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function test() {
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var list = [
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{"name": "John", "Average": 15, "High": 10, "DtmStamp": 1358226000000},
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{"name": "Jane", "Average": 16, "High": 92, "DtmStamp": 1358226000000},
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{"name": "Jane", "Average": 17, "High": 45, "DtmStamp": 1358226000000},
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{"name": "John", "Average": 18, "High": 87, "DtmStamp": 1358226000000},
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{"name": "Jane", "Average": 15, "High": 10, "DtmStamp": 1358226060000},
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{"name": "John", "Average": 16, "High": 87, "DtmStamp": 1358226060000},
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{"name": "John", "Average": 17, "High": 45, "DtmStamp": 1358226060000},
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{"name": "Jane", "Average": 18, "High": 92, "DtmStamp": 1358226060000}
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];
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var sorted = this.groupBy(list, function (item) {
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return [item.name + item.High];
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});
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console.log(sorted);
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}
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test();
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function groupBy(array, f) {
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debugger;
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var groups = {};
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array.forEach(function (o) {
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var group = JSON.stringify(f(o));
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groups[group] = groups[group] || [];
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groups[group].push(o);
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});
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return Object.keys(groups).map(function (group) {
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return groups[group];
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});
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}
具體實現思路:
- 函數groupBy有兩個形參,一爲對象數組,二爲匿名函數(該函數功能:返回對象的某個指定屬性的屬性值並存放在數組中);
- groupBy函數內,先創建一個空對象;
- 然後forEach遍歷對象數組,遍歷時要執行的函數中只有一個形參o(數組中的每個元素);
- 由於下面函數調用是想用name來分組,因此let group = JSON.stringify( f(o) ),相當於先獲取到對象數組list中的name屬性對應的屬性值並放入數組中:["John"],然後再將屬性值轉換爲json字符串:'["John"]';
- groups[group] = groups[group] || [],在js中對象也是關聯數組,因此這裏相當於做了兩件事,一是把group作爲groups的key,二是將對應的value初始化,第一次執行爲空數組,循環執行時找到相同的name時保持不變;
- groups[group].push( o ),這句相當於把list中每個對象壓入數組中作爲value;
- 最後,Object.keys(groups)是取出groups對象中的所有key,然後遍歷一個個key組成的新數組,返回分好了組的二維數組
代碼中,我把分組參數多加了一個,結果顯示也是可以分組成功的。