1, 1:1000);gc();system.time(print(object.size(x.2)))
used (Mb) gc trigger (Mb)
Ncells 711006 19.02235810 59.8
Vcells 4300288 32.9 14409257 110.0
[1] 145860
[1] 0.01 0.00 0.01 NA NA
# trying list of 10,000
x.2 - list.subset(x.1,
1:1);gc();system.time(print(object.size(x.2)))
used (Mb) gc trigger (Mb)
Ncells 711006 19.02235810 59.8
Vcells 4381288 33.5 14409257 110.0
[1] 1491948
[1] 0.28 0.00 0.28 NA NA
# list of 40,000
x.2 - list.subset(x.1,
1:4);gc();system.time(print(object.size(x.2)))
used (Mb) gc trigger (Mb)
Ncells 711006 19.02235810 59.8
Vcells 4651288 35.5 14409257 110.0
[1] 5988460
[1] 7.15 0.00 7.15 NA NA
# list of 60,000
x.2 - list.subset(x.1,
1:6);gc();system.time(print(object.size(x.2)))
used (Mb) gc trigger (Mb)
Ncells 711006 19.02235810 59.8
Vcells 4831288 36.9 14409257 110.0
[1] 9001556
[1] 17.33 0.00 17.32NANA
# list of 100,000
x.2 - list.subset(x.1,
1:10);gc();system.time(print(object.size(x.2)))
used (Mb) gc trigger (Mb)
Ncells 711006 19.02235810 59.8
Vcells 5191288 39.7 14409257 110.0
[1] 15044780
[1] 51.85 0.00 51.86NANA
# list structure of the last object
str(x.2)
List of 10
$ : chr [1:10] sadc sar date ksh ...
$ : chr [1:10] aprperf aprperf aprperf aprperf ...
$ : num [1:10] 23 23 0 23 23 0 0 0 0 23 ...
$ : num [1:10] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:10] 3600 3600 0.01 3600 3600 0.01 0.01 0.01 0.01 3600 ...
$ : num [1:10] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:10] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:10] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:10] 62608 6796829 10208 13128 ...
$ : num [1:10] 0 1 0 0 1 0 0 0 0 0 ...
# with the first two items on the list converted to factors,
# 'object.size' performs a lot better
str(x.1)
List of 10
$ : Factor w/ 175 levels #bpbkar,#bpcd,..: 132 133 60 93 13 160 60 84
60 132 ...
$ : Factor w/ 8 levels apra3g,aprperf,..: 2 2 2 2 2 2 2 2 2 2 ...
$ : num [1:227299] 23 23 0 23 23 0 0 0 0 23 ...
$ : num [1:227299] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:227299] 3600 3600 0.01 3600 3600 0.01 0.01 0.01 0.01 3600 ...
$ : num [1:227299] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:227299] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:227299] 0.01 0 0.01 0 0.01 0 0.01 0 0 0.01 ...
$ : num [1:227299] 62608 6796829 10208 13128 ...
$ : num [1:227299] 0 1 0 0 1 0 0 0 0 0 ...
system.time(print(object.size(x.1))) # now it is fast
[1] 16374176
[1] 0 0 0 NA NA
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor0.1
year 2004
month11
day 15
language R
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
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Re: [R] how can I get the coefficients of x^0, x^1, x^2, . , x^6 from expansion of (1+x+x^2)^3
Use the 'polynom' library: p - as.polynomial(c(1,1,1)) p 1 + x + x^2 p^3 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6 unclass(p^3) [1] 1 3 6 7 6 3 1 __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 Peter Yang [EMAIL PROTECTED]To: [EMAIL PROTECTED] cc: Sent by: Subject: [R] how can I get the coefficients of x^0, x^1, x^2, . ,x^6 from [EMAIL PROTECTED] expansion of (1+x+x^2)^3 ath.ethz.ch 12/03/2004 14:56 Hi, I would like to get the coefficients of x^0, x^1, x^2, . , x^6 from expansion of (1+x+x^2)^3. The result should be 1, 3, 6, 7, 6, 3, 1; How can I calculate in R? You help will be greatly appreciated. Peter [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] scatterplot of 100000 points and pdf file format
