similar to: are arithmetic comparison operators binary?

i wonder about the following examples showing incoherence in how type conversions are done in r: x = TRUE x[2] = as.raw(1) # Error in x[2] = as.raw(1) : # incompatible types (from raw to logical) in subassignment type fix it seems that there is an attempt to coerce the raw value to logical here, which fails, even though as.logical(as.raw(1)) # TRUE likewise, x[2]

i wonder about the following examples showing incoherence in how type conversions are done in r: x = TRUE x[2] = as.raw(1) # Error in x[2] = as.raw(1) : # incompatible types (from raw to logical) in subassignment type fix it seems that there is an attempt to coerce the raw value to logical here, which fails, even though as.logical(as.raw(1)) # TRUE likewise, x[2]

Dear R Gurus: I read somewhere that functions are considered vectors. Is this true, please? thanks Edna Bell

somewhat related to a previous discussion [1] on how 'names<-' would sometimes modify its argument in place, and sometimes produce a modified copy without changing the original, here's another example of how it becomes visible to the user when r makes or doesn't make a copy of an object: x = NULL dput(x) # NULL class(x) = 'integer' # error: invalid

given what ?identical says, i find the following odd: x = 1:10 y = 1:10 all.equal(x,y) [1] TRUE identical(x,y) [1] TRUE y[11] = 11 y = y[1:10] all.equal(x,y) [1] TRUE identical(x,y) [1] FALSE y [1] 1 2 3 4 5 6 7 8 9 10 length(y) [1] 10 looks like a bug. platform i686-pc-linux-gnu arch i686 os linux-gnu system

the following is a trivialized version of some functional code i tried to use in r: (funcs = lapply(1:5, function(i) function() i)) # a list of no-parameter functions, each with its own closure environment, # each supposed to return the corresponding index when applied to no arguments sapply(funcs, function(func) func()) # supposed to return c(1,2,3,4,5) there is absolutely nothing unusual in

m = matrix(1:4, 2) apply(m, 1, cat, '\n') # 1 2 # 3 4 # NULL why the null? vQ

Following Duncan's suggestion, I forward the below to R-devel. vQ -------- Original Message -------- Subject: Re: [R] Randomly remove condition-selected rows from a matrix Date: Fri, 02 Jan 2009 10:34:52 -0500 From: Duncan Murdoch <murdoch at stats.uwo.ca> To: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no> CC: R help <R-help at stat.math.ethz.ch>

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48365 OS: Ubuntu 8.04 Linux 32bit Submission from: (NULL) (129.241.110.141) sprintf has a documented limit on strings included in the output using the format '%s'. It appears that there is a limit on the length of strings included with, e.g., the format '%d' beyond which surprising things happen (output modified for conciseness):

Dear R Gurus: How do I find the functions which are primitives, please? Thanks, Edna Bell

I am wondering if people on the list could recommend books that they have found helpful about programming concepts and style? I often find that students write R programs by copying existing code but could really benefit from the understanding of more general programming ideas. An example would be to avoid writing functions which attempt to modify their parameters. Another principle would be

Hi all I have the following regular expression problem: I want to find complete elements of a vector that end in a repeated character but where the repetition doesn't make up the whole word. That is, for the vector vec: vec<-c("aaaa", "baaa", "bbaa", "bbba", "baamm", "aa") I would like to get "baaa" "bbaa"

I was wondering why the following doesn't work: > a=c(1,2) > names(a)=c("one","two") > a one two 1 2 > > names(a[2]) [1] "two" > > names(a[2])="too" > names(a) [1] "one" "two" > a one two 1 2 I must not be understanding some basic concept here. Why doesn't the 2nd name change to

I was wondering why the following doesn't work: > a=c(1,2) > names(a)=c("one","two") > a one two 1 2 > > names(a[2]) [1] "two" > > names(a[2])="too" > names(a) [1] "one" "two" > a one two 1 2 I must not be understanding some basic concept here. Why doesn't the 2nd name change to

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48269 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.199.164) In the following example (and many other cases): quote(a=1) # 1 the argument matching is apparently incorrect wrt. the documentation (The R Language Definition, v 2.8.1, sec. 4.3.2, p. 23), which specifies the following algorithm for argument matching: 1. Attempt to

Full_Name: Wacek Kusnierczyk Version: 2.8.0 and 2.10.0 r48242 OS: Ubuntu 8.04 Linux 32 bit Submission from: (NULL) (129.241.110.161) In the following code: duplicated(data.frame(), incomparables=NA) # Error in if (!is.logical(incomparables) || incomparables) .NotYetUsed("incomparables != FALSE") : # missing value where TRUE/FALSE needed the raised error is clearly not the

Dear all, let's assume I have a vector of character strings: x <- c("abcdef", "defabc", "qwerty") What I would like to find is the following: all elements where the word 'abc' does not appear (i.e. 3 in this case of 'x'). Since I am not really experienced with regular expressions, I started slowly and thought I find all word were

Hello! I have the problem that in my function the passed variable is not used, but the variable name of the dataframe itself?- difficult to explain, but an easy example: TestFunc<-function(df, group) { ??? print(names(subset(df, select=group))) } df1<-data.frame(group="G1", visit="V1", value=0.9) TestFunc(df1, c("group", "visit")) Result: [1]

Hi, How do I tell an if statement to generate two seperate outputs. E.g If X>5 I want to create df1 and df2: if (X>5) {df1<-c(4,5,6,7,8) AND df2<-c(9,10,11,12,13)} Thanks, James -- View this message in context: http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22650844.html Sent from the R help mailing list archive at Nabble.com.

Hello, I am wondering about the behaviour of strsplit. When the pattern matches the beginning of the search string, the mepty string is added to the result, but that's not the case when the pattern matches the end of the search string: strsplit(" hello dolly ") [1] "" "hello" "dolly" The man for strsplit explains the algorithm: " The algorithm