Fast way to convert a list of character vectors to a list of numeric vectors

Question

I want to quickly convert a list of character vectors to a list of numeric vectors. I'm trying to avoid purrr::map() , lapply() and the like. I'm ending up with a list of character vectors from the output of a stringr operation. I'm willing to use Rcpp or R's internal C stuff. It's for the filesstrings package. The C++ standard lib provides stod() defined in <string> but it doesn't behave like as.numeric() , for example it converts "12a" to the number 12, but I like how as.numeric() returns NA for that. Here's how I'm doing it now.

nums_as_chars <- stringr::str_extract_all(c("a1b2", "c3d4e5", "xyz"), "\\d")
nums_as_chars
#> [[1]]
#> [1] "1" "2"
#> 
#> [[2]]
#> [1] "3" "4" "5"
#> 
#> [[3]]
#> character(0)
nums <- purrr::map(nums_as_chars, as.numeric)
nums
#> [[1]]
#> [1] 1 2
#> 
#> [[2]]
#> [1] 3 4 5
#> 
#> [[3]]
#> numeric(0)

Created on 2018-04-04 by the reprex package (v0.2.0).

Answer 1

OK, so I created a more realistic example with 3000 strings to do some profiling, and tried every way I could think of to do this.

character_vector <- rep(c("a1b2", "c3d4e5", "xyz"), 1000)


try_1 <- function(chars){
  extracted_numbers <- stringr::str_extract_all(chars, "\\d")
  lapply(extracted_numbers, as.numeric)
}

try_2 <- function(chars){
  extracted_numbers <- stringr::str_extract_all(chars, "\\d")
  purrr::map(extracted_numbers, as.numeric)
}

try_3 <- function(chars){
  extracted_numbers <- stringr::str_extract_all(chars, "\\d")
  relist(as.numeric(unlist(extracted_numbers)), extracted_numbers)
}

try_4 <- function(chars){
  convert_fun <- function(x){as.numeric(stringr::str_extract_all(x, "\\d")[[1]])}
  lapply(chars, convert_fun)
}

# if you don't need to keep the list ...
try_5 <- function(chars){
  extracted_numbers <- stringr::str_extract_all(chars, "\\d")
  suppressWarnings(as.numeric(unlist(extracted_numbers)))
}


microbenchmark::microbenchmark(try_1(character_vector),
                               try_2(character_vector),
                               try_3(character_vector),
                               try_4(character_vector),
                               try_5(character_vector))
#> Unit: milliseconds
#>                     expr        min         lq       mean     median
#>  try_1(character_vector)   2.701769   2.866486   3.304917   3.005177
#>  try_2(character_vector)   3.936557   4.295735   4.872892   4.391737
#>  try_3(character_vector)  12.441844  13.317455  15.759840  14.250013
#>  try_4(character_vector) 183.180143 187.789907 191.298661 190.073565
#>  try_5(character_vector)   1.846848   1.964761   2.090801   2.026860
#>          uq        max neval
#>    3.275250  10.569255   100
#>    4.726425  17.007687   100
#>   16.995679  49.983457   100
#>  193.012754 215.532544   100
#>    2.105214   4.396379   100

Notice that the units are milliseconds, and for 3000 entries, it took lapply 3 milliseconds to do the list of 3000. That doesn't seem that unreasonable to me.

The purrr::map solution was very close to lapply , and then @roland solution was longer, and then my first thought was a lot worse. If you don't care about the list structure (which I imagine you do), then you can get down to 2 milliseconds.

Answer 2

You don't provide anything suitable for reasonable benchmarks. So, test this for yourself:

relist(as.numeric(unlist(nums_as_chars)),
       nums_as_chars)
#[[1]]
#[1] 1 2
#
#[[2]]
#[1] 3 4 5
#
#[[3]]
#numeric(0)

Answer 3

The C++ standard lib provides stod() defined in <string> but it doesn't behave like as.numeric() , for example it converts "12a" to the number 12 , but I like how as.numeric() returns NA for that.

This part is easily solved: Simply use the second parameter of the function to verify that the input was consumed:

Rcpp::NumericVector as_numeric(std::string const& str) {
    std::size_t pos;
    double value = std::stod(&str[0], &pos);
    return NumericVector::create(pos == str.size() ? value : NA_REAL);
}

〉 as_numeric('12')
[1] 12

〉 as_numeric('12a')
[1] NA

… obviously this should be vectorised for performance.

Answer 4

Inspired by @Konrad's answer, I coded up the following using Rcpp .

