按特定条件查找数组内元素的最后一个索引 [英] Find last index of element inside array by certain condition

问题描述

假设我有一个对象数组:

Let's say I have an array of objects:

[
  {'a': 'something', 'b':12},
  {'a': 'something', 'b':12},
  {'a': 'somethingElse', 'b':12},
  {'a': 'something', 'b':12},
  {'a': 'somethingElse', 'b':12}
]

获取元素的最后一个索引的最简洁的方法是什么，其中 a 的值为 'something'.在这种情况下 3. 有没有办法避免循环...

What would be the most cleanest way to get the last index of the element where a has the value 'something'. In this case 3. Is there any way to avoid loops...

推荐答案

这是一个可重用的打字稿版本，它反映了 ES2015 findIndex 函数的签名:

Here's a reusable typescript version which mirrors the signature of the ES2015 findIndex function:

/**
* Returns the index of the last element in the array where predicate is true, and -1
* otherwise.
* @param array The source array to search in
* @param predicate find calls predicate once for each element of the array, in descending
* order, until it finds one where predicate returns true. If such an element is found,
* findLastIndex immediately returns that element index. Otherwise, findLastIndex returns -1.
*/
export function findLastIndex<T>(array: Array<T>, predicate: (value: T, index: number, obj: T[]) => boolean): number {
    let l = array.length;
    while (l--) {
        if (predicate(array[l], l, array))
            return l;
    }
    return -1;
}

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