Have you tried plot(...,pch='.') This will use the period as the plotting character instead of the 'circle' which is drawn. This should reduce the size of the PDF file. I have done scatter plots with 2M points and they are typically meaningless with that many points overlaid. Check out 'hexbin' on Bioconductor (you can download the package from the RGUI window. This is a much better way of showing some information since it will plot the number of points that are within a hexagon. I have found this to be a better way of looking at some data. __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 Witold Eryk Wolski [EMAIL PROTECTED]To: R Help Mailing List [EMAIL PROTECTED] cc: Sent by: Subject: [R] scatterplot of 10 points and pdf file format [EMAIL PROTECTED] ath.ethz.ch 11/24/2004 10:34 Hi, I want to draw a scatter plot with 1M and more points and save it as pdf. This makes the pdf file large. So i tried to save the file first as png and than convert it to pdf. This looks OK if printed but if viewed e.g. with acrobat as document figure the quality is bad. Anyone knows a way to reduce the size but keep the quality? /E -- Dipl. bio-chem. Witold Eryk Wolski MPI-Moleculare Genetic Ihnestrasse 63-73 14195 Berlin tel: 0049-30-83875219 __(_ http://www.molgen.mpg.de/~wolski \__/'v' http://r4proteomics.sourceforge.net||/ \ mail: [EMAIL PROTECTED]^^ m m [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] timeDate
You might want to check out 'chron'. This stores the time as days and fractions of a day. If you take the current date, as.numeric(chron(dates.=11/23/2004)) [1] 12745 you get the value above. If you change this to millisecond, you get as.numeric(chron(dates.=11/23/2004)) * 86400 * 1000 [1] 1.101168e+12 this value requires 46 bits and since a floating point number has 54 bits of value, it should be enough to give you millisecond resolution and still maintain the 'date' __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 Yasser El-Zein [EMAIL PROTECTED]To: [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED]Subject: Re: [R] timeDate ath.ethz.ch 11/23/2004 09:55 Please respond to Yasser El-Zein I am looking for up to the millisecond resolution. Is there a package that has that? On Mon, 22 Nov 2004 21:48:20 + (UTC), Gabor Grothendieck [EMAIL PROTECTED] wrote: Yasser El-Zein abu3ammar at gmail.com writes: From the document it is apparent to me that I need as.POSIXct (I have a double representing the number of millis since 1/1/1970 and I need to construct a datetime object). I see it showing how to construct the time object from a string representing the time but now fro a double of millis. Does anyone know hoe to do that? If by millis you mean milliseconds (i.e. one thousandths of a second) then POSIXct does not support that resolution, but if rounding to seconds is ok then structure(round(x/1000), class = c(POSIXt, POSIXct)) should give it to you assuming x is the number of milliseconds. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to extract data?
By 'ignore', can we delete those from the list of data? I would then assume that if you have a sequence of +0+0+ that you would want the last + for the increase of three. If that is the case, then do a 'diff' and delete the entries that are 0. Then create a new 'diff' and then use 'rle' to see what the length of the sequences are: x - c(1,2,2,3,3,4,3,3,2,2,2,1) x [1] 1 2 2 3 3 4 3 3 2 2 2 1 x.d - diff(x) x.d [1] 1 0 1 0 1 -1 0 -1 0 0 -1 x.new - x[c(x.d,1) != 0] x.new [1] 1 2 3 4 3 2 1 x.d1 - diff(x.new) x.d1 [1] 1 1 1 -1 -1 -1 rle(x.d1) Run Length Encoding lengths: int [1:2] 3 3 values : num [1:2] 1 -1 you can check the results of 'rle' to determine where the changes are. __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 ebashi [EMAIL PROTECTED] To: [EMAIL PROTECTED], [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED]Subject: [R] How to extract data? ath.ethz.ch 11/23/2004 15:54 I appreciate if anyone can help me, I have a table as follow, rate DATE VALUE 1 1997-01-10 5.30 2 1997-01-17 5.30 3 1997-01-24 5.28 4 1997-01-31 5.30 5 1997-02-07 5.29 6 1997-02-14 5.26 7 1997-02-21 5.24 8 1997-02-28 5.26 9 1997-03-07 5.30 10 1997-03-14 5.30 ... ... ... ... ... ... I want to extract the DATE(s) on which the VALUE has already dropped twice and the DATE(s) that VALUE has already increased for three times,( ignore where VALUE(i+1)-VALUE(i)=0),I try to use diff() function, however that works only for one increase or decrease. Sincerely, Sean __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] RE : Create sequence for dataset
I think this might do it.
