On Fri, 7 Oct 2011, Petr PIKAL wrote:
Hi
Is it necessary to use sapply? With lapply you will get what you want.
Another possibility is to use vapply.
sapply's simplification is only 'unexpected' by those who have not
studied its help page. See the posting guide ....
Regards
Petr
Dear folks--
The function below is a snippet of a larger function that is not doing
what
it is supposed to do, and I do not understand its behavior. The larger
function is supposed to produce an array containing the results of a
user-specified function applied to groups of data defined by the
intersection of one or more factors, and return them in an array with a
dimension for each factor and a dimension level for each factor level.
This
snippet is supposed to take a data frame, a vector of column numbers
containing factors, and a column number for the data, and return (in the
test function below, just print) a list of character vectors of the
level
names (one vector per dimension) and the length of those vectors.
It works fine so long as I give it more than one factor column, but if I
give it a vector of factor columns of length 1, it behave differently
and
when I try to assign the names from test.levels to the dimnames of the
array, I end up with an error message:
Error in dimnames(data) <- dimnames :
length of 'dimnames' [1] not equal to array extent
The example below shows the function output for a test data frame
(“test.df”) when run first of a vector of two column number for factors
and
then on just one. You can see how the structure of the output shifts. I
can
not understand what is happening. What I want it to do when given just
factor cols =c(1) is to give me back exactly what it gives me bact for
factor colum 1 in factor.cols = c(1,2).
Any help or suggestions would be greatly appreciated.
Sincerely,
andrewH
# Test Data
test.df <- data.frame(AA=rep(LETTERS[1:2], c(6,6)),BB=rep(LETTERS[3:5],
c(4,4,4)),
CC=rep(LETTERS[6:9],c(3,3,3,3)),
DD=c(1:12))
# The function
getLevels <- function(data.df, factor.cols, data.col){
test.levels <- sapply(test.df[,factor.cols, drop=F], levels)
cat("test.levels:\n"); print(test.levels)
no.levels <- sapply(sapply(data.df[,factor.cols, drop=F], levels),
length)
cat("no.levels:\n"); print(no.levels)
}
# Run it with two factors and again with 1, Output below
cat("\nTest 2 factors:\n")
getLevels(test.df, c(1,2), 4)
cat("\nTest 1 factor:\n")
getLevels(test.df, c(1), 4)
Test 2 factors:
getLevels(test.df, c(1,2), 4)
test.levels=
$AA
[1] "A" "B"
$BB
[1] "C" "D" "E"
no.levels=AA BB
2 3
cat("\nTest 1 factor:\n")
Test 1 factor:
getLevels(test.df, c(1), 4)
test.levels= AA
[1,] "A"
[2,] "B"
no.levels=A B
1 1
--
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behavior-of-extract-or-sapply-functions-tp3881176p3881176.html
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______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
