我有一个数组:
var arr = [
{class: 'class-1', team: 'abcdef'},
{class: 'class-21', team: 'xmsqmd'},
{class: 'class-4', team: 'xaksjs'},
{class: 'class-21', team: 'xmsqmd'},
{class: 'class-5', team: 'asioda'},
{class: 'class-44', team: 'csdcsw'}
];如您所见,这个对象:
{class: 'class-21', team: 'xmsqmd'}被重复了2次。无论如何,这只是一个例子,但我需要创建另一个数组,它应该如下所示:
newArr = [
[{class: 'class-1', team: 'abcdef'}, 1],
[{class: 'class-21', team: 'xmsqmd'}, 2],
[{class: 'class-4', team: 'xaksjs'}, 1],
[{class: 'class-5', team: 'asioda'}, 1],
[{class: 'class-44', team: 'csdcsw'}, 1]
]因此,新的数组应该是一个2d数组,第一个值是对象,第二个值是重复的时间。
预先谢谢,我已经搜索过了,但我只找到了一个解决方案,结果是一个数组,其中一个对象的键是已压缩的对象。
发布于 2017-04-27 22:21:12
可能的解决办法。
Array#forEach创建一个哈希对象,每个元素都作为key,它在arr数组中的表象数是value。Array#map映射哈希对象,以从原始数组中获取指定的对象及其从哈希对象中出现的次数。
var arr = [{
class: 'class-1',
team: 'abcdef'
}, {
class: 'class-21',
team: 'xmsqmd'
}, {
class: 'class-4',
team: 'xaksjs'
}, {
class: 'class-21',
team: 'xmsqmd'
}, {
class: 'class-5',
team: 'asioda'
}, {
class: 'class-44',
team: 'csdcsw'
}],
hash = arr.map(v => v.class),
obj = {};
hash.forEach(v => !obj[v] ? obj[v] = 1 : obj[v]++);
var res = Object.keys(obj).map(v => [Object.assign({}, arr.find(c => c.class == v)), obj[v]])
.sort((a,b) => b[1] - a[1]);
document.write('<pre>' + JSON.stringify(res, null, 2) + '</pre>');
发布于 2017-04-27 22:58:51
我们可以在两个循环中简化解决方案:
var arr = [
{class: 'class-1', team: 'abcdef'},
{class: 'class-21', team: 'xmsqmd'},
{class: 'class-4', team: 'xaksjs'},
{class: 'class-21', team: 'xmsqmd'},
{class: 'class-5', team: 'asioda'},
{class: 'class-44', team: 'csdcsw'}
];
var res = [];
var obj = {};
//construct the object with key and recurrence count as value
arr.forEach(function(value, i) {
var str = JSON.stringify(value);
if(!obj[str]) {
obj[str] = 1;
} else {
obj[str]++;
}
});
//Format the data in the desired output
for(var i in obj) {
var tempArr = [JSON.parse(i)];
tempArr.push(obj[i]);
res.push(tempArr);
}
console.log(res);
发布于 2017-04-27 22:20:29
您可以使用forEach
newArr = {}
arr.forEach(function(obj) {
var key = JSON.stringify(obj)
newArr[key] = (newArr[key] || 0) + 1
})
https://stackoverflow.com/questions/43668713
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