Re: [R] creating unique and sorted values by combining columns
apply(tmp, 1, function(x) paste(sort(unique(x[x!=0])),collapse=_))
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
[EMAIL PROTECTED] a écrit :
Hi List,
I am combining column values of a dataframe to create a new variable
that is the sorted and unqiue combination of the columns (and excluding
0). The following works, but surely there is a more elegant way of doing
this?
t1-NULL
for(i in 1:nrow(tmp)) {
if(i == 1){
sort(c(tmp[i,1], tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),
decreasing=TRUE)-t1
}
else {
rbind(t1,sort(c(tmp[i,1],
tmp[i,2],tmp[i,3],tmp[i,4],tmp[i,5]),decreasing=TRUE))-t1
}
}
t2-NULL
for(i in 1:nrow(t1)){
if(i == 1) paste(unique(t1[i,t1[i,]0]),collapse=_)-t2
else cbind(t2,paste(unique(t1[i,t1[i,]0]),collapse=_))-t2
}
tmp
t1
t2
Any hints appreciated.
Thanks
Herry
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Re: [R] correlation and matrix
it should be smth like that:
apply(sapply(seq(1, 204, by=12), seq, length=4), 2, function(x)
{
M - dta[,x]
z - sapply(M, nlevels) # if dta is a dataframe
if (sum(z==1)3) cor(as.matrix(M[,z!=0]), use=comp, method=spear)
else NA
})
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
[EMAIL PROTECTED] a écrit :
Dear everyone,
I am new in R and I've got difficulties in realizing the following
tasks:
-I have variables (factors) with different numbers of levels, either 1,
2 or 3.
-I have a matrix containing these 204 factors and I have to correlate
them by groups of 4 variables.
-I have to delete the factors just having one level ( because when
correlating one-level factors, the output is NA)
here is my code:
lst-seq(1, 204, by=12) % there are 12 factors for 17 natural resources
for (n in lst)
{
Mx- matrix(0, byrow = F, ncol = 4, nrow=nrow(dta)) % I extract the 4
factors I have to correlate and I'd like to do it for each n
{if (nlevels(dta[,n+4])!=1)
Mx[,1]-dta[,n+4]
else
Mx[,1]-NA}
{if (nlevels(dta[,n+5])!=1)
Mx[,2]-dta[,n+5]
else
Mx[,2]-NA}
{if (nlevels(dta[,n+7])!=1)
Mx[,3]-dta[,n+7]
else
Mx[,3]-NA}
{if (nlevels(dta[,n+8])!=1)
Mx[,4]-dta[,n+8]
else
Mx[,4]-NA}
p-0% I compute the number of non - NA columns and I'd
like to delete the Na columns from that matrix
for (i in 1:4)
{
if(!is.na(sum(Mx[,i])0)) p-p+1
}
print(p)
{if (p==0 | p==1) stop(computation impossible)
else {
r-0
for (i in 1:4)
{
if(is.na(sum(Mx[,i])0)) r-i
}
print(r)
print(cor((as.matrix(Mx[,-r])), use=complete.obs, method=spearman))
}
}
} %The problem is the last step doesn't work for p==2.
In fact, it seems the loop for doesn't work either.
I hope it is clear enough and I thank you in advance for your help.
Nathalie
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Re: [R] regular expressions : extracting numbers
gsub( , , gsub(%, , gsub([a-z], , c(tr3,jh40%qs dqd [1] 3 40 Jacques VESLOT INRA - Biostatistique Processus Spatiaux Site Agroparc 84914 Avignon Cedex 9, France Tel: +33 (0) 4 32 72 21 58 Fax: +33 (0) 4 32 72 21 84 GOUACHE David a écrit : Hello all, I have a vector of character strings, in which I have letters, numbers, and symbols. What I wish to do is obtain a vector of the same length with just the numbers. A quick example - extract of the original vector : lema, rb 2% rb 2% rb 3% rb 4% rb 3% rb 2%,mineuse rb rb rb 12 rb rj 30% rb rb rb 25% rb rb rb rj, rb and the type of thing I wish to end up with : 2 2 3 4 3 2 12 30 25 or, instead of , NA would be acceptable (actually it would almost be better for me) Anyways, I've been battling with gsub() and things of the sort, but I'm drowning in the regular expressions, despite a few hours of looking at Perl tutorials... So if anyone can help me out, it would be greatly appreciated!! In advance, thanks very much. David Gouache Arvalis - Institut du Végétal Station de La Minière 78280 Guyancourt Tel: 01.30.12.96.22 / Port: 06.86.08.94.32 __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to add elements to a table
z - read.table(clipboard) z V1 V2 1 1990 20 2 1991 15 3 1995 17 4 1997 9 5 1998 10 6 1999 11 7 2000 5 zz - merge(data.frame(V1=1990:2000), z, by=V1, all.x=T) plot(zz, type=l) Jacques VESLOT INRA - Biostatistique Processus Spatiaux Site Agroparc 84914 Avignon Cedex 9, France Tel: +33 (0) 4 32 72 21 58 Fax: +33 (0) 4 32 72 21 84 [EMAIL PROTECTED] a écrit : Hi, I've been using R for a few weeks now, but consider myself still a newbie, especially in what concerns the basics of data handling in R. My situation is this: I read in data from a database using RODBC, and then use table to produce a table of counts per years, table which I give to plot. All is well, as long as there are no gaps in the years sequence, but now I have the following result (example): 1990 20 1991 15 1995 17 1997 9 1998 10 1999 11 2000 5 The plot function is quite intelligent, in the sense that it draws appropriate gaps in the x-axis between years, but not intelligent enough to interrupt the data line during the gap. This gives the impression that, although no year is marked on the x-axis between 1991 and 1995, there has been data for this period, which is not correct. What I tried to do is convert the table to a matrix, insert zeros for the missing years using rbind and cbind, and convert the result back to table. But the structure of this resulting table is not the same as for the originating table, so that I need to pass tab[1,] to plot. It's no longer a contingency table in fact. I've seen in the mailing list archives that there is an issue on using tables when matrixes or other structures would be more appropriate. I like the table, because plot automatically plots the corresponding years on the x-axis, which I find less error-prone than adding the tick labels later by hand, i.e. the association between years and counts is stronger. Also, as I tabulate counts of cases per gender, or per age categories, I think a contingency table is the right thing to use, isn't it? I'd be glad on any advice about what would be the best data structure to handle my situation, and I'd also like to know how I could automagically check a list of year, cases rows against a fixed list of years, insert zero or NA values for missing years, and get an easily usable result that I can forward to plot or barplot. Thanks a lot in advance, Anne-Marie __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Repeat if
sapply(1:85, function(i) eval(parse(text=paste(range(V, i, ,
na.rm=T), sep=
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
Birgit Lemcke a écrit :
Hello,
(Power Book G4, Mac OS X, R 2.5.0)
I would like to repeat the function range for 85 Vectors (V1-V85).
I tried with this code:
i-0
repeat {
+ i-i+1
+ if (i85) next
+ range (Vi, na.rm = TRUE)
+ if (i==85) break
+ }
I presume that the Vi is wrong, because in this syntax i is not known
as a variable. But I don´t know how to say that it is a variable here.
Would be nice if somebody could help me.
Perhaps I´m thinking too complicated and there is an easier way to do
this.
Thanks in advance
Greetings
Birgit
Birgit Lemcke
Institut für Systematische Botanik
Zollikerstrasse 107
CH-8008 Zürich
Switzerland
Ph: +41 (0)44 634 8351
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Repeat if
you may have a vector with only NA values in it...
max(c(NA,NA), na.rm=T)
[1] -Inf
Warning message:
aucun argument pour max ; -Inf est renvoyé
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
Birgit Lemcke a écrit :
Thanks that was really a quick answer.
It works but I get this warning message anyway:
1: kein nicht-fehlendes Argument für min; gebe Inf zurück (None
not-lacking argument for min; give Inf back)
2: kein nicht-fehlendes Argument für max; gebe -Inf zurück
what does this mean?
Greeting and thanks for your help!
Birgit
Am 28.06.2007 um 12:05 schrieb Jacques VESLOT:
sapply(1:85, function(i) eval(parse(text=paste(range(V, i, ,
na.rm=T), sep=
Jacques VESLOT
INRA - Biostatistique Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France
Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84
Birgit Lemcke a écrit :
Hello,
(Power Book G4, Mac OS X, R 2.5.0)
I would like to repeat the function range for 85 Vectors (V1-V85).
I tried with this code:
i-0
repeat {
+ i-i+1
+ if (i85) next
+ range (Vi, na.rm = TRUE)
+ if (i==85) break
+ }
I presume that the Vi is wrong, because in this syntax i is not
known as a variable. But I don´t know how to say that it is a
variable here.
Would be nice if somebody could help me.
Perhaps I´m thinking too complicated and there is an easier way to
do this.
Thanks in advance
Greetings
Birgit
Birgit Lemcke
Institut für Systematische Botanik
Zollikerstrasse 107
CH-8008 Zürich
Switzerland
Ph: +41 (0)44 634 8351
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailto: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
and provide commented, minimal, self-contained, reproducible code.
Birgit Lemcke
Institut für Systematische Botanik
Zollikerstrasse 107
CH-8008 Zürich
Switzerland
Ph: +41 (0)44 634 8351
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to replace some objects?
you can use as.numeric(factor( )); in your example: ex - sample(letters[1:3], 10, T) ex [1] b b c b a a b b a a as.numeric(factor(ex)) [1] 2 2 3 2 1 1 2 2 1 1 if the order is different, use levels: as.numeric(factor(ex, levels=letters[3:1])) [1] 2 2 1 2 3 3 2 2 3 3 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Zhang Jian a écrit : I want to replace some objects in one row or column.For example, One colume: a,b,a,c,b,b,a,a,c. I want to replace a with 1, b with 2, and c with 3. Like this: 1,2,1,3,2,2,1,1,3. How to do it? I donot know how to do it. Thanks. [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to replace some objects?
similarly: as.character(factor(c(abca,coma),labels=c(aaa,bbb))) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Zhang Jian a écrit : The example is simply.But the other example: One column: acba,coma,acmo,acmo,acba,coma I want to replace acba with aaa, coma with bbb, and acmo with ddd. Like this: aaa,bbb,ddd,ddd,aaa,bbb How to do it? Thanks. On 12/19/06, *Jacques VESLOT* [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: you can use as.numeric(factor( )); in your example: ex - sample(letters[1:3], 10, T) ex [1] b b c b a a b b a a as.numeric(factor(ex)) [1] 2 2 3 2 1 1 2 2 1 1 if the order is different, use levels: as.numeric(factor(ex, levels=letters[3:1])) [1] 2 2 1 2 3 3 2 2 3 3 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Zhang Jian a écrit : I want to replace some objects in one row or column.For example, One colume: a,b,a,c,b,b,a,a,c. I want to replace a with 1, b with 2, and c with 3. Like this: 1,2,1,3,2,2,1,1,3. How to do it? I donot know how to do it. Thanks. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailto: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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] barplot help needed
tab - do.call(rbind, list(data1, data2, data3, data4)) etype - rep(c(sd1, sd2, sd3, sd4), length(data1)) b - barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named position 1) and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named position 2) with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo - barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = Percentage) arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] barplot help needed
thought sd1, sd2... were scalars but if not just do: etype - c(sd1, sd2, sd3, sd4) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : Thank you very much for your help. I just don't understand the following line (which also gives me a dimension error later in the arrows command) etype - rep(c(sd1, sd2, sd3, sd4), length(data1)) Antje (I don't see my emails to the mailinglist anymore... just the answers from other people... I don't understand???) Jacques VESLOT schrieb: tab - do.call(rbind, list(data1, data2, data3, data4)) etype - rep(c(sd1, sd2, sd3, sd4), length(data1)) b - barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named position 1) and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named position 2) with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo - barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = Percentage) arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Error Message saying .Call(R_lazyLoadDBfetch, etc.
