React JS concat数组并转换为新的二维数组并按二维数组数据和排序
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【中文标题】React JS concat数组并转换为新的二维数组并按二维数组数据和排序【英文标题】：React JS concate array and transform into new two dimension array and sort by the two dimension array data sum 【发布时间】：2019-07-15 07:35:04 【问题描述】：我正在开发一个反应原生应用程序 假设我有这些数组列表：
let soldList1 = [
"itemCode":"X001" , "soldRate":0.0789,
"itemCode":"5555" , "soldRate":0.0543,
"itemCode":"3141" , "soldRate":0.0112,
"itemCode":"Mix-001" , "soldRate":0.01,
"itemCode":"7689-L" , "soldRate":0.005,
"itemCode":"1111" , "soldRate":0.003
]
let soldList2 = [
"itemCode":"3141" , "soldRate":0.0712,
"itemCode":"7689-L" , "soldRate":0.03,
"itemCode":"5555" , "soldRate":0.0234,
"itemCode":"1111" , "soldRate":0.011,
"itemCode":"X001" , "soldRate":0.008,
"itemCode":"Mix-001" , "soldRate":0.004
]
let soldList3 = [
"itemCode":"5555" , "soldRate":0.0339,
"itemCode":"X001" , "soldRate":0.0221,
"itemCode":"3141" , "soldRate":0.0111,
"itemCode":"1111" , "soldRate":0.0089,
"itemCode":"Mix-001" , "soldRate":0.0077,
"itemCode":"7689-L" , "soldRate":0.0032
]
let soldList4 =[
"itemCode":"8888" , "soldRate":0.13,
"itemCode":"9999" , "soldRate":0.11,
"itemCode":"3141" , "soldRate":0.08,
"itemCode":"1111" , "soldRate":0.07
]
let soldList5 =[
"itemCode":"3141" , "soldRate":0.044,
"itemCode":"1111" , "soldRate":0.011,
"itemCode":"8888" , "soldRate":0.0011,
"itemCode":"9999" , "soldRate":0.0001
]
let soldList6 =[
"itemCode":"Mix-001" , "soldRate":0.5678,
"itemCode":"7689-L" , "soldRate":0.546
"itemCode":"8888" , "soldRate":0.323,
"itemCode":"9999" , "soldRate":0.0032,
"itemCode":"Mix-001" , "soldRate":0.0022,
"itemCode":"UV-007" , "soldRate":0.0012
"itemCode":"TT-08" , "soldRate":0.0011,
"itemCode":"PP-03" , "soldRate":0.0009
]
如您所见，一些列表获得了其他列表没有的元素，并且每个列表项顺序为desc order by 'soldRate'，每个列表长度也可能不同。
所以目标是连接这些数组列表和处理以构建一个新的数组列表，如下所示：
let finalAllConcateAndSortedByDataSumList = [
"itemCode":"Mix-001" , "data": [0.01, 0.004, 0.0077, 0, 0, 0.5678], "dataSum":0.5895,
"itemCode":"7689-L" , "data": [0.005, 0.03, 0.0032, 0, 0, 0.546], "dataSum":0.5842,
"itemCode":"8888" , "data": [0, 0, 0, 0.13, 0.0011, 0.323], "dataSum":0.4541,
"itemCode":"3141" , "data": [0.0112, 0.0712, 0.0111, 0.08, 0.044, 0], "dataSum":0.2175,
"itemCode":"X001" , "data": [0.0789, 0.008, 0.0221, 0, 0, 0.0221], "dataSum":0.1311,
"itemCode":"9999" , "data": [0, 0, 0, 0.11, 0.0001, 0.0032], "dataSum":0.1133,
"itemCode":"5555" , "data": [0.0543, 0.0234, 0.0339, 0, 0, 0], "dataSum":0.1116,
"itemCode":"1111" , "data": [0.003, 0.011, 0.0089, 0.07, 0.011, 0], "dataSum":0.1039,
"itemCode":"UV-007" , "data": [0, 0, 0, 0, 0, 0.0012], "dataSum":0.0012,
"itemCode":"TT-08" , "data": [0, 0, 0, 0, 0, 0.0011], "dataSum":0.0011,
"itemCode":"PP-03" , "data": [0, 0, 0, 0, 0, 0.0009], "dataSum":0.0009,
]
如您所见，上面列表 1 2 3 4 5 6 中的所有'soldRate' 元素都已连接并生成新的数组列表'data' 属性，每个数据顺序的位置与上面完全相同original list 1 2 3 4 5 6....如果position中的元素值不存在于任何原始列表中，则它是'0'值
最终将 finalAllConcateAndSortedByDataSumList 中的每个对象，数据数组汇总到属性'dataSum'
那么新列表 finalAllConcateAndSortedByDataSumList 是desc order by dataSum
这是我个人尝试让它工作，但它不起作用......
