JavaScript how to filter two arrays at the same time - Stack Overflow

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Say I have two arrays:

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

If i were to filter the first array so that it returns this result:

 var resultOne = one.filter(v => v == 2 || v == 3 );

The result would return [2,3] but how would I filter the other array to return [B,C]... based on the result of this first one ?

Say I have two arrays:

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

If i were to filter the first array so that it returns this result:

 var resultOne = one.filter(v => v == 2 || v == 3 );

The result would return [2,3] but how would I filter the other array to return [B,C]... based on the result of this first one ?

• javascript

Share Improve this question Follow edited Jul 23, 2019 at 14:53 Fraction 13k55 gold badges3232 silver badges5252 bronze badges asked Jul 23, 2019 at 14:43 ShaneOG97ShaneOG97 53022 gold badges1111 silver badges2727 bronze badges 5

• 1 Is it depending on Index? – Christheoreo Commented Jul 23, 2019 at 14:45
• Yes that is exactly what im looking for here – ShaneOG97 Commented Jul 23, 2019 at 14:46
• Will the two arrays always be the same length? – DBS Commented Jul 23, 2019 at 14:46
• Yes they will be the same – ShaneOG97 Commented Jul 23, 2019 at 14:47
• What about just using two.splice(1,2)? – leonheess Commented Jul 23, 2019 at 14:47

Add a ment  | 

6 Answers 6

Sorted by: Reset to default 4

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

var resultOne = two.filter((v,i) => one[i] == 2 || one[i] == 3 );

console.log(resultOne)

You can implement filter on the array by passing the index, return if index + 1 includes in the resulted array.

Please Note: Since the solution is based on the index, it will not work for the random numbers of array (not sequential).

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

var resultOne = one.filter(v => v == 2 || v == 3 );
var resultTwo = two.filter((v,i) => resultOne.includes(i+1));
console.log(resultOne);
console.log(resultTwo);

OR: In Single line:

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

var result = two.filter((v,i) => one.filter(v => v == 2 || v == 3).includes(i+1));
console.log(result);

You can use Array.reduce():

var one = [1,2,3,4,5];

var two = ["A","B","C","D","E"];

const [resultOne, resultTwo] = one.reduce((acc, v, i) => (v == 2 || v == 3) ? (acc[0].push(v), acc[1].push(two[i]), acc) : acc, [[], []]); 
 
 console.log(resultOne);
 console.log(resultTwo);

Or use this version if the two arrays haven't the same length:

var one = [1,2,3,4,5,6];

var two = ["A","B","C","D","E"];

const [resultOne, resultTwo] = one.reduce((acc, v, i) => (v == 2 || v == 6) ? (acc[0].push(v), (i in two) ? acc[1].push(two[i]) : '', acc) : acc, [[], []]); 
 
 console.log(resultOne);
 console.log(resultTwo);

A good old for loop or use the second parameter of the filter callback which gives you the current array index. Simply push the corresponding element of two onto resultTwo by index.

var resultTwo = [];
var resultOne = one.filter((v, i) => { 
  var isMatch = v == 2 || v == 3;
  if (isMatch) resultTwo.push(two[i]);
  return isMatch;
});

EDIT: Just realized somone posted the same answer... The only thing I can add is that you need to subtract 1 from your original array to arrive at the correct index.

If you use map you can map the results from your filtering. This is assuming the result will always be an index for your second collection.

https://developer.mozilla/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map

var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
var resultOne = one.filter(v => v == 2 || v == 3 );
var resultTwo = resultOne.map((indexFromResultOne) => two[indexFromResultOne - 1]); 
// Because you aren't using a zero based index.
console.log(resultTwo);

Not the most efficient, but this is the most general and readable way I can think of to filter multiple corresponding arrays:

var a = [1, 2, 3, 4, 5];
var b = ["A", "B", "C", "D", "E"];

var indices = [];

for (var i = 0; i < a.length; i++) {
  var v = a[i];
  
  if (v == 2 || v == 3) {
    indices.push(i);
  }
}

console.log(indices.map(i => a[i]));
console.log(indices.map(i => b[i]));

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