我正在开发应用程序,我需要对不同的javascript对象进行分组,这些将基于月,日和年。
我正在做的事情如下
var calculateByDay = function(inputList){
var outPutList = [];
var result = {}
var item = null, key = null;
for(i=0; c<inputList.length; i++) {
item=inputList[c];
key = Object.keys(item)[0];
item=item[key];
if(!result[key]) {
result[key] = item;
}
else {
result[key] += item;
}
for (r in result)
{
var docs = {};
docs["date"] = r;
docs["amount"] = result[r];
outPutList.push(docs);
}
}
return outPutList;
}
如何改进上述代码并将其用于月份和年度计算? 我彻底使用了underscore.js并且它有一个groupBy方法。但似乎不符合我的要求。 我想按月和年分组, 为了
var inputList = [{"2012-12-02T00:00": 2000}, {"2013-01-01T00:00": 1200},{"2013-02-02T00:00": 550}, {"2013-02-02T00:00": 1000}];
输出应为:
Monthly :
[{"December 2012": 2000}, {"January 2013": 1200},{"February 2013": 1550}];
Yearly
[{"year 2012": 2000}, {"year 2013": 2750}];
似乎我需要这种地图,减少大数据(数组集)的方法,是否有其他库或实践可以使代码稳固?
提前致谢。
答案 0 :(得分:16)
鉴于数据结构略有不同:
var data = [{
"date": "2011-12-02T00:00",
"value": 1000
}, {
"date": "2013-03-02T00:00",
"value": 1000
}, {
"date": "2013-03-02T00:00",
"value": 500
}, {
"date": "2012-12-02T00:00",
"value": 200
}, {
"date": "2013-04-02T00:00",
"value": 200
}, {
"date": "2013-04-02T00:00",
"value": 500
}, {
"date": "2013-03-02T00:00",
"value": 500
}, {
"date": "2013-04-12T00:00",
"value": 1000
}, {
"date": "2012-11-02T00:00",
"value": 600
}];
你可以使用下划线:
var grouped = _.groupBy(data, function(item) {
return item.date;
});
var groupedByYear = _.groupBy(data, function(item) {
return item.date.substring(0,4);
});
var groupedByMonth = _.groupBy(data, function(item) {
return item.date.substring(0,7);
});
console.log(groupedByYear);
参见相关回答:Javascript - underscorejs map reduce groupby based on date
答案 1 :(得分:2)
请查看以下重构是否对您有用 http://jsfiddle.net/wkUJC/
var dates = [{"2012-12-02T00:00": 2000}, {"2013-01-01T00:00": 1200},{"2013-02-02T00:00": 550}, {"2013-02-02T00:00": 1000}];
function calc(dates) {
var response = {};
dates.forEach(function(d){
for (var k in d) {
var _ = k.split("-");
var year = _[0]
var month = _[1]
if (!response[year]) response[year] = {total: 0}
response[year][month] = response[year][month] ? response[year][month]+d[k] : d[k]
response[year].total+= d[k]
}
});
console.log(response);
return response;
}
calc(dates);