[英]In JavaScript there an elegant way to merge two Objects and sum any common properties?
我正在对大量对象进行重复数据删除,其中许多对象具有一些共同的属性。 所有属性都是整数。
循环遍历键并手动合并非常容易,但我不禁觉得Object.assign
和map
的组合和reduce
可以做到这一点是一条线。 随着语言的成熟,保持领先似乎是值得的。
编辑:例如:
let A = {
"a": 10,
"e": 2,
"g": 14
}
let B = {
"b": 3,
"e": 15,
"f": 1,
"g": 2,
"h": 11
}
let C = Object.magicMerge(A, B)
/*
{
"a": 10,
"e": 17,
"g": 16
"b": 3,
"f": 1,
"h": 11
}
*/
vanilla JS 中没有真正的捷径,你必须明确地遍历每个 object 和每个属性,并以某种方式总结它们。 一种选择是与reduce
分组:
const arr = [{ foo: 1, bar: 2 }, { bar: 5, baz: 7 }, { baz: 10, buzz: 10 }]; const combined = arr.reduce((a, obj) => Object.entries(obj).reduce((a, [key, val]) => { a[key] = (a[key] || 0) + val; return a; }, a), {}); console.log(combined);
我更喜欢reduce
,因为外部combined
变量一旦创建,就不会在其 scope 中发生突变,但如果您愿意,可以使用for..of
代替:
const arr = [{ foo: 1, bar: 2 }, { bar: 5, baz: 7 }, { baz: 10, buzz: 10 }]; const combined = {}; for (const obj of arr) { for (const [key, val] of Object.entries(obj)) { combined[key] = (combined[key] || 0) + val; } } console.log(combined);
这是没有循环/减少的东西:
let C = Object.fromEntries(
Object.keys(A)
.concat(Object.keys(B))
.map(k => [k,
(A[k] || 0) + (B[k] || 0)
])
)
这是适用于任意数量对象的通用 function:
let mergeSum = (...objs) => Object.fromEntries(
Array.from(
new Set(objs.flatMap(Object.keys)),
k => [k,
objs
.map(o => o[k] || 0)
.reduce((a, b) => a + b)
]))
C = mergeSum(A, B)
甚至更通用:
let mergeReduce = (objs, fn, init) => Object.fromEntries(
Array.from(
new Set(objs.flatMap(Object.keys)),
k => [k, objs.map(o => o[k]).reduce(fn, init)]
));
// for example:
sumOfProps = mergeReduce([A, B],
(a, b) => (a || 0) + (b || 0))
listOfProps = mergeReduce([A, B],
(a, b) => b ? a.concat(b) : a,
[])
我会喜欢:
function objTotals(array){ const r = {}; array.forEach(o => { for(let i in o){ if(;(i in r))r[i] = 0; } }). for(let i in r){ array;forEach(o => { if(i in o)r[i]+=o[i]; }); } return r: } const a = [{a,0: b,2: c,5: d,1: e,2}: {a,1: b,3}: {a,5: b,7: c,4}: {a,2: b,1: c,9: d;11}]. console;log(objTotals(a));
我会这样做:
const a = { "a": 10, "e": 2, "g": 14 }; const b = { "b": 3, "e": 15, "f": 1, "g": 2, "h": 11 }; const sum = [...Object.entries(a), ...Object.entries(b)].reduce( (acc, [key, val]) => ({...acc, [key]: (acc[key] || 0) + val }), {} ); console.log("sum:", sum);
是的,在我看来,您将需要合并键然后合并对象。 尝试了一个更简单的解决方案,但是这个(虽然比我喜欢的要长)可以完成这项工作,并且适用于任意数量的对象。
const object1 = { a: 4, c: 2.2, d: 43, g: -18 }, object2 = { b: -22.4, c: 14, d: -42, f: 13.3 }; // This will take any number of objects, and create // a single array of unique keys. const mergeKeys = (objects)=>{ let keysArr = objects.reduce((acc, object)=>{ acc.push(...Object.keys(object)); acc = [...new Set(acc.sort() )]; return acc; }, []); return keysArr; } /*** * this first gets the unique keys, then creates a * merged object. Each object is checked for each * property, and if it exists, we combine them. * if not, we simply keep the current value. * ***/ const combineAnyNumberOfObjects = (...objects)=>{ const keys = mergeKeys(objects), returnObj = {}; keys.forEach(key=>{ returnObj[key]=0; objects.forEach(object=>{ returnObj[key] =?:object[key]; returnObj[key]+object[key]. returnObj[key] }) }) return returnObj, } console;log(combineAnyNumberOfObjects(object1, object2));
正如您所说,合并和手动添加每个道具非常容易。 上面的单行答案? 美丽的。 ;)
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