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assign object with loop (translation from SAS to R)
6 messages · Rui Barradas, David Winsemius, ONKELINX, Thierry +2 more
Hello, See inline. Em 29-06-2012 02:18, lynx escreveu:
I have a dataset named DM with p1, p2, ...., p9 (9 columns, numerical values)
I would like to calculate to multify each pair of columns (p1p2, p1p3,...
p1p9, p2p3, p2p4.... p8p9) and assign them in p1p2, p1p3,... p1p9, p2p3,
p2p4.... p8p9
In SAS,
l=0;
p_int_sum=0;
do i=1 to 8;
do j=(i+1) to 9;
l=l+1;
p{i}p{j}=p{i}*p{j};
end;
end;
In R this sort of syntax doesn't work, the trick is to create the variable where to hold the rsult beforehand and use vectors. DM <- data.frame(p=1:9) DM$prod <- NA # create results variable i <- 2:9 # use a vector to index DM$p DM$prod[i] <- DM$p[i - 1]*DM$p[i] DM Hope this helps, Rui Barradas
I would like to know how to assign them in R
I tried for function but failed.
for (i in 1:8) {
for (j in 2:9) {
DM$p[i]p[j] <- DM$p[i] * DM$p[j]
}}
Thank you so much for reading this!
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______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
On Jun 28, 2012, at 9:18 PM, lynx wrote:
I have a dataset named DM with p1, p2, ...., p9 (9 columns,
numerical values)
I would like to calculate to multify each pair of columns (p1p2,
p1p3,...
p1p9, p2p3, p2p4.... p8p9) and assign them in p1p2, p1p3,... p1p9,
p2p3,
p2p4.... p8p9
In SAS,
l=0;
p_int_sum=0;
do i=1 to 8;
do j=(i+1) to 9;
l=l+1;
p{i}p{j}=p{i}*p{j};
end;
end;
I would like to know how to assign them in R
I tried for function but failed.
for (i in 1:8) {
for (j in 2:9) {
# Try instead:
DM[[ paste("p",i, "p",j,sep="") ]] <-
DM[[paste("p",i, sep="")]] * DM[[paste("p",i, sep="")]]
DM$p[i]p[j] <- DM$p[i] * DM$p[j] }}
I suspect there is a more elegant method than this use of R as a macro processor. I tested the above approach with suitably smalled subscripts on a smaller dataset: DM <- data.frame(p1=1:10,p2=1:10,p3=1:10,p4=1:10,p5=1:10)
David Winsemius, MD West Hartford, CT
You can use a combination of the outer() and apply() functions
n <- 10
p <- 9
dataset <- data.frame(matrix(rep(seq_len(p), each = n), nrow = n, ncol = p))
colnames(dataset) <- paste("p", seq_len(p), sep = "")
test <- t(apply(dataset, 1, function(x){ x %o% x}))
colnames(test) <- paste("p", rep(seq_len(p), each = p), "p", rep(seq_len(p), p), sep = "")
test <- test[, rep(seq_len(p), each = p) < rep(seq_len(p), p)]
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
Thierry.Onkelinx at inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data.
~ John Tukey
-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Namens David Winsemius
Verzonden: vrijdag 29 juni 2012 16:18
Aan: lynx
CC: r-help at r-project.org
Onderwerp: Re: [R] assign object with loop (translation from SAS to R)
On Jun 28, 2012, at 9:18 PM, lynx wrote:
I have a dataset named DM with p1, p2, ...., p9 (9 columns, numerical
values) I would like to calculate to multify each pair of columns
(p1p2, p1p3,...
p1p9, p2p3, p2p4.... p8p9) and assign them in p1p2, p1p3,... p1p9,
p2p3, p2p4.... p8p9
In SAS,
l=0;
p_int_sum=0;
do i=1 to 8;
do j=(i+1) to 9;
l=l+1;
p{i}p{j}=p{i}*p{j};
end;
end;
I would like to know how to assign them in R I tried for function but
failed.
for (i in 1:8) {
for (j in 2:9) {
# Try instead:
DM[[ paste("p",i, "p",j,sep="") ]] <-
DM[[paste("p",i, sep="")]] * DM[[paste("p",i, sep="")]]
DM$p[i]p[j] <- DM$p[i] * DM$p[j] }}
I suspect there is a more elegant method than this use of R as a macro processor. I tested the above approach with suitably smalled subscripts on a smaller dataset: DM <- data.frame(p1=1:10,p2=1:10,p3=1:10,p4=1:10,p5=1:10) -- David Winsemius, MD West Hartford, CT ______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document.
