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Substitute list value
8 messages · Charles Stangor, michael.weylandt at gmail.com (R. Michael Weylandt, Jessica Streicher +2 more
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It should. Reproducible example of it not? Michael
On Jul 11, 2012, at 3:06 PM, Charles Stangor <cstangor at gmail.com> wrote:
I can't seem to determine how to get the name of a list member to
substitute as a variable name:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[1], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
--
Charles Stangor
Professor
--
Charles Stangor
Professor
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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.
1 day later
two things: - R always counts from 1, not from 0 - listmembers are accessed by using [[ ]] , not [ ] try t1[t==ll[[1]], "v"] <- 99 greetings Jessi
On 11.07.2012, at 15:47, Charles Stangor wrote:
I can't seem to determine how to get the name of a list member to
substitute:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[0], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
--
Charles Stangor
Professor
[[alternative HTML version deleted]]
______________________________________________ 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.
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On 13/07/2012 9:50 AM, Bert Gunter wrote:
Jessica: On Fri, Jul 13, 2012 at 1:35 AM, Jessica Streicher <j.streicher at micromata.de
wrote:
two things: - R always counts from 1, not from 0 - listmembers are accessed by using [[ ]] , not [ ]
FALSE! -- or at least not clearly stated:
> x <- list(a=letters[1:3],b=1:4) x[[2]]
[1] 1 2 3 4
x[2]
$b [1] 1 2 3 4 Note that the first is a vector, the second component of x; while the second is a list whose only component is the second component of x.
I think Jessica was right, and clear. List members are accessed using [[ ]]. Lists are subsetted using [ ]. Your first example extracts the second member of x. Your second example constructs a new list with a subset of the members that are in x. Duncan Murdoch
-- Bert
try t1[t==ll[[1]], "v"] <- 99 greetings Jessi On 11.07.2012, at 15:47, Charles Stangor wrote:
I can't seem to determine how to get the name of a list member to
substitute:
ll <- list("a1" = "a","a2" = "b")
t1[t==ll[0], "v"] <- 99
why doesn't this substitute to:
t1[t=="a", "v"] <- 99
Thank you!
--
Charles Stangor
Professor
[[alternative HTML version deleted]]
______________________________________________ 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. ______________________________________________ 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.