Hi, I got the following error message when running a function of mine doing intensive computations: Erreur dans .Call(R_lazyLoadDBfetch, key, file, compressed, hook, PACKAGE = base) : référence d'argument par défaut récursive I haven't found neither where the problem lies nor what could be recursive. Besides, I wonder whether it might or not be a problem of memory size. Could you please give me any suggestion on how to interpret this message? Thanks in advance, jacques --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] barplot help needed
tab - do.call(rbind, list(data1, data2, data3, data4)) etype - do.call(rbind, list(sd1, sd2, sd3, sd4)) b - barplot(tab, beside=T, ylim=c(0,max(tab+etype))) arrows(as.vector(b), as.vector(tab) - as.vector(etype), as.vector(b), as.vector(tab) + as.vector(etype), code=3) unlist() is not correct - sorry - since all are matrices - not data frames ! so use as.vector() --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : Still, there is one problem. The SD-Values don't fit to the bar they belong to. I made the following experiment: data1 - c(2,4,6,2,5) data2 - data1 sd1 - c(0.5,1,1.5,1,2) sd2 - sd1 tab - do.call(rbind, list(data1, data2)) etype - c(sd1,sd2) b - barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) I expect the bars with the same height and the same stddev. The height is okay, but the stddev is messed up... if I do it like this: etype - matrix(c(sd1,sd2), nrow=2, byrow=TRUE) it works (but maybe there is an easier way...) Antje Jacques VESLOT schrieb: thought sd1, sd2... were scalars but if not just do: etype - c(sd1, sd2, sd3, sd4) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : Thank you very much for your help. I just don't understand the following line (which also gives me a dimension error later in the arrows command) etype - rep(c(sd1, sd2, sd3, sd4), length(data1)) Antje (I don't see my emails to the mailinglist anymore... just the answers from other people... I don't understand???) Jacques VESLOT schrieb: tab - do.call(rbind, list(data1, data2, data3, data4)) etype - rep(c(sd1, sd2, sd3, sd4), length(data1)) b - barplot(tab, beside=T) arrows(unlist(b), unlist(tab) - etype, unlist(b), unlist(tab) + etype, code=3) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Antje a écrit : hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named position 1) and each bar in different colors. data1[2],data2[2],data3[2],data4[2] at the next x-axis label (named position 2) with the same color scheme and so on over the whole length. I managed to plot one set in the following way: par(mai=c(1.5,1,1,0.6)) plotInfo - barplot(data1, las=2, ylim = c(0,plotMax+1), ylab = Percentage) arrows(plotInfo,data1,plotInfo, data1 + sd1, length=0.1, angle=90) arrows(plotInfo,data1,plotInfo, data1 - sd1, length=0.1, angle=90) could anybody give me a help on this? Antje __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorise a for loop?
tt$fold - ifelse(tt$M 0, 1/(2^tt$M), 2^tt$M)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
john seers (IFR) a écrit :
Hi R guru coders
I wrote a bit of code to add a new column onto a topTable dataframe.
That is a list of genes processed using the limma package. I used a for
loop but I kept feeling there was a better way using a more vector
oriented approach. I looked at several commands such as apply, by
etc but could not find a good way to do it. I have this feeling there is
a command or technique eluding me. (Is there an expr:value1?value2
construction in R?)
Can anybody suggest an elegant solution?
Details:
So, the topTable looks like this:
topa1[1:5,c(1,2,3,4)]
IDName GB_accession M
11195 245828 SIGKEC9 AX135029 -7.670197
10966107FHL1 B14446 -5.089926
6287 25744 M90LL137340 -4.531744
777 2288 VSNL1 LF039555 -4.035472
11310 272294 M98LL031650 3.866422
I want to add a fold column so it will look like this:
topa1[1:5,c(1,2,3,4,10)]
IDName GB_accession M fold
11195 245828 SIGKEC9 AX135029 -7.670197 203.68521
10966107FHL1 B14446 -5.089926 34.05810
6287 25744 M90LL137340 -4.531744 23.13082
777 2288 VSNL1 LF039555 -4.035472 16.39828
11310 272294 M98LL031650 3.866422 14.58508
The fold values is calculated from the M column which is a log2 value.
The calculation is different depending on whether the M value is
negative or positive. That is if the gene is down regulated the
reciprocal value has to be used to calculate a fold value.
Here is my clunky, not vectorised code :
# Function to add a fold column to the toptable
ttfold-function(tt) {
fold-NULL
for (i in 1:length(tt$M)) {
if (tt$M[i] 0 ) {
fold[i]-1/(2^tt$M[i])
} else {
fold[i]-2^tt$M[i]
}
}
tt-cbind(tt, fold)
}
# Add fold column to top tables
topa1-ttfold(topa1)
Regards
J
---
John Seers
Institute of Food Research
Norwich Research Park
Colney
Norwich
NR4 7UA
tel +44 (0)1603 251497
fax +44 (0)1603 507723
e-mail [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
e-disclaimer at http://www.ifr.ac.uk/edisclaimer/
http://www.ifr.ac.uk/edisclaimer/
Web sites:
www.ifr.ac.uk http://www.ifr.ac.uk/
www.foodandhealthnetwork.com http://www.foodandhealthnetwork.com/
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Re: [R] Compiling a contingency table of counts by case
dat - read.delim(clipboard, sep=;) dat - dat[order(dat$case, dat$name), ] res - apply(combinations(nlevels(dat$name), 2), 1, function(x) with(dat[dat$name %in% levels(dat$name)[x],], table(unlist(sapply(split(x, case), function(y) ifelse(length(y) == 2, paste(y, collapse=), NA)) names(res) - apply(combinations(nlevels(dat$name), 2), 1, function(x) paste(levels(dat$name)[x], collapse=.)) res $Joe.John 11 1 $Joe.Karl character(0) $Joe.Mike 10 11 1 1 $Joe.Zoe 11 2 $John.Karl character(0) $John.Mike 10 1 $John.Zoe 11 1 $Karl.Mike 01 1 $Karl.Zoe 00 1 $Mike.Zoe 01 10 11 1 1 1 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Serguei Kaniovski a écrit : I have asked a similar question before but this time the problem is somewhat more involved. I have the following data: case;name;x 1;Joe;1 1;Mike;1 1;Zoe;1 2;Joe;1 2;Mike;0 2;Zoe;1 2;John;1 3;Mike;1 3;Zoe;0 3;Karl;0 I would like to count the number of case in which any two name a. both have x=1, b. the first has x=0 - the second has x=1, c. the first has x=1 - the second has x=0, d. both have x=0, The difficulty is that the number of names and their identity changes from case to case. Thanks a lot for you help, Serguei Kaniovski __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compiling a contingency table of counts by case
what's different from: with(dat, tapply(x, list(name,case), sum)) 1 2 3 Joe 1 1 NA John NA 1 NA Karl NA NA 0 Mike 1 0 1 and how to deal with this table ? --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- hadley wickham a écrit : I have asked a similar question before but this time the problem is somewhat more involved. I have the following data: case;name;x 1;Joe;1 1;Mike;1 1;Zoe;1 2;Joe;1 2;Mike;0 2;Zoe;1 2;John;1 3;Mike;1 3;Zoe;0 3;Karl;0 I would like to count the number of case in which any two name a. both have x=1, b. the first has x=0 - the second has x=1, c. the first has x=1 - the second has x=0, d. both have x=0, One way is to use the reshape package: dat - read.delim(clipboard, sep=;) dat - rename(dat, c(x=value)) cast(dat, name ~ case) (which doesn't give you exactly what you want, but I think might be more illuminating) Hadley __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] coerce matrix to number
if only 2 letters: (z==v)*1 else: lapply(z, function(x) as.numeric(as.character(factor(x,levels= c(d,v,w),labels=c(0,1,2) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Simone Gabbriellini a écrit : Dear List, how can I coerce a matrix like this [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0 1 1 0 0 0 [2,] 1 0 1 0 0 0 [3,] 1 1 0 0 0 0 [4,] 0 0 0 0 1 0 [5,] 0 0 0 1 0 0 [6,] 0 0 0 0 0 0 to be filled with numbers? this is the result of replacing some character (v, d) with 0 and 1, using the code I found with RSiteSearch() z[] - lapply(z, factor, levels = c(d, v), labels = c(0, 1)); thank you, Simone |-| dott. Simone Gabbriellini PhD Student Dipartimento di Scienze Sociali Università di Pisa via Colombo 35 - 56100 Pisa mail: [EMAIL PROTECTED] mobile: +39 3475710037 |-| Please avoid sending me Word or PowerPoint attachments. See http://www.gnu.org/philosophy/no-word-attachments.html [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Export data
res - as.data.frame(t(c(A=1,B=2,C=3))) res A B C 1 1 2 3 write.table(res, file=file.csv, sep=\t, row.names=F, col.names=T) write.table(t(4:6), append=T, file=file.csv, sep=\t, row.names=F, col.names=F) read.csv(file.csv, sep=\t) A B C 1 1 2 3 2 4 5 6 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Marie-Noëlle Rolland a écrit : Hi there, I would like to write a data output. In first time, a data frame (called res) is written in a file *csv such as: ABC# names of data 1.5 4.5 7# values So I use write.table which prints res to a file file.csv write.table(res,file=file.csv,sep=,row.names=TRUE,col.names=TRUE,eol = \n) After, I would like to append other data to the same file, ABC 1.5 4.5 7 369 # appended data I use then the same function but with different arguments: write.table(res,file=file.csv,append=TRUE,sep=,row.names=TRUE,col.names=FALSE,eol = \n) Actually, it doesn't work, the file is overwritten by the new one. I got something like that: ABC 369 Coul you help me? Thank you in advance Kind regards, Marie-Noëlle __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] newbie question about index
(a==1)*1 or ifelse(a == 1, 1, 0) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- z s a écrit : Hi, I am trying to convert a variable a = sample(1:3,100,rep = T) represents choices into a 3X100 dummy varible b with corresponding element set to 1 otherwise 0. eg. a: 1 3 2 1 2 3 1 1 b: 1 0 0 1 0 0 1 1.. 0 0 1 0 1 0 0 0... 0 1 0 0 0 1 0 0... Is there something like b[a] =1 existing? I could not figure this out myself. - Mp3·è¿ñËÑ-иèÈȸè¸ßËÙÏ [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Barplot
barplot(t(sapply(split(z1[,1:8], z1$V9),colSums)), beside=T) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Muhammad Subianto a écrit : Dear all, I have a dataset. I want to make barplot from this data. Zero1 - V1 V2 V3 V4 V5 V6 V7 V8 V9 1 1 0 0 0 1 0 0 0 Positive 2 0 0 1 0 1 0 1 1 Negative 3 0 0 1 0 0 0 1 1 Positive 4 0 1 0 1 1 1 0 1 Negative 5 0 0 1 0 1 1 0 0 Positive 6 0 1 0 0 1 1 1 1 Negative 7 1 0 1 1 1 1 1 1 Negative 8 0 0 0 0 1 0 0 1 Negative 9 0 1 1 1 1 0 0 1 Negative 10 0 0 0 1 1 0 1 0 Positive 11 0 0 0 0 1 0 0 1 Negative 12 0 0 1 1 1 1 1 0 Positive 13 0 1 1 0 1 1 1 1 Negative z1 - read.table(textConnection(Zero1), header=TRUE) z1 str(z1) A simple way I can use mosaic plot mosaicplot(table(z1)) library(vcd) mosaic(table(z1)) I have tried to learn ?xtabs ?table and ?ftable but I can't figure out. I need a barplot for all variables and the result maybe like | | | | | | | | | || | | |pos|neg| |pos|neg||pos|neg| | | | | | || | | - -- v1v2v3 v7 v8 Thanks you for any helps. Regards, Muhammad Subianto __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] First elements of a list.
sapply(a, [, 1) [1] John Jane koda gunner sapply(a, [, 2) [1] Smith Doe NA NA --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- cory a écrit : Suppose I have the following list: a - strsplit(c(John;Smith, Jane;Doe, koda, gunner), ;) I want to get to these two vectors without looping... firstNames:c(John, Jane, koda, gunner) lastNames:c(Jane, Doe, NA, NA) Thanks cn [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] [Fwd: Trend test and test for homogeneity of odd-ratios]
I partly answered my question since independence_test() function in coin package apparently do Cochran-Armitage trend test just like Eric Lecoutre's function tabletrend() - slightly modified here: independence_test(pheno ~ geno, data = dat2, teststat = quad, scores = list(geno = c(0, 1, 2))) Asymptotic General Independence Test data: pheno by groups 1 2 3 chi-squared = 0.2268, df = 1, p-value = 0.6339 tabletrend(with(dat2, table(pheno, geno))) [1] 0.6338308 Message original Sujet: Trend test and test for homogeneity of odd-ratios Date: Wed, 16 Aug 2006 17:39:33 +0200 De: Jacques VESLOT [EMAIL PROTECTED] Pour: R-Help r-help@stat.math.ethz.ch Dear r-users, I am looking for some R functions to do Cochran-Armitage trend test for 2*3 tables (binary phenotype vs. genotypes) and for testing the homogeneity of odds ratios within 2*3*k tables (binary phenotype vs. genotypes vs. strata). In R-Help archives, I've found a 2003 script by Eric Lecoutre for Cochran-Armitage trend test and a script for Breslow-Day test for 2*2*k tables. Could someone please tell we if there were some functions available on CRAN to do such tests ? Thanks, jacques __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to remove similar successive objects from a vector?
VECTOR=c(3,2,2,3,4,4,5,5,5,3,3,3,5,1,6,6)
NEWVECTOR - ifelse(VECTOR[-length(VECTOR)]==VECTOR[-1],NA,VECTOR)
NEWVECTOR[!is.na(NEWVECTOR)]
[1] 3 2 3 4 5 3 5 1
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Atte Tenkanen a écrit :
Is there some (much) more efficient way to do this?
VECTOR=c(3,2,4,5,5,3,3,5,1,6,6);
NEWVECTOR=VECTOR[1];
for(i in 1:(length(VECTOR)-1))
{
if((identical(VECTOR[i], VECTOR[i+1]))==FALSE){
NEWVECTOR=c(NEWVECTOR,VECTOR[i+1])}
}
VECTOR
[1] 3 2 4 5 5 3 3 5 1 6 6
NEWVECTOR
[1] 3 2 4 5 3 5 1 6
___
Atte Tenkanen
University of Turku, Finland
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[R] Trend test and test for homogeneity of odd-ratios
Dear r-users, I am looking for some R functions to do Cochran-Armitage trend test for 2*3 tables (binary phenotype vs. genotypes) and for testing the homogeneity of odds ratios within 2*3*k tables (binary phenotype vs. genotypes vs. strata). In R-Help archives, I've found a 2003 script by Eric Lecoutre for Cochran-Armitage trend test and a script for Breslow-Day test for 2*2*k tables. Could someone please tell we if there were some functions available on CRAN to do such tests ? Thanks, jacques __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combinations question
library(gtools)
cb - function(n,r) t(apply(combinations(n, r), 1, function(x) ifelse(1:n %in%
x, 1, 0)))
cb(6,3)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Martin a écrit :
I need to generate a {0,1} matrix wifht nCr rows and n columns. The rows of
the matrix will consist of all possible combinations containing r ones.
