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
>
> __
> 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.
>
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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.
>
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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
> __
> 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.
>
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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
> "table"s 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.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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 (i<85) 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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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 (i<85) 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]]
>
>
>
>
> __
> 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.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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.
> >
>
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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] 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,plo
[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
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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.
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] Reshape into a contingency table/Fisher's test
> test <- read.table("clipboard",sep=";",row=1,header=T, as.is=T)
> test
X00 X01 X10 X11
John.Mike 123 313 12 31
John.Jim54 57 39 36
John.Steve 135 47 47 74
Mike.Jim63 37 27 16
Mike.Steve 15 15 5 61
Jim.Steve6 10 34 35
> list.mat <- mapply(function(x,y) structure(matrix(x,2,2),
> dimnames=list(unlist(strsplit(y, ".",
fixed=T)), 0:1)), as.data.frame(t(test)), rownames(test), SIMPLIFY=F)
>
> list.mat
[[1]]
0 1
John 123 12
Mike 313 31
[[2]]
0 1
John 54 39
Jim 57 36
[[3]]
0 1
John 135 47
Steve 47 74
[[4]]
0 1
Mike 63 27
Jim 37 16
[[5]]
0 1
Mike 15 5
Steve 15 61
[[6]]
0 1
Jim6 34
Steve 10 35
>
> cbind(rownames(test), sapply(list.mat, function(x) sprintf("%.G",
> fisher.test(x)$p.value)))
[,1] [,2]
[1,] "John.Mike" "1"
[2,] "John.Jim" "0.8"
[3,] "John.Steve" "1E-09"
[4,] "Mike.Jim" "1"
[5,] "Mike.Steve" "7E-06"
[6,] "Jim.Steve" "0.4"
---
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Serguei Kaniovski a écrit :
> Dear all,
>
> how can I "reshape"/"cast" the following matrix
>
> 00;01;10;11
> John.Mike;123;313;12;31
> John.Jim;54;57;39;36
> John.Steve;135;47;47;74
> Mike.Jim;63;37;27;16
> Mike.Steve;15;15;5;61
> Jim.Steve;6;10;34;35
>
> into a set of stacked 2x2 contingency tables
>
> 0;1
> John;123;12
> Mike;313;31
> John;54;39
> Jim;57;36
> John;135;47
> Steve;47;16
> ...
>
> so that the "fisher.test" and "chisq.test" can be "applied" to the
> pairs of names. As as result I would like to have the original format, ie
>
> John.Mike
> John.Jim
> John.Steve
> Mike.Jim
> Mike.Steve
> Jim.Steve
>
> with two columns containing the p-values of the tests.
>
> Thanks a lot,
> Serguei
> --
> ___
>
> Austrian Institute of Economic Research (WIFO)
>
> Name: Serguei Kaniovski P.O.Box 91
> Tel.: +43-1-7982601-231 Arsenal Objekt 20
> Fax: +43-1-7989386 1103 Vienna, Austria
> Mail: [EMAIL PROTECTED]
>
> http://www.wifo.ac.at/Serguei.Kaniovski
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorise a for loop?
tt$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/>
>
>
> [[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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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] 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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] 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.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] [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 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.
[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] 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
>
> __
> 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.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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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
>
>
> [[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.
>
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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] 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] 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] 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] 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] 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] Permutation Distribution
> data1 <- expand.grid(var1=1:15, var2=1:2) > test <- replicate(1000, with(data.frame(var1=data1$var1, > var2=sample(data1$var2)), diff(tapply(var1, var2, mean > hist(test) ----------- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Jacob van Wyk a écrit : > Hallo > > Is there an elegant way to do the following: > > Dataset consists of 2 variables: var1: some measurements, and var2: a > grouping variable with two values, 1 and 2. > There are (say) 10 measurements from group 1 and 15 measurements from group 2. > The idea is to study the permutation distribution of mean(group 1) * > mean(group2). > One way would be to permute 1s and 2s and select the corresponding > measurements; calculate the difference in means. > Redo this 1000 times, say. Etc. > > Any help is much appreciated. > Thanks > Jacob > > > Jacob L van Wyk > Department of Statistics > University of Johannesburg, APK > P O Box 524 > Auckland Park 2006 > South Africa > Tel: +27 11 489 3080 > Fax: +27 11 489 2832 > > > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A contingency table of counts by case
sorry, answered to quickly...
actually it's easier using paste():
df <- df[order(df$case),]
apply(combinations(9,2), 1, function(y) table(factor(do.call(paste,
c(with(df[df$id %in% y, ],
split(x, id)), sep="")), levels=c("00","01","10","11"
-----------
Jacques VESLOT
CNRS UMR 8090
I.B.L (2ème étage)
1 rue du Professeur Calmette
B.P. 245
59019 Lille Cedex
Tel : 33 (0)3.20.87.10.44
Fax : 33 (0)3.20.87.10.31
http://www-good.ibl.fr
---
Serguei Kaniovski a écrit :
> Here is an example of the data.frame that I have,
>
> df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9)
>
> "case" represents the cases,
> "id" the persons, and
> "x" is the binary state.
>
> I would like to know in how many cases any two persons
>
> a. both have "1",
> b. the first has "0" - the second has "1",
> c. the first has "0" - the second has "0",
> d. both have "0".
>
> There will be choose(9,2) sums, denoted s_ij for 1<=i where i and j are running indices for an "id"-pair.
>
> Please help, this is way beyond my knowledge of R!
