Reputation: 253
I'd like to replace a character in a string with another character, by first sampling by the character. I'm having trouble having it print out the character instead of the index.
Example data, is labelled "try":
L 0.970223325 - 0.019851117 X 0.007444169
K 0.962779156 - 0.027295285 Q 0.004962779
P 0.972704715 - 0.027295285 NA 0
C 0.970223325 - 0.027295285 L 0.00248139
V 0.970223325 - 0.027295285 T 0.00248139
I'm trying to sample a character for a given row using weighted probabilities.
samp <- function(row) {
sample(try[row,seq(1, length(try), 2)], 1, prob = try[row,seq(2, length(try), 2)])
}
Then, I want to use the selected character to replace a position in a given string.
subchar <- function(string, pos, new) {
paste(substr(string, 1, pos-1), new , substr(string, pos+1, nchar(string)), sep='')
}
My question is - if I do, for example
> subchar("KLMN", 3, samp(4))
[1] "KL1N"
But I want it to read "KLCN". As.character(samp(4)) doesn't work either. How do I get it to print out the character instead of the index?
Upvotes: 0
Views: 471
Reputation: 66844
The problem arises because your letters are stored as factors
rather than characters
, and samp
is returning a data.frame
.
C
is the first level in your factor so that is stored as 1
internally, and as.character
(which gets invoked by the paste
statement) pulls this out when working on the mini-data.frame:
samp(4)
V1
4 C
as.character(samp(4))
[1] "1"
You can solve this in 2 ways, either dropping the data.frame
of the samp
output in your call to subchar
, or modifying samp
to do so:
subchar("KLMN", 3, samp(4)[,1])
[1] "KLCN"
samp2 <- function(row)
{ sample(try[row,seq(1, length(try), 2)], 1, prob = try[row,seq(2, length(try), 2)])[,1]
}
subchar("KLMN",3,samp2(4))
[1] "KLCN
You may also find it easier to sample within your subsetting, and you can drop the data.frame from there:
samp3 <- function(row){
try[row,sample(seq(1,length(try),2),1,prob=try[row,seq(2,length(try),2)]),drop=TRUE]
}
Upvotes: 1