reduce的几种用法

[leefinder]

求最大子集

const subcollection = (arr) => arr.reduce((t, v) => t.concat(t.map(item => item.concat(v))), [[]])
subcollection([1, 2, 3]) // [] [1] [2] [1, 2] [3] [1, 3] [2, 3] [1, 2, 3]

累加

const Accumulation = (...nums) => nums.reduce((t, v) => t + v, 0)

Accumulation(1, 2, 3, 4, 5); // 15

累乘

const Multiplication = (...nums) => nums.reduce((t, v) => t * v, 1)

Multiplication(1, 2, 3, 4, 5); // 120

实现数组reverse

const Reverse = (arr = []) => arr.reduceRight((t, v) => (t.push(v), t), [])

Reverse([1, 2, 3, 4, 5]); // [5, 4, 3, 2, 1]

reduce实现map

[1, 2, 3, 4].map(t => t * 2)
[1, 2, 3, 4].reduce((t, v) => [...t, t * 2], [])

##reduce实现filter

[1, 2, 3, 4].filter(t => t > 2)
[1, 2, 3, 4].reduce((t, v) => t > 2 ? [...t, v] : t, [])

reduce实现some

[1, 2, 3, 4].some(t => t > 2)
[1, 2, 3, 4].reduce((t, v) => t || v > 2, false)

reduce实现every

[1, 2, 3, 4].every(t => t > 2)
[1, 2, 3, 4].reduce((t, v) => t && v > 2, true)

数组分割

const Chunk = (arr, size) => arr.length > 0 ? arr.reduce((t, v) => (t[t.length - 1].length === size ? t.push([v]) : t[t.length - 1].push(v), t), [[]]) : []
Chunk([1, 2, 3, 4, 5, 6, 7, 8, 9], 2) // [[1, 2], [3, 4], [5, 6], [7, 8], [9]]

数组过滤

const Difference = (arr1, arr2) => arr1.reduce((t, v) => (!arr2.includes(v) && t.push(v), t), [])
Difference([1, 2, 3, 4], [3, 4, 5]) // [1, 2] 

数组填充

const Fill = (arr, start, end, text) => {
    if (start < 0 || start >= end || end > arr.length) return arr;
    return [
        ...arr.slice(0, start),
        ...arr.slice(start, end).reduce((t, v) => (t.push(text || v), t), []),
        ...arr.slice(end, arr.length)
    ]
}  
const arr = [0, 1, 2, 3, 4, 5, 6];
Fill(arr, 'aaa', 2, 5); // [0, 1, 'aaa', 'aaa', 'aaa', 5, 6]

数组拍平

const Flat = arr => arr.reduce((t, v) => t.concat(Array.isArray(v) ? Flat(v) : v), [])
Flat([[1,2,3], [4,5], [5,6, [7, 8]]]) // [1, 2, 3, 4, 5, 6, 7, 8]

数组去重

const Dupli = arr => arr.reduce((t, v) => (!t.includes(v) && t.push(v), t), [])
Dupli([1,2,1,2,3,4,5]) // [1, 2, 3, 4, 5]

参考链接