使用谓词查找数组中的所有元素/索引 - Typescript

Find All elements / indexes in an array with predicate - Typescript

我想找到列表/数组中某项所有出现的 indexes,最好使用 PREDICATE

我使用 IONIC - ANGULAR 框架,因此在 TYPESCRIPT.

这是我想要的具体示例:

        const myList = [0, 2, 1, 1, 3, 4, 1];
        // what already exists:
        myList.findIndex(x => x === 1); // return 2

        // what I would like it to be:
        myList.findAllIndexes(x => x === 1); // should return [2, 3, 6]

在此先感谢您的帮助。

解决方案 :

    /**
     * Returns the indexes of all elements in the array where predicate is true, [] 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,
     * it is added to indexes and the functions continue..
     */
    findAllIndexes<T>(array: Array<T>, predicate: (value: T, index: number, obj: T[]) => boolean): number[] {
        const indexes = [];
        let l = array.length;
        while (l--) {
            if (predicate(array[l], l, array)) {
                indexes.push(l);
            }
        }
        return indexes;
    }

并使用它:

const myList = [0, 2, 1, 1, 3, 4, 1];
const indexes = this.findAllIndexes(myList, x => x === 1);
// return [6, 3, 2]

其他方法:

有点不同但很有用(允许获取所有元素而不是索引):

const myList = [0, 2, 1, 1, 3, 4, 1];
const allElements = myList.filter(x => x === 1);

PS :我选择从头到尾迭代循环,可以将其反转得到 [2, 3, 6] 而不是 [6, 3, 2]。

祝大家编码愉快!