过滤，直到第一次满足某些条件

Question

It's not a real world example, I over-simplified it. Giving this array:

const a = [1,2,3,4,5,6,7,8,4,5]; // Etc. Random numbers after.

I want to filter it to have only those matching a pattern (let's say greater than 3 for this trivial example) until something appends for the first time (let's say element is greater than 7)

So for this example, I just want: [4,5,6,7] . But with filter , I would have the trailing 4 and 5 :

const a = [1,2,3,4,5,6,7,8,4,5].filter((v) => v > 3)
// returns: [4, 5, 6, 7, 8, 4, 5]

So I want to get item from an array and definitively stop after a condition. How can I filter then stop after the first time a condition is not met? (without for loop, I want to keep it "functional-like")

const a = [1,2,3,4,5,6,7,8,4,5,1,2,976,-1].awsome_function();
// returns: [4, 5, 6, 7, 8] because it stopped after the first 8.

Answer 1

You could use Array#some and combine both conditions.

 var array = [1,2,3,4,5,6,7,8,4,5], result = []; array.some(a => (a > 3 && result.push(a), a > 7)); console.log(result); 

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ES5

 var array = [1,2,3,4,5,6,7,8,4,5], result = []; array.some(function (a) { if (a > 3) { result.push(a); } return a > 7; }); console.log(result); 

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Answer 2

If you want to keep the functional style you can use this :

Array.prototype.filterUntil = function(predicate, stop){

  let shouldStop = false;

  return this.filter(function filter(value, index){
    if(stop(value)){
      shouldStop = true;
    }

    return shouldStop && predicate(value);
  });
}

In your case you can call it like this :

data.filterUntil(value => value > 3, value => value < 7)