x.1 - data.frame(x=sample(1:3,20,T), y=sample(10:12,20,T)) # create
test data
x.1 # print it out
x y
1 2 11
2 3 11
3 2 10
4 1 12
5 3 11
6 1 10
7 3 10
8 1 11
9 1 12
10 1 11
11 1 12
12 1 12
13 2 11
14 3 11
15 3 10
16 3 10
17 2 12
18 2 10
19 3 11
20 2 11
# split the data by the numbers in 'x' (would be your 'amnl_key)
# and add a column containing the sequence number
x.s - by(x.1, x.1$x, function(x){x$seq - seq(along=x$x); x})
# the result in 'x.s' is a list and the rows have to be recombined (rbind)
to form the result
x.s # print out the data
x.1$x: 1
x y seq
4 1 12 1
6 1 10 2
8 1 11 3
9 1 12 4
10 1 11 5
11 1 12 6
12 1 12 7
x.1$x: 2
x y seq
1 2 11 1
3 2 10 2
13 2 11 3
17 2 12 4
18 2 10 5
20 2 11 6
x.1$x: 3
x y seq
2 3 11 1
5 3 11 2
7 3 10 3
14 3 11 4
15 3 10 5
16 3 10 6
19 3 11 7
do.call('rbind', x.s) # bind the rows and print out the result
x y seq
1.4 1 12 1
1.6 1 10 2
1.8 1 11 3
1.9 1 12 4
1.10 1 11 5
1.11 1 12 6
1.12 1 12 7
2.1 2 11 1
2.3 2 10 2
2.13 2 11 3
2.17 2 12 4
2.18 2 10 5
2.20 2 11 6
3.2 3 11 1
3.5 3 11 2
3.7 3 10 3
3.14 3 11 4
3.15 3 10 5
3.16 3 10 6
3.19 3 11 7
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
[EMAIL PROTECTED]
Sent by: To:
[EMAIL PROTECTED]
[EMAIL PROTECTED]cc:
ath.ethz.ch Subject: [R] RE : Create
sequence for dataset
11/21/2004 16:28
Dear members,
I want to create a sequence of numbers for the multiple records of
individual animal in my dataset. The SAS code below will do the trick, but
I want to learn to do it in R. Can anyone help ?
data htssn;
set htssn;
by anml_key;
if first.anml_key then do;
seq_ht_rslt=0;
end;
seq_ht_rslt+1;
Thanks in advance.
Stella
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Re: [R] scan or source a text file into a list
use an 'environment' to read in the values; e.g.,
with(e1 - new.env(),source('/tempxx.txt', local=T)) # read in the file
to a new environment
myList - list() # define empty list
for (i in ls(e1)){ # process each element
+ myList[i] - get(i, e1)
+ }
ls(e1) # show objects in the list
[1] fnrnYears qe year0
myList # output my list
$fnr
[1] 0.3
$nYears
[1] 50
$qe
[1] 0.04
$year0
[1] 1970
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Andy Bunn
[EMAIL PROTECTED] To: R-Help
[EMAIL PROTECTED]
Sent by: cc:
[EMAIL PROTECTED]Subject: [R] scan or source a
text file into a list
ath.ethz.ch
11/11/2004 10:57
I've ported somebody else's rather cumbersome Matlab model to R for
colleagues that want a free build of the model with the same type of I/O.
The Matlab model reads a text file with the initial parameters specified
as:
C:\Data\Carluc\Rportmore Params.R
# Number of years to simulate
nYears = 50;
# Initial year for graphing purposes
year0 = 1970;
# NPP/GPP ratio (cpp0 unitless)
fnr = 0.30;
# Quantum efficency
qe = 0.040;
That is, there are four input variables (for this run - there can be many
more) written in a way that R can understand them. In R, I can have the
model source the parameter text file easily enough and have the objects in
the workspace. The model function in R takes a list at runtime. How can I
have R read that file and put the contents into the list I need?
E.g.,
rm(list = ls())
source(Params.R)
ls()
[1] fnrnYears qe year0
fnr
[1] 0.3
nYears
[1] 50
foo.list - list(fnr = fnr, nYears = nYears)
foo.list
$fnr
[1] 0.3
$nYears
[1] 50
The model is then run with
CarlucR(inputParamList = foo.list, ...)
I can't build inputParamList by hand as above because the number of
initial parameters changes with the model run and this runs in a wrapper.
Any thoughts? Some combination of paste with scan or parse?