My clumsy attempt for n = 6 and r = 3 is
X - expand.grid(c(1,0),c(1,0),c(1,0),c(1,0),c(1,0),c(1,0))
Y - X[rowSums(X)==3,]
I can genralize this in a function but the result is quite ugly. Any
suggestions?
Thank you in advance.
Martin
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] questions on aggregate data
data.frame(x = with(df1, rep(x, freq))) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- zhijie zhang a écrit : Dear friends, my question is how to aggregate dataset and the inverse manipulation. e.g.My dataset data structure1: x 1 1 2 3 3 data structure2: x freq 1 2 2 1 3 2 Then how to generate dataset2 from dataset1 and generate dataset1 from dataset2? e.g. dataset2 from dataset1 : x-c(1,1,2,3,3) a-tab(x) as.data.frame(a) *But i can't do the inverse manipulation:generate dataset1 from dataset2*, anybody can help me on the two different manipulations? Thanks a lot! __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] missing value
dat[dat==9] - NA because the result of mean() is real and summary()'s output is a vector. --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Mauricio Cardeal a écrit : Hi all # I have this data set and how can I assign NA´s in just one command ? And why the summary(dat) function preserves the value 9 as real. ? x - c(1,2,3,9,4) y - c(3,6,9,2,3) z - c(9,9,2,2,8) w - c(6,5,3,0,9) dat - cbind(x,y,z,w) summary(dat) x[x==9] - NA y[y==9] - NA z[z==9] - NA w[w==9] - NA summary(dat) summary(x) summary(y) summary(z) summary(w) Thank you all, Mauricio __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with allp ossible combination.
diffs - do.call(expand.grid, dt) diffs$delta - rowSums(expand.grid(dt$a, -dt$b)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Arun Kumar Saha a écrit : Dear R Users, Suppose I have a dataset like this: a b 39700 485.00 39300 485.00 39100 480.00 38800 487.00 38800 492.00 39300 507.00 39500 493.00 39400 494.00 39500 494.00 39100 494.00 39200 490.00 Now I want get a-b for all possible combinations of a and b. Using two 'for' loop it is easy to calculate. But problem arises when row length of the data set is large eg. 1000 or more. Then R takes lot of time to do that. Can anyone please tell me whether there is any R-function to do such kind of job quickly? Thanks and regards, [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] mirror vector?
mat - matrix(1:16,4,4) mat [,1] [,2] [,3] [,4] [1,]159 13 [2,]26 10 14 [3,]37 11 15 [4,]48 12 16 apply(mat,2,rev) [,1] [,2] [,3] [,4] [1,]48 12 16 [2,]37 11 15 [3,]26 10 14 [4,]159 13 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- [EMAIL PROTECTED] a écrit : Hello, I'm an absolut beginner with R and now I got a 2D vector with numbers. I would like to mirror this vector now by the rows (so that the first row becomes last, second becomes one before last, ...). I don't know if there is any method I can use to do this. Could you please help me? Antje - Was Sie schon immer wissen wollten aber nie zu Fragen trauten? Yahoo! Clever hilft Ihnen. [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster alternative to by?
table(mapped$col2) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- michael watson (IAH-C) a écrit : Hi I have a data.frame, two columns, 12304 rows. Both columns are factors. I want to do an equivalent of an SQL group by statement, and count the number of rows in the data frame for each unique value of the second column. I have: countl - by(mapped, mapped$col2, nrow) Now, mapped$col2 has 10588 levels, so this statement takes a really long time to run. Is there a more efficient way of doing this in R? Thanks Mick __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] the first and last case
do.call(rbind,lapply(split(dat, dat$ind), function(x) x[c(1,nrow(x)),])) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Mauricio Cardeal a écrit : Hi all Sometime ago I asked for a solution about how to aggregate data and the help was wonderful. Now, I´d like to know how to extract for each individual case below the first and the last observation to obtain this: ind y 18 19 27 2 11 39 3 10 4 8 4 5 # Below the example: ind - c(1,1,1,2,2,3,3,3,4,4,4,4) y - c(8,10,9,7,11,9,9,10,8,7,6,5) dat - as.data.frame(cbind(ind,y)) dat attach(dat) mean.ind - aggregate(dat$y, by=list(dat$ind), mean) mean.ind Thanks Mauricio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Permutation Distribution
data1 - expand.grid(var1=1:15, var2=1:2) test - replicate(1000, with(data.frame(var1=data1$var1, var2=sample(data1$var2)), diff(tapply(var1, var2, mean hist(test) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Jacob van Wyk a écrit : Hallo Is there an elegant way to do the following: Dataset consists of 2 variables: var1: some measurements, and var2: a grouping variable with two values, 1 and 2. There are (say) 10 measurements from group 1 and 15 measurements from group 2. The idea is to study the permutation distribution of mean(group 1) * mean(group2). One way would be to permute 1s and 2s and select the corresponding measurements; calculate the difference in means. Redo this 1000 times, say. Etc. Any help is much appreciated. Thanks Jacob Jacob L van Wyk Department of Statistics University of Johannesburg, APK P O Box 524 Auckland Park 2006 South Africa Tel: +27 11 489 3080 Fax: +27 11 489 2832 [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] hist or barplot
s1 - runif(10,0,10) s1 [1] 8.328396 2.840870 7.401377 9.998165 5.045539 9.568728 5.372493 5.232439 [9] 5.774790 4.224103 s2 - runif(10,0,10) s2 [1] 1.230750 3.855060 8.652698 7.846725 9.100171 7.309179 9.235562 1.581741 [9] 6.979521 1.918997 ss - cbind(s1,s2) smin - apply(ss,1,min) smax - apply(ss,1,max) barplot(smax) barplot(smin, add=T, col=white) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- COMTE Guillaume a écrit : Hi all, I wish to draw 2 hist (or barplot) on the same graph. I can do it simply by using par(new=TRUE) , but it overlap with the first drawn, how to tell R to put in front of the graph the min value of the two graph in order for it to be seen and don't hide each other. I've been looking at help for barplot or hist but didn't find anything... (or my english is too poor to understand) thks COMTE Guillaume [[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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] A contingency table of counts by case
sorry, answered to quickly... actually it's easier using paste(): df - df[order(df$case),] apply(combinations(9,2), 1, function(y) table(factor(do.call(paste, c(with(df[df$id %in% y, ], split(x, id)), sep=)), levels=c(00,01,10,11 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Serguei Kaniovski a écrit : Here is an example of the data.frame that I have, df-data.frame(case=rep(1:5,each=9),id=rep(1:9,times=5),x=round(runif(length(rep(1:5,each=9) case represents the cases, id the persons, and x is the binary state. I would like to know in how many cases any two persons a. both have 1, b. the first has 0 - the second has 1, c. the first has 0 - the second has 0, d. both have 0. There will be choose(9,2) sums, denoted s_ij for 1=ij=9, where i and j are running indices for an id-pair. Please help, this is way beyond my knowledge of R! Thank you, Serguei Kaniovski __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Object name and Strings?
deparse(substitute(a)) [1] a --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Stéphane Cruveiller a écrit : Hi all, Is there a simple way to convert an object name to a characters string? Stéphane. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] A contingency table of counts by case
library(gtools) apply(combinations(9,2), 1, function(x) with(df[df$id %in% x, ], table(x, id))) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Serguei Kaniovski a écrit : Here is an example of the data.frame that I have, df-data.frame(case=rep(1:5,each=9),id=rep(1:9,times=5),x=round(runif(length(rep(1:5,each=9) case represents the cases, id the persons, and x is the binary state. I would like to know in how many cases any two persons a. both have 1, b. the first has 0 - the second has 1, c. the first has 0 - the second has 0, d. both have 0. There will be choose(9,2) sums, denoted s_ij for 1=ij=9, where i and j are running indices for an id-pair. Please help, this is way beyond my knowledge of R! Thank you, Serguei Kaniovski __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple question about variables....
see ?$ 'x$name' is equivalent to 'x[[name]]' so you need use : eval(parse(text = paste(OBJ$, varname))) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Stéphane Cruveiller a écrit : Dear R users, I have a simple question on variable manipulation. Imagine I have an object OBJ that has toto as one of its variables. I would like to understand why if I do varname - toto OBJ$varname returns no results whereas OBJ[varname]returns the column entitled toto Thanks for your help. Stéphane. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] simple question about variables....
OBJ$toto works to... b - as.data.frame(matrix(1:4,2)) b V1 V2 1 1 3 2 2 4 b$V1 [1] 1 2 but varname is not evaluated in OBJ$varname. --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Joerg van den Hoff a écrit : Stéphane Cruveiller wrote: Dear R users, I have a simple question on variable manipulation. Imagine I have an object OBJ that has toto as one of its variables. I would like to understand why if I do varname - toto OBJ$varname returns no results whereas OBJ[varname]returns the column entitled toto Thanks for your help. Stéphane. because if the value of `varname' is substituted in the expressions, in the first case that yields OBJ$toto and in the second OBJ[toto] the latter is valid, the former is not (you'd need `OBJ$toto' there), read ` ?$ ': ...Both '[[' and '$' select a single element of the list. The main difference is that '$' does not allow computed indices, whereas '[[' does. 'x$name' is equivalent to 'x[[name]]'... not, too, the difference between `[' (sublist) and `[[' (single element extraction) joerg __ 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] Problems plotting a function defined as a product
plot(t, sapply(t,g))
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Alessandro Antonucci a écrit :
In order to define a function f as:
f - function(x) (x+1)*(x+2)
I want to use the notation:
v = c(1,2)
g - function(x) prod((v+x))
That apparently works and, for instance,
the loop:
for (i in 1:100) { print(f(i)-g(i)) }
Produces a sequence of zeros.
Nevertheless, if I try to plot
the function g by:
t = seq(0,100,1)
plot(t,g(t),type=l)
I obtain the following errors/warning:
Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ
In addition: Warning message:
longer object length
is not a multiple of shorter object length in: v + x
Execution halted
Any idea about that?
Kind regards,
Alessandro
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Re: [R] colors on graph
?image Cf. See Also --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- COMTE Guillaume a écrit : Hy all, I need to draw something in 2 dimension that has 3 dimension, the choice has been made to use colors to display the third dimension into the graph. Has someone done something like that, i can't figure out how to parametize the colors. Thks for all ideas, COMTE Guillaume [[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] Table of P values for Fisher's exact test
apply(yourdata[, c(X2N_CHB,X2N_AA,Counts_CHB,Counts_AA)], 1, function(x) fisher.test(cbind(x[3:4], x[1:2]-x[3:4]))$p.value) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- jenny tan a écrit : Hi, I have a table of observed counts for various genetic markers. Instead of doing Fisher's exact test for each marker one at a time and recording the P value manually, is there a script to go through the whole list and generate the P value column automatically? An example of my data: Counts_CHB and Counts_AA are the observed counts for one allele. 2N_CHB and 2N_AA are the total number of alleles. gene Local_Pos 2N_CHB 2N_AA Counts_CHB Counts_AA Exact_P B2M 247590 46 0 5 B2M 353290 44 0 1 FCN2 220388 46 0 1 FCN2 353690 46 0 9 FCN2 402784 46 0 9 FCN2 403690 46 0 9 FCN2 431890 46 0 9 FCN2 439290 46 0 9 FCN2 957590 46 0 1 __ 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] new object
names(summary(fit)) [1] surv time n.risk n.event conf.int std.err [7] lowerupperstrata call summary(fit)$n.risk [1] 11 10 8 7 5 4 2 12 10 8 6 5 4 3 2 1 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Mauricio Cardeal a écrit : Hi ! Please, how can I extract n.event and n.risk as a new object from example below ? Thanks in advance. Mauricio require(survival) fit - survfit(Surv(time, status) ~ x, data=aml) summary(fit) Call: survfit(formula = Surv(time, status) ~ x, data = aml) x=Maintained time n.risk n.event survival std.err lower 95% CI upper 95% CI 9 11 10.909 0.0867 0.75411.000 13 10 10.818 0.1163 0.61921.000 18 8 10.716 0.1397 0.48841.000 23 7 10.614 0.1526 0.37690.999 31 5 10.491 0.1642 0.25490.946 34 4 10.368 0.1627 0.15490.875 48 2 10.184 0.1535 0.03590.944 x=Nonmaintained time n.risk n.event survival std.err lower 95% CI upper 95% CI 5 12 2 0.8333 0.1076 0.64701.000 8 10 2 0.6667 0.1361 0.44680.995 12 8 1 0.5833 0.1423 0.36160.941 23 6 1 0.4861 0.1481 0.26750.883 27 5 1 0.3889 0.1470 0.18540.816 30 4 1 0.2917 0.1387 0.11480.741 33 3 1 0.1944 0.1219 0.05690.664 43 2 1 0.0972 0.0919 0.01530.620 45 1 1 0. NA NA NA [[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] How to include NA's of a factor in table?
fcv - factor(c('a', NA, 'c'), exclude=NULL)
table(fcv, exclude=NULL)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Heinz Tuechler a écrit :
Dear All,
Is there a better way to include NA's of a factor in the output of table()
than using as.character()?
