How to filter array based on nested array data? [duplicate]

Question

I have an array with a few objects:

const x = [
  {label: 'a', data: [0,0]},
  {label: 'b', data: [0,0]},
  {label: 'c', data: [0,1]}
  {label: 'd', data: [0,1]},
]

I want to filter out the objects where the data values are equal to 0 and end up with:

const x = [
  {label: 'c', data: [0,1]}
  {label: 'd', data: [0,1]},
]

I tried:

const result = x.filter((x) => {
    return x.data.filter((value) => value > 0);
});

expecting the .filter() to only return objects where the value would be higher than 0, but this is not working as expected.

From the documentation: "It should return a truthy to keep the element in the resulting array, and a falsy value otherwise". You are returning the result of .filter(), which is an array. — Ivar
– Ivar, Commented Dec 5, 2022 at 12:31

Nick Parsons · Accepted Answer · 2022-12-05 12:24:15Z

2

Your inner .filter() method always returns an array, which is considered a truthy value. So the outer .filter() keeps all objects. Instead, you can use .some() to return true if any of the values in the data array are > 0 to keep the object in the resulting array:

const x = [ {label: 'a', data: [0,0]}, {label: 'b', data: [0,0]}, {label: 'c', data: [0,1]}, {label: 'd', data: [0,1]}, ];

const result = x.filter((x) => {
  return x.data.some((value) => value > 0);
});
console.log(result);

answered Dec 5, 2022 at 12:24

Nick Parsons

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Atena Dadkhah · Accepted Answer · 2022-12-05 12:41:09Z

0

You can filter based on if some values in data object inside the main array are not 0 and then keep them in the array.

const x = [
  { label: 'a', data: [0, 0] },
  { label: 'b', data: [0, 0] },
  { label: 'c', data: [0, 1] },
  { label: 'd', data: [0, 1] },
]
const res = x.filter(el => el.data.some(y => y!=0))
console.log(res)

answered Dec 5, 2022 at 12:41

Atena Dadkhah

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