-Andy
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor0.0
year 2004
month10
day 04
language R
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[R] Error in PDF output in R 2.0.0
The following script works fine in R 1.9.1. It was creating a PDF file
with the graphs in it. In R 2.0.0, I got the error message below. I tried
the same script just outputting to Windows and postscript and the output
was OK. The error message only showed up when trying to create a PDF file.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor0.0
year 2004
month10
day 04
language R
## output to windows -- OK
print(xyplot(I(usr/sys) ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
print(xyplot(csw ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
## ouput to postscript file -- OK
postscript('out.ps')
print(xyplot(I(usr/sys) ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
print(xyplot(csw ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
dev.off()
windows
2
## output to PDF file -- ERRORS
pdf('out.pdf')
print(xyplot(I(usr/sys) ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
Error in [-(`*tmp*`, pos.heights[[nm]], value = numeric(0)) :
nothing to replace with
print(xyplot(csw ~ time|factor(cpu), memIn,
+ panel=function(x,y)panel.xyplot(x,y,type='l')))
Error in [-(`*tmp*`, pos.heights[[nm]], value = numeric(0)) :
nothing to replace with
dev.off()
windows
2
## traceback on error
traceback()
3: calculateGridLayout(x, rows.per.page, cols.per.page, number.of.cond,
panel.height, panel.width, main, sub, xlab, ylab, x.alternating,
y.alternating, x.relation.same, y.relation.same, xaxis.rot,
yaxis.rot, xaxis.cex, yaxis.cex, par.strip.text, legend)
2: print.trellis(xyplot(csw ~ time | factor(cpu), memIn, panel =
function(x,
y) panel.xyplot(x, y, type = l)))
1: print(xyplot(csw ~ time | factor(cpu), memIn, panel = function(x,
y) panel.xyplot(x, y, type = l)))
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
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Re: [R] Error with repeat lines() in function
The problem was is that you were not return a value from the apply
function. It was trying to store the result of the apply into an array and
there was no value.
See the line I added in your function.
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Sean Davis
[EMAIL PROTECTED]To: Uwe Ligges [EMAIL
PROTECTED]
cc: r-help [EMAIL
PROTECTED]
Sent by: Subject: Re: [R] Error with repeat
lines() in function
[EMAIL PROTECTED]
ath.ethz.ch
09/24/2004 13:09
Here is an example that seems to reproduce the error:
rf1 - matrix(sort(abs(round(runif(4)*100))),nrow=1)
annot1 - sort(abs(round(runif(193)*100)))
annot2 - annot1 + 70
annot3 - cbind(annot1,annot2)
rat2 - rnorm(193)
rat1 - rnorm(193)
plotter -
function(annot,rat1,rat2,rf1,...) {
par(las=2)
xmax - max(annot[,2])
xmin - min(annot[,1])
par(mfrow=c(2,1))
plot(annot[,1],rat1,type=l,xlab=,ylab=log2 Ratio,...)
points(annot[,1],rat1)
apply(rf1,1,function(z) {
if (z[4]==+) {
color - 'green'
yoffset=1
} else {
color - 'red'
yoffset=-1
}
lines(list(x=c(z[1],z[4]),y=c(-2-yoffset/10,-2-yoffset/
10)),lwd=2,col=color)
lines(list(x=c(z[2],z[3]),y=c(-2-yoffset/10,-2-yoffset/
10)),lwd=4,col=color)
1 # fake a return value
})
abline(h=0,lty=2)
}
plotter(annot3,rat1,rat2,rf1)
Error in ans[[1]] : subscript out of bounds
Enter a frame number, or 0 to exit
1:plotter(annot3, rat1, rat2, rf1)
2:apply(rf1, 1, function(z) {
Selection: 0
On Sep 24, 2004, at 12:05 PM, Uwe Ligges wrote:
Sean Davis wrote:
I have a function that does some plotting. I then add lines to the
plot. If executed one line at a time, there is not a problem. If I
execute the function, though, I get:
Error in ans[[1]] : subscript out of bounds
This always occurs after the second lines command, and doesn't happen
with all of my data points (some do not have errors). Any ideas?
Please give an example how to produce the error,
i.e. specify a very small toy example (including generated data and
the call to your function).
Many people on this list are quite busy these days and don't want to
think about how to call your function and invent an example ...
