Hi all, there is a vector v with several NAs. I want to create a new vector n of the same length as v and the same NAs as in v and tried this: n <- vector(length=length(v), mode="numeric") replace(n, which(is.na(v)), NA) but this does't work, all values in n are 0. What went wrong? Thanks, Sven -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
replacing values in a vector
11 messages · Sven Garbade, Jens Nieschulze, Kaspar Pflugshaupt +7 more
On Thu, 26 Jul 2001, Sven Garbade wrote:
Hi all, there is a vector v with several NAs. I want to create a new vector n of the same length as v and the same NAs as in v and tried this: n <- vector(length=length(v), mode="numeric") replace(n, which(is.na(v)), NA) but this does't work, all values in n are 0. What went wrong?
replace
function (x, list, values)
{
x[list] <- values
x
}
from the function body I gues you should use
replace(n, which(is.na(v)), NA)->n
instead because I assume that x is a local variable so the global n wouldn't be changed by replace() J
Thanks, Sven -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
*********************************************************************** Jens Nieschulze Institute for Forest Biometrics & Phone: ++49-551-39-12107 Applied Computer Science Fax : ++49-551-39-3465 Buesgenweg 4 37077 Goettingen E-mail: jniesch at uni-forst.gwdg.de GERMANY http://www.uni-forst.gwdg.de/~jniesch -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
On Thursday 26 July 2001 13:42, Sven Garbade wrote:
Hi all, there is a vector v with several NAs. I want to create a new vector n of the same length as v and the same NAs as in v and tried this: n <- vector(length=length(v), mode="numeric") replace(n, which(is.na(v)), NA) but this does't work, all values in n are 0. What went wrong?
You have to assign the result of the replacement, like in n <- replace(n, which(is.na(v)), NA) Or, somewhat simpler: n[is.na(v)] <- NA Cheers Kaspar Pflugshaupt
Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Jens Nieschulze wrote:
On Thu, 26 Jul 2001, Sven Garbade wrote:
Hi all, there is a vector v with several NAs. I want to create a new vector n of the same length as v and the same NAs as in v and tried this: n <- vector(length=length(v), mode="numeric") replace(n, which(is.na(v)), NA) but this does't work, all values in n are 0. What went wrong?
replace
function (x, list, values)
{
x[list] <- values
x
}
from the function body I gues you should use
replace(n, which(is.na(v)), NA)->n
instead because I assume that x is a local variable so the global n wouldn't be changed by replace()
oh, of course... Thanks! Sven -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
n <- rep(0, length(v)) #Or replace 0 by the initial value you want in the
vector n
n[is.na(v)] <- NA
Best Regards,
Philippe
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-----Message d'origine-----
De : owner-r-help at stat.math.ethz.ch
[mailto:owner-r-help at stat.math.ethz.ch]De la part de Sven Garbade
Envoye : jeudi 26 juillet 2001 13:43
A : R-Help
Objet : [R] replacing values in a vector
Hi all,
there is a vector v with several NAs. I want to create a new vector n of
the same length as v and the same NAs as in v and tried this:
n <- vector(length=length(v), mode="numeric")
replace(n, which(is.na(v)), NA)
but this does't work, all values in n are 0. What went wrong?
Thanks, Sven
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You should assign the result of replace back to n: n <- replace(n, which(is.na(v)), NA) Giovanni
__________________________________________________ [ ] [ Giovanni Petris GPetris at uark.edu ] [ Department of Mathematical Sciences ] [ University of Arkansas - Fayetteville, AR 72701 ] [ Ph: (501) 575-6324, 575-8630 (fax) ] [ http://definetti.uark.edu/~gpetris/ ] [__________________________________________________] > Date: Thu, 26 Jul 2001 13:42:44 +0200 > From: Sven Garbade <garbade at psy.uni-muenchen.de> > X-Accept-Language: en > Sender: owner-r-help at stat.math.ethz.ch > Precedence: SfS-bulk > Content-Type: text/plain; charset=us-ascii > Content-Length: 649 > > Hi all, > > there is a vector v with several NAs. I want to create a new vector n of > the same length as v and the same NAs as in v and tried this: > > n <- vector(length=length(v), mode="numeric") > replace(n, which(is.na(v)), NA) > > but this does't work, all values in n are 0. What went wrong? > > Thanks, Sven > -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- > r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html > Send "info", "help", or "[un]subscribe" > (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch > _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._ > -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Actually, if you just want a vector as v with 0 except for the NA, the easiest is just: v*0 which will let be the NA if present. Agus Dr. Agustin Lobo Instituto de Ciencias de la Tierra (CSIC) Lluis Sole Sabaris s/n 08028 Barcelona SPAIN tel 34 93409 5410 fax 34 93411 0012 alobo at ija.csic.es -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Note that the below does not work for Inf/-Inf (gives NaN) Giles Dr. G.T. Innocent Comparative Epidemiology and Informatics Group, Dept. of Veterinary Clinical Studies, Glasgow University Vet. School. Bearsden Road, Glasgow G61 1QH Tel. 0141 339 8855 Ext. 0531. Fax. 0141 330 5729 e-mail G.Innocent at vet.gla.ac.uk
Actually, if you just want a vector as v with 0 except for the NA, the easiest is just: v*0 which will let be the NA if present. Agus Dr. Agustin Lobo Instituto de Ciencias de la Tierra (CSIC) Lluis Sole Sabaris s/n 08028 Barcelona SPAIN tel 34 93409 5410 fax 34 93411 0012 alobo at ija.csic.es
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I have a dataset (of say values of X), each value associated with a level of a factor F (25 levels in all), in a dataframe. I want to make a single plot which contains 25 histograms of X, one for each level of F and each one appropriately labelled. (And, since there are 25 levels, a 5x5 array would be nice). What's the trick in R, please? With thanks, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 167 1972 Date: 27-Jul-01 Time: 20:47:11 ------------------------------ XFMail ------------------------------ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
On Fri, 27 Jul 2001 Ted.Harding at nessie.mcc.ac.uk wrote:
I have a dataset (of say values of X), each value associated with a level of a factor F (25 levels in all), in a dataframe. I want to make a single plot which contains 25 histograms of X, one for each level of F and each one appropriately labelled. (And, since there are 25 levels, a 5x5 array would be nice). What's the trick in R, please?
Use the histogram function in the new 'lattice' package -thomas Thomas Lumley Asst. Professor, Biostatistics tlumley at u.washington.edu University of Washington, Seattle -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
2 days later
Ted.Harding at nessie.mcc.ac.uk writes:
I have a dataset (of say values of X), each value associated with a level of a factor F (25 levels in all), in a dataframe. I want to make a single plot which contains 25 histograms of X, one for each level of F and each one appropriately labelled. (And, since there are 25 levels, a 5x5 array would be nice). What's the trick in R, please?
The trick is to the par() function. Here is an example ...
#
# generate some dummy data
#
x <- rnorm(n = 1000, mean = 12, sd = 1)
f <- round(runif(n = 1000, min = 1, max = 25))
#
# set device for 5 * 5 array of charts
#
par(mfrow = c(5,5))
#
# plot histograms
#
tapply(x, f, hist, ylab="n", xlab="x")
I think you'll need a for() loop to get better control over the plots.
Something like:
for(i in min(f):max(f))
{
hist(x[f==i], xlim=range(x), ylab="n", xlab="x", main = i)
}
But I might be wrong.
I hope that helps.
Mark
--
Mark Myatt
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