如何有條件地基於多個鍵對數組進行分組？

Question

我有條件地將一個簡單的問題分組。

const aInput = [
{"date":"2018/9/10", "code":"A"},
{"date":"2018/9/10", "code":"B"},
{"date":"2018/9/11", "code":"B"},
{"date":"2018/9/11", "code":"B"}];

我想按日期和代碼對數組進行分組，但如果A和B都存在一天，那么只有B停留

 const aInput = [ {"date":"2018/9/10", "code":"A"}, {"date":"2018/9/10", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/11", "code":"B"}]; // I want to group the array above by day & code BUT if there is both A & B then only B stays const aOutput = [ {"date":"2018/9/10", "code":"B"}, {"date":"2018/9/11", "code":"B"}] 

Answer 1

您可以使用reduce類似的操作：

創建一個以每個date為鍵的累加器。 如果累加器的上下文中沒有當前日期，或者當前對象具有首選代碼，則將該對象添加為值。

（添加了一個額外的A代碼對象來演示）

 const aInput = [ {"date":"2018/9/10", "code":"A"}, {"date":"2018/9/10", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/12", "code":"A"}] const preferredCode = "B"; const merged = aInput.reduce((r,{date, code}) =>{ if(!r[date] || code === preferredCode) r[date] = {date, code}; return r; },{}) const final = Object.values(merged) console.log(final) 

上面的代碼假定只有2個code 。 如果您有兩個以上的code並且您希望每天獲取所有唯一代碼（除非有B ，那么可以使用reduce和some代碼來完成以下操作

 const aInput = [ {"date":"2018/9/10", "code":"A"}, {"date":"2018/9/10", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/12", "code":"A"}, {"date":"2018/9/12", "code":"C"}] const preferredCode = "B"; const merged = aInput.reduce((r,{date, code}) =>{ r[date] = r[date] || []; if(code === preferredCode) r[date] = [{date, code}] else if(!r[date].some(e => e.code === preferredCode || e.code === code)) r[date].push({date, code}) return r; },{}) const final = [].concat.apply([],Object.values(merged)) console.log(final) 

Answer 2

您可以使用.reduce在此數組中按代碼分組

 const aInput = [ {"date":"2018/9/10", "code":"A"}, {"date":"2018/9/10", "code":"B"}, {"date":"2018/9/11", "code":"B"}, {"date":"2018/9/11", "code":"B"}]; const result = aInput.reduce((r, a) => { r[a.code] = [...r[a.code] || [], a]; return r; }, {}); console.log(result)