replace values in vector from a replacement table

[Liviu Andronic]

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,] "a"    "A"
[3,] "b"    "B"
[4,] NA     NA
[5,] "d"    "D"
[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


--
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[jholtman]

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,] "a"    "A"
[3,] "b"    "B"
[4,] NA     NA
[5,] "d"    "D"
[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 <[hidden email]> 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,] "a"    "A"
> [3,] "b"    "B"
> [4,] NA     NA
> [5,] "d"    "D"
> [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
>
> ______________________________________________
> [hidden email] 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.

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[Liviu Andronic]

On Mon, Jul 30, 2012 at 6:00 PM, jim holtman <[hidden email]> 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

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[Liviu Andronic]

On Mon, Jul 30, 2012 at 6:00 PM, jim holtman <[hidden email]> 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,] "a"    "A"
> [3,] "b"    "B"
> [4,] NA     NA
> [5,] "d"    "D"
> [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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