我有一系列具有相同属性的对象。如何将具有相同period属性的对象合并到新的对象数组中。我在stackoverflow中看到了这些示例，但它们主要使用外部库，如JQuery或lodash。我可以在vanilla.js中制作这个吗？
Here 是Plunkr，带有代码和预期值的标记。谢谢你的帮助
答案 0 :(得分：0)
您可以使用forEach()循环并添加到新阵列。
var a = [{"period":"01.07.2016","count":11,"scale":"d"},{"period":"01.07.2016","count":22,"scale":"n"},{"period":"01.08.2016","count":33,"scale":"n"},{"period":"01.08.2016","count":44,"scale":"d"},{"period":"01.09.2016","count":55,"scale":"d"},{"period":"01.09.2016","count":66,"scale":"n"}]
var result = []
a.forEach(function(e) {
if (!this[e.period]) {
this[e.period] = {
date: e.period.split('.')[1],
day: 0,
night: 0
}
result.push(this[e.period])
}
if (e.scale == 'd') {
this[e.period].day = (this[e.period].day || 0) + e.count
} else if (e.scale == 'n') {
this[e.period].night = (this[e.period].night || 0) + e.count
}
}, {})
console.log(result)
答案 1 :(得分：0)
您可以将其与日期分组，并使用对象作为哈希表的引用。另一个对象用于访问day / nicht属性。
var array = [{ period: '01.07.2016', count: 11, scale: 'd' }, { period: '01.07.2016', count: 22, scale: 'n' }, { period: '01.08.2016', count: 33, scale: 'n' }, { period: '01.08.2016', count: 44, scale: 'd' }, { period: '01.09.2016', count: 55, scale: 'd' }, { period: '01.09.2016', count: 66, scale: 'n' }],
grouped = [];
array.forEach(function (a) {
var day = a.period.slice(3, 5);
if (!this[day]) {
this[day] = { date: day, day: 0, night: 0 };
grouped.push(this[day]);
}
this[day][{ d: 'day', n: 'night' }[a.scale]] += a.count;
}, Object.create(null));
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }
您也可以在哈希表上使用闭包，而不使用thisArg。
var array = [{ period: '01.07.2016', count: 11, scale: 'd' }, { period: '01.07.2016', count: 22, scale: 'n' }, { period: '01.08.2016', count: 33, scale: 'n' }, { period: '01.08.2016', count: 44, scale: 'd' }, { period: '01.09.2016', count: 55, scale: 'd' }, { period: '01.09.2016', count: 66, scale: 'n' }],
grouped = [];
array.forEach(function (hash) {
return function (a) {
var day = a.period.slice(3, 5);
if (!hash[day]) {
hash[day] = { date: day, day: 0, night: 0 };
grouped.push(hash[day]);
}
hash[day][{ d: 'day', n: 'night' }[a.scale]] += a.count;
};
}(Object.create(null)));
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }
在不滥用thisArg和IFFE
var array = [{ period: '01.07.2016', count: 11, scale: 'd' }, { period: '01.07.2016', count: 22, scale: 'n' }, { period: '01.08.2016', count: 33, scale: 'n' }, { period: '01.08.2016', count: 44, scale: 'd' }, { period: '01.09.2016', count: 55, scale: 'd' }, { period: '01.09.2016', count: 66, scale: 'n' }],
grouped = function () {
var result = [],
hash = Object.create(null);
array.forEach(function (a) {
var day = a.period.slice(3, 5);
if (!hash[day]) {
hash[day] = { date: day, day: 0, night: 0 };
result.push(hash[day]);
}
hash[day][{ d: 'day', n: 'night' }[a.scale]] += a.count;
});
return result;
}();
console.log(grouped);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 2 :(得分：0)
您可以使用 forEach 循环遍历数组，然后使用 some 来检查是否已存在具有相同键的对象来合并对象。
P.S：您预期的输出有一段错误：“01.08.2016”。它已经改变了白天和黑夜的数量。
var a = [{
'period': "01.07.2016",
'count': 11,