Admittedly, I do not understand the help page for table concerning the
exclude argument applied to factors. I tried in different ways, but could
not get NA to be included in the table, if not using as.character() (see
example).
Greetings,
Heinz
## example
fcv - factor(c('a', NA, 'c'))
table(fcv)# shows a, c
table(fcv, exclude='a') # shows c
table(fcv, exclude=)# shows a, c
table(fcv, exclude=NULL) # shows a, c
table(as.character(fcv), exclude=NULL) # shows a, c, NA
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status Patched
major 2
minor 3.1
year 2006
month 07
day01
svn rev38471
language R
version.string Version 2.3.1 Patched (2006-07-01 r38471)
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Re: [R] questions on data management
merge(mn[sample(1:nrow(mn), 3, rep=F),], xy, by.x=c(m,n), by.y=c(x,y)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- zhijie zhang a écrit : Dear friends, suppose i have two datasets: A and B A: id-1:6 x-c(1,2,3,4,5,6) y-c(2,4,6,8,3,2) xy-data.frame(id,x,y) B m-c(1,1,3,3,5,5) n-c(2,2,6,6,3,3) mn-data.frame(m,n) Now, i want to perfomr two tasks: 1. get a subset of B,no duplicate values,: C: m n 1 2 3 6 5 3 2.Extract the values in A on the conditions that x=m and y=n the results should be: id x y 1 1 2 3 3 6 5 5 3 Thanks very much! __ 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] questions on data management
mnu - unique(mn[order(mn$m,mn$n),]) mnu$p - table(paste(mn$m, mn$n)) merge(mnu[sample(1:nrow(mnu), size=3, prob=mnu$p, replace=F), c(m,n)], xy, by.x=c(m,n), by.y=c(x,y)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- zhijie zhang a écrit : Dear friends, suppose i have two datasets: A and B A: id-1:6 x-c(1,2,3,4,5,6) y-c(2,4,6,8,3,2) xy-data.frame(id,x,y) B m-c(1,1,3,3,5,5) n-c(2,2,6,6,3,3) mn-data.frame(m,n) Now, i want to perfomr two tasks: 1. get a subset of B,no duplicate values,: C: m n 1 2 3 6 5 3 2.Extract the values in A on the conditions that x=m and y=n the results should be: id x y 1 1 2 3 3 6 5 5 3 Thanks very much! __ 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] tapply question
i think you can't have column with the same names. data.frame(AAA=1:3, AAA=4:6) AAA AAA.1 1 1 4 2 2 5 3 3 6 but you could subset the data frame by names using substring(): sapply(unique(substring(names(data1), 1, 3)), function(x) rowMeans(data1[, substring(names(data1), 1, 3) == x]) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- [EMAIL PROTECTED] a écrit : I think I understand tapply but i still can't figure out how to do the following. I have a dataframe where some of the column names are the same and i want to make a new dataframe where columns that have the same name are averaged by row. so, if the data frame, DF, was AAABBB CCC AAA DDD 1 07 11 13 20 8 12 14 30 6 0 15 then the resulting data frame would be exactly the same except that the AAA column would be 6 comes from (11 + 1)/2 7comes from (12 + 2)/2 3 stays 3 because the element in the other AAA is zero so i don't want to average that one. it shoulsd just stay 3. So, I do DF[DF == 0]-NA rowaverage-function(x) x[rowMeans(forecastDf[x],na.rm=TRUE) revisedDF-tapply(seq(DF),names(DF),rowmeans) there are two problems with this : 1) i need to go through the rows of the same name, not the columns so i don't think seq(DF) is right because that goes through the columns but i want to go through rows. 2) BBB will come back with ALL NA's ( since it was unique and there was nothing else to average ( and I don't know how to transform that BB column to all zero's. thanks and i'm sorry for so many questions. i'm getting bettter with this stuff and my questions will decrease soon. my guess is that i no longer should be using tapply ? and should be using some other version of apply. thanks mark __ 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] IMPORTING FILE FROM EXCEL
if you don't need to import several dataset, you can directly change current
directory from the file
menu and move to your working directory instead of substuting all backslashes
for slashes in import
function argument.
you can also copy your dataset in excel and do :
data1 - read.delim(clipboard)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Rashmi Pant a écrit :
Hi
I am a beginner with R.
I have been trying to import a tab delimited excel file but i get the
following error message
file.show('C:\Documents and Settings\stats\Desktop\SUMI\plasma2.txt')
Warning message:
file.show(): file 'C:Documents and SettingsstatsDesktopSUMIplasma2.txt' does
not exist
I have understood the programming part but i cannot go ahead unless i have
imported the file. I have consulted the R-help archive without success.
Any help will be appreciated
Thanks in advance
-
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Re: [R] random sampling problems?
with replacement or not ? without replacement: data1 - cbind(id=1:9, expand.grid(x=1:3,y=1:3)) merge(data1, sapply(data1[,c(x,y)], sample, 3), all.y=T) why not: data1[sample(data1$id, 3),] --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- zhijie zhang a écrit : Dear friends, suppose my dataset is the following data: id-1:9 x-c(1,2,3,1,2,3,1,2,3) y-c(1,1,1,2,2,2,3,3,3) data-data.frame(id,x,y) id x y 1 1 1 1 2 2 2 1 3 3 3 1 4 4 1 2 5 5 2 2 6 6 3 2 7 7 1 3 8 8 2 3 9 9 3 3 i want to do sampling like this:say the sample size is 3. First: random sampling from x; Next ,random sampling from y ;and combing sampled x and sampled y; Finally, output the samples: id x and y. I think i could call it two-dimension sampling. Thanks very much! __ 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] Query : Chi Square goodness of fit test
chisq.test(counts, p=Expected/sum(Expected), simulate.p.value =FALSE, correct = FALSE) Chi-squared test for given probabilities data: counts X-squared = 40.5207, df = 13, p-value = 0.0001139 Warning message: l'approximation du Chi-2 est peut-être incorrecte in: chisq.test(counts, p = Expected/sum(Expected), simulate.p.value = FALSE, but the use of Chi2 test is incorrect since some of Expected frequencies are lower than 5. --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- priti desai a écrit : I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R run the script The R script is as follows ## start # No_of_Frouds- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N - length(No_of_Frouds) # Estimation of Parameter lambda- sum(No_of_Frouds)/N lambda pmf - dpois(i, lambda, log = FALSE) step_function - ppois(i, lambda, lower.tail = TRUE, log.p = FALSE) # Chi-Squared Goodness of Fit Test # Ho: The data follow a Poisson distribution Vs H1: Not Ho Frauds - c(1:13) counts- c(2,3,3,5,7,2,1,1,2,3,2,1,1,0) # Observed frequency Expected -c(0.251005528,1.224602726,2.987288468,4.85811559,5.925428863,5.7817821 03,4.701348074,3.276697142,1.998288788,1.083247457,0.528493456,0.2344006 79,0.095299266,0.035764993) chisq.test(counts, Expected, simulate.p.value =FALSE, correct = FALSE) # end The result of R is as follows Pearson's Chi-squared test data: counts and poisson_fit X-squared = 70, df = 65, p-value = 0.3135 Warning message: Chi-squared approximation may be incorrect in: chisq.test(counts, poisson_fit, simulate.p.value = FALSE, correct = FALSE) But I have done calculations in Excel. I am getting different answer. Observed = 2,3,3,5,7,2,1,1,2,3,2,1,1,0 Expected=0.251005528,1.224602726,2.987288468,4.85811559,5.925428863,5.78 1782103,4.701348074,3.276697142,1.998288788,1.083247457,0.528493456,0.23 4400679,0.095299266,0.035764993 Estimated Parameter =4.878788 Chi square stat = 0.000113 My excel answer tally with the book which I have refer for excel. Please tell me the correct calculations in R. ## Awaiting your positive reply. Regards. Priti. __ 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] Passing arguments to glm()
f.myglm - function(y=y, subset=x2 == 'yes', data=d.d.mydata)
eval(parse(text=glm(,
deparse(substitute(y)), ~ x1, family=binomial, data=,
deparse(substitute(data)), , subset =,
subset, )))
f.myglm()
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Christian Bieli a écrit :
Hi there
I want to pass arguments (i.e. the response variable and the subset
argument) in a self-made function to glm.
Here is one way I can do this:
f.myglm - function(y,subfact,subval) {
glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval)
}
str(d.mydata)
`data.frame':15806 obs. of 3 variables:
$ y : Factor w/ 2 levels no,yes: 1 1 1 1 1 1 1 NA 1 1 ...
$ x1: Factor w/ 2 levels no,yes: 2 2 1 2 2 2 2 2 2 2 ...
$ x2: Factor w/ 2 levels no,yes: 1 1 1 1 1 2 2 1 2 2 ...
f.myglm('y','x2','yes')
But is there a way I can pass the arguments and use the data argument of
glm()?
In a naive way of thinking I'd like to something like this:
f.myglm - function(y,sub) {
glm(y~x1,family=binomial,data=d.mydata,subset=sub)
}
f.myglm(y=y,sub=x2=='yes')
I know that's not possible, because the objects y and x2 are not defined
in the user workspace.
So, something like passing the arguments as an expression and evaluate
it in the glm function should work, but I didn't manage to do it.
I'd appreciate your advice.
Christian
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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Re: [R] Passing arguments to glm()
i forgot a paste():
f.myglm - function(y=y, subset=x2 == 'yes', data=d.d.mydata)
eval(parse(text=paste(glm(,
deparse(substitute(y)), ~ x1, family=binomial, data=,
deparse(substitute(data)), , subset =,
subset, ), sep=)))
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Jacques VESLOT a écrit :
f.myglm - function(y=y, subset=x2 == 'yes', data=d.d.mydata)
eval(parse(text=glm(,
deparse(substitute(y)), ~ x1, family=binomial, data=,
deparse(substitute(data)), , subset =,
subset, )))
f.myglm()
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Christian Bieli a écrit :
Hi there
I want to pass arguments (i.e. the response variable and the subset
argument) in a self-made function to glm.
Here is one way I can do this:
f.myglm - function(y,subfact,subval) {
glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval)
}
str(d.mydata)
`data.frame':15806 obs. of 3 variables:
$ y : Factor w/ 2 levels no,yes: 1 1 1 1 1 1 1 NA 1 1 ...
$ x1: Factor w/ 2 levels no,yes: 2 2 1 2 2 2 2 2 2 2 ...
$ x2: Factor w/ 2 levels no,yes: 1 1 1 1 1 2 2 1 2 2 ...
f.myglm('y','x2','yes')
But is there a way I can pass the arguments and use the data argument of
glm()?
In a naive way of thinking I'd like to something like this:
f.myglm - function(y,sub) {
glm(y~x1,family=binomial,data=d.mydata,subset=sub)
}
f.myglm(y=y,sub=x2=='yes')
I know that's not possible, because the objects y and x2 are not defined
in the user workspace.
So, something like passing the arguments as an expression and evaluate
it in the glm function should work, but I didn't manage to do it.
I'd appreciate your advice.
Christian
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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Re: [R] help with table partition
do.call(cbind, split(as.data.frame(test_table), rep(1:170,each=366)))
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Wong, Kim a écrit :
Hi,
I have a test_table where the dim is 62220 by 73 (row by col)
I would like to partition the rows into 170 equal parts (170 tables
where each is of dim 366 by 73), and rearrange them horizontally. The
source codes I have:
for (i in 1:170) {
c = cbind(c,test_table[(367*i+1):(367*(i+1)),2:73]);
}
Unfortunately, using for loop and cbind for a table of this size causes
long running time. What is the most efficient way to get the table that
I want?
Thanks for any help.
K.
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Re: [R] plotting gaussian data
curve(coef$A * dnorm(x, coef$mu, coef$sig), -4, 4) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- H. Paul Benton a écrit : Ok I guess it's time to ask. So I want to plot my data. It's my data from a frequency table, temp. My formula is just a Gaussian eq. I have done the nls function to get my parameters and now I want to do the whole plot (...) and then lines(..) This is what I have done. temp bin x 1 -4.0 0 2 -3.9 0 3 -3.8 0 4 -3.7 0 5 -3.6 0 6 -3.5 0 and so on fo x ~ (A/(sig * sqrt(2 * pi))) * exp(-1 * ((bin - mu)^2/(2 * sig^2))) fo.v x ~ (335.48/(0.174 * sqrt(2 * pi))) * exp(-1 * ((bin - (-0.0786))^2/(2 * 0.174^2))) coef $A [1] 335.4812 $mu [1] -0.07863746 $sig [1] 0.1746473 plot(fo, temp, coef) Error in eval(expr, envir, enclos) : object sig not found plot(fo, coef, temp) Error in eval(expr, envir, enclos) : object x not found plot(fo, temp, list=coef) If someone could point me in the right direction I would be very grateful. Cheers, __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Date calculation
test
V1V2
1 5/2/2006 36560
2 5/3/2006 36538
3 5/4/2006 36452
4 5/5/2006 36510
5 5/8/2006 36485
6 5/9/2006 36502
7 5/10/2006 36584
8 5/11/2006 36571
dates - strptime(test$V1, %d/%m/%Y)
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
stat stat a écrit :
Dear all R users,
Suppose I have a data frame data like this:
5/2/2006 36560
5/3/2006 36538
5/4/2006 36452
5/5/2006 36510
5/8/2006 36485
5/9/2006 36502
5/10/2006 36584
5/11/200636571
Now I want to create a for loop like this:
date = 5/10/2006
for (i in 1: 8)
{
if (data[i,1] date) break
}
But I get error while executing this. Can anyone tell me the right way to
do this?
