Re: [R] replace values in vector from a replacement table
On Mon, Jul 30, 2012 at 6:00 PM, jim holtman jholt...@gmail.com wrote:
try this:
(x - rep(letters,2))
[1] a b c d e f g h i j k l m n o p
q r s t u v w
[24] x y z a b c d e f g h i j k l m
n o p q r s t
[47] u v w x y z
values - c(aa, a, b, NA, d, zz)
repl - c(aa, A, B, NA, D, zz)
(repl.tab - cbind(values, repl))
values repl
[1,] aa aa
[2,] aA
[3,] bB
[4,] NA NA
[5,] dD
[6,] zz zz
indx - match(x, repl.tab[, 1], nomatch = 0)
x[indx != 0] - repl.tab[indx, 2]
x
[1] A B c D e f g h i j k l m n o p
q r s t u v w
[24] x y z A B c D e f g h i j k l m
n o p q r s t
[47] u v w x y z
Based on this code I came up with the following function.
replace2 - function(x, ind, repl){
if(any(is.na(ind))) ind[is.na(ind)] - 0
if(is.vector(x) is.vector(repl)) {
(x[ind != 0] - repl[ind])
return(x)
} else if(identical(ncol(x), ncol(repl))){
(x[ind != 0, ] - repl[ind, ])
return(x)
}
}
Whereas replicate() can be used only on vectors of same dimension,
replicate2() can be used on vectors and matrices/dataframes, and the
replacement data can have different nr of rows. It also works with
index vectors containing NAs.
##for vectors
(indx - match(x, repl.tab[, 1], nomatch = 0))
[1] 2 3 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 5 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
[46] 0 0 0 0 0 0 0
head(replace2(x, indx, repl.tab[, 2]))
[1] A B c D e f
(indx - match(x, repl.tab[, 1])) ##index vector with NAs
[1] 2 3 NA 5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA NA NA NA 2 3 NA 5
[31] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
head(replace2(x, indx, repl.tab[, 2]))
[1] A B c D e f
##for matrices/dataframes
head(xx - cbind(x, x))
x x
[1,] a a
[2,] b b
[3,] c c
[4,] d d
[5,] e e
[6,] f f
(repl.tab2 - cbind(repl.tab[, 2], repl.tab[, 2]))
[,1] [,2]
[1,] aa aa
[2,] A A
[3,] B B
[4,] NA NA
[5,] D D
[6,] zz zz
head(replace2(xx, indx, repl.tab2))
x x
[1,] A A
[2,] B B
[3,] c c
[4,] D D
[5,] e e
[6,] f f
Does this function have any generic value? Are there obvious
implementation mistakes?
Regards
Liviu
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[R] replace values in vector from a replacement table
Dear all I've got stuck when trying to replace values in a vector by selecting replacements from a replacement table. I'm trying to use only base functions. Here's a dummy example: (x - rep(letters,2)) [1] a b c d e f g h i j k l m n o p q r s t u v [23] w x y z a b c d e f g h i j k l m n o p q r [45] s t u v w x y z values - c(aa, a, b, NA, d, zz) repl - c(aa, A, B, NA, D, zz) (repl.tab - cbind(values, repl)) values repl [1,] aa aa [2,] aA [3,] bB [4,] NA NA [5,] dD [6,] zz zz Now I can easily compute all four combinations of 'match' and '%in%': (ind - match(x, repl.tab[ ,1])) [1] 2 3 NA 5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA 2 3 NA [30] 5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA (ind - match(repl.tab[ ,1], x)) [1] NA 1 2 NA 4 NA (ind - x %in% repl.tab[ ,1]) [1] TRUE TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [15] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE [29] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [43] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE (ind - repl.tab[ ,1] %in% x) [1] FALSE TRUE TRUE FALSE TRUE FALSE But how do I actually proceed to obtain the following vector? Can it be done without an explicit apply() or loop? res [1] A B c D e f g h i j k l m n o p q r s t u v [23] w x y z A B c D e f g h i j k l m n o p q r [45] s t u v w x y z Regards Liviu -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@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.
Re: [R] replace values in vector from a replacement table
try this: (x - rep(letters,2)) [1] a b c d e f g h i j k l m n o p q r s t u v w [24] x y z a b c d e f g h i j k l m n o p q r s t [47] u v w x y z values - c(aa, a, b, NA, d, zz) repl - c(aa, A, B, NA, D, zz) (repl.tab - cbind(values, repl)) values repl [1,] aa aa [2,] aA [3,] bB [4,] NA NA [5,] dD [6,] zz zz indx - match(x, repl.tab[, 1], nomatch = 0) x[indx != 0] - repl.tab[indx, 2] x [1] A B c D e f g h i j k l m n o p q r s t u v w [24] x y z A B c D e f g h i j k l m n o p q r s t [47] u v w x y z On Mon, Jul 30, 2012 at 11:53 AM, Liviu Andronic landronim...@gmail.com wrote: Dear all I've got stuck when trying to replace values in a vector by selecting replacements from a replacement table. I'm trying to use only base functions. Here's a dummy example: (x - rep(letters,2)) [1] a b c d e f g h i j k l m n o p q r s t u v [23] w x y z a b c d e f g h i j k l m n o p q r [45] s t u v w x y z values - c(aa, a, b, NA, d, zz) repl - c(aa, A, B, NA, D, zz) (repl.tab - cbind(values, repl)) values repl [1,] aa aa [2,] aA [3,] bB [4,] NA NA [5,] dD [6,] zz zz Now I can easily compute all four combinations of 'match' and '%in%': (ind - match(x, repl.tab[ ,1])) [1] 2 3 NA 5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA 2 3 NA [30] 5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA (ind - match(repl.tab[ ,1], x)) [1] NA 1 2 NA 4 NA (ind - x %in% repl.tab[ ,1]) [1] TRUE TRUE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [15] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE [29] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [43] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE (ind - repl.tab[ ,1] %in% x) [1] FALSE TRUE TRUE FALSE TRUE FALSE But how do I actually proceed to obtain the following vector? Can it be done without an explicit apply() or loop? res [1] A B c D e f g h i j k l m n o p q r s t u v [23] w x y z A B c D e f g h i j k l m n o p q r [45] s t u v w x y z Regards Liviu -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@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.
Re: [R] replace values in vector from a replacement table
On Mon, Jul 30, 2012 at 6:00 PM, jim holtman jholt...@gmail.com wrote: try this: indx - match(x, repl.tab[, 1], nomatch = 0) x[indx != 0] - repl.tab[indx, 2] x [1] A B c D e f g h i j k l m n o p q r s t u v w [24] x y z A B c D e f g h i j k l m n o p q r s t [47] u v w x y z This is excellent! Thank you Liviu __ R-help@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.