'scale': 'd'
}, {
'period': "01.07.2016",
'count': 22,
'scale': 'n'
}, {
'period': "01.08.2016",
'count': 33,
'scale': 'n'
}, {
'period': "01.08.2016",
'count': 44,
'scale': 'd'
}, {
'period': "01.09.2016",
'count': 55,
'scale': 'd'
}, {
'period': "01.09.2016",
'count': 66,
'scale': 'n'
}]
function createMatrix(list) {
var arr = [];
list.forEach((item) => {
var key = item["period"];
var result = arr.some(function(currentValue, index, arr) {
if (arr[index].date == key.split(".")[1]) {
var dayNight = (item.scale == "d" ? "day" : "night");
arr[index][dayNight] = item.count;
return true;
}
});
if (result === false) {
var dayNight = (item.scale == "d" ? "day" : "night");
var obj = {};
obj[dayNight] = item.count;
obj["date"] = key.split(".")[1];
arr.push(obj);
}
});
return arr;
}
console.log(createMatrix(a));
答案 3 :(得分：0)
首先，我们可以通过“scale”对原始数据列表进行排序，然后在一个reduce内获取收集的目标/期望列表...例如.e.g。 ...
var rawData = [{
'period': '01.07.2016',
'count': 11,
'scale': 'd'
}, {
'period': '01.07.2016',
'count': 22,
'scale': 'n'
}, {
'period': '01.08.2016',
'count': 33,
'scale': 'n'
}, {
'period': '01.08.2016',
'count': 44,
'scale': 'd'
}, {
'period': '01.09.2016',
'count': 55,
'scale': 'd'
}, {
'period': '01.09.2016',
'count': 66,
'scale': 'n'
}];
// expecting
/*
result = [{
date : '07',
day : 11,
night : 22
}, {
date : '08',
day : 44,
night : 33
}, {
date : '09',
day : 55,
night : 66
}]
*/
function sortByDay(a, b) {
return ((a.scale < b.scale) ? -1 : ((a.scale > b.scale) ? 1 : 0));
}
function collectSortedDayAndNightTotals(collector, item/*, idx, list*/) {
var
dateRegistry = collector.dateRegistry,
dateKey = item.period,
registeredItem = dateRegistry[dateKey];
if (registeredItem == null) {
dateRegistry[dateKey] = item;
} else {
collector.totalList.push({
date : dateKey.split('.')[1],
day : registeredItem.count,
night : item.count
});
}
return collector;
}
var result = rawData.sort(sortByDay).reduce(collectSortedDayAndNightTotals, {
dateRegistry: {},
totalList : []
}).totalList;
console.log('result : ', result);
答案 4 :(得分：0)
我用reduce
var arr = [{
'period': '01.07.2016',
'count': 11,
'scale': 'd'
},
{
'period': '01.07.2016',
'count': 22,
'scale': 'n'
},
{
'period': '01.07.2016',
'count': 22,
'scale': 'n'
},
{
'period': '01.08.2016',
'count': 11,
'scale': 'd'
},
{
'period': '01.08.2016',
'count': 22,
'scale': 'n'
},
{
'period': '01.08.2016',
'count': 22,
'scale': 'n'
},
];
function functionName(a) {
var b = [];
var c = 0;
var d = a.reduce(function(first, second, i){
if(first.period == second.period){
b[c] = b[c] || [{
day: '',
night: '',
date: '',
}];
if (first.scale == 'd') {
b[c].day = b[c].day + first.count || first.count;
}else {
b[c].night = b[c].day + first.count || first.count
}
b[c].date = b[c].date || first.period;
}else {
c= c + 1;
}
return second;
});
b[c] = b[c] || [{
day: '',
night: '',
date: '',
}];
if (d.scale == 'd') {
b[c].day = b[c].day + d.count || d.count;
}else {
b[c].night = b[c].day + d.count || d.count;
}
b[c].date = b[c].date || d.period;
return b;
}
console.log(functionName(arr));