Sincerely yours
stat
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Re: [R] expand only one of variable
toy.df$sign - ifelse(toy.df$size == 0, negative, positive) toy.df[rep(1:nrow(toy.df), ifelse(toy.df$size==0, 1, toy.df$size)),] --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Muhammad Subianto a écrit : Dear all, I want to expand only one of variable in data frame and the others variable will be following with the expand variable. Here my toy example: toy.df - data.frame(size=c(3,1,2,0,3,5,1,0), group=LETTERS[1:8], country=c(Germany,England,Argentina,Mexico,Italy,Brazil,France,Spain), w=rep(0,8), d=rep(0,8), l=rep(0,8)) toy.df The size variable is the number of expand and signed by positive but size with 0 (zero) must be signed by negative. The result something like: size group country w d l sign 3 A Germany 0 0 0 positive 3 A Germany 0 0 0 positive 3 A Germany 0 0 0 positive 1 B England 0 0 0 positive 2 C Argentina 0 0 0 positive 2 C Argentina 0 0 0 positive 0 DMexico 0 0 0 negative 3 E Italy 0 0 0 positive 3 E Italy 0 0 0 positive 3 E Italy 0 0 0 positive 5 FBrazil 0 0 0 positive 5 FBrazil 0 0 0 positive 5 FBrazil 0 0 0 positive 5 FBrazil 0 0 0 positive 5 FBrazil 0 0 0 positive 1 GFrance 0 0 0 positive 0 H Spain 0 0 0 negative I would be very happy if anyone could help me. Thank you very much in advance. Kindly regards, Muhammad Subianto __ 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] about string
?nchar sapply(strsplit(f:\\JPCS_signal.txt, _), function(x) substring(x[1], 4)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- XinMeng a écrit : Hello sir: There are 2 questions about string. 1 How to calculate the width of a string? e.g string abc's width is 3; 2 How can I get the substring in such kind of condition: f:\\JPCS_signal.txt f:\\PC1_signal.txt f:\\PC2_signal.txt What I wanna get is JPCS PC1 PC2.How can I achieve them by R cammand? Thanks a lot! My best -- *** Xin Meng Capitalbio Corporation National Engineering Research Center for Beijing Biochip Technology BioPharma-informatics Software Dept. Research Engineer Tel: +86-10-80715888/80726868-6438 Fax: +86-10-80726790 [EMAIL PROTECTED] Address:18 Life Science Parkway, Changping District, Beijing 102206, China __ 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] SPSS variable lables import
data1 - read.spss(file1.sav, F, T) # works ! --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Frank Thomas a écrit : Hi, I try to get the variable labels of a SPSS data file into R but don't find this mentioned in the help file for foreign. Is there another way to get them ? BTW: An SPSS variable name is like: VAR001, whereas the variable label might be 'Identification no.' Thanks in advance, F. Thomas __ 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] How to create list of objects?
lapply(f, summary) sapply(f, AIC) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Rainer M Krug a écrit : Hi I am doing several mle and want to store them in a list (or whatever is the right construct) to be able to analyse them later. at the moment I am doing: f - list() f$IP - mle(...) f$NE - mle(...) but when I say: summary(f) I get: Length Class Mode IP 0 mle list NE 0 mle list I don't get the output I would have, i.e. the one from summary(f$IP) summary(f$IP) Maximum likelihood estimation Call: mle(minuslogl = IPNeglogPoisL, method = L-BFGS-B, fixed = list(), control = list(maxit = 1e+08, factr = 1e-20)) Coefficients: Estimate Std. Error a 1242.0185506 44.92341097 b0.8802538 0.01685811 -2 log L: 145.3509 What I want to do is something like: AICs - AIC(logLik(f)) and then have all the AICs in the vector AICs. It must be possible or is this again a namespace issue? Rainer __ 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] Function (x) as consecutive values
?cumsum
system.time({ z - NULL ; for (i in 1:1000) z - c(z, sum((1:i)**2)) })
[1] 0.04 0.00 0.04 NA NA
system.time( zz - cumsum((1:1000)**2) )
[1] 0 0 0 NA NA
all.equal(z,zz)
[1] TRUE
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Paul Chatfield a écrit :
Hi - I'm trying to avoid using a 'for' loop due to inefficiency and instead
use a function (and ultimately tapply as I'm working on a matrix) but I can't
figure out how to get 'function' to take the variables as anything other than
vectors for example
aa-0
x-1:4
test.fun-function(x)
{aa-(x*x +aa)
return(aa)}
test.fun(1:4)
This code returns 'aa' as 1 4 9 16, but I'd like it to return aa as 1 5 14
30 taking into consideration that I've just calculated aa for x=1. Aside
from using loops, is there not a simple way of telling R to work out x for
consecutive values?
thanks
Paul Chatfield
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Re: [R] Function (x) as consecutive values
sorry, don't understand your problem...
i think it's better to use matrices directly or faster funtions;
but sapply(1:4, function(x) ...) can do the job easily instead of 'for' loops.
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Paul Chatfield a écrit :
Thanks for your reply, though this still wouldn't work with a function
for example, starting code like below fails because x is read as a
vector as opposed to doing it for x=1 then x=2 - is there any way of
tweaking the code easily, or do I just resign myself to for loops to do
that?
x-1:4
trial- function(x)
{xx-matrix(runif(20), 2, 10)
if (xx[1,x]0.5) { ...}
Thanks
Paul
*/Jacques VESLOT [EMAIL PROTECTED]/* wrote:
?cumsum
system.time({ z - NULL ; for (i in 1:1000) z - c(z,
sum((1:i)**2)) })
[1] 0.04 0.00 0.04 NA NA
system.time( zz - cumsum((1:1000)**2) )
[1] 0 0 0 NA NA
all.equal(z,zz)
[1] TRUE
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Paul Chatfield a écrit :
Hi - I'm trying to avoid using a 'for' loop due to inefficiency
and instead use a function (and ultimately tapply as I'm working on
a matrix) but I can't figure out how to get 'function' to take the
variables as anything other than vectors for example
aa-0
x-1:4
test.fun-function(x)
{aa-(x*x +aa)
return(aa)}
test.fun(1:4)
This code returns 'aa' as 1 4 9 16, but I'd like it to return aa
as 1 5 14 30 taking into consideration that I've just calculated aa
for x=1. Aside from using loops, is there not a simple way of
telling R to work out x for consecutive values?
thanks
Paul Chatfield
Send instant messages to your online friends
http://uk.messenger.yahoo.com
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Re: [R] Break Matrix
split(as.data.frame(matrix(rnorm(331*12),331,12)), gl(ceiling(331/12),12,331)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- SUMANTA BASAK a écrit : Hi All, I have a (331*12) matrix. I wan t to braek it into 28 parts each window having 12 rows, so that each matrix become (12*12) matrix. How can i do this. Thanks, Sumanta. - Send instant messages to your online friends - NOW [[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] Break Matrix
mat - matrix(rnorm(331*12),331,12) z - rep(seq(0,331,by=11)+1, each=2) zz - z[-c(1,length(z))] ind - as.data.frame(matrix(zz, nr=2)) lapply(ind, function(x) mat[x[1]:x[2],]) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- SUMANTA BASAK a écrit : Hi, Thanks for reply. I'm sorry that i forgot to mention that from 2nd matrix the first row values of the second matrix (and so on..) will be the first row (and so on..). Please help me. Thanks, SB. */Jacques VESLOT [EMAIL PROTECTED]/* wrote: split(as.data.frame(matrix(rnorm(331*12),331,12)), gl(ceiling(331/12),12,331)) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- SUMANTA BASAK a écrit : Hi All, I have a (331*12) matrix. I wan t to braek it into 28 parts each window having 12 rows, so that each matrix become (12*12) matrix. How can i do this. Thanks, Sumanta. - Send instant messages to your online friends - NOW [[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 Yahoo! India Answers: Share what you know. Learn something new. Click here http://us.rd.yahoo.com/mail/in/mailcricket/*http://in.answers.yahoo.com Send instant messages to your online friends - NOW http://messenger.yahoo.com/download.php;_ylt=Ah5_.LTcbbJtYrNKnfM5e6xwMMIF __ 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] Survival analysis
Dear r-users, I have longitudinal data with measures and a binary trait (status) made on hundreds of people every 3 years during nine years. I tried different models but I wonder if it is possible to use survival analysis. If I take Surv(time, status), I have hundreds of ties and I am not sure it is a nice case to do survival analysis. I wonder if I could take Surv(age, status) in a cox model with coxph() or a logrank test for a genotype factor with survdiff() since people enter the study at different ages which implies different survival time and thus very few ties. Actually I can't find similar examples in documentation. Thanks in advance. jacques --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr __ 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] Finding ties in data (be)for(e) BradleyTerry
z - apply(dat, 1, function(x) paste(sort(x), collapse=))
zz - outer(z,z,==)
data.frame(first=col(zz)[lower.tri(zz)][zz[lower.tri(zz)]],second=row(zz)[lower.tri(zz)][zz[lower.tri(zz)]])
there should be a bettter solution...
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Paul Lemmens a écrit :
Dear all,
I have carried out a pairwise comparison study that I want to analyze
using the BradleyTerry package to establish a rank order of my
stimuli. However, BT does not handle ties between stimuli, so I need
to find those in my data before I can use that model.
The code below goes from the format of my result file(s) to a data
frame suitable for BT, but as you can see, there are some ties. I need
to find the rows with identical (but swapped) winner and loser and
with the same frequency. How can I accomplish that using the R-way
(not looping through the entire thing; in reality, I have approx 40
stimuli with around 180 observations of an odd 40 subjects)?
# $lp was left picture; $rp, right one; $wr was the winner/chosen one
by subject.
dat - data.frame(subjno=gl(4,3),
lp=factor(c(1,3,2,2,3,1,3,1,2,1,2,3), labels=c('a','b','c')),
rp=factor(c(2,1,3,3,1,2,1,2,3,3,1,2), labels=c('a', 'b', 'c')),
wr=factor(c(1,1,2,1,2,2,1,1,2,2,1,2), labels=c('lp', 'rp')))
dat.lp - subset(dat, wr=='lp')
dat.rp - subset(dat, wr=='rp')
names(dat.rp)[c(2,3)] - c('loser', 'winner')
names(dat.lp)[c(2,3)] - c('winner', 'loser')
(dat - with(merge(dat.lp, dat.rp, all=TRUE),
data.frame(table(winner,loser
Thank you for your help!
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Re: [R] Breaking a matrix into parts
lapply(split(mat, gl(nrow(mat)/4, 4, nrow(mat)), cov) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- SUMANTA BASAK a écrit : Hi, I've a matrix in 20*11 order. There are 11 variables, i.e 11 columns and each variable have 20 row data. Now i want to calculate covariance between any variable with others taking 4 rows at a time, so that there will be 5 blocks. How can i do this using any R-function? If i want to do it in any 'loop' function? Thanks a lot, SB. __ 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] Giving Error
sorry (need a data.frame) : lapply(split(as.data.frame(mat), gl(nrow(mat)/4, 4, nrow(mat))), cov) --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- SUMANTA BASAK a écrit : I tried your code, but it's giving the following error.. Error in match.fun(FUN) : argument FUN is missing, with no default __ Yahoo! India Answers: Share what you know. Learn something new. http://in.answers.yahoo.com __ 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] www.r-project.org
think it could be quite useful for most r-users to (keep r-project web site as functional as today but to) offer a links page complementary to other Documentation pages that could link to some of the numerous resources (scripts, etc) on the internet in personal home pages or academic sites impossible to be catched by a single person. These links could be classified by theme such as those in task views or rather according to the different areas of statistics. Indeed many answers in rhelp ended by you should have a look at www Adaikalavan Ramasamy a écrit : I am coming late into the discussion, so apologies if the following points are redundant. 1) IMHO, the most important feature that would make life a lot easier for everyone is having search engines on the main webpage. I know you can click on the Search on the left hand side pane but putting it on the main webpage is much more useful. We can also have a targets section for the search (c.f. http://finzi.psych.upenn.edu/nmz.html) where one can search mailing list, html Manual, FAQ, user-inputted package name etc. 2) About having explicit URL print, may I suggest using http://maps.google.com approach of using the Link to this page (top right hand of the page) ? 3) I understand that R is restricted in terms of priority and human resources. But given that Asia (e.g. India, Singapore, China) has low labour costs and abundant computing personals, would it not make sense for some Asian research group to offer to spearhead and maintain the website ? From a marketing point of view some nice graphics, search functions and navigation etc would be useful to attract newcomers. There could be a simple version alternative (as it is now) for those who prefer or those who have trouble accessing the site. Just my £0.02. Regards, Adai On Tue, 2006-04-25 at 12:33 -0700, Spencer Graves wrote: Hi, Gabor: inline Gabor Grothendieck wrote: On Windows, right click the web page, choose Properties and copy the url there. That works, and I will use it in the future. Thanks. However, if the subject is not educating Spencer Graves but how to make www.r-project.org more user friendly, then it still might help to display as Address the actual web address of the archive page rather than www.r-project.org. It may not look as pretty, but I'm for function first and cosmetics only if they don't interfere with functionality. Best Wishes, spencer graves On 4/25/06, Spencer Graves [EMAIL PROTECTED] wrote: __ 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 -- --- [EMAIL PROTECTED] CNRS UMR 8090 - http://www-good.ibl.fr Génomique et physiologie moléculaire des maladies métaboliques I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 __ 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] ordered boxplots
boxplot(count ~ spray, data = InsectSprays, col = lightgray, at=with(InsectSprays, rank(tapply(count, spray, median Chuck Cleland a écrit : ?reorder.factor shows the follwing example: bymedian - with(InsectSprays, reorder(spray, count, median)) boxplot(count ~ bymedian, data = InsectSprays, xlab = Type of spray, ylab = Insect count, main = InsectSprays data, varwidth = TRUE, col = lightgray) Thomas Hoffmann wrote: Dear List-Members, I would like to produce a ordered boxplot in which the categories with the smallest median are plotted at the left end and the box with the largest median at the right. Thanks in advance for any advices Thomas H. __ 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 -- --- [EMAIL PROTECTED] CNRS UMR 8090 - http://www-good.ibl.fr Génomique et physiologie moléculaire des maladies métaboliques I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 __ 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] www.r-project.org
what about improving the Links page, with for instance a list of links - such as thoses in task views - grouped by theme, instead of modifying r-project site ? -- --- [EMAIL PROTECTED] CNRS UMR 8090 - http://www-good.ibl.fr Génomique et physiologie moléculaire des maladies métaboliques I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 __ 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] Problem with data frame
as.data.frame(replicate(20, rnorm(500)))
Arun Kumar Saha a écrit :
Dear r-users,
suppose I have n normal distributions with parameter N(0,i) i=1,2,...,n
respectively.
