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Map and filter in one method?

The following code processes a list of file paths and should return only the file names (without extension) of XML files. Currently I got to this:

const filteredFiles = files
  .map(f => f.match(/.*\/(.*)\.xml/)) // map to regex match with capture
  .filter(v => v)                     // non-matches returned null and will be filtered out here
  .map(m => m[1])                     // map out the regex capture

I find this code quite cumbersome. Is there no way to combine the matching and filtering in a more "efficient" way? And by "efficient" I mean code-readable-efficient and not time-efficient as the input array holds 100 values at most but most of the time between 10 and 20.

about 4 years ago · Juan Pablo Isaza

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Answer 1 · 2022-07-07T01:10:58.761Z

This doesn't solve your need of mapping and filtering out the non matching values in one shot... but it makes one step easier by using the optional chaining operator ?.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Optional_chaining

I also slightly changed the regex to allow the filename with no path specifed.

const files = [
  './path/to/filename_01.xml',
  'non_matching_value',
  './path/to/filename_02.xml',
  './path/to/filename_03.xml',
  './path/to/filename_04.xml',
  'filename_05.xml',
];

const filteredFiles = files
  .map(filename => filename.match(/^(.*\/)?(.*)\.xml/)?.[2])
  .filter(filename => filename);
  
console.log(filteredFiles);

Answer 2 · 2022-07-07T01:10:58.770Z

You can (ab)use flat map:

const filteredFiles = files.flatMap((f)=>{
  let match = f.match('...');
  if (match) {
      return [match[1]]
  } else {
      return []
  }
})

Not sure if it's actually better than the original though.

Answer 3 · 2022-07-07T01:10:58.777Z

Map and filter, otherwise known as reduce

const rx = /\/(.*)\.xml$/;

const filteredFiles = files.reduce((arr, f) => {
  const match = f.match(rx);
  return match ? [...arr, match[1]] : arr;
}, []);

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Map and filter in one method?

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Andres GPT