Uwe Ligges
Thanks,
Sean
function(x,annot,rat1,rat2,rf,...) {
par(las=2)
wh - which(annot[,5]==x)
xmax - max(annot[wh,4])
xmin - min(annot[wh,3])
chr - annot[wh,2][1]
wh.rf - rf$chrom==as.character(chr) rf$txStartxmin
rf$txEndxmax
par(mfrow=c(2,1))
plot(annot[wh,3],rat1[wh],type=l,xlab=,ylab=log2
Ratio,main=x,...)
points(annot[wh,3],rat1[wh])
apply(rf[wh.rf,],1,function(z) {
browser()
if (z[4]==+) {
color - 'green'
yoffset=1
} else {
color - 'red'
yoffset=-1
}
lines(list(x=c(z[5],z[6]),y=c(-2-yoffset/10,-2-yoffset/
10)),lwd=2,col=color)
lines(list(x=c(z[5],z[6]),y=c(-2-yoffset/10,-2-yoffset/
10)),lwd=2,col=color)
})
abline(h=0,lty=2)
}
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Re: [R] Unique lists from a list
Try this: l.1 - list(list(name='a', addr='123'),list(name='b', addr='234'), list(name='b', addr='234'), list(name='a', addr='123')) # create a list l.names - unlist(lapply(l.1, '[[', 'name')) # get the 'name' l.u - unique(l.names) # make unique new.list - l.1[match(l.u, l.names)] # create new list with just one 'name' __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 michael watson (IAH-C) To: [EMAIL PROTECTED] [EMAIL PROTECTED]cc: .ac.uk Subject: [R] Unique lists from a list Sent by: [EMAIL PROTECTED] ath.ethz.ch 09/01/2004 10:31 Hi I have a list. Two of the elements of this list are Name and Address, both of which are character vectors. Name and Address are linked, so that the same Name always associates with the same Address. What I want to do is pull out the unique values, as a new list of the same format (ie two elements of character vectors). Now I've worked out that unique(list$Name) will give me a list of the unique names, but how do I then go and link those to the correct (unique) addresses so I end up with a new list which is the same format as the rest, but now unique? Cheers Mick __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] naive question
It is amazing the amount of time that has been spent on this issue. In most cases, if you do some timing studies using 'scan', you will find that you can read some quite large data structures in a reasonable time. If you initial concern was having to wait 10 minutes to have your data read in, you could have read in quite a few data sets by now. When comparing speeds/feeds of processors, you also have to consider what it being done on them. Back in the dark ages we had a 1 MIP computer with 4M of memory handling input from 200 users on a transaction system. Today I need a 1GHZ computer with 512M to just handle me. Now true, I am doing a lot different processing on it. With respect to I/O, you have to consider what is being read in and how it is converted. Each system/program has different requirements. I have some applications (running on a laptop) that can read in approximately 100K rows of data per second (of course they are already binary). On the other hand, I can easily slow that down to 1K rows per second if I do not specify the correct parameters to 'read.table'. So go back and take a look at what you are doing, and instrument your code to see where time is being spent. The nice thing about R is that there are a number of ways of approaching a solution and it you don't like the timing of one way, try another. That is half the fun of using R. __ James HoltmanWhat is the problem you are trying to solve? Executive Technical Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] +1 (513) 723-2929 [EMAIL PROTECTED] mple.eduTo: [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED], [EMAIL PROTECTED] [EMAIL PROTECTED], [EMAIL PROTECTED] ath.ethz.ch Subject: Re: [R] naive question 06/30/2004 16:25 [EMAIL PROTECTED] writes: I did not use R ten years ago, but reasonable RAM amounts have multiplied by roughly a factor of 10 (from 128Mb to 1Gb), CPU speeds have gone up by a factor of 30 (from 90Mhz to 3Ghz), and disk space availabilty has gone up probably by a factor of 10. So, unless the I/O performance scales nonlinearly with size (a bit strange but not inconsistent with my R experiments), I would think that things should have gotten faster (by the wall clock, not slower). Of course, it is possible that the other components of the R system have been worked on more -- I am not equipped to comment... I think your RAM calculation is a bit off. in late 1993, 4MB systems were the standard PC, with 16 or 32 MB on high-end workstations. I beg to differ. In 1989, Mac II came standard with 8MB, NeXT came standard with 16MB. By 1994, 16MB was pretty much standard on good quality (= Pentium, of which the 90Mhz was the first example) PCs, with 32Mb pretty common (though I suspect that most R/S-Plus users were on SUNs, which were somewhat more plushly equipped). Comparable figures today are probably 256MB for the entry-level PC and a couple GB in the high end. So that's more like a factor of 64. On the other hand, CPU's have changed by more than the clock speed; in particular, the number of clock cycles per FP calculation has decreased considerably and is currently less than one in some circumstances. I think that FP performance has increased more than integer performance, which has pretty much kept pace with the clock speed. The compilers have also improved a bit... Igor __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] binding rows from different matrices
Try:
veca=matrix(1:25,5,5)
vecb=matrix(letters[1:25],5,5)
vecc=matrix(LETTERS[1:25],5,5)
x.1 - lapply(1:5,function(x)rbind(veca[x,],vecb[x,],vecc[x,]))
do.call('rbind',x.1)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] a f k p u
[3,] A F K P U
[4,] 2 7 12 17 22
[5,] b g l q v
[6,] B G L Q V
[7,] 3 8 13 18 23
[8,] c h m r w
[9,] C H M R W
[10,] 4 9 14 19 24
[11,] d i n s x
[12,] D I N S X
[13,] 5 10 15 20 25
[14,] e j o t y
[15,] E J O T Y
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
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Stephane DRAY
[EMAIL PROTECTED]To: [EMAIL PROTECTED]
eal.ca cc:
Sent by: Subject: [R] binding rows from
different matrices
[EMAIL PROTECTED]
ath.ethz.ch
06/29/2004 11:00
Hello list,
I have 3 matrices with same dimension :
veca=matrix(1:25,5,5)
vecb=matrix(letters[1:25],5,5)
vecc=matrix(LETTERS[1:25],5,5)
I would like to obtain a new matrix composed by alternating rows of these
different matrices (row 1 of mat 1, row 1 of mat 2, row 1 of mat 3, row 2
of mat 1.)
I have found a solution to do it but it is not very pretty and I wonder if
I can do it in an other way (perhaps with apply ) ?
res=matrix(0,1,5)
for(i in 1:5)
+ res=rbind(res,veca[i,],vecb[i,],vecc[i,])
res=res[-1,]
res
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] a f k p u
[3,] A F K P U
[4,] 2 7 12 17 22
[5,] b g l q v
[6,] B G L Q V
[7,] 3 8 13 18 23
[8,] c h m r w
[9,] C H M R W
[10,] 4 9 14 19 24
[11,] d i n s x
[12,] D I N S X
[13,] 5 10 15 20 25
[14,] e j o t y
[15,] E J O T Y
Thanks in advance !
Stéphane DRAY
--
Département des Sciences Biologiques
Université de Montréal, C.P. 6128, succursale centre-ville
Montréal, Québec H3C 3J7, Canada
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E-mail : [EMAIL PROTECTED]
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Re: [R] Specifying suitable PC to run R
If you are running Windows, do you have the Performance Monitor running?
This will help identify the reasons that programs are running slow. Most
likely, you are low on memory and are paging a lot. I alway have it
running and when I am running a large R script, if I am not using 100% of
the CPU, then I must be paging (assuming that I am not reading in my data).
You can also sprinkle the following function throughout your code to see
how much CPU and memory you are using. I bracket all my major
computational sections with it:
my.stats - function(text = stats)
{
cat(text, -,sys.call(sys.parent())[[1]], :, proc.time()[1:3], :
, round(
memory.size()/2.^20., 1.), MB\n)
invisible(flush.console())
}
This prints out a message like:
my.stats('Begin Reading')
Begin Reading - my.stats : 5.61 3.77 22309.67 : 18.7 MB
This says that I have used 5.61 CPU seconds of 'user' time, 3.77 CPU
seconds of 'system' time and the R session has been running for 22309
seconds (I always have one waiting for simple calculation) and I have
18.7MB of memory allocated to objects.
My first choice is get as much memory on your machine as you can; 1GB since
this the most that R can use. I noticed a big difference in upgrading from
256M - 512M. I also watch the Performance Monitor and when memory gets
low and I want to run a large job, I restart R. Most of my scripts are
setup to run R without saving any data in the .Rdata file. If I need to
save a large object, I do it explicitly since memory is key performance
limiting factor and Windows is not that good at freeing up memory after you
have used a lot of it.
A faster CPU will also help, but it would be the second choice, since if
you are paging, most of your time is spent on data transfer and not
computation.
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James HoltmanWhat is the problem you are trying to solve?
