[R] using cut
Suppose I have some data x - rnorm(1000); y - x*x; then try to cut it into 2 chunks, c - cut(y, breaks=2); summary(y) Min. 1st Qu.Median Mean 3rd Qu. Max. 6.879e-06 9.911e-02 3.823e-01 9.499e-01 1.297e+00 8.342e+00 summary(c) (-0.00833,4.17] (4.17,8.35] 958 42 Is that the correct behavior? Why is the left hand side of the interval negative? How would I split a data vector into groups/intervals such that each interval contained the same number of points? Thanks much! Scott [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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.
Re: [R] using cut
You can split a vector into groups with equal numbers using the quantcut function in the gtools package. For example, to split into two groups, use: c - quantcut(y,c(0,.5,1)) On 26/05/07, J. Scott Olsson [EMAIL PROTECTED] wrote: Suppose I have some data x - rnorm(1000); y - x*x; then try to cut it into 2 chunks, c - cut(y, breaks=2); summary(y) Min. 1st Qu.Median Mean 3rd Qu. Max. 6.879e-06 9.911e-02 3.823e-01 9.499e-01 1.297e+00 8.342e+00 summary(c) (-0.00833,4.17] (4.17,8.35] 958 42 Is that the correct behavior? Why is the left hand side of the interval negative? How would I split a data vector into groups/intervals such that each interval contained the same number of points? Thanks much! Scott [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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. -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP __ R-help@stat.math.ethz.ch 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] using cut on matrices
Dear list, I'd like to use the function cut() on matrices, ie that when I apply it to a matrix, it would return a matrix of the same dimensions instead of a vector. I wonder if there is a better (more elegant) solution than matrix(cut(a, ...), ncol=ncol(a), nrow=nrow(a)) because I would like to use cut on both vectors and matrices and avoid testing whether a is a matrix. Thanks, Tamas -- Tamás K. Papp E-mail: [EMAIL PROTECTED] (preferred, especially for large messages) [EMAIL PROTECTED] Please try to send only (latin-2) plain text, not HTML or other garbage. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] using cut on matrices
Tamas Papp [EMAIL PROTECTED] writes: Dear list, I'd like to use the function cut() on matrices, ie that when I apply it to a matrix, it would return a matrix of the same dimensions instead of a vector. I wonder if there is a better (more elegant) solution than matrix(cut(a, ...), ncol=ncol(a), nrow=nrow(a)) because I would like to use cut on both vectors and matrices and avoid testing whether a is a matrix. Will this not work?: ac - cut(a) dim(ac) - dim(a) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] using cut on matrices
Peter's solution led me to an apparent bug in R.
a - 1:9
cut(a)
Error in cut.default(a) : Argument breaks is missing, with no default
##
That didn't work, so I read the documentation, found that a second
argument was required. Result:
cut(a, 2)
[1] (0.992,5] (0.992,5] (0.992,5] (0.992,5] (5,9.01] (5,9.01] (5,9.01]
[8] (5,9.01] (5,9.01]
Levels: (0.992,5] (5,9.01]
# LOOK VERY CAREFULLY:
# R 1.6.2 coded 5 as (5, 9.01].
# (I know I need to upgrade to R 1.7.1.)
# S-Plus 6.1 for Windows 2000 produced the following:
cut(a, 2)
[1] 1 1 1 1 1 2 2 2 2
attr(, levels):
[1] 0.92+ thru 5.00 5.00+ thru 9.08
# NOTE: S-Plus 6.1 did it correctly.
# I'm sorry to report an error without a fix,
# but I'm out of time for this now.
# Before I found the bug, I wrote a function
# to retain dimnames:
Cut -
function(a, ...){
ac - cut(a, ...)
if(is.array(a)){
dim(ac) - dim(a)
dimnames(ac) - dimnames(a)
}
else names(ac) - names(a)
ac
}
Cut(a, 2)
[1] (0.992,5] (0.992,5] (0.992,5] (0.992,5] (5,9.01] (5,9.01] (5,9.01]
[8] (5,9.01] (5,9.01]
Levels: (0.992,5] (5,9.01]
# That worked fine. What about a vector with names?
a1 - a
names(a1) - letters[1:9]
Cut(a1, 2)
[1] (0.992,5] (0.992,5] (0.992,5] (0.992,5] (5,9.01] (5,9.01] (5,9.01]
[8] (5,9.01] (5,9.01]
Levels: (0.992,5] (5,9.01]
# What happened to the names?
names(Cut(a1,2))
[1] a b c d e f g h i
# The names were there, but R chose not to display them.
# S-Plus 6.1 under Win2000 produced the following:
Cut(a1, 2)
a b c d e f g h i
1 1 1 1 1 2 2 2 2
attr(, levels):
[1] 0.92+ thru 5.00 5.00+ thru 9.08
# The names appear with codes and a translate table
# Something similar happens with an array:
# In R 1.6.2:
a2 - a
dim(a2) - c(3,3)
dimnames(a2) - list(LETTERS[1:3], c(ab,bc,cd))
Cut(a2, 2)
[1] (0.992,5] (0.992,5] (0.992,5] (0.992,5] (5,9.01] (5,9.01] (5,9.01]
[8] (5,9.01] (5,9.01]
Levels: (0.992,5] (5,9.01]
dimnames(Cut(a2,2))
[[1]]
[1] A B C
[[2]]
[1] ab bc cd
# S-Plus produced the following:
Cut(a2, 2)
ab bc cd
A 1 1 2
B 1 1 2
C 1 2 2
attr(, levels):
[1] 0.92+ thru 5.00 5.00+ thru 9.08
#
hope this helps.
spencer graves
Peter Dalgaard BSA wrote:
Tamas Papp [EMAIL PROTECTED] writes:
Dear list,
I'd like to use the function cut() on matrices, ie that when I apply
it to a matrix, it would return a matrix of the same dimensions
instead of a vector.
I wonder if there is a better (more elegant) solution than
matrix(cut(a, ...), ncol=ncol(a), nrow=nrow(a))
because I would like to use cut on both vectors and matrices and avoid
testing whether a is a matrix.
Will this not work?:
ac - cut(a)
dim(ac) - dim(a)
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