NumericVector char_to_num(CharacterVector x) {
  std::size_t n = x.size();
  if (n == 0) return NumericVector(0);
  NumericVector out(n);
  for (std::size_t i = 0; i != n; ++i) {
    std::string x_i(x[i]);
    double number = NA_REAL;
    try {
      std::size_t pos;
      number = std::stod(x_i, &pos);
      number = ((pos == x_i.size()) ? number : NA_REAL);
    } catch (const std::invalid_argument& e) {
      ;  // do nothing
    }
    out[i] = number;
  }
  return out;
}

// [[Rcpp::export]]
List lst_char_to_num(List x) {
  std::size_t n = x.size();
  List out(n);
  for (std::size_t i = 0; i != n; ++i)
    out[i] = char_to_num(x[i]);
  return out;
}

This lst_char_to_num() turns out to be the best answer. I compare it to my favourite answers so far which are try1 , try2 and try3 from @rmflight. try1 was fastest so far (on a big dataset, which is what I'm worried about). I've taken the stringr operation out of the timings because I want to purely evaluate the speed of the list conversion.

character_vector <- rep(c("a1b2", "c3d4e5", "xyz"), 1000)
extracted_numbers <- stringr::str_extract_all(character_vector, "\\d")

try_1 <- function(char_list) {
  lapply(char_list, as.numeric)
}

try_2 <- function(char_list) {
  purrr::map(char_list, as.numeric)
}

try_3 <- function(char_list) {
  relist(as.numeric(unlist(char_list)), char_list)
}

microbenchmark::microbenchmark(try_1(extracted_numbers),
                               try_2(extracted_numbers),
                               try_3(extracted_numbers),
                               lst_char_to_num(extracted_numbers),
                               times = 1000)

Unit: microseconds
                              expr       min         lq       mean     median        uq        max neval  cld
          try_1(extracted_numbers)  1068.823  1334.9060  1518.7589  1477.7825  1559.791   5318.318  1000  b  
          try_2(extracted_numbers)  2029.832  2581.6655  2974.4126  2856.8560  3057.930   9846.862  1000   c 
          try_3(extracted_numbers) 10015.929 12261.6405 14043.5922 13188.8465 14802.795 165217.152  1000    d
lst_char_to_num(extracted_numbers)   500.858   681.5895   827.5021   765.9505   830.311   6744.985  1000 a   

Answer 5

Here is a base solution, faster for the small example, but slower for @rmflight's extended benchmark :

mm <- function(chars){
  lapply(strsplit(chars,"[a-z]"),function(x) as.numeric(x[x!=""]))
}

character_vector <- c("a1b2", "c3d4e5", "xyz")

mm(character_vector)
# [[1]]
# [1] 1 2
# 
# [[2]]
# [1] 3 4 5
# 
# [[3]]
# numeric(0)

microbenchmark::microbenchmark(try_1(character_vector),
                               try_2(character_vector),
                               try_3(character_vector),
                               try_4(character_vector),
                               try_5(character_vector),
                               mm(character_vector))

# Unit: microseconds
#                    expr     min       lq      mean   median       uq      max neval
# try_1(character_vector)  74.007  83.0690  96.18563  93.2630 102.1365  158.962   100
# try_2(character_vector) 375.694 409.4875 455.99043 444.7915 489.5345  853.335   100
# try_3(character_vector)  97.416 110.6315 128.83893 128.5670 141.9715  218.620   100
# try_4(character_vector) 211.823 229.7590 251.82474 244.6740 257.5115  412.696   100
# try_5(character_vector)  71.363  94.0180 121.96305 113.4635 141.4050  255.245   100
#    mm(character_vector)  14.726  23.0330  62.70177  26.4315  29.2630 3652.345   100

character_vector <- rep(c("a1b2", "c3d4e5", "xyz"), 1000)

microbenchmark::microbenchmark(try_1(character_vector),
                               try_2(character_vector),
                               try_3(character_vector),
                               try_4(character_vector),
                               try_5(character_vector),
                               mm(character_vector))

# Unit: microseconds
#                    expr        min         lq       mean      median         uq        max neval
# try_1(character_vector)   3286.091   3466.764   4080.628   3579.2825   3836.604  30358.295   100
# try_2(character_vector)   5013.525   5290.859   6081.001   5526.0920   5878.942  32019.653   100
# try_3(character_vector)  18456.932  19466.395  22987.824  20040.6965  21318.998  65432.958   100
# try_4(character_vector) 215098.270 224078.287 245975.607 237691.9815 259353.446 356781.135   100
# try_5(character_vector)    906.951   1044.013   1106.771   1063.8365   1105.559   1649.276   100
#    mm(character_vector)   7550.495   7872.383   9204.357   8215.6040   8733.268  36034.853   100

FYI another solution, but slower:

mm2 <- function(chars){
  lapply(gsub("[a-z]"," ",vec),function(x) scan(text=x,what=double(),quiet=TRUE))
}

mm2(character_vector)
# [[1]]
# [1] 1 2
# 
# [[2]]
# [1] 3 4 5
# 
# [[3]]
# numeric(0)