因此,我正在尝试使用一些现有数据来创建对象列表的列表。我拥有的数据包含2个值,即``a''和``b'',其中许多相似的``a''值在数据集中重复多次。 我正在尝试将数据转换为以下格式:对于任何“ a”值,该列表将在单个位置包含对应于“ a”值的所有“ b”值。 数据集为: const tempData = [ {a:1557637200025,b:“ aaaa”}, {a:1557637200025,b:“ bbbb”}, {a:1557637200025,b:“ cccc”}, {a:1557637200025,b:“ dddd”}, {a:1557637200025,b:“ eeee”}, {a:1557637200025,b:“ ffff”}, {a:1557637200025,b:“ gggg”}, {a:1555563600017,b:“ hhhh”}, {a:1555563600017,b:“ iiii”}, {a:1555563600017,b:“ jjjj”}, {a:1555563600017,b:“ kkkk”}, {a:1555563600017,b:“ llll”}, {a:1555563600017,b:“ mmmm”}, {a:1555563600017,b:“ nnnn”}, {a:1555563600017,b:“ oooo”}, {a:1555563600017,b:“ pppp”}, {a:1555563600017,b:“ qqqq”}, {a:1555563600017,b:“ rrrr”}, {a:1555563600017,b:“ ssss”}}, {a:1555563600017,b:“ tttt”}, {a:1555563600017,b:“ uuuu”}, {a:1557982800028,b:“ vvvv”}, ] 我希望它看起来像这样: const newTempData = [ { a:1557637200025,b:[ “ aaaa”, “ bbbb”, ... ]}, { a:1555563600017,b:[ “啊……”, “ iiii”, ]}, ... ] 我已经尝试过基本嵌套的for循环,但是我无法在任何地方实现我想要的结果。我没有任何办法来更新row2的for循环内的row1值,因此我可以停止循环以再次匹配值然后再次。 var newTempData = []; for(让tempData中的row1){ console.log(row1); var some = []; var row2 = Number(row1)+1; var x = tempData [row1] .a; for(tempData中的第2行){ var y = tempData [row2] .a; var instance = tempData [row2] .b; if(x === y){ some.push({ins:instance}); } } newTempData.push({a:x,b:some}); } console.log(newTempData); 我得到的输出是这个。 0:{a:1557637200025,b:Array(7)} 1:{a:1557637200025,b:Array(7)} 2:{a:1557637200025,b:Array(7)} 3:{a:1557637200025,b:Array(7)} 4:{a:1557637200025,b:Array(7)} 5:{a:1557637200025,b:Array(7)} 6:{a:1557637200025,b:Array(7)} 7:{a:1555563600017,b:Array(14)} 8:{a:1555563600017,b:Array(14)} 9:{a:1555563600017,b:Array(14)} 10:{a:1555563600017,b:Array(14)} 11:{a:1555563600017,b:Array(14)} 12:{a:1555563600017,b:Array(14)} 13:{a:1555563600017,b:Array(14)} 14:{a:1555563600017,b:Array(14)} 15:{a:1555563600017,b:Array(14)} 16:{a:1555563600017,b:Array(14)} 17:{a:1555563600017,b:Array(14)} 18:{a:1555563600017,b:Array(14)} 19:{a:1555563600017,b:Array(14)} 20:{a:1555563600017,b:Array(14)} 21:{a:1558972800028,b:Array(1)} 我只是想获得所需的数据而无需重复。现在,我能够获取数据,但是对于每个匹配的'a'值都将重复该数据。我想知道更好的方法,而不是业余的方法。谢谢。
2021-01-31 08:16:27
尝试减少。我已经注释掉b是否已经在数组中的检查,这取决于您是否要重复b值。 const tempData = [ {a:1557637200025,b:“ aaaa”}, {a:1557637200025,b:“ bbbb”}, {a:1557637200025,b:“ cccc”}, {a:1557637200025,b:“ dddd”}, {a:1557637200025,b:“ eeee”}, {a:1557637200025,b:“ ffff”}, {a:1557637200025,b:“ gggg”}, {a:1555563600017,b:“ hhhh”}, {a:1555563600017,b:“ iiii”}, {a:1555563600017,b:“ jjjj”}, {a:1555563600017,b:“ kkkk”}, {a:1555563600017,b:“ llll”}, {a:1555563600017,b:“ mmmm”}, {a:1555563600017,b:“ nnnn”}, {a:1555563600017,b:“ oooo”}, {a:1555563600017,b:“ pppp”}, {a:1555563600017,b:“ qqqq”}, {a:1555563600017,b:“ rrrr”}, {a:1555563600017,b:“ ssss”}, {a:1555563600017,b:“ tttt”}, {a:1555563600017,b:“ uuuu”}, {a:1557982800028,b:“ vvvv”} ]; const result = tempData.reduce((acc,{a,b})=> { const element = acc.find(el => el.a === a); //如果元素存在 如果(element / * &&!element.b.includes(b)* /){ element.b.push(b); 返回acc; } / * else if(element){return acc; } * / acc.push({a,b:[b]}); 返回acc; },[]); | 如果您关心性能,则可以使用Map来代替使用数组的.find(): 函数add(map,{a,b}){ 如果(!map.has(a)){ map.set(a,{a:a,b:[]}); 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