Now I want to generate 500 random number for each distribution. And want to
put all 500*n random numbers
in a single data frame.
I tried with following code:
n=20
random = data.frame(n)
for ( i in 2: length)
{
random[,i] = random(500,mean=0,sd=i)
}
but while executing this I am getting errors.
Can anyone give me any suggestion?
Thanks and regards
Arun
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Génomique et physiologie moléculaire des maladies métaboliques
I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex
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Re: [R] aov contrasts residual error calculation
why not using lme() ? first, you need transform data: dat2 - as.data.frame(lapply(subset(dat, sel=-c(A,B,C)), rep, 3)) dat2$y - unlist(subset(dat, sel=c(A,B,C)), F, F) dat2$cond - factor(rep(c(A,B,C), each=nrow(dat))) dat2$inter - factor(dat2$map):factor(dat2$cond) lme1 - lme(fixed = y ~ mapping + cond + inter + other fixed effects, random = ~ 1 |subj, data=dat2, contrast=list(inter=poly(nlevels(dat2$inter)[,1:4])) Steven Lacey a écrit : Hi, I am using aov with an Error component to model some repeated measures data. By repeated measures I mean the data look something like this... subjABC 1 411 15 2 312 17 3 5914 4 610 18 For each subject I have 3 observations, one in each of three conditions (A, B, C). I want to test the following contrast (1, 0, -1). One solution is to apply the contrast weights at the subject level explicitly and then call t.test on the difference scores. However, I am looking for a more robust solution as I my actual design has more within-subjects factors and one or more between subjects factors. A better solution is to specify the contrast in an argument to aov. The estimated difference of the contrast is the same as that in the paired t-test, but the residual df are double. While not what I expected, it follows from the documentation, which explicitly states that these contrasts are not to be used for any error term. Even though I specify 1 contrast, there are 2 df for a 3 level factor, and I suspect internally the error term is calculated by pooling across multiple contrasts. While very useful, I am wondering if there is way to get aov to calculate the residual error term only based on the specified contrasts (i.e., not assume homogeneity of variance and sphericity) for that strata? If not, I could calculate them directly using model.matrix, but I've never done that. If that is the preferred solution, I'd also appreciate coding suggestions to do it efficiently. How would I do the same thing with a two factor anova where one factor is within-subjects and one is between... Condition Mapping SubjectABC 11 411 15 12 13 14 15 16 17 18 29 210 Mapping is a between-subject factor. Condition is a within-subject factor. There are 5 levels of mapping, 8 subjects nested in each level of mapping. For each of the 40 combinations of mapping and subject there are 3 observations, one in each level of the condition factor. I want to estimate the pooled error associated with the following set of 4 orthogonal contrasts: condition.L:mapping.L condition.L:mapping.Q condition.L:mapping.C condition.L:mapping^4 What is the best way to do this? One way is to estimate the linear contrast for condition for each subject, create a 40 row matrix where the measure for each combination of mapping and subject is the linear contrast on condition. If I pass this dataframe to aov, the mse it returns is the value I am looking for. If possible, I would like to obtain the estimate without collapsing the dataframe, but am not sure how to proceed. Suggestions? Thanks, Steve __ 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 -- --- [EMAIL PROTECTED] CNRS UMR 8090 - http://www-good.ibl.fr Génomique et physiologie moléculaire des maladies métaboliques I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 __ 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] aov contrasts residual error calculation
with error strata, aov() becomes difficult too... a nice and brief presentation of mixed models with r by John Fox at: http://cran.r-project.org/doc/contrib/Fox-Companion/appendix.html Steven Lacey a écrit : Jacques, Thanks for the reply. I am not using lme because I don’t have the time to understand how it works and I have a balanced design, so typcial linear modelling in aov should be sufficient for my purposes. Down the road I plan to learn lme, but I'm not there yet. So any suggestions with respect to aov would be greatly appreciated. Steve -Original Message- From: Jacques Veslot [mailto:[EMAIL PROTECTED] Sent: Friday, April 21, 2006 11:58 AM To: Steven Lacey Cc: r-help@stat.math.ethz.ch Subject: Re: [R] aov contrasts residual error calculation why not using lme() ? first, you need transform data: dat2 - as.data.frame(lapply(subset(dat, sel=-c(A,B,C)), rep, 3)) dat2$y - unlist(subset(dat, sel=c(A,B,C)), F, F) dat2$cond - factor(rep(c(A,B,C), each=nrow(dat))) dat2$inter - factor(dat2$map):factor(dat2$cond) lme1 - lme(fixed = y ~ mapping + cond + inter + other fixed effects, random = ~ 1 |subj, data=dat2, contrast=list(inter=poly(nlevels(dat2$inter)[,1:4])) Steven Lacey a écrit : Hi, I am using aov with an Error component to model some repeated measures data. By repeated measures I mean the data look something like this... subjABC 1 411 15 2 312 17 3 5914 4 610 18 For each subject I have 3 observations, one in each of three conditions (A, B, C). I want to test the following contrast (1, 0, -1). One solution is to apply the contrast weights at the subject level explicitly and then call t.test on the difference scores. However, I am looking for a more robust solution as I my actual design has more within-subjects factors and one or more between subjects factors. A better solution is to specify the contrast in an argument to aov. The estimated difference of the contrast is the same as that in the paired t-test, but the residual df are double. While not what I expected, it follows from the documentation, which explicitly states that these contrasts are not to be used for any error term. Even though I specify 1 contrast, there are 2 df for a 3 level factor, and I suspect internally the error term is calculated by pooling across multiple contrasts. While very useful, I am wondering if there is way to get aov to calculate the residual error term only based on the specified contrasts (i.e., not assume homogeneity of variance and sphericity) for that strata? If not, I could calculate them directly using model.matrix, but I've never done that. If that is the preferred solution, I'd also appreciate coding suggestions to do it efficiently. How would I do the same thing with a two factor anova where one factor is within-subjects and one is between... Condition Mapping SubjectABC 11 411 15 12 13 14 15 16 17 18 29 210 Mapping is a between-subject factor. Condition is a within-subject factor. There are 5 levels of mapping, 8 subjects nested in each level of mapping. For each of the 40 combinations of mapping and subject there are 3 observations, one in each level of the condition factor. I want to estimate the pooled error associated with the following set of 4 orthogonal contrasts: condition.L:mapping.L condition.L:mapping.Q condition.L:mapping.C condition.L:mapping^4 What is the best way to do this? One way is to estimate the linear contrast for condition for each subject, create a 40 row matrix where the measure for each combination of mapping and subject is the linear contrast on condition. If I pass this dataframe to aov, the mse it returns is the value I am looking for. If possible, I would like to obtain the estimate without collapsing the dataframe, but am not sure how to proceed. Suggestions? Thanks, Steve __ 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 -- --- [EMAIL PROTECTED] CNRS UMR 8090 - http://www-good.ibl.fr Génomique et physiologie moléculaire des maladies métaboliques I.B.L 2eme etage - 1 rue du Pr Calmette, B.P.245, 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 __ 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] How to generate a list of lists recursively (for bayesm)
1) see in ?[[:
The most important distinction between '[', '[[' and '$' is that the
'[' can select more than one element whereas the other two select a
single element.
2) what are Sk, att, attq8, regdata1 and regdata2 ?
[EMAIL PROTECTED] a écrit :
Dear all,
I need to generate a list of lists as required by the bayesm-package. This
means in my application that I have to generate a list which consists of 2000
elements, which are lists themselves:
list(list(y1,X1),...,list(y2000,X2000)).
The y are vectors and the X are matrices of different dimensions. I tried to
solve this problem iteratively by the following code, but received an error
message, which I do not understand:
for(i in 1:1067){
+ c-0
+ for(j in 1:300){
+ if(Sk[i,j]!=0){
+ c-c+1
+ if(c==1){
+ X1-att[j,]
+ X2-attq8[j,]
+ y-Sk[i,j]
+ }
+ else{
+ X1-rbind(X1,att[j,])
+ X2-rbind(X2,attq8[j,])
+ y-rbind(y,Sk[i,j])
+ }
+ }}
+ regdata1[[c(i,1)]]-y
+ regdata1[[c(i,2)]]-X1
+ regdata2[[c(i,1)]]-y
+ regdata2[[c(i,2)]]-X2
+ }
Fehler: objekt regdata1 nicht gefunden
??? So, does anybody know what I did wrong or how I can generate my list of
lists?
Regards,
Moritz
Moritz Marienfeld
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Re: [R] which function to use to do classification
if you want to classify rows or columns, read: ?hclust ?kmeans library(cluster) ?pam Baoqiang Cao a écrit : Dear All, I have a data, suppose it is an N*M matrix data. All I want is to classify it into, let see, 3 classes. Which method(s) do you think is(are) appropriate for this purpose? Any reference will be welcome! Thanks! Best, Baoqiang Cao __ 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] which function to use to do classification
try this (suppose mat is your matrix): hc - hclust(dist(mat,manhattan), ward) plot(hc, hang=-1) (x - identify(hc)) # rightclick to stop cutree(hc, 3) km- kmeans(mat, 3) km$cluster km$centers pam(daisy(mat, metric = manhattan), k=3, diss=T)$clust Baoqiang Cao a écrit : Thanks! I tried kmeans, the results is not very positive. Anyway, thanks Jacques! Please let me know if you have any other thoughts! Best regards, Baoqiang Cao === At 2006-03-29, 00:08:44 you wrote: === if you want to classify rows or columns, read: ?hclust ?kmeans library(cluster) ?pam Baoqiang Cao a écrit : Dear All, I have a data, suppose it is an N*M matrix data. All I want is to classify it into, let see, 3 classes. Which method(s) do you think is(are) appropriate for this purpose? Any reference will be welcome! Thanks! Best, Baoqiang Cao __ 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 . = = = = = = = = = = = = = = = = = = = = Baoqiang Cao [EMAIL PROTECTED] 2006-03-29 __ 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] apply(ing) to sum subset of a vector
apply(cbind(from,to), 1, function(x) sum(g[x[1]:x[2]])) Fred J. a écrit : Dear R users I am trying to sum selective elements of a vector but my solution is not cutting it. Example: g - 1:5; from - 1:3; to - 3:5; from to 1 3 2 4 3 5 so I expect 3 sums from g 1+2+3 that is 1 to 3 of g 2+3+4 that is 2 to 4 of g 3+4+5 that is 3 to 5 of g my solution will not work. sum.em - function(g, c1, c2) sum(g[c1:c2]) apply(g, 1, sum.em, ...) I don't think so because apply is not aware of the from and to. and if I f - list(g, from, to) that will not fit with the second arg of apply. thank you - [[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] data management on R
if A, B and C are of matrix class: C - cbind(x=A[, 1], n=B[, 2]) if A, B and C are of data.frame class: C - cbind.data.frame(x=A$x, n=B$n) C - data.frame(x=A$x, n=B$n) linda.s a écrit : what is the difference between the two matrix B and C? B m n 1 1 2 2 7 8 C - cbind(x=A[, 1], n=B[, 2]) C x n [1,] 1 2 [2,] 3 8 For B, it use 1 and 2 to indicate rows while C use[1,] and [2,] On 3/23/06, Jacques VESLOT [EMAIL PROTECTED] wrote: C - cbind(x=A[, 1], n=B[, 2]) __ 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] How to avoid for or while loops when one index depends on another
mat - matrix(rnorm(1),100)
system.time({res1 - NULL ;
for (i in 1:(ncol(mat)-1)) for (j in (i+1):ncol(mat))
res1 - rbind(res1, c(i, j))})
[1] 1.51 0.01 1.53 NA NA
system.time(res2 - which(upper.tri(mat), T))
[1] 0.02 0.00 0.02 NA NA
all.equal(res1,res2[order(res2[,row]),])
[1] TRUE
Daniel Goldstein a écrit :
Dear R Community,
I'm trying to exploit the elegance of R by doing the following
pseudocode routine without a WHILE or FOR loop in R:
for i = 1 to length-1
for j = (i+1) to length
print a[i], a[j]
That is, I want i and j to be the indices of a half-matrix
1 2, 1 3, 1 4, ..., 1 length,
2 3, 2 4, ..., 2 length,
3 4, ..., 3 length
1. Can this be done with the 'whole object' approach (Introduction to R
v2.1.1 section 9.2.2) and not while loops?
