CODEWITHSUNDEEP
rstringstringr
M. Schumacher
M. Schumacher

Reputation: 150

zeros disappear using str_c()

I got a problem which I just don't understand.

I would like to put together strings, depending on the number of characters:

the vector c("1", "23", "302", "2004") should be transformed into c("1900001", "1900023", "1900302", "1902004").

Therefore I used this code:

test <- c("1", "23", "302", "2004")
fun_function <- function(test){

  if(nchar(test == 4)){
    str_c("190",test, sep="")
  } else {
    if(nchar(test == 3)){
      str_c("1900", test, sep="")
    } else{
      if(nchar(test == 2)){
        str_c("19000", test, sep="")
      } else{
        if(nchar(test == 1)){
          str_c("190000", test, sep="")
        }
      }
    }
  }
}

res <- map(test, 
    ~fun_function(.x)) %>% 
  unlist()

However, instead of getting c("1900001", "1900023", "1900302", "1902004") I get c("1901", "19023", "190302", "1902004").

Now, why I get that vector? If I'm using this function for a single case, it works:

nchar("1") == 1
str_c("190000", "1", sep="")

So, where is the problem? I just don't understand the reason for my outcome. Can someone help me with this? I also tried a For Loop using the same logic, but that does not work either.

Upvotes: 0

Views: 106

Answers (1)

astrofunkswag
astrofunkswag

Reputation: 2718

You're condition checking in the if statements is wrong. Its nchar(test) == 4, not nchar(test == 4)

But the str_pad function from stringr is a much more concise way to approach the problem. First I pad with zeros to 5 digits, and then paste a '19' onto each.

test2 <- str_pad(test, 5, 'left', pad = '0')
test3 <- paste0('19', test2)
test3
[1] "1900001" "1900023" "1900302" "1902004"

Upvotes: 2

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