similar to: Programming Concepts and Philosophy

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

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

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>

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

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):

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"

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

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

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

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

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]

the man page for relational operators (see, e.g., ?'<') says: " Binary operators which allow the comparison of values in atomic vectors. Arguments: x, y: atomic vectors, symbols, calls, or other objects for which methods have been written. " it is somewhat surprizing that the following works: '<'(1) # logical(0) '<'() #

the man page for relational operators (see, e.g., ?'<') says: " Binary operators which allow the comparison of values in atomic vectors. Arguments: x, y: atomic vectors, symbols, calls, or other objects for which methods have been written. " it is somewhat surprizing that the following works: '<'(1) # logical(0) '<'() #

?ginv provides 'Modern Applied Statistics with S' (MASS), 3rd, by Venables and Ripley as the sole reference. I happen to have this book (4th ed) on loan from our library, and as far as I can see, ginv is mentioned there twice, and it is *used*, not *explained* in any way. (It is used on p. 148 in the 4th edition.) ginv does not appear in the index of MASS. ginv is an implementation of

Hi group! Suppose I have 2 matrices A and B of equal dimensions. I want to apply a function f to all corresponding pairs of rows from A and B in an efficient manner. Basically, I want mapply(f, data.frame(A), data.frame(B)) but for rows. How do I do it? Thanks, Andrey

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

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

Tested in: R version 2.8.1 (2008-12-22) / Windows Recursive default argument references normally give nice clear errors. In the first set of examples, you get the error: Error in ... : promise already under evaluation: recursive default argument reference or earlier problems? (function(a = a) a ) () (function(a = a) c(a) ) () (function(a = a) a[1] ) () (function(a = a)