2. (Extra credit) Is your solution likely to be more efficient than a loop?
3. (Extra credit) How could once do this with FOR as opposed to WHILE in
R? Clearly if you attempt j in i+1:length you're in trouble when i
exceeds length.
Thanks for your help with this excellent open-source resource,
Dan
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Re: [R] sqlSave
Dear Tobias,
I finally succeeded in exporting my dataframes from R into Access with
this script:
library(RODBC)
canal - odbcConnectAccess(D:/Floristique.mdb)
sqlSave(channel=canal, dat=flore, tablename=Floristique, rownames=F,
safer=F, fast=F,
varTypes=c(dates=Date))
odbcClose(canal)
My problem in exporting my dataframe flore seems to be closely linked
with the name of the column of dates in my dataframe, which was formerly
date and which I changed into dates. Since my exporting worked (?).
The varTypes argument works as described in the sqlSave() help page: an
optional named character vector giving the DBMSs datatypes to be used
for some (or all) of the columns if a table is to be created. However,
I have only been able to export my dataframe with the dates column as
Date; not as POSIX, though dates are in POSIX when importing into R with
sqlQuery (?).
I have still questions about the functionning of these functions, but I
have to acknowledge that I haven't yet reviewed many things about the
subject.
Best regards,
jacques
Brandt, T. (Tobias) a écrit :
Dear Jacques
I refer to your post on r-help on 20 March 2006.
I see that you had some trouble with sqlSave. I also noticed that you
said you tried to use varTypes but didn't understand it properly.
I've had the same problem in that I'm trying to use sqlSave to save my
dataframes which contain dates to a database but have run into
problems. My solution until now has been to save the dates as
characters and then convert them to dates with a query. However I
would prefer to import them as dates in the first place. Perhaps this
can be done with varTypes but I have been unable to figure out how to
use this properly.
(...)
I look forward to your response.
Kind regards
Tobias Brandt
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jacques VESLOT
Sent: 20 March 2006 12:30 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] sqlSave
OK, I finally found what's wrong - date column name.
Jacques VESLOT a écrit :
Dear R-users,
I tried to export a dataframe form R to Access, that way:
library(RODBC)
channel - odbcConnectAccess(d:/test.mdb) sqlSave(channel=channel,
flore, rownames=F)
odbcClose(channel)
But I always got this error message:
Erreur dans sqlSave(channel = channel, flore, Florist) :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][Pilote ODBC Microsoft Access] Erreur
de syntaxe
dans la définition de champ.
Note that I succeeded in exporting this dataframe into Excel and then
import Excel file from Access.
I tried to find where the problem comes from by exporting columns one
by one, as follows:
library(RODBC)
canal - odbcConnectAccess(d:/test.mdb) for (i in names(flore)) {
cat(i)
sqlSave(channel=canal, dat=flore[i]) }
odbcClose(canal)
I could export all columns but one, named date, which
consists of dates.
I tried to export this column as POSIX, as Date and even as
character,
but without success. I still had the same error message:
dateErreur dans sqlSave(channel = canal, dat = flore[i]) :
[RODBC] ERROR: Could not SQLExecDirect 37000 -3553
[Microsoft][Pilote ODBC Microsoft Access] Erreur de syntaxe dans la
définition de champ.
I also tried with varTypes, though I am not sure how to use this
argument correctly. I did:
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date))
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date/Heure))
But still the same error message.
Maybe it's in Windows, but I don't understand...
Thanks for helping,
jacques
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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R
Re: [R] data management on R
C - cbind(x=A[, 1], n=B[, 2]) linda.s a écrit : On 3/23/06, Jim Porzak [EMAIL PROTECTED] wrote: C - cbind(A[, 1], B[, 2]) The result is: [,1] [,2] [1,]12 [2,]38 How to keep x and n as the column title? Linda __ 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] ordering boxplots according to median
boxplot(count~spray, InsectSprays) spray2 - with(InsectSprays, factor(spray, levels=levels(spray)[order(tapply(count,spray,median))])) boxplot(count~spray2, InsectSprays) Talloen, Willem [PRDBE] a écrit : Dear R-users, Does anyone knows how I can order my serie of boxplots from lowest to highest median (which is much better for visualization purposes). thanks in advance, willem [[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
[R] sqlSave
Dear R-users,
I tried to export a dataframe form R to Access, that way:
library(RODBC)
channel - odbcConnectAccess(d:/test.mdb)
sqlSave(channel=channel, flore, rownames=F)
odbcClose(channel)
But I always got this error message:
Erreur dans sqlSave(channel = channel, flore, Florist) :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][Pilote ODBC Microsoft Access] Erreur de syntaxe
dans la définition de champ.
Note that I succeeded in exporting this dataframe into Excel and then
import Excel file from Access.
I tried to find where the problem comes from by exporting columns one by
one, as follows:
library(RODBC)
canal - odbcConnectAccess(d:/test.mdb)
for (i in names(flore)) {
cat(i)
sqlSave(channel=canal, dat=flore[i]) }
odbcClose(canal)
I could export all columns but one, named date, which consists of dates.
I tried to export this column as POSIX, as Date and even as character,
but without success. I still had the same error message:
dateErreur dans sqlSave(channel = canal, dat = flore[i]) :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][Pilote ODBC Microsoft Access] Erreur de syntaxe
dans la définition de champ.
I also tried with varTypes, though I am not sure how to use this
argument correctly. I did:
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date))
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date/Heure))
But still the same error message.
Maybe it's in Windows, but I don't understand...
Thanks for helping,
jacques
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] sqlSave
OK, I finally found what's wrong - date column name.
Jacques VESLOT a écrit :
Dear R-users,
I tried to export a dataframe form R to Access, that way:
library(RODBC)
channel - odbcConnectAccess(d:/test.mdb)
sqlSave(channel=channel, flore, rownames=F)
odbcClose(channel)
But I always got this error message:
Erreur dans sqlSave(channel = channel, flore, Florist) :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][Pilote ODBC Microsoft Access] Erreur de syntaxe
dans la définition de champ.
Note that I succeeded in exporting this dataframe into Excel and then
import Excel file from Access.
I tried to find where the problem comes from by exporting columns one by
one, as follows:
library(RODBC)
canal - odbcConnectAccess(d:/test.mdb)
for (i in names(flore)) {
cat(i)
sqlSave(channel=canal, dat=flore[i]) }
odbcClose(canal)
I could export all columns but one, named date, which consists of dates.
I tried to export this column as POSIX, as Date and even as character,
but without success. I still had the same error message:
dateErreur dans sqlSave(channel = canal, dat = flore[i]) :
[RODBC] ERROR: Could not SQLExecDirect
37000 -3553 [Microsoft][Pilote ODBC Microsoft Access] Erreur de syntaxe
dans la définition de champ.
I also tried with varTypes, though I am not sure how to use this
argument correctly. I did:
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date))
sqlSave(channel=canal, dat=flore, varTypes=c(date=Date/Heure))
But still the same error message.
Maybe it's in Windows, but I don't understand...
Thanks for helping,
jacques
R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
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
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to use the result of hclust?
?cutree ?plot.hclust ?identify.hclust hc- hclust(dist(tab, manhattan), ward) plot(hc, hang=-1) (x - identify(hc)) cutree(hc, 2) Michael a écrit : Hi all, Does hclust provide concrete clustered results? I could not see how to use it to make 6 clusters... and it does not give the 6 cluster labels... How to use the result of hclust? thanks a lot, Michael. [[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] combinatorics again
library(gtools)
combinations(5,3)
[,1] [,2] [,3]
[1,]123
[2,]124
[3,]125
[4,]134
[5,]135
[6,]145
[7,]234
[8,]235
[9,]245
[10,]345
Robin Hankin a écrit :
Hi
I want to enumerate all vectors of length J, whose elements are
integers in the range 1 to S, without regard to ordering.
With J=S=3, the combinations are as follows:
[,1] [,2] [,3]
[1,]111
[2,]112
[3,]113
[4,]122
[5,]123
[6,]133
[7,]222
[8,]223
[9,]233
[10,]333
Note that (eg) c(1,2,1) is not on the list because we already have
c(1,1,2) which would be equivalent [because the problem is to
enumerate the cases without regard to ordering] and I do not want
repeats.
The best I can do is to create all S^J possibilities and weed out the
repeats, using unique() ; code below.
Why is this no good? Well, even for the tiny case of J=S=10, this
would require a matrix of 10^10 rows, and my little linux machine
refuses to cooperate, complaining about allocating a vector of
length 1410065408. For these values of J and S, I happen to know
that the are 6360 distinct combinations, which is eminently handleable.
Anyone got any better ideas?
allcomb - function(J,S){
f - function(...) {
1:S
}
out - as.matrix(do.call(expand.grid, lapply(1:J, FUN = f)))
out - t(apply(out,1,sort))
unique(out)
}
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] combinatorics again
combinations(5,3,rep=T)
Robin Hankin a écrit :
Thank you Jacques
but your solution misses (eg) c(1,1,2) which I need.
best wishes
Robin
On 6 Mar 2006, at 09:17, Jacques VESLOT wrote:
library(gtools)
combinations(5,3)
[,1] [,2] [,3]
[1,]123
[2,]124
[3,]125
[4,]134
[5,]135
[6,]145
[7,]234
[8,]235
[9,]245
[10,]345
Robin Hankin a écrit :
Hi
I want to enumerate all vectors of length J, whose elements are
integers in the range 1 to S, without regard to ordering.
With J=S=3, the combinations are as follows:
[,1] [,2] [,3]
[1,]111
[2,]112
[3,]113
[4,]122
[5,]123
[6,]133
[7,]222
[8,]223
[9,]233
[10,]333
Note that (eg) c(1,2,1) is not on the list because we already have
c(1,1,2) which would be equivalent [because the problem is to
enumerate the cases without regard to ordering] and I do not want
repeats.
The best I can do is to create all S^J possibilities and weed out the
repeats, using unique() ; code below.
Why is this no good? Well, even for the tiny case of J=S=10, this
would require a matrix of 10^10 rows, and my little linux machine
refuses to cooperate, complaining about allocating a vector of
length 1410065408. For these values of J and S, I happen to know
that the are 6360 distinct combinations, which is eminently
handleable.
Anyone got any better ideas?
allcomb - function(J,S){
f - function(...) {
1:S
}
out - as.matrix(do.call(expand.grid, lapply(1:J, FUN = f)))
out - t(apply(out,1,sort))
unique(out)
}
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
__
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
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
__
R-help@stat.math.ethz.ch mailing list
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Re: [R] Contingency table and zeros
myvector - c(1, 2, 3, 2, 1, 3, 5) myf - factor(myvector, levels=1:5) table(myf) myf 1 2 3 4 5 2 2 2 0 1 cumsum(table(myf)) 1 2 3 4 5 2 4 6 6 7 Nicolas Perot a écrit : Hello, Let's assume I have a vector of integers : myvector - c(1, 2, 3, 2, 1, 3, 5) My purpose is to obtain the cumulative distribution of these numerical data, i.e. something like : value nb_occur. =12 =24 =36 =46 =57 For this, I create a table with ; mytable - table(myvector) 1 2 3 5 2 2 2 1 However, table() returns an array of integers, mytable[4] returns the occurence number of the 5 item, which makes this table hard to index. I would prefer to have a data structure where mytable[4] could return 0, as there is no 4 in my vector. table() may not be the proper way to do it. For the moment, I use a loop which scans the vector from its lowest value to its highest, and counts the number of times each value appears. Is there a built-in function, or a table() option to do such a task ? Thanks. Regards, Nicolas __ 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] Averaging over columns
t(as.matrix(aggregate(t(as.matrix(DF)), rep(1:12,each=3), mean)[,-1])) michael watson (IAH-C) a écrit : Hi I've been reading the help for by and aggregate but can't get my head round how to do this. I have a data frame - the first three columns are replicate measurements, then the next 3 are replicates etc up to 36 (so 12 variables with 3 replicate measurements each). I want to compute the mean for each of the 12 variables, so that, for each row, I have 12 means. A grouping variable across columns can easily be created by rep(1:12,each=3), but I can't figure out which function to use to get R to calculate the means I want. Thanks in advance Mick __ 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] Averaging over columns
sorry, i forgot list(): t(as.matrix(aggregate(t(as.matrix(DF)), list(rep(1:12,each=3)), mean)[,-1])) michael watson (IAH-C) a écrit : Hi I've been reading the help for by and aggregate but can't get my head round how to do this. I have a data frame - the first three columns are replicate measurements, then the next 3 are replicates etc up to 36 (so 12 variables with 3 replicate measurements each). I want to compute the mean for each of the 12 variables, so that, for each row, I have 12 means. A grouping variable across columns can easily be created by rep(1:12,each=3), but I can't figure out which function to use to get R to calculate the means I want. Thanks in advance Mick __ 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] break a vector into classes
see: ?cut ?findInterval [EMAIL PROTECTED] a écrit : Hi, I'm looking for a function which divides a vector into n classes and returns the breaks as well as the number of values in each class. This is actually what hist(vector, breaks=n) does, but in hist() n is a suggestion only, and is a suggestion only and cannot be enforced (as far as I know...) It is not so difficult to do it yourself, but a ready made function would be nice... cheers wouter __ 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] returning the largest element in an array/matrix?
