React JS连接数组并转换为新的二维数组并按二维数组数据总和排序

Question

我正在研究本机应用程序 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，某些列表获得了其他列表没有的元素，并且每个列表项的顺序都是desc order by 'soldRate' ，并且每个 list.length 也可能不同。

所以目标是连接这些数组列表并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到新的数组列表'data'属性中，每个数据顺序的位置与上面原始列表 1 完全相同2 3 4 5 6.... 如果 position 中的元素值不存在于任何原始列表中，则它是一个 '0' 值

最终在 finalAllConcateAndSortedByDataSumList 中的每个对象，数据数组将汇总到属性'dataSum'

然后新列表 finalAllConcateAndSortedByDataSumList 是desc order by dataSum

这是我个人尝试使其工作，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，代码示例会非常有帮助谢谢

Answer 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itemCode存在于每个数组中，则向结果中添加soldRate或0 。

 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);

Answer 2

我将您的流程分解为不同的步骤，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。

这是一个代码沙箱，显示了这一点（请参阅控制台输出）。

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)

Answer 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);