从数组中获取所有元素均为真的子数组的所有索引

Question

This is an example of an array I can have:

let testArray = [
  [true, true, false],
  [false, true, false],
  [true, true, true],
  [false, true, false],
  [true, true, true]
]

How can I check if all the values in a sub-array are true and then get all indexes of sub-arrays which pass the check?

Answer 1

Use map to create objects that hold the index and a boolean whether all sub-array elements are true (with every and Boolean as the callback), filter to select only those where all elements are true , then another map to get only the indexes.

 let testArray = [ [true, true, false], [false, true, false], [true, true, true], [false, true, false], [true, true, true] ]; console.log(testArray .map((subArray, index) => ({ index, allTrue: subArray.every(Boolean) })) .filter((entry) => entry.allTrue) .map((entry) => entry.index)); 

Answer 2

You could reduce the outer array by checking the inner array and take the index or nothing.

 var array = [[true, true, false], [false, true, false], [true, true, true], [false, true, false]], allTrue = array.reduce((r, a, i) => r.concat(a.every(Boolean) ? i : []), []); console.log(allTrue); 

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