How can I replace numeric values by characters in R? - r

Question

I have a file like this.
"1" 10 2 0 0 0 0 0 0 0 0 0 0 0 4 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
"2" 10 3 6 17 11 15 8 17 14 1 42 21 22 15 9 9 17 12 9 16 4 8 12 29 23 11 0 0 0 0
"3" 10 4 39 39 14 33 16 23 37 21 29 22 46 26 16 26 21 22 21 10 16 3 10 14 20 12 6 0 0 0
"4" 100 18 0 0 0 1 0 0 0 0 0 0 2 0 0 1 0 2 8 5 2 1 2 4 9 6 4 3 0 0
.....................
What I want to do is, replace the values from column 4 onwards by characters, i.e. if value is between 0 to 10, then it will be replaced by character 'a' and if it is between 10 to 20, it will be replaced by character b and so on.
For example, the output file will be of the form,
"1" 10 2 0 0 0 0 0 0 0 0 0 0 0 a 0 0 a 0 0 0 0 0 a 0 0 0 0 0 0 0
.............................
How can I do it in R? Is there someway I can automate the assigning of characters because currently I am using two for loops and harcoding the values by the range.
Edit: My approach:
> for ( i in 1:nrow(x) )
+ for ( j in j:ncol(x) )
+ {
+ if (x[i,j] < 10 && x[i,j] > 0 )
+ x[i,j] = a
+ else if ( x[i,j] < 20 && x[i,j] > 10 )
+ x[i,j] = b
+ }
The above is my approach. This is showing an error in conditions, and I know will take a lot of time since it involves usage of two for loops.

Answer 1

One possible solution is to create a dummy data set to match against, and then match all non zero values to it (assuming df is your data set)
matchData <- data.frame(lets = c(0, rep(letters, each = 10)),
nums = c(0, seq_len(length(letters)*10)))
df[, -seq_len(3)] <- sapply(df[, -seq_len(3)], function(x) matchData$lets[match(x, matchData$nums)])
df
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 V25
# 1 1 10 2 0 0 0 0 0 0 0 0 0 0 0 a 0 0 a 0 0 0 0 0 a 0
# 2 2 10 3 a b b b a b b a e c c b a a b b a b a a b c
# 3 3 10 4 d d b d b c d c c c e c b c c c c a b a a b
# 4 4 100 18 0 0 0 a 0 0 0 0 0 0 a 0 0 a 0 a a a a a a a
# V26 V27 V28 V29 V30 V31
# 1 0 0 0 0 0 0
# 2 c b 0 0 0 0
# 3 b b a 0 0 0
# 4 a a a a 0 0

Answer 2

I think the following will be close, just a quick answer that hopefully helps you along. You'd have to apply over this method to do it for the entire dataframe. Also there's coercion that I didn't handle here, so when testing on a single row everything got coerced into a char.
The basic thought is that if you want 1-10 to correspond to "a", 11-20 to correspond to "b", then we can get that by dividing the number by 10, then calling ceiling. 1-10 then maps to 1, 11-20 then maps to 2, and so forth. letters[1] maps to "a", letters[2] maps to "b", ect so we get the desired functionality.
#everything coerced to char, I know
testVect<-c("2", 10, 3, 6, 17, 11, 15, 8 ,17, 14, 1, 42, 21, 22, 15, 9, 9, 17, 12, 9, 16, 4, 8, 12 ,29, 23, 11, 0, 0 ,0 ,0)
testAfter4<-sapply(testVect[4:length(testVect)],
function(entry) {
ifelse(entry==0, 0, letters[ceiling(as.numeric(entry)/10)])
} )
#need to cast entry back to numeric as it was coerced to char when initializing testVect
testVect[4:length(testVect)]<-testAfter4
testVect
#[1] "2" "10" "3" "a" "b" "b" "b" "a" "b" "b" "a" "e" "c" "c" "b"
#[16] "a" "a" "b" "b" "a" "b" "a" "a" "b" "c" "c" "b" "0" "0" "0"
#[31] "0"

Answer 3

You can use the ascii codes and an offset based on your value/10 (without remainder)...
mydat = c(10,2,0,19,20,19,0,0)
# Convert a number divided by 10 to its offset (hat tip to MrFlick for `letters`
# this uses the cryptic looking %/% operator for division without remainder
char10 = letters[1+(md %/% 10)]
# convert zeroes, and if desired replace column 1:4 with original data
char10[md==0] = 0
Output:
> char10
[1] "b" "a" "0" "b" "c" "b" "0" "0"