查找两个数组是否在数组中重复,然后选择它们 [英] Find if two arrays are repeated in array and then select them
问题描述
我在主/父数组中有多个数组,如下所示:
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
为了更简单的阅读,这里是数组:
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
我想选择重复 3 次或更多次 (> 3) 的数组并将其分配给一个变量.所以在这个例子中,var repeatArrays
将是 [1, 17]
和 [2, 12]
.
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays
would be [1, 17]
and [2, 12]
.
所以这应该是最终的结果:
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
我在这里找到了类似的东西,但它使用了 underscore.js 和 lodash.
I found something similar here but it uses underscore.js and lodash.
如何使用 javascript 甚至 jquery(如果需要)?
How could I it with javascript or even jquery (if need be)?
推荐答案
您可以参考 Map
,然后按计数过滤并恢复数组.
You could take a Map
with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
使用所需数量的地图进行过滤.
Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
独一无二!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
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