Executive Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
(513) 723-2929
Michael Dewey
[EMAIL PROTECTED]To: [EMAIL PROTECTED]
uk cc:
Sent by: Subject: [R] Specifying suitable
PC to run R
[EMAIL PROTECTED]
ath.ethz.ch
10/09/2003 14:04
If I am buying a PC where the most compute intensive task will be running R
and I do not have unlimited resources what trade-offs should I make?
Specifically should I go for
1 - more memory, or
2 - faster processor, or
3 - something else?
If it makes a difference I shall be running Windows on it and I am thinking
about getting a portable which I understand makes upgrading more difficult.
Extra background: the tasks I notice going slowly at the moment are fitting
models with lme which have complex random effects and bootstrapping. By the
standards of r-help posters I have small datasets (few thousand cases, few
hundred variables). In order to facilitate working with colleagues I need
to stick with windows even if linux would be more efficient
Michael Dewey
[EMAIL PROTECTED]
http://www.aghmed.fsnet.co.uk/home.html
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Re: [R] timezones
Part of the problem is that 'now' is POSIXct and 'now.gmt' is POSIXlt. If you use as.POSIXct, you get the right answer. (now - Sys.time()) [1] 2003-08-03 18:29:38 EDT str(now) `POSIXct', format: chr 2003-08-03 18:29:38 (now.gmt - as.POSIXlt(now,tz=GMT)) [1] 2003-08-03 22:29:38 GMT str(now.gmt) `POSIXlt', format: chr 2003-08-03 22:29:38 (now.gmt - as.POSIXct(now,tz=GMT)) [1] 2003-08-03 18:29:38 EDT str(now.gmt) `POSIXct', format: chr 2003-08-03 18:29:38 now-now.gmt Time difference of 0 secs __ James HoltmanWhat is the problem you are trying to solve? Executive Consultant -- Office of Technology, Convergys [EMAIL PROTECTED] (513) 723-2929 Jerome Asselin [EMAIL PROTECTED]To: [EMAIL PROTECTED], [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED]Subject: Re: [R] timezones ath.ethz.ch 07/31/2003 12:30 I share your concerns regarding Problems 1 and 2. However, I am unable to provide help on those at this moment. As for Problem 3, an alternative for the time being would be to use another package such as chron or date, although it would be preferable to use the classes of the base package if possible. Sorry I can't be more helpful. Jerome On July 30, 2003 09:19 pm, Gabor Grothendieck wrote: I have some questions and comments on timezones. Problem 1. # get current time in current time zone (now - Sys.time()) [1] 2003-07-29 18:23:58 Eastern Daylight Time # convert this to GMT (now.gmt - as.POSIXlt(now,tz=GMT)) [1] 2003-07-29 22:23:58 GMT # take difference now-now.gmt Time difference of -5 hours Note that the difference between the times displayed by the first two R expressions is -4 hours. Why does the last expression return -5 hours? Problem 2. Why do the two expressions below give different answers? I take the difference between two dates in GMT and then repeat it in the current time zone (EDT). # days since origin in GMT julian(as.POSIXct(2003-06-29,tz=GMT),origin=as.POSIXct(1899-12-30 ,tz=GMT)) Time difference of 37801 days # days since origin in current timezone julian(as.POSIXct(2003-06-29),origin=as.POSIXct(1899-12-30)) Time difference of 37800.96 days I thought this might be daylight savings time related but even with standard time I get: julian(as.POSIXct(2003-06-29,tz=EST),origin=as.POSIXct(1899-12-30 ,tz=EST)) Time difference of 37800.96 days Problem 3. What is the general strategy of dealing with dates, as opposed to datetimes, in R? I have had so many problems and a great deal of frustration, mostly related to timezones. The basic problem is that various aspects of the date such as the year, the month, the day of the month, the day of the week can be different depending on the timezone you use. This is highly undesirable since I am not dealing with anything more granular than a day yet timezones, which are completely extraneous to dates and by all rights should not have to enter into my problems, keep fowling me up. A lesser problem is that I find myself using irrelevant constants such as the number of seconds in a day which you would think would be something I would not have to deal with since I am concerned with daily data. With R have nifty object oriented features I think a good project would be to implement a class in the base that handled dates without times (or timezones!) P.S. I have sent an earlier version of this but did not see it posted so if both get posted please ignore the prior one since this one has more info in it. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing
[R] Problem reading a PDF output
I generated a PDF output file of 10 plots. When I try to view it with
Adobe reader (R4 R5), it will lockup the reader (it is consuming 100% of
the CPU) after presenting the 4th plot. I can generate the plots just fine
in Windows and as a postscript file reading it with GSview.
