javascript 从键值对创建对象数组
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Creating object array from key-value pairs
提问by tpolyak
From a service I receive a JSON object with key-value pairs, and I need to dynamically create objects from them, grouped by one column.
我从服务中收到一个带有键值对的 JSON 对象，我需要从它们动态创建对象，按一列分组。
The JSON looks like this:
JSON 如下所示：
[
{ "Group" : "A", "Key" : "Name", "Value" : "John" },
{ "Group" : "A", "Key" : "Age", "Value" : "30" },
{ "Group" : "A", "Key" : "City", "Value" : "London" },
{ "Group" : "B", "Key" : "Name", "Value" : "Hans" },
{ "Group" : "B", "Key" : "Age", "Value" : "35" },
{ "Group" : "B", "Key" : "City", "Value" : "Berlin" },
{ "Group" : "C", "Key" : "Name", "Value" : "José" },
{ "Group" : "C", "Key" : "Age", "Value" : "25" },
{ "Group" : "C", "Key" : "City", "Value" : "Madrid" }
]
I would need to transform it to the following array of objects:
我需要将其转换为以下对象数组：
[
{ Group : "A", Name : "John", Age : 30, City : "London" },
{ Group : "B", Name : "Hans", Age : 35, City : "Berlin" },
{ Group : "C", Name : "José", Age : 25, City : "Madrid" }
]
Each group can contain any number of key-value pairs.
每个组可以包含任意数量的键值对。
Currently I have a working solution for this, but I don't know if it's optimal:
目前我有一个可行的解决方案，但我不知道它是否是最佳的：
var items = [
{ "Group" : "A", "Key" : "Name", "Value" : "John" },
{ "Group" : "A", "Key" : "Age", "Value" : "30" },
{ "Group" : "A", "Key" : "City", "Value" : "London" },
{ "Group" : "B", "Key" : "Name", "Value" : "Hans" },
{ "Group" : "B", "Key" : "Age", "Value" : "35" },
{ "Group" : "B", "Key" : "City", "Value" : "Berlin" },
{ "Group" : "C", "Key" : "Name", "Value" : "José" },
{ "Group" : "C", "Key" : "Age", "Value" : "25" },
{ "Group" : "C", "Key" : "City", "Value" : "Madrid" }
], item, record, hash = {}, results = [];
// Create a "hash" object to build up
for (var i = 0, len = items.length; i < len; i += 1) {
item = items[i];
if (!hash[item.Group]) {
hash[item.Group] = {
Group : item.Group
};
}
hash[item.Group][item.Key] = item.Value;
}
// Push each item in the hash to the array
for (record in hash) {
if(hash.hasOwnProperty(record)) {
results.push(hash[record]);
}
}
You can check the fiddle here: http://jsbin.com/ozizom/1/
你可以在这里检查小提琴：http: //jsbin.com/ozizom/1/
Do you have a better solution for this?
你有更好的解决方案吗？
回答by sp00m
Assuming that the JSON records will always be sorted by Group, here is another approach:
假设 JSON 记录将始终按 Group 排序，这是另一种方法：
var json = [
{ "Group" : "A", "Key" : "Name", "Value" : "John" },
{ "Group" : "A", "Key" : "Age", "Value" : "30" },
{ "Group" : "A", "Key" : "City", "Value" : "London" },
{ "Group" : "B", "Key" : "Name", "Value" : "Hans" },
{ "Group" : "B", "Key" : "Age", "Value" : "35" },
{ "Group" : "B", "Key" : "City", "Value" : "Berlin" },
{ "Group" : "C", "Key" : "Name", "Value" : "José" },
{ "Group" : "C", "Key" : "Age", "Value" : "25" },
{ "Group" : "C", "Key" : "City", "Value" : "Madrid" }
];
var array = [];
var previousGroup = null;
for(var i=0; i<json.length; i++) {
var group = json[i].Group;
if(previousGroup != group) {
array.push({Group: group});
previousGroup = group;
}
array[array.length-1][json[i].Key] = json[i].Value;
}
Hereis a working example.