let finalAllConcateAndSortedByDataSumList = soldList1.concat(soldList2)
.concat(soldList3)
.concat(soldList4)
.concat(soldList5)
.concat(soldList16).map((item) =>
let newItem = itemCode: item.itemCode, data: [item.soldRate], dataSum: Math.sum([item.soldRate])
return newItem
).sort((item) => return item.dataSum)
所以在这里寻求帮助，代码示例真的很有帮助 谢谢
【问题讨论】：
【参考方案1】：您可以首先提取数组中所有唯一的 itemCode 值，然后循环遍历它们，如果每个数组中都存在 itemCode，则将 soldRate 或 0 添加到结果中。
let soldList1 = [
itemCode: "X001", soldRate: 0.0789 ,
itemCode: "5555", soldRate: 0.0543 ,
itemCode: "3141", soldRate: 0.0112 ,
itemCode: "Mix-001", soldRate: 0.01 ,
itemCode: "7689-L", soldRate: 0.005 ,
itemCode: "1111", soldRate: 0.003
];
let soldList2 = [
itemCode: "3141", soldRate: 0.0712 ,
itemCode: "7689-L", soldRate: 0.03 ,
itemCode: "5555", soldRate: 0.0234 ,
itemCode: "1111", soldRate: 0.011 ,
itemCode: "X001", soldRate: 0.008 ,
itemCode: "Mix-001", soldRate: 0.004
];
let soldList3 = [
itemCode: "5555", soldRate: 0.0339 ,
itemCode: "X001", soldRate: 0.0221 ,
itemCode: "3141", soldRate: 0.0111 ,
itemCode: "1111", soldRate: 0.0089 ,
itemCode: "Mix-001", soldRate: 0.0077 ,
itemCode: "7689-L", soldRate: 0.0032
];
let soldList4 = [
itemCode: "8888", soldRate: 0.13 ,
itemCode: "9999", soldRate: 0.11 ,
itemCode: "3141", soldRate: 0.08 ,
itemCode: "1111", soldRate: 0.07
];
let soldList5 = [
itemCode: "3141", soldRate: 0.044 ,
itemCode: "1111", soldRate: 0.011 ,
itemCode: "8888", soldRate: 0.0011 ,
itemCode: "9999", soldRate: 0.0001
];
let soldList6 = [
itemCode: "Mix-001", soldRate: 0.5678 ,
itemCode: "7689-L", soldRate: 0.546 ,
itemCode: "8888", soldRate: 0.323 ,
itemCode: "9999", soldRate: 0.0032 ,
itemCode: "Mix-001", soldRate: 0.0022 ,
itemCode: "UV-007", soldRate: 0.0012 ,
itemCode: "TT-08", soldRate: 0.0011 ,
itemCode: "PP-03", soldRate: 0.0009
];
function sumSoldRates(...arrs)
let itemCodes = [].concat(...arrs).reduce((acc, obj) =>
if (!acc.includes(obj.itemCode))
acc.push(obj.itemCode);
return acc;
, []);
let result = itemCodes.map(code =>
let obj = itemCode: code, data: [], dataSum: 0 ;
arrs.forEach(arr =>
let item = arr.find(el => el.itemCode === code);
if (item)
obj.data.push(item.soldRate);
obj.dataSum += item.soldRate;
else
obj.data.push(0);
);
return obj;
);
result.sort((a, b) => b.dataSum - a.dataSum);
return result;
let result = sumSoldRates(
soldList1,
soldList2,
soldList3,
soldList4,
soldList5,
soldList6
);
console.log(result);【讨论】：
【参考方案2】：我将您的流程分解为多个步骤，因为它可以让您更清楚地了解您在做什么（并且更易于调试）。代码可能不像其他答案那样紧凑或“优雅”，但我相信这样更容易理解。
这是一个code sandbox，它显示了这一点（参见控制台输出）。
const getItemCodes = (list) =>
return list.map(item => item.itemCode)
const getDataValue = (list, itemCode) =>
let index = list.findIndex(item => item.itemCode === itemCode)
return (index === -1 ? 0 : list[index].soldRate)
const sumArray = (array) =>
const sum = array.reduce((prev, curr) =>
return (parseFloat(prev) + parseFloat(curr))
)
return sum.toFixed(4)
const sortByDataSum = (array) =>
array.sort((a, b) =>
if (a.dataSum > b.dataSum)
return -1
else if (a.dataSum < b.dataSum)
return 1
else
return 0
)
return array
// concat all item codes together, which will result in duplicates
let allItemCodes = [...getItemCodes(soldList1),
...getItemCodes(soldList2),
...getItemCodes(soldList3),
...getItemCodes(soldList4),
...getItemCodes(soldList5),
...getItemCodes(soldList6)]
// remove duplicate codes
allItemCodes = [...new Set(allItemCodes)]
let finalList = []
allItemCodes.forEach((code) =>
let item =
'itemCode': code,
'data': [getDataValue(soldList1, code),
getDataValue(soldList2, code),
getDataValue(soldList3, code),
getDataValue(soldList4, code),
getDataValue(soldList5, code),
getDataValue(soldList6, code)]
item.dataSum = sumArray(item.data)
finalList.push(item)
)
sortByDataSum(finalList)
console.log(finalList)
【讨论】：
【参考方案3】：我会在字典中使用数组的组合：
let result =
soldList1.concat(soldList2)
.concat(soldList3)
.concat(soldList4)
.concat(soldList5)
.concat(soldList6)
.forEach(item =>
if(result[item.itemCode])
result[item.itemCode].data.push(item.soldRate);
result[item.itemCode].dataSum += item.soldRate;
else
result[item.itemCode] = data: [item.soldRate], itemCode: item.itemCode, dataSum: item.soldRate;
);
let finalAllConcateAndSortedByDataSumList = Object.values(result).sort((item) => return item.dataSum)
console.log(finalAllConcateAndSortedByDataSumList);
【讨论】：
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