How to remove all characters from a string, except the integers?

Ali Çehreli acehreli at
Fri Mar 4 10:34:29 UTC 2022

On 3/3/22 04:14, BoQsc wrote:

 > and if it contains integers, remove all the regular string characters.

Others assumed you wanted integer values but I think you want the digits 
of the integers. It took me a while to realize that chunkBy can do that:

// Convenience functions to tuple members of the result
// of chunkBy when used with a unary predicate.
auto isMatched(T)(T tuple) {
   return tuple[0];

// Ditto
auto chunkOf(T)(T tuple) {
   return tuple[1];

auto numbers(R)(R range) {
   import std.algorithm : chunkBy, filter, map;
   import std.uni : isNumber;

   return range

unittest {
   import std.algorithm : equal, map;
   import std.conv : text;

   // "٤٢" is a non-ASCII number example.
   auto r = "123 ab ٤٢ c 456 xyz 789".numbers;
   assert(!text.equal(["123", "٤٢", "456", "789"]));

void main() {

isMatched() and chunkOf() are not necessary at all. I wanted to use 
readable names to fields of the elements of chunkBy instead of the 
cryptic t[0] and t[1]:

   return range
          .filter!(t => t[0])   // Not pretty
          .map!(t => t[1]);     // Not pretty

Those functions could not be nested functions because otherwise I would 
have to write e.g.

   return range
          .filter!(t => isMatched(t))   // Not pretty
          .map!(t => chunkOf(t));       // Not pretty

To get integer values, .to!int would work as long as the numbers consist 
of ASCII digits. (I am removing ٤٢.)

   import std.stdio;
   import std.algorithm;
   import std.conv;
   writeln("123 abc 456 xyz 789"!(to!int));


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