Is there anyway to tell what might be wrong with the PDF output? The file
is 890KB in size if anyone would like to look at it. The postscript file
is 783KB in size.
I was able to isolate it to a single plot in a PDF file and it had 38,000
lines of the following that composed 99% of the file: (I was plotting out
individual events, which were about that many)
86.66 88.67 m
86.66 92.81 l
86.66 96.95 l
86.66 101.09 l
86.66 105.23 l
86.66 109.37 l
86.66 113.51 l
: 38,000 more of the same
388.87 96.95 l
388.87 92.81 l
388.97 88.67 l
389.07 84.53 l
S
Q q
0.000 0.000 0.000 RG
0.75 w
[] 0 d
Is this breaking some limit in PDF?
I am running:
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major1
minor7.1
year 2003
month06
day 16
language R
__
James Holtman What is the problem you are trying to solve?
Executive Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
(513) 723-2929
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Re: [R] Convert char vector to numeric table
use 'textConnection':
x.1 - c('1 2 3','4 5 6','7 8 9','8 7 6','6 5 4') # create character
vector
x.in - textConnection(x.1) # setup connection
x.data - read.table(x.in) # read in the character vector
x.data
V1 V2 V3
1 1 2 3
2 4 5 6
3 7 8 9
4 8 7 6
5 6 5 4
Nurnberg-LaZerte
[EMAIL PROTECTED] To: R's help mailing
list [EMAIL PROTECTED]
Sent by: cc:
[EMAIL PROTECTED]Subject: [R] Convert char vector to
numeric table
ath.ethz.ch
03/31/03 17:09
Please respond to
Nurnberg-LaZerte
I'm a great fan of read.table(), but this time the data had a lot of cruft.
So I used readLines() and editted the char vector to eventually get
something like this:
23.4 1.5 4.2
19.1 2.2 4.1
and so on. To get that into a 3 col numeric table, I first just used:
writeLines(data,tempfile)
read.table(tempfile,col.names=c(A,B,C))
Works fine, but writing to a temporary file seems ... inelegant? And
read.table() doesn't take a char vector as a file or connection argument.
The following works but it seems like a lot of code:
data - sub( +,,data)# remove leading blanks
for strsplit
data - strsplit(data, +)# strsplit returns a
list of char vectors
ndata - character(0) # vectorize the list
of char vectors
for (ii in 1:length(data)) ndata - c(ndata,data[[ii]])
ndata - as.numeric(ndata)
dim(ndata) - c(3,length(data))
data - t(ndata)
data.frame(A=data[,1],B=data[,2],C=data[,3])
Am I missing something?
Thanks,
Bruce L.
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Re: [R] overlapping pattern match (errata 2.0)
Another way to find all the multiple occurances of a character in a string is to use 'rle': x.s - 'aaabbcdeeeggiijjysbbddeffghjjjsdk' x - unlist(strsplit(x.s, NULL)) x [1] a a a b b c d e e e f f f f g g i i j [20] j y s b b d d e f f g h j j j s d k k [39] k k k rle(x) Run Length Encoding lengths: int [1:21] 3 2 1 1 3 4 2 2 2 1 ... values : chr [1:21] a b c d e f g i j y s b d e f g ... When the lengths are 1, the corresponding 'values' are the repeated characters. FMGCFMGC [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent by: cc: [EMAIL PROTECTED] [EMAIL PROTECTED]Subject: Re: [R] overlapping pattern match (errata 2.0) ath.ethz.ch 03/28/03 17:36 well! excuse me again but... your.string - aaacdf nc1 - nchar(your.string)-1 x - unlist(strsplit(your.string, NULL)) CORRECT x2 - c() for (i in 1:nc1) x2 - c(x2, paste(x[i], x[i+1], sep=)) ERRATA 2 cat(ocurrences of aa in your.string: , length(grep(aa, x2)), sep=, fill=TRUE) Fran PD: sorry again __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- NOTICE: The information contained in this electronic mail transmission is intended by Convergys Corporation for the use of the named individual or entity to which it is directed and may contain information that is privileged or otherwise confidential. If you have received this electronic mail transmission in error, please delete it from your system without copying or forwarding it, and notify the sender of the error by reply email or by telephone (collect), so that the sender's address records can be corrected. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
