在指定位置之后替换字符串中的所有元素

Question

I would like to replace all elements in a string after a specified position that varies among strings. Ideally the solution would use regex in base R .

Here is a worked example and the desired result:

my.last.position <- c(5, 7, 3, NA, 10)

my.data <- read.table(text='
                         my.string
                         .1.222.2.2
                         ..1..1..2.
                         1.1.2.2...
                         .222.232..
                         ..1..1...1
', header=TRUE, stringsAsFactors = FALSE)
my.data

desired.result <- read.table(text='
                         my.string
                         .1.22.....
                         ..1..1....
                         1.1.......
                         .222.232..
                         ..1..1...1
', header=TRUE, stringsAsFactors = FALSE)
desired.result

The vector my.last.position specifies the last position to retain in each string. The data.frame desired.result contains the desired result.

Thanks for any advice. Sorry if this is a duplicate.

Answer 1

Hopefully someone can come up with a more elegant solution than this, but here goes:

mapply(
    function(s,i) paste(collapse='',if (is.na(i)) s else c(s[seq_len(i)],rep('.',length(s)-i))),
    strsplit(my.data$my.string,''),
    my.last.position
);
## [1] ".1.22....." "..1..1...." "1.1......." ".222.232.." "..1..1...1"

Answer 2

Here is another method. Not sure that it is any more elegant:

# as in the example output, any NAs are treated as do not mess with this
# vector element
my.last.position.NoNA <- ifelse(is.na(my.last.position),
                                nchar(my.data$my.string), my.last.position)

# perform the replacement
paste0(substr(my.data$my.string, 1, my.last.position.NoNA), 
  sapply(1:nrow(my.data), function(i) paste0(rep(".", 
                length=(nchar(my.data$my.string[i])-my.last.position[i])),
                collapse="")))