我有两个要合并的数组，但也添加了数组的“Count”和“Total”部分。
数组一是:(较小）
[ { ReasonCode: '', Count: 2, Total: 15.63 },
{ ReasonCode: '01', Count: 13, Total: -144 },
{ ReasonCode: '03', Count: 7, Total: -394.87 },
{ ReasonCode: '04', Count: 128, Total: -3556.1 },
{ ReasonCode: '07', Count: 2, Total: -4.83 },
{ ReasonCode: '09', Count: 192, Total: -20826.25 } ]
数组二是:(更大）
[ { ReasonCode: '', Count: 6, Total: 412.21 },
{ ReasonCode: '01', Count: 7, Total: -9.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 162, Total: -1199.16 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 76, Total: -507.23 },
{ ReasonCode: '09', Count: 41, Total: -743.07 } ]
我正在使用此函数合并它们并创建新数组。
function CombineArrays(BiggerArray, SmallerArray, NewArray) {
BiggerArray.forEach(function (BA) {
let match = false;
SmallerArray.forEach(function (SA) {
if(BA.ReasonCode === SA.ReasonCode){
match = true;
BA.Count += SA.Count;
BA.Total += SA.Total;
BA.ReasonCode = SA.ReasonCode;
NewArray.push(BA);
}
});
if(!match) NewArray.push(BA);
});
}
从该函数生成的数组缺少第一个数组中的“ReasonCode：'03'”。我怎样才能这样做，它将添加两个数组中的所有数组对象，并将所需的两列相加。
[ { ReasonCode: '', Count: 8, Total: 427.84 },
{ ReasonCode: '01', Count: 20, Total: -153.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 290, Total: -4755.26 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 78, Total: -512.0600000000001 },
{ ReasonCode: '09', Count: 233, Total: -21569.32 } ]
提前谢谢。
答案 0 :(得分：1)
我会循环遍历最小的数组，如果该项不在大数组中，则添加它，否则添加到Count/Total
const arr1 = [{ ReasonCode: '', Count: 2, Total: 15.63 },
{ ReasonCode: '01', Count: 13, Total: -144 },
{ ReasonCode: '03', Count: 7, Total: -394.87 },
{ ReasonCode: '04', Count: 128, Total: -3556.1 },
{ ReasonCode: '07', Count: 2, Total: -4.83 },
{ ReasonCode: '09', Count: 192, Total: -20826.25 }]
const arr2 = [{ ReasonCode: '', Count: 6, Total: 412.21 },
{ ReasonCode: '01', Count: 7, Total: -9.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 162, Total: -1199.16 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 76, Total: -507.23 },
{ ReasonCode: '09', Count: 41, Total: -743.07 }]
function CombineArrays(biggest, smallest) {
// Create a new instance of the bigger array
let result = [].concat(biggest)
smallest.forEach(smallItem => {
// Try to get the item from the bigger list
var bigItem = result.find(item => item.ReasonCode == smallItem.ReasonCode)
// If it is not in the bigger list, add it
if (!bigItem) result.push(smallItem)
// If it is in the bigger list, then increment count/total by the small item amounts
else {
bigItem.Count += smallItem.Count
bigItem.Total += smallItem.Total
}
})
return result
}
console.log(CombineArrays(arr2, arr1))
答案 1 :(得分：1)
我会通过concat数组执行此操作，然后reduce匹配正确的ReasonCode：
let arr1 = [ { ReasonCode: '', Count: 2, Total: 15.63 },
{ ReasonCode: '01', Count: 13, Total: -144 },
{ ReasonCode: '03', Count: 7, Total: -394.87 },
{ ReasonCode: '04', Count: 128, Total: -3556.1 },
{ ReasonCode: '07', Count: 2, Total: -4.83 },
{ ReasonCode: '09', Count: 192, Total: -20826.25 } ];
let arr2 = [ { ReasonCode: '', Count: 6, Total: 412.21 },
{ ReasonCode: '01', Count: 7, Total: -9.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 162, Total: -1199.16 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 76, Total: -507.23 },
{ ReasonCode: '09', Count: 41, Total: -743.07 } ];
// Merge the arrays to a single one - Can be more if you need, just append them in concat()
let source = arr1.concat(arr2);
// Reduce the array holding all separated values to a new one
// with distinct ReasonCodes - the last argument `[]` is the initial accumulator value
let merged = source.reduce((accumulator, candidate) => {
// Check if this candidate is already in the accumulator
