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,] "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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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"

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?
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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

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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.
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
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