这是一个工作示例。
回答by dharcourt
Here's a solution where the code size is reduced (for better or for worse :-) by the use of JavaScript idioms. This solution doesn't depend on the order of input values:
这是一个解决方案，通过使用 JavaScript 习语来减少代码大小（无论好坏:-）。此解决方案不依赖于输入值的顺序：
var values = [
{Group: 'A', Key: 'Name', Value: 'John'},
{Group: 'A', Key: 'Age', Value: '30'},
{Group: 'A', Key: 'City', Value: 'London'},
{Group: 'B', Key: 'Name', Value: 'Hans'},
{Group: 'B', Key: 'Age', Value: '35'},
{Group: 'B', Key: 'City', Value: 'Berlin'},
{Group: 'C', Key: 'Name', Value: 'José'},
{Group: 'C', Key: 'Age', Value: '25'},
{Group: 'C', Key: 'City', Value: 'Madrid'}
];
var map = {};
values.forEach(function(value) {
map[value.Group] = map[value.Group] || {Group: value.Group};
map[value.Group][value.Key] = value.Value;
});
var results = Object.keys(map).map(function(key) { return map[key]; });
A working example is at http://jsfiddle.net/arQww.
一个工作示例位于http://jsfiddle.net/arQww。
Here's the fastest solution I can find, which assumes that the values will always be sorted by Group:
这是我能找到的最快的解决方案，它假设值将始终按组排序：
var group, results = [];
for (var i = 0; i < values.length; ) {
results.push({Group: group = values[i].Group});
do {
results.push[results.length - 1][values[i].Key] = values[i].Value;
} while (++i < values.length && values[i].Group == group);
}
A performance comparison is at http://jsperf.com/vnmzc. While this second solutions is faster, the performance of both is O(n) and the real-world difference between them will be inconsequential, so the first solution is probably preferable since it's simpler and more general.
性能比较位于http://jsperf.com/vnmzc。虽然第二个解决方案更快，但两者的性能都是 O(n) 并且它们之间的实际差异将无关紧要，因此第一个解决方案可能更可取，因为它更简单，更通用。
回答by Ivan Klass
If you have to manipulate data a lot I would recommend underscoreframework. This is how solution will look in it:
如果您必须大量操作数据，我会推荐下划线框架。这是解决方案的外观：
/*
We group items into object that looks like {group: attributes, ..}
Then for each group we create result object denoting group,
and extend result with object created from keys and values of attributes
*/
_.map(_.groupBy(items, function (item) {return item.Group}),
function (attributes, group) {
return _.extend({Group: group},
_.object(_.pluck(attributes, 'Key'),
_.pluck(attributes, 'Value')))
})
回答by user3771408
I need help with a similar problem, buy I want to sum values. I have this array with objects
我需要帮助解决类似的问题，购买我想求和的值。我有这个带有对象的数组
var myArrWithObj = [
{DateToSort: "Jul2014", ValueOneToSum: "60", ValueTwoToSum: "15"},
{DateToSort: "Jul2014", ValueOneToSum: "30", ValueTwoToSum: "50"},
{DateToSort: "Jul2014", ValueOneToSum: "12", ValueTwoToSum: "22"},
{DateToSort: "Aug2014", ValueOneToSum: "65", ValueTwoToSum: "25"},
{DateToSort: "Aug2014", ValueOneToSum: "13", ValueTwoToSum: "10"},
{DateToSort: "Aug2014", ValueOneToSum: "90", ValueTwoToSum: "20"},
{DateToSort: "Sep2014", ValueOneToSum: "60", ValueTwoToSum: "15"},
{DateToSort: "Sep2014", ValueOneToSum: "60", ValueTwoToSum: "18"},
{DateToSort: "Sep2014", ValueOneToSum: "75", ValueTwoToSum: "18"}
];
and I want the user to choose what month to sum from an option menu.
我希望用户从选项菜单中选择要汇总的月份。
So If the user selects August 2014from the select menu, I want to sum all the ValueOneToSumvalues and ValueTwoToSumvalues based on Aug2014, how could I do that?
因此，如果用户August 2014从选择菜单中进行选择，我想将所有ValueOneToSum值和ValueTwoToSum基于的值相加，我该Aug2014怎么做？
For example: the totalSumOnefor Aug2014would be 168and totalSumTwowould be 55.
例如：totalSumOneforAug2014将是168并且totalSumTwo将是55。
