提问人：Chris 提问时间：10/23/2023 最后编辑：Chris 更新时间：10/23/2023 访问量：123
一个属性的 JavaScript 数组总数
Total JavaScript Array by one Property
问：
我看了很多例子，但无法让它起作用。充其量，我最终会得到一个现有数组的版本，该版本复制了应该在摘要中丢失的属性。
我想为每个位置输出一个新的总计数组，并且不包括特定期间、期间属性等的任何详细信息。 我认为我的问题是 Object.assign（{}， o）;我正在创建一个完整的副本，但我看不到如何创建一个仅具有位置的对象，可以将总数添加到该位置。
这是一个简化的示例，在原版中，我将多个列相加并删除多个列。
let helper = {};
let locationPeriodCounts = [
{"period": "2023-10-21", "location": "228", "countIn": 6, "countOut": 1},
{"period": "2023-10-22", "location": "228", "countIn": 8, "countOut": 2},
{"period": "2023-10-23", "location": "228", "countIn": 3, "countOut": 3},
{"period": "2023-10-24", "location": "228", "countIn": 1, "countOut": 4},
{"period": "2023-10-21", "location": "229", "countIn": 5, "countOut": 6},
{"period": "2023-10-22", "location": "229", "countIn": 18, "countOut": 6},
{"period": "2023-10-23", "location": "229", "countIn": 8, "countOut": 6},
{"period": "2023-10-24", "location": "230", "countIn": 3, "countOut": 6},
{"period": "2023-10-25", "location": "230", "countIn": 4, "countOut": 6}
];
let locationCounts = locationPeriodCounts.reduce(function(r, o) {
let key = o.location;
if (!helper[key]) {
helper[key] = Object.assign({}, o); // create a copy of o
helper[key].totalCountIn = 0;
helper[key].totalCountOut = 0;
r.push(helper[key]);
} else {
helper[key].totalCountIn += o.countIn;
helper[key].totalCountOut += o.countOut;
}
return r;
}, []);
console.log(locationCounts);这就是我正在寻找的输出......
[
{"location": "228", "totalCountIn": 18, "totalCountOut": 10},
{"location": "229", "totalCountIn": 31, "totalCountOut": 18},
{"location": "230", "totalCountIn": 7, "totalCountOut": 12}
];
答：
4赞
mplungjan
10/23/2023
#1
我添加了一个 Object.entries 以转换为对象数组
现在，我们需要将嵌套对象展开，使两个计数器与位置位于同一级别
let locationCounts = Object.entries(
locationPeriodCounts
.reduce(function (acc, { location, countIn, countOut }) {
acc[location] ??= { totalCountIn: 0, totalCountOut: 0 }; // create if it does not exist
acc[location].totalCountIn += countIn; // always add to the total
acc[location].totalCountOut += countOut; // always add to the total
return acc;
}, {})
)
.map(([location, { totalCountIn, totalCountOut }]) => ({ location, totalCountIn, totalCountOut }));
console.log(locationCounts);<script>
let locationPeriodCounts = [
{"period": "2023-10-21", "location": "228", "countIn": 6, "countOut": 1},
{"period": "2023-10-22", "location": "228", "countIn": 8, "countOut": 2},
{"period": "2023-10-23", "location": "228", "countIn": 3, "countOut": 3},
{"period": "2023-10-24", "location": "228", "countIn": 1, "countOut": 4},
{"period": "2023-10-21", "location": "229", "countIn": 5, "countOut": 6},
{"period": "2023-10-22", "location": "229", "countIn": 18, "countOut": 6},
{"period": "2023-10-23", "location": "229", "countIn": 8, "countOut": 6},
{"period": "2023-10-24", "location": "230", "countIn": 3, "countOut": 6},
{"period": "2023-10-25", "location": "230", "countIn": 4, "countOut": 6}
];
</script>评论
0赞
Chris
10/23/2023
谢谢@mplungjan。totakCount 具体是如何设置的？如果我想将两个值添加到两个总数中，我该怎么做（出于此处的目的，我们可以将 countIn 添加到两个总数）？
0赞
mplungjan
10/23/2023
我的代码创建一个对象：并按位置添加到每个对象中。然后，它将该对象转换为对象数组，并替换为 和 替换为 。同样，当您要求转换某些输入时，请显示预期的输出{ "228":0,"229":0,"230":0}228"location":"228"18"totalCount":18
0赞
Chris
10/23/2023
很抱歉@mplungjan，我已经澄清了代码以证明我需要它来处理多个属性。
0赞
mplungjan
10/23/2023
请参阅更新的代码。我现在正在展开嵌套计数器，以将它们与位置放在同一水平上
1赞
Filo.
