多种数组去重的方法

原创已于 2025-08-11 14:22:57 修改 · 150 阅读

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#java #前端 #javascript

于 2025-08-11 14:21:52 首次发布

1、第三方库lodash方法

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = _.uniq(arryData); // [1, 3, 4, 6, 9, 2]

2、new Set 方法

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = [...new Set(arryData)]; // 或者Array.from(new Set(arrayData))

3、includes方法

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = [];
arryData.forEach(e => {
    if(!result.includes(e)){
        result.push(e);
    }
});

4、indexOf方法

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = [];
arryData.forEach(e => {
	if (result.indexOf(e) === -1) {
        result.push(e);
    }
});

5、filter 过滤出实际的索引值与找到元素的索引值相同的值

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = arryData.filter((e, index) => arryData.indexOf(e) === index);

6、reduce 方法

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const result = arryData.reduce((pre, item) => {
	return pre.includes(item) ? pre : [...pre, item];
}, []);

7、使用对象key具有唯一性的特点

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
const obj = {};
arryData.forEach(e => (obj[e] = 11));// 给value 赋予任意的值
const result = Object.keys(obj).map(e => +e);

8、sort 排序后对比前后两值是否相同，不同则push

const arryData = [1, 3, 1, 4, 6, 3, 9, 2];
arryData.sort((a, b) => a - b);
const result = [];
for (let i = 0; i < arryData.length; i++) {
	if (arryData[i] !== arryData[i + 1]) {
		result.push(arryData[i]);
	}
}