如何按月对一组对象进行分组？

Question

我正在使用JavaScript。我有一个包含这种格式数据的数组：

[
        {"USER_NAME":"User1","LAST_SUCCESSFUL_CONNECT":"1373978337642"},
        {"USER_NAME":"User2","LAST_SUCCESSFUL_CONNECT":"1374515704026"},
        {"USER_NAME":"User3","LAST_SUCCESSFUL_CONNECT":"1374749782479"}
]

（上面的数字代表UTC日期/时间，以毫秒为单位。

我想按月分组（统计）数据。像这样：

[
    {"Month":"January, 2014","User_Count": 2},
    {"Month":"February, 2014","User_Count": 1},
]

如果它简化了问题，我可以使用jQuery。

Answer 1

这看起来像map reduce问题。高级解决方案如下：

1. 重新组织名单成员。
2. 算上他们。

以下是实现此目标的逐步操作方法：

的地图

1. 遍历词典列表
2. 将datetime字符串转换为javascript datetime对象。
3. 使用月份作为键和词典列表作为值。

现在按月份分组。

示例：

var l = [...];
var o = {};
var f = function(x){
    var dt_object = Date(x["LAST_SUCCESSFUL_CONNECT"]); // convert to datetime object
    var key = dt_object.year + '-' + dt_object.month;

    if (o[key] === undefined) {
        var o[key] = [];
    };

    o[key].push(x)
}

_.map(l, f(x)) //apply f to each member of l

的减少

1. 遍历包含列表字典的新对象。
2. 计算每个字典列表的长度。
3. 使用count作为关键字和列表长度作为其值。

示例：

var g = function(member_count){
    //extra logic may go here
    return member_count
}

for member in o {
    count = _.reduce(l, g(member))
    member['count'] = count
}

生成的API

o['month-year'] //for list of dictionaries in that month
o['month-year']['count'] //for the count of list of dictionaries in that month.

的参考文献：

对于javascript中的map和reduce函数，请参阅underscore.js：
http://underscorejs.org/#map
http://underscorejs.org/#reduce

Javascript日期对象：
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date

有关Date和DateTime对象的更多信息：
https://en.wikipedia.org/wiki/ISO_8601

有关map reduce的更多信息：
https://en.wikipedia.org/wiki/MapReduce

Answer 2

使用Map-reduce。以下是使用underscore.js的一个示例。它虽然有点冗长但很简单。

var data = [{
    "USER_NAME": "User1",
        "LAST_SUCCESSFUL_CONNECT": "1373978337642"
}, {
    "USER_NAME": "User2",
        "LAST_SUCCESSFUL_CONNECT": "1374515704026"
}, {
    "USER_NAME": "User3",
        "LAST_SUCCESSFUL_CONNECT": "1374749782479"
}, {
    "USER_NAME": "User4",
        "LAST_SUCCESSFUL_CONNECT": "1274749702479"
}];

var monthNames = ["January", "February", "March", "April", "May", "June",
    "July", "August", "September", "October", "November", "December"];

var map_result = _.map(data, function (item) {
    var d = new Date(new Number(item.LAST_SUCCESSFUL_CONNECT));
    var month = monthNames[d.getMonth()] + ", " + d.getFullYear();
    return {
        "Month": month,
        "User_Count": 1
    };
});

var result_temp = _.reduce(map_result, function (memo, item) {
    if (memo[item.Month] === undefined) {
        memo[item.Month] = item.User_Count;
    }else{
        memo[item.Month] += item.User_Count;
    }
    return memo;
},{});

//then wrap the result to the format you expected.
var result = _.map(result_temp, function(value, key){
    return {
        "Month": key,
        "User_Count": value
    };
});

console.log(result); 

Answer 3

也许这会很有用

LINQ for JavaScript
http://linqjs.codeplex.com/

Answer 4

优化方案

b = 1:3
m = 5

for(j in 1:2){
  for(i in 1:5){
    print((1-i/m)* b[j] + (i/m)* b[j+1])
  }
}