我有一些数据需要做一些计算/操作:
data=[{"sex":"M","age":"0","pop":"310"},
{"sex":"M","age":"5","pop":"306"},
{"sex":"M","age":"10","pop":"313"},
{"sex":"M","age":"15","pop":"332"},
....
{"sex":"M","age":"100","pop":"4"},
{"sex":"W","age":"0","pop":"294"},
{"sex":"W","age":"5","pop":"291"},
{"sex":"W","age":"10","pop":"300"},
{"sex":"W","age":"15","pop":"318"},
....
{"sex":"W","age":"100","pop":"1"}
]
我提取男性(M)和女性(W)的子数据:
var male=data.filter(function(d){
if (data.sex=="M"){return d.pop;}
})
var female=data.filter(function(d){
if (data.sex=="F"){return d.pop;}
})
现在,我想计算两性的人口(pop),即创建一个名为“both”的新变量,其中包含变量“age”:
both=[{"sex":"MW","age":"0","pop":"604"},
{"sex":"MW","age":"5","pop":"597"},
{"sex":"MW","age":"10","pop":"613"},
{"sex":"MW","age":"15","pop":"650",
....
{"sex":"MW","age":"100","pop":"5"}
]
我该怎么做?
答案 0 :(得分:2)
听起来您希望将M和W条目组合成一个具有合并pop
值的单个条目,按年龄。如果是这样,您需要按年龄创建条目的地图并将pop
添加到一起,您可以使用对象在ES5中轻松完成(在ES2015中 - 也就是“ES6” - 您可能会使用一个Map
)。完成后将其转换回数组。见评论:
// Your data -- note that I've fixed the pop values, if they're
// meant to be numbers, they shouldn't be in quotes
var data = [
{"sex":"M","age":"0","pop":310},
{"sex":"M","age":"5","pop":306},
{"sex":"M","age":"10","pop":313},
{"sex":"M","age":"15","pop":332},
{"sex":"M","age":"100","pop":4},
{"sex":"W","age":"0","pop":294},
{"sex":"W","age":"5","pop":291},
{"sex":"W","age":"10","pop":300},
{"sex":"W","age":"15","pop":318},
{"sex":"W","age":"100","pop":1}
];
// Create the "map" of entries by age:
var byAge = Object.create(null);
data.forEach(function(entry) {
// Get the entry for this age, if any
var ageEntry = byAge[entry.age];
if (!ageEntry) {
// None, create and add it with an initial pop of 0
ageEntry = byAge[entry.age] = {sex: "MW", age: entry.age, pop: 0};
}
// Add in the pop for this entry
ageEntry.pop += entry.pop;
});
// Convert it back into an array (if you want):
var result = Object.keys(byAge).map(function(key) {
return byAge[key];
});
// Show result:
console.log(result);
答案 1 :(得分:2)
var data = [
{"sex":"M","age":"0","pop":310},
{"sex":"M","age":"5","pop":306},
{"sex":"M","age":"10","pop":313},
{"sex":"M","age":"15","pop":332},
{"sex":"M","age":"100","pop":4},
{"sex":"W","age":"0","pop":294},
{"sex":"W","age":"5","pop":291},
{"sex":"W","age":"10","pop":300},
{"sex":"W","age":"15","pop":318},
{"sex":"W","age":"100","pop":1}
];
var ages=[]
for(var i=0;i<data.lenght;i++){
if(!ages[data[i]['age']]){
ages[data[i]['age']].push({"sex": "MW", "age": data[i]['age'], "pop":data[i]['pop'] })
}else{
ages[data[i]['age']].pop +=data[i]['pop']
}
}
答案 2 :(得分:0)
听起来,reduce方法可以满足您的需求。
`var population = ARR.reduce(function(a, b) {
return Number(a.pop) + Number(b.pop);
});`
请注意,您必须将填充的字符串表示形式转换为数字,否则它将被连接。