動態平均計算我有一個JSON陣列等：經由lodash
[{
"x": {
"x1": 1
},
"y": {
"yt": 0,
"zt": 4,
"qa": 3,
"ft": 0,
...
}
},
{
"x": {
"x1": 5
},
"y": {
"yt": 10,
"zt": 2,
"qa": 0,
"ft": 0,
...
}
}]
我想爲每個字段來計算平均值。結果結構應該相同。像：
{
"x": {
"x1": 3
},
"y": {
"yt": 5,
"zt": 3,
"qa": 1.5,
"ft": 0,
...
}
}
感謝
您可以合併使用傳播語法和lodash的_.mergeWith()的對象。 合併時，如果第二個參數（b）是一個數字，則將其除以原始數組中的項目數，以獲得它對總平均值的各自貢獻。如果第一個參數（a）是一個數字，只需加上它就可以不分割（避免多次分割），如果未定義則加0。
我已經添加了2個對象數組和3個對象數組的示例。
const getAvg = (data) => _.mergeWith({}, ...data, (a, b) => {
if(_.isNumber(b)) {
return ((b || 0)/data.length) + (_.isNumber(a) ? (a || 0) : 0);
}
});
const data1 = [
{"x":{"x1":1},"y":{"yt":0,"zt":4,"qa":3,"ft":0}},
{"x":{"x1":5},"y":{"yt":10,"zt":2,"qa":0,"ft":0}}
];
const data2 = [
{"x":{"x1":1},"y":{"yt":0,"zt":4,"qa":3,"ft":0}},
{"x":{"x1":5},"y":{"yt":10,"zt":2,"qa":0,"ft":0}},
{"x":{"x1":3},"y":{"yt":2,"zt":6,"qa":3,"ft":0}}
];
const result1 = getAvg(data1);
console.log('2 objects in the array: ', result1);
const result2 = getAvg(data2);
console.log('3 objects in the array: ', result2);<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>你可以先收集，並在相同的數據結構總結所有的值，然後用給定數組的長度calculkate由一個部門的平均值。
function getParts(array, result) {
function iter(o, r) {
Object.keys(o).forEach(function (k) {
if (o[k] && typeof o[k] === 'object') {
return iter(o[k], r[k] = r[k] || {});
}
r[k] = (r[k] || 0) + o[k];
});
}
function avr(o) {
Object.keys(o).forEach(function (k) {
if (o[k] && typeof o[k] === 'object') {
return avr(o[k]);
}
o[k] = o[k] /data.length;
});
}
data.forEach(function (a) {
iter(a, result);
});
avr(result);
}
var data = [{ x: { x1: 1 }, y: { yt: 0, zt: 4, qa: 3, ft: 0, } }, { x: { x1: 5 }, y: { yt: 10, zt: 2, qa: 0, ft: 0, } }],
result = {};
getParts(data, result);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }let objectArray = [{
"x": {
"x1": 1
},
"y": {
"yt": 0,
"zt": 4,
"qa": 3,
"ft": 0,
}
},
{
"x": {
"x1": 5
},
"y": {
"yt": 10,
"zt": 2,
"qa": 0,
"ft": 0,
}
}];
function findAverage(array) {
let counter = {},
result = {},
i,
obj,
key,
subKey;
// Iterate through array
for (i = 0; i < array.length; i++) {
obj = array[i];
// Copy each key in array element to counter object
for (key in obj) {
counter[key] = counter[key] || {};
// Increment and keep count of key-values of counter based on values in array element
for (subKey in obj[key]) {
counter[key][subKey] = counter[key][subKey] || {total: 0, numElements: 0};
counter[key][subKey].total += obj[key][subKey];
counter[key][subKey].numElements += 1;
}
}
}
// Go back through counter to find average of all existing subkeys (based on incremented total and the number of elements recorded) and throw it into result object
for (key in counter) {
result[key] = result[key] || {};
for (subKey in counter[key]) {
result[key][subKey] = counter[key][subKey].total/counter[key][subKey].numElements;
}
}
return result;
}
console.log(findAverage(objectArray));設計不是絕對最佳，和複製對象可以遞歸不預先知道它們的結構來完成的，但我想保持的步驟儘可能清楚。
編輯爲允許測試爲片段。不知道你甚至可以這麼做！
var array = [{
"x": {
"x1": 1
},
"y": {
"yt": 0,
"zt": 4,
"qa": 3,
"ft": 0
}
},
{
"x": {
"x1": 5
},
"y": {
"yt": 10,
"zt": 2,
"qa": 0,
"ft": 0
}
}];
function aintob(){
var o = {};
var first = array[0],
second = array[1];
var result = {x:{},y:{}};
var each = function(letter, oa, ob){
var i,
letter = {};
for(i in oa){
letter[i] = (oa[i]+ob[i])/2;
}
return letter;
}
o.x = each("x", first.x, second.x);
o.y = each("y", first.y, second.y);
return o;
}
console.log(aintob());
_ 「有一個問題，如果數據爲0的代碼被太複雜了。」 _？ – guest271314
好吧，也許是我的代碼。沒關係那部分。我正在刪除它。 – serkan