indcol - rep(1:ncol(mat), each=nrow(mat))[which.max(mat)] indrow - rep(1:nrow(mat), ncol(mat))[which.max(mat)] indrow - which(apply(mat==max(mat),1,sum)!=0) indcol - which(apply(mat==max(mat),2,sum)!=0) Michael a écrit : Hi all, I want to use which.max to identify the maximum in a 2D array/matrix and I want argmin and return the row and column indices. But which.max only works for vector... Is there any convinient way to solve this problem? Thanks a lot! [[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] Applying strptime() to a data set or array
maybe you need to transform A$date to character: A$date - strptime(as.character(A$date), ...) see also: ?ISOdatetime Andrew Athan a écrit : I'm sure this is just the result of a basic misunderstanding of the syntax of R, but I am stumped. A - read.table(file=sumByThirtyMinute.csv,sep=,,col.names=c(date,pandl)) A now consists of thousands of rows, but A$date is a string... ... 3183 2006-02-28 12:00:00548.470 3184 2006-02-28 12:30:00515.240 3185 2006-02-28 13:00:00140.120 3186 2006-02-28 13:30:00450.940 3187 2006-02-28 14:00:00318.570 ... So, I try to convert A$date to a POSIXlt ... A[,1]-strptime(A$date,%Y-%m-%d %H:%M:%s) A$date-strptime(A$date,%Y-%m-%d %H:%M:%s) which gives me a warning that the length of the array I am trying to replace A$date with is 9 ... but if I print strptime(A$date,%Y-%m-%d %H:%M:%s), it clearly has thousands of rows. Yet, if I ask for length(strptime(A$date,%Y-%m-%d %H:%M:%s)), I get 9. What am I doing wrong? Do I need to convert the return value of strptime(A$date,%Y-%m-%d %H:%M:%s) to some array/vector/matrix datatype before attempting to assign it? Thanks, Andrew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Applying strptime() to a data set or array
it's probably a factor, not a string vector... so i would do as.vector or as.character - but it may be not necessary ! as.POSIXct(strptime(as.vector(A$date),...)) or: seq( from = ISOdate(2006,02,28, 12, 0, 0, tz=), to = ISOdate(...), by=30 min) Andrew Athan a écrit : A$date is already a string, as read from the file. I tried it anyway, for you... A$date-strptime(as.character(A$date),%Y-%m-%d %H:%M:%s) Error in $-.data.frame(`*tmp*`, date, value = list(sec = c(0, 0, : replacement has 9 rows, data has 3198 A. Jacques VESLOT wrote: maybe you need to transform A$date to character: A$date - strptime(as.character(A$date), ...) see also: ?ISOdatetime Andrew Athan a écrit : I'm sure this is just the result of a basic misunderstanding of the syntax of R, but I am stumped. A - read.table(file=sumByThirtyMinute.csv,sep=,,col.names=c(date,pandl)) A now consists of thousands of rows, but A$date is a string... ... 3183 2006-02-28 12:00:00548.470 3184 2006-02-28 12:30:00515.240 3185 2006-02-28 13:00:00140.120 3186 2006-02-28 13:30:00450.940 3187 2006-02-28 14:00:00318.570 ... So, I try to convert A$date to a POSIXlt ... A[,1]-strptime(A$date,%Y-%m-%d %H:%M:%s) A$date-strptime(A$date,%Y-%m-%d %H:%M:%s) which gives me a warning that the length of the array I am trying to replace A$date with is 9 ... but if I print strptime(A$date,%Y-%m-%d %H:%M:%s), it clearly has thousands of rows. Yet, if I ask for length(strptime(A$date,%Y-%m-%d %H:%M:%s)), I get 9. What am I doing wrong? Do I need to convert the return value of strptime(A$date,%Y-%m-%d %H:%M:%s) to some array/vector/matrix datatype before attempting to assign it? Thanks, Andrew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Two quick questions
DF$date - as.Date(paste(DF$day,DF$month,DF$year), %d %m %Y) # if 4-figure year aggregate(DF[c(var1, var2, var3)], DF[c(date, sector)], sum, is.na=T) Serguei Kaniovski a écrit : Hi all, 1. How to construct a date from three variables year, month, and day, where all three are integers? 2. I have a dataframe by date and sector. I would like to add-up all entries for all variable with identical date and sector, replacing the original entries, i.e. emulate the STATA command collapse (sum) var1 var2 var3, by(date sector). Thank you, Serguei Kaniovski __ 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] Maximally independent variables
library(gtools) z - combinations(ncol(DF), 3) maxcor - function(x) max(as.vector(as.dist(cor(DF[,x] names(DF)[z[which.min(apply(z, 1, maxcor)),]] Gabor Grothendieck a écrit : Are there any R packages that relate to the following data reduction problem fo finding maximally independent variables? Currently what I am doing is solving the following minimax problem: Suppose we want to find the three maximally independent variables. From the full n by n correlation matrix, C, of all n variables chooose three variables and form their 3 by 3 correlation submatrix, C1, finding the offdiagonal entry of C1 which is largest in absolute value. Call that z. Thus for each set of 3 variables we can associate such a z. Now for each possible set of three variables find the one for which its value of z is least. I only give the above formulation because that is what I am doing now but I would be happy to consider other different formulations. __ 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] Help with barplot
par(mar=par()$mar+c(0,3,0,0)) with(x, barplot(structure(Median, names=as.character(City)),horiz=T,las=2)) [EMAIL PROTECTED] a écrit : Hello all, I need to create a horizontal barplot with the following data: City Median Springfield 34 Worcester 66 Fitchburg 65 Lowell 63 Quincy 62 Boston 36 Cambridge54 Waltham 42 Medford 52 Pittsfield 65 Rensselaer 60 Schenectady 56 Glens Falls 64 The code below produces the bars for the Median, but how or where do I specify the corresponding city name on the y-axis for the bars. x-data.frame(City=ozone.ne.trim$City,Median=ozone.ne.trim$Median) with (x, barplot(Median,horiz=TRUE)) Please help. Mathangi __ 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] rotated labels in barplot with beside=T and multiple groups
why not: colnames(lsu) - seq(0.1,1,by=0.1) barplot(lsu, bes=T) Karin Lagesen a écrit : I have a data set that I display using barplot. I don't know what you call it, but when I look at it, it looks like this: lsu (0,0.1] (0.1,0.2] (0.2,0.3] (0.3,0.4] (0.4,0.5] (0.5,0.6] A 0.052631579 0.0 0.0 0.0 0.0 0.0 B 0.0 0.0 0.001007049 0.003021148 0.0 0.0 E 0.2 0.0 0.0 0.0 0.1 0.0 (0.6,0.7] (0.7,0.8] (0.8,0.9] (0.9,1] A 0.0 0.0 0.0 0.947368421 B 0.0 0.004028197 0.005035247 0.986908359 E 0.1 0.0 0.1 0.5 Now, trying the examples shown via the r-help mailing list I am trying to make a plot where each of the groups gets displayed in a histogram-like fashion upwards with the number 0.1, 0.2 and so forth underneath the group. What I do is the following: par(mar = c(6, 4, 4, 2) + 0.1) bplot = barplot(lsu, beside=TRUE, col=colors[1:length(lsu[,1])], ylim = c(0,1.0), xaxt = n, xlab = ) axis(side=1,at=bplot, labels=FALSE, tick=TRUE) NULL nam=rep(a,10) text(bplot, par(usr)[3] - 1.5, srt = 45, adj = 1, labels = nam, xpd = TRUE) NULL The result is the bars pointing upwards, like I want, but I get one tickmark per bar, and no labels underneath. I want no tickmark, and one label per group. Any ideas as to what I am doing wrong? TIA, Karin __ 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] How to Import Data
select the directory with setwd() and then import data: setwd(d:/.../yourdirectory) x111 - read.table(x111.csv,...) or indicate path behind filename: x111 - read.table(d:/.../yourdirectory/x111.csv,...) besides, there are other functions to import data. see ?read.table Carl Klarner a écrit : Hello, I am a very new user of R. I've spent several hours trying to import data, so I feel okay asking the list for help. I had an Excel file, then I turned it into a csv file, as instructed by directions. My filename is x111.csv. I then used the following commands to read this (fairly small) dataset in. x111 -read.table(file='x111.csv', sep=,header=T, quote=,comment.char=,as.is=T) I then get the following error message. Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'x111.csv', reason 'No such file or directory' I would imagine I'm not putting my csv file in the right location for R to be able to read it. If that's the case, where should I put it? Or is there something else I need to do to it first? Thanks for your help, Carl __ 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] creating 3-way tables for mantelhaen.test
library(gtools) index - cbind(combinations(7,2),8) lapply(as.data.frame(t(index)), function(x) mantelhaen.test(table(mydata[,x]))) Taka Matzmoto a écrit : Hi R users I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and one continuous variable (e.g., Y1). I combined these variables using data.frame() mydata - data.frame(X1,X2,X3,X4,X5,X6,X7,Y1) after that, I sorted this data.frame rank.by.Y1-order(mydata[,8]) sorted.mydata-mydata[rank.by.Y1,] after that, I replaced Y1's values with values ranging from 1 to 10 ( 1 represents the lowest group on Y1 and 10 presents the hight group on Y1). Now Y1 becomes a grouping variable. What I like to do is to apply mantelhaen.test for each binary variable pair (e.g, X1 and X2, X1 and X3, X1 and X4, , X6 and X7) In order to apply mantelhaen.test, a 3-dimensional contingency table is required. Could you provide some advice on how to create a 3-dimensional contingency table (first dimension represents the first variable of variable pair, second dimension the second variable of variable pair, and third dimension represents 1 to 10 ) and apply mantelhaen.test ? I looked at arrary, xtabs, table commands but I couldn't figure out yet. Thank you __ 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] aggregate data.frame using column-specific functions
you can use mapply()...
z - as.data.frame(matrix(1:3,3,3,T))
mapply(function(x,y) x(y), c(sum,prod,sum), z)
Markus Preisetanz a écrit :
Dear Colleagues,
does anybody know how to aggregate a data.frame using different functions for
different columns?
Sincerely
___
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Re: [R] lattice histogram finetuning
it looks like a barchart rather than an histogram (?). if it is, you can do something like that: CT.flow$age - factor(CT.flow$age) CT.flow$succession - factor(CT.flow$succession,levels=c(early,mid,late)) ct - expand.grid( succession=levels(CT.flow$succession), site=levels(CT.flow$site), age=levels(CT.flow$age) ct$eff - as.vector(ftable(xtabs(~.,CT.flow))) barchart(eff~age|site*succession,horiz=F,ct) Patrick Kuss a écrit : Dear list, I have some difficulties fine-tuning a lattice conditional histogram plot and found little help in the documentation. My dataset consists of plant flowering ages from 3 different altitudes and 3 successional sites at each altitudinal level. My questions are: How do I avoid that lattice automatically sorts the different $condlevels alphabetically? I need: hist$condlevels$succession - list(early,mid,late) # not early,late,mid hist$condlevels$site - list(FU,SP,JU) Additionally, how do I add a vertical line indicating the mean flowering age at each panel? And how do I assure breaks to happen for each year? hist$panel.args.common$breaks Happy for any suggestions Patrick # Flowering age at different altitudes and successional site FU.e - round(rnorm(20,mean=9,sd=1),dig=0) FU.m - round(rnorm(20,mean=12,sd=1),dig=0) FU.l - round(rnorm(20,mean=15,sd=1),dig=0) SP.e - round(rnorm(20,mean=8,sd=1),dig=0) SP.m - round(rnorm(20,mean=11,sd=1),dig=0) SP.l - round(rnorm(20,mean=14,sd=1),dig=0) JU.e - round(rnorm(20,mean=7,sd=1),dig=0) JU.m - round(rnorm(20,mean=10,sd=1),dig=0) JU.l - round(rnorm(20,mean=13,sd=1),dig=0) age - c(FU.e,FU.m,FU.l,SP.e,SP.m,SP.l,JU.e,JU.m,JU.l) site - rep(c(FU,SP,JU),each=60) succession - rep(c(early,mid,late),each=20,times=3) CT.flow - data.frame(site,succession,age) library(lattice) trellis.device(color=F) hist - histogram(~age|succession*site,data=CT.flow) hist -- Patrick Kuss PhD-student Institute of Botany University of Basel Schönbeinstr. 6 CH-4056 Basel +41 61 267 2976 __ 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