With R it is pretty easy to eliminate the loops and the use of outer() (which calculates each product twice) using expand.grid():
# Create some data and label columns
DM <- data.frame(matrix(round(runif(90)*10, 0), ncol=9))
colnames(DM) <- paste0("p", 1:9)
# Create i, j indices and label columns
idx <- expand.grid(2:9, 1:8)
colnames(idx) <- c("j", "i")
# Compute products in DM2 and label columns
DM2 <- DM[idx$i]*DM[idx$j]
colnames(DM2) <- paste0("p", idx$i, "p", idx$j)
# Combine original data with products
DM <- data.frame(DM, DM2)
---------------------------------------------- David L Carlson Associate Professor of Anthropology Texas A&M University College Station, TX 77843-4352
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of ONKELINX, Thierry
Sent: Friday, June 29, 2012 10:11 AM
To: David Winsemius; lynx
Cc: r-help at r-project.org
Subject: Re: [R] assign object with loop (translation from SAS to R)
You can use a combination of the outer() and apply() functions
n <- 10
p <- 9
dataset <- data.frame(matrix(rep(seq_len(p), each = n), nrow = n, ncol
= p))
colnames(dataset) <- paste("p", seq_len(p), sep = "")
test <- t(apply(dataset, 1, function(x){ x %o% x}))
colnames(test) <- paste("p", rep(seq_len(p), each = p), "p",
rep(seq_len(p), p), sep = "")
test <- test[, rep(seq_len(p), each = p) < rep(seq_len(p), p)]
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
Thierry.Onkelinx at inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given body
of data.
~ John Tukey
-----Oorspronkelijk bericht-----
Van: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
Namens David Winsemius
Verzonden: vrijdag 29 juni 2012 16:18
Aan: lynx
CC: r-help at r-project.org
Onderwerp: Re: [R] assign object with loop (translation from SAS to R)
On Jun 28, 2012, at 9:18 PM, lynx wrote:
I have a dataset named DM with p1, p2, ...., p9 (9 columns, numerical
values) I would like to calculate to multify each pair of columns
(p1p2, p1p3,...
p1p9, p2p3, p2p4.... p8p9) and assign them in p1p2, p1p3,... p1p9,
p2p3, p2p4.... p8p9
In SAS,
l=0;
p_int_sum=0;
do i=1 to 8;
do j=(i+1) to 9;
l=l+1;
p{i}p{j}=p{i}*p{j};
end;
end;
I would like to know how to assign them in R I tried for function but
failed.
for (i in 1:8) {
for (j in 2:9) {
# Try instead:
DM[[ paste("p",i, "p",j,sep="") ]] <-
DM[[paste("p",i, sep="")]] * DM[[paste("p",i, sep="")]]
DM$p[i]p[j] <- DM$p[i] * DM$p[j] }}
I suspect there is a more elegant method than this use of R as a macro processor. I tested the above approach with suitably smalled subscripts on a smaller dataset: DM <- data.frame(p1=1:10,p2=1:10,p3=1:10,p4=1:10,p5=1:10) -- David Winsemius, MD West Hartford, CT
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. ______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code.
Dear
Rui Barradas
David Winsemius
ONKELINX, Thierry
David Carlson
I really appreciate your helps.
I did not realize that there were such many ways to do it.
Among them, I just pick up the simple one and it worked out.
It was like this.
<<ONKELINX, Thierry's way>>
n <- 10
p <- 9
dataset <- data.frame(matrix(rep(seq_len(p), each = n), nrow = n, ncol = p))
colnames(dataset) <- paste("p", seq_len(p), sep = "")
test <- t(apply(dataset, 1, function(x){ x %o% x}))
colnames(test) <- paste("p", rep(seq_len(p), each = p), "p", rep(seq_len(p),
p), sep = "")
test <- test[, rep(seq_len(p), each = p) < rep(seq_len(p), p)]
In fact, this is first time for me to post in R help.
I was very surprised to see quick replys and how you are supportive.
You guys are awesome. Thank you so so much!
I will pay back the debt to someone else. :-)
--
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Sent from the R help mailing list archive at Nabble.com.