>
> Thank you,
> Serguei Kaniovski
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] A contingency table of counts by case
library(gtools)
apply(combinations(9,2), 1, function(x) with(df[df$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<=i 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
__
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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] 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] 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] 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] 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
>
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Re: [R] new object
> names(summary(fit)) [1] "surv" "time" "n.risk" "n.event" "conf.int" "std.err" [7] "lower""upper""strata" "call" > summary(fit)$n.risk [1] 11 10 8 7 5 4 2 12 10 8 6 5 4 3 2 1 --- Jacques VESLOT CNRS UMR 8090 I.B.L (2ème étage) 1 rue du Professeur Calmette B.P. 245 59019 Lille Cedex Tel : 33 (0)3.20.87.10.44 Fax : 33 (0)3.20.87.10.31 http://www-good.ibl.fr --- Mauricio Cardeal a écrit : > Hi ! > > Please, how can I extract n.event and n.risk as a new object from > example below ? Thanks in advance. > Mauricio > > require(survival) > fit <- survfit(Surv(time, status) ~ x, data=aml) > > summary(fit) > > Call: survfit(formula = Surv(time, status) ~ x, data = aml) > > x=Maintained > time n.risk n.event survival std.err lower 95% CI upper 95% CI > 9 11 10.909 0.0867 0.75411.000 >13 10 10.818 0.1163 0.61921.000 >18 8 10.716 0.1397 0.48841.000 >23 7 10.614 0.1526 0.37690.999 >31 5 10.491 0.1642 0.25490.946 >34 4 10.368 0.1627 0.15490.875 >48 2 10.184 0.1535 0.03590.944 > > x=Nonmaintained > time n.risk n.event survival std.err lower 95% CI upper 95% CI > 5 12 2 0.8333 0.1076 0.64701.000 > 8 10 2 0.6667 0.1361 0.44680.995 >12 8 1 0.5833 0.1423 0.36160.941 >23 6 1 0.4861 0.1481 0.26750.883 >27 5 1 0.3889 0.1470 0.18540.816 >30 4 1 0.2917 0.1387 0.11480.741 >33 3 1 0.1944 0.1219 0.05690.664 >43 2 1 0.0972 0.0919 0.01530.620 >45 1 1 0. NA NA NA > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Table of P values for Fisher's exact test
apply(yourdata[, c("X2N_CHB","X2N_AA","Counts_CHB","Counts_AA")], 1,
function(x)
fisher.test(cbind(x[3:4], x[1:2]-x[3:4]))$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
>
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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)
>
> __
> 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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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] 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!
>
>
>
>
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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!
>
>
>
>
__
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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!
>
>
>
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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
>
>
> -
>
> [[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
>
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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
>>
>>__
>>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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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] 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] 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.
>
>
>
>
> -
> CONFIDENTIALITY NOTICE: This message and any attachments rel...{{dropped}}
>
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>
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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
>
> Send instant messages to your online friends http://in.messenger.yahoo.com
>
> Stay connected with your friends even when away from PC. Link:
> http://in.mobile.yahoo.com/new/messenger/
> [[alternative HTML version deleted]]
>
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> 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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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
>
> __
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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] 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
>
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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
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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
> > [[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
> >
>
>
> Send instant messages to your online friends http://uk.messenger.yahoo.com
>
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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
>
>
> Send instant messages to your online friends http://uk.messenger.yahoo.com
> [[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
>
__
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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] 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
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] 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!
>
> __
> 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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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] 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] 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
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: >>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] 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
>
> [[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
>
--
---
[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
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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
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
__
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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] 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] 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
>
>Viel oder wenig? Schnell oder langsam? Unbegrenzt surfen + telefonieren
>ohne Zeit- und Volumenbegrenzung? DAS TOP ANGEBOT JETZT bei Arcor: günstig
>und schnell mit DSL - das All-Inclusive-Paket für clevere Doppel-Sparer,
>nur 44,85 € inkl. DSL- und ISDN-Grundgebühr!
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>PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
>
>
>
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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] 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] 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] 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] 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
> >>
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
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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[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
__
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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] 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
>>>
>>>
>>>
>>
>
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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
>
>__
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>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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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] 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] 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] 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] 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] 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
>
>
__
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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
>
>__
>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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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] 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
>
>__
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>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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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
>
>
__
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Re: [R] labels inside the bars of horizontal barplot
b <- barplot(t(as.matrix(mydf)),horiz=TRUE,beside=TRUE,space=c(-0.9,0.1)) text(myvect1, b[1,], mylabels, col="white", adj=1) Albert Vilella a écrit : >Dears, > >I would like to add labels inside the bars of a horizontal bar >plot like this: > >mylabels = c( > "long text goes here first", > "long text goes here second", > "long text goes here third", > "long text goes here fourth", > "long text goes here fourth", > "long text goes here fifth", > "long text goes here sixth", > "long text goes here seventh", > "long text goes here bla bla", > "long text goes here and more" >) >myvect1 = c(0.25,0.21,0.20,0.22,0.20,0.22,0.20,0.22,0.20,0.22) >myvect2 = c(0.13,0.08,0.09,0.11,0.13,0.08,0.09,0.11,0.13,0.08) >mydf = data.frame(myvect1,myvect2) # Forgive this df<->matrix mess, I >use the df for other stuff >barplot(t(as.matrix(mydf)), >horiz=TRUE, >beside=TRUE, >space=c(-0.9,0.1)) > >What I would like is to have the labels inside the upper bar. > >Any ideas? > >Bests, > >Albert. > >__ >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