let index = accumulator.findIndex(
// Make it short with a fat arrow function that checks for an existing ReasonCode
existing => existing.ReasonCode === candidate.ReasonCode
);
if(index > -1) {
// If found in results, just increment the existing values with the ones from the candidate
accumulator[index].Count += candidate.Count;
accumulator[index].Total += candidate.Total;
} else {
// If candidate was not present yet, push it to the accumulator
accumulator.push(candidate);
}
return accumulator;
}, []);
console.log(merged);
使用此方法，您可以保证合并所有ReasonCodes与合并的数组。有关reduce：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce
答案 2 :(得分：1)
尝试以下方法：
var arr1 = [ { ReasonCode: '', Count: 2, Total: 15.63 },
{ ReasonCode: '01', Count: 13, Total: -144 },
{ ReasonCode: '03', Count: 7, Total: -394.87 },
{ ReasonCode: '04', Count: 128, Total: -3556.1 },
{ ReasonCode: '07', Count: 2, Total: -4.83 },
{ ReasonCode: '09', Count: 192, Total: -20826.25 } ];
var arr2 = [ { ReasonCode: '', Count: 6, Total: 412.21 },
{ ReasonCode: '01', Count: 7, Total: -9.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 162, Total: -1199.16 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 76, Total: -507.23 },
{ ReasonCode: '09', Count: 41, Total: -743.07 } ];
var map = {};
for(var i = 0; i < arr1.length; i++){
map[arr1[i].ReasonCode] = {
"index" : i
};
}
var result = [];
for(var i = 0; i < arr2.length; i++){
if(map[arr2[i].ReasonCode]){
arr2[i].Count += arr1[map[arr2[i].ReasonCode].index].Count;
arr2[i].Total += arr1[map[arr2[i].ReasonCode].index].Total;
map[arr2[i].ReasonCode].found = true;
}
result.push(arr2[i]);
}
Object.keys(map).forEach(function(key){
if(!map[key].found)
result.push(arr1[map[key].index]);
});
console.log(result);
答案 3 :(得分：1)
var array1= [ { ReasonCode: '', Count: 2, Total: 15.63 },
{ ReasonCode: '01', Count: 13, Total: -144 },
{ ReasonCode: '03', Count: 7, Total: -394.87 },
{ ReasonCode: '04', Count: 128, Total: -3556.1 },
{ ReasonCode: '07', Count: 2, Total: -4.83 },
{ ReasonCode: '09', Count: 192, Total: -20826.25 } ];
var array2 = [ { ReasonCode: '', Count: 6, Total: 412.21 },
{ ReasonCode: '01', Count: 7, Total: -9.75 },
{ ReasonCode: '02', Count: 5, Total: -37.03 },
{ ReasonCode: '04', Count: 162, Total: -1199.16 },
{ ReasonCode: '05', Count: 1, Total: -3.8 },
{ ReasonCode: '06', Count: 2, Total: -58.83 },
{ ReasonCode: '07', Count: 76, Total: -507.23 },
{ ReasonCode: '09', Count: 41, Total: -743.07 } ];
function CombineArrays(BiggerArray, SmallerArray) {
var NewObject = {};
BiggerArray.forEach(function(BA) {
var temp = NewObject[BA.ReasonCode] || {};
temp.Count = temp.Count ? temp.Count + BA.Count : BA.Count;
temp.Total = temp.Total ? temp.Total + BA.Total : BA.Total;
temp.ReasonCode = BA.ReasonCode;
NewObject[BA.ReasonCode] = temp;
})
SmallerArray.forEach(function(SA) {
var temp = NewObject[SA.ReasonCode] || {};
temp.Count = temp.Count ? temp.Count + SA.Count : SA.Count;
temp.Total = temp.Total ? temp.Total + SA.Total : SA.Total;
temp.ReasonCode = SA.ReasonCode;
NewObject[SA.ReasonCode] = temp;
});
return NewObject;
}
console.log(CombineArrays(array1, array2));
答案 4 :(得分：1)
你的algorythm效率不高。如果找到forEach，您的内部ReasonCode循环将遍历数组的每个元素。如果您使用了常规数组，则可以在break内执行if。
这是我的答案，一种不同的方法。
将对象用作地图非常适合您的情况，因为您可以轻松存储和访问每个ReasonCode，您还可以将实际对象作为值附加。
let map = {};
function execute(array) {
for(let i = 0; i < array.length; i++) {
let SA = array[i];
if( SA.ReasonCode in map === false ) {
map[SA.ReasonCode] = SA;
}
else {
let obj = map[SA.ReasonCode];
obj.Count += SA.Count;
obj.Total += SA.Total;
}
}
}
execute(SmallerArray);
execute(BiggerArray);
console.log(Object.values(map));