10/23/2023
#2
我发现@mplungjan的反应比我的更聪明、更紧凑。 然而，这是我通常做的：
let locationPeriodCounts = [
{ "period": "2023-10-21", "location": "228", "countIn": 6 },
{ "period": "2023-10-22", "location": "228", "countIn": 8 },
{ "period": "2023-10-23", "location": "228", "countIn": 3 },
{ "period": "2023-10-24", "location": "228", "countIn": 1 },
{ "period": "2023-10-21", "location": "229", "countIn": 5 },
{ "period": "2023-10-22", "location": "229", "countIn": 18 },
{ "period": "2023-10-23", "location": "229", "countIn": 8 },
{ "period": "2023-10-24", "location": "230", "countIn": 3 },
{ "period": "2023-10-25", "location": "230", "countIn": 4 }
]
let locationCounts = locationPeriodCounts.reduce((acc, { location, countIn }) => {
let currLocationCount = acc.find(count => count.location === location)
if (currLocationCount) {
currLocationCount.totalCount += countIn
} else {
acc.push({
location: location,
totalCount: countIn
})
}
return acc;
}, []);
//If you also need sorting
locationCounts.sort((a, b) => {
//return a.totalCount - b.totalCount // asc
return b.totalCount - a.totalCount //desc
})
console.log(locationCounts);
这是输出：
[
{location: "229", totalCount: 31},
{location: "228", totalCount: 18},
{location: "230", totalCount: 7}
]
评论
1赞
Konrad
10/23/2023
这要慢得多，因为find
0赞
Filo.
10/23/2023
是的，这就是为什么我说@mplungjan比我的聪明得多！
0赞
Filo.
10/23/2023
但是，一个问题：这样做是像 ~ O（1） 中的哈希图一样访问数组还是进行隐式查找？非常感谢！acc[location]
2赞
Konrad
10/23/2023
在另一个答案中，它不是一个数组，而是一个对象，所以很可能是 O（1）accacc[location]
0赞
Filo.
10/23/2023
呃，很好，谢谢。绝对有趣！
2赞
pilchard
10/23/2023
#3
在原始代码中使用“helper”对象并不是一个坏的倾向，而常见的分组模式通过使用在传递给 的累加器对象中保存一个结果数组来做到这一点。这样做的好处包括：在reduce调用完成后不需要序列化对象，并且在reduce调用完成并检索结果数组后丢弃“helper”对象。reducereduce
let locationCounts = locationPeriodCounts.reduce(
(acc, { location, countIn, countOut }) => {
if (acc[location] === undefined) {
acc[location] = { location, totalCountIn: 0, totalCountOut: 0 };
acc._.push(acc[location]); // push a reference to the result array
}
acc[location].totalCountIn += countIn;
acc[location].totalCountOut += countOut;
return acc;
},
{ _: [] } // declare a result array in the accumulator
)._; // retrieve the result array
console.log(locationCounts);<script>
let locationPeriodCounts = [
{"period": "2023-10-21", "location": "228", "countIn": 6, "countOut": 1},
{"period": "2023-10-22", "location": "228", "countIn": 8, "countOut": 2},
{"period": "2023-10-23", "location": "228", "countIn": 3, "countOut": 3},
{"period": "2023-10-24", "location": "228", "countIn": 1, "countOut": 4},
{"period": "2023-10-21", "location": "229", "countIn": 5, "countOut": 6},
{"period": "2023-10-22", "location": "229", "countIn": 18, "countOut": 6},
{"period": "2023-10-23", "location": "229", "countIn": 8, "countOut": 6},
{"period": "2023-10-24", "location": "230", "countIn": 3, "countOut": 6},
{"period": "2023-10-25", "location": "230", "countIn": 4, "countOut": 6}
];
</script>无需解构
let locationCounts = locationPeriodCounts.reduce(
(acc, obj) => {
const key = obj.location;
if (acc[key] === undefined) {
acc[key] = {
location: obj.location,
totalCountIn: 0,
totalCountOut: 0,
};
acc._.push(acc[key]);
}
acc[key].totalCountIn += obj.countIn;
acc[key].totalCountOut += obj.countOut;
return acc;
},
{ _: [] }
)._;
console.log(locationCounts);<script>
let locationPeriodCounts = [
{"period": "2023-10-21", "location": "228", "countIn": 6, "countOut": 1},
{"period": "2023-10-22", "location": "228", "countIn": 8, "countOut": 2},
{"period": "2023-10-23", "location": "228", "countIn": 3, "countOut": 3},
{"period": "2023-10-24", "location": "228", "countIn": 1, "countOut": 4},
{"period": "2023-10-21", "location": "229", "countIn": 5, "countOut": 6},
{"period": "2023-10-22", "location": "229", "countIn": 18, "countOut": 6},
{"period": "2023-10-23", "location": "229", "countIn": 8, "countOut": 6},
{"period": "2023-10-24", "location": "230", "countIn": 3, "countOut": 6},
{"period": "2023-10-25", "location": "230", "countIn": 4, "countOut": 6}
];
</script>您现有的代码
您所缺少的只是创建一个仅具有相关属性的新对象（在您的问题中正确指出为问题）。要么使用上面的解构，要么像现在这样手动。Object.assign()
要使您自己的代码按原样工作，只需替换为 ，然后将求和移出块即可。这仍然在全局范围内徘徊，但通过将其作为闭包传递，可以很容易地解决：helper[key] = Object.assign({}, o);helper[key] = { location: o.location };elsehelper
let locationCounts = (helper =>
locationPeriodCounts.reduce(function (r, o) {
let key = o.location;
if (!helper[key]) {
helper[key] = { location: o.location };
helper[key].totalCountIn = 0;
helper[key].totalCountOut = 0;
r.push(helper[key]);
}
helper[key].totalCountIn += o.countIn;
helper[key].totalCountOut += o.countOut;
return r;
}, []))({});
或使用上述设置。
评论