最終目的是為以下數(shù)組中的每個(gè)名稱獲取一個(gè)。（類似于下面代碼片段底部注釋掉的行）如果有同名，則只取數(shù)量最多的，如果有重復(fù)，則只取版本較高的。我嘗試使用map、reduce的組合。請(qǐng)幫我解決這個(gè)問(wèn)題。//Original arrayvar arr = [  {'name': 'b', 'version': '103', 'count': 10},   {'name': 'a', 'version': '0.1b', 'count': 73},   {'name': 'b', 'version': '109', 'count': 67},   {'name': 'a', 'version': '1.0.3', 'count': 100},   {'name': 'a', 'version': '2.1.0', 'count': 100},   {'name': 'c', 'version': 'a103', 'count': 30},   {'name': 'c', 'version': 'a111', 'count': 100}];//expected console output// { name: 'a', version: '2.1.0', count: 100 }// { name: 'b', version: '109', count: 67 }// { name: 'c', version: 'a111', count: 100 }首先，作為該過(guò)程的“一部分”，我嘗試從數(shù)組中獲取“a”。var filteredArr = arr.filter(item => item.name === 'a');var m = Math.max(...(filteredArr.map(({count}) => count))),  maxes = arr.map(({count}) => count).reduce((p,c,i,a) => c ==  m ? p.concat(i) : p,[]);console.log(maxes);我期望從上面得到 [3, 4]，但它打印出 [3, 4, 6]。=====附加問(wèn)題：我該如何處理這另一個(gè)原始的預(yù)期。//Another original array.var arr2 = [   { name: 'a', version: '0.1', count: 1 },  { name: 'b', version: '216', count: 1 },  { name: 'b', version: '209', count: 1 },  { name: 'b', version: '216', count: 1 },  { name: 'b', version: '216', count: 1 },  { name: 'b', version: '213', count: 1 },  { name: 'b', version: '213', count: 1 },  { name: 'b', version: '209', count: 1 },  { name: 'b', version: '216', count: 1 },  { name: 'b', version: '216', count: 1 }];//expected output// { name: 'a', version: '0.1', count: 1 },// { name: 'b', version: '216', count: 5 },// { name: 'b', version: '209', count: 2 },// { name: 'b', version: '213', count: 2 }