针对 ID 存在根据特定性别数据拆分组合列表 [英] Split a combined List based on particular gender data is present for ID
本文介绍了针对 ID 存在根据特定性别数据拆分组合列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这是一个组合列表,具有唯一的年龄代码.
Here is a combined list, with unique ageCode.
[
{
"ageCode": 1,
"ageDesc": "0-4",
"qM": 358,
"sM": 158,
"qF": 328,
"sF": 258
},
{
"ageCode": 3,
"ageDesc": "15-59",
"qM": 525,
"sM": 125
},
{
"ageCode": 4,
"ageDesc": "60+",
"qF": 458,
"sF": 358
}
]
组合列表需要根据是否 M & 拆分为单个对象年龄代码存在 F 值.转换后的列表如下所示.如果没有M"日期,那么将没有带有性别"的对象:M"同样适用于F"
Combined list need to be split into individual object based on if both M & F value is present for an ageCode. The converted list look like this. If there is No "M" date then there will be no object with "gender": "M" same applied to "F"
[
{
"ageCode": 1,
"ageDesc": "0-4",
"q": 358,
"s": 158,
"gender": "M"
},
{
"ageCode": 1,
"ageDesc": "0-4",
"q": 328,
"s": 258,
"gender": "F"
},
{
"ageCode": 3,
"ageDesc": "15-59",
"q": 525,
"s": 125,
"gender": "M"
},
{
"agCode": 4,
"ageDesc": "60+",
"q": 458,
"s": 358,
"gender": "F"
}
]
尝试的解决方案:
for(let item of this.ageData) {
if (this.ageData.find((i) => { i.agCode=== item.agCode})){
//}
}
这里需要重复和多个 for 循环等问题,有没有什么有效的方法来实现这一点.
Here issues like duplicate and multiple for loop is required, Is there any efficient way to achieve this.
推荐答案
尝试下一个代码
const combined = [
{
"ageCode": 1,
"ageDesc": "0-4",
"qM": 358,
"sM": 158,
"qF": 328,
"sF": 258
},
{
"ageCode": 3,
"ageDesc": "15-59",
"qM": 525,
"sM": 125
},
{
"ageCode": 4,
"ageDesc": "60+",
"qF": 458,
"sF": 358
}
];
const original = combined.reduce((result, item) => {
if (item.qM !== undefined) {
result.push({
"ageCode": item.ageCode,
"ageDesc": item.ageDesc,
"q": item.qM,
"s": item.sM,
"gender": "M"
});
}
if (item.qF !== undefined) {
result.push({
"ageCode": item.ageCode,
"ageDesc": item.ageDesc,
"q": item.qF,
"s": item.sF,
"gender": "F"
});
}
return result;
}, []);
console.log(original);
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