Question

使用托儿所-我需要在两个数组的两个列上匹配两个数组：

部门出勤：[日期，部门，出席情况]

    0: (3) ["09-30-2019", "00_Infants", 22]
    1: (3) ["09-30-2019", "01_Ones", 38]
    2: (3) ["09-30-2019", "02_Twos", 40]
    3: (3) ["09-30-2019", "03_Threes", 42]
    4: (3) ["09-30-2019", "04_Fours", 19]
    5: (3) ["10-01-2019", "00_Infants", 27]
    6: (3) ["10-01-2019", "01_Ones", 42]
    7: (3) ["10-01-2019", "02_Twos", 51]

ETC ...

部门费用[日期，部门，费用]

    0: (3) ["09-30-2019", "00_Infants", "584.56"]
    1: (3) ["09-30-2019", "01_Ones", "701.51"]
    2: (3) ["09-30-2019", "02_Twos", "614.02"]
    3: (3) ["09-30-2019", "03_Threes", "442.50"]
    4: (3) ["09-30-2019", "04_Fours", "166.65"]
    5: (3) ["09-30-2019", "06_Floater", "141.37"]
    6: (3) ["09-30-2019", "07_Office", "246.91"]
    7: (3) ["09-30-2019", "08_Administration", "0.00"]
    8: (3) ["09-30-2019", "09_Director", "0.00"]
    9: (3) ["09-30-2019", "12_Kitchen", "0.00"]
    10: (3) ["10-01-2019", "00_Infants", "760.37"]
    11: (3) ["10-01-2019", "01_Ones", "756.48"]
    12: (3) ["10-01-2019", "02_Twos", "640.23"]
    13: (3) ["10-01-2019", "03_Threes", "552.66"]

-

我只想将.department出现在出勤情况中，将Expense.date && Expense.department与Attendance.date && Attendance.department进行匹配然后，将Expense.expense附加到出勤中的匹配记录中

我尝试过映射，过滤，d3.js nest（），Object.assign（），ifs，$。each ...

这是我还没有删除并从头开始的最后一件非常糟糕的事情-我知道仅此一项看起来就很糟糕，而且还没有结束；但是，需要的代码。

    let ffs = attended.map(function (x, i) {
    return {
        "date": emp[x],
        "other": emp[i][0]
    }
    }.bind(this));

let mfer = attended.map(x => Object.assign(x, emp.find(y => y[0] === x[0])));

预期结果：[日期，部门，出席情况，费用]

    0: (3) ["09-30-2019", "00_Infants", 22, 584.56]
    1: (3) ["09-30-2019", "01_Ones", 38, 701.51]
    2: (3) ["09-30-2019", "02_Twos", 40, 613.02]

实际结果：除了一切之外的一切。

我的浏览器搜索记录显示昨天19:00至现在08:30之间的115条相关查询。

我认为这应该是简单的解决方案，但是我处于学习状态，我所面临的挑战超出了我应该花的时间。

- 我认为这与我需要的最接近，但是当尝试将第二个过滤器应用于捕获部门时，一切都变得不清楚：

Applying filter within a filter in JavaScript

一些打开的标签...

Find all matching elements with in an array of objects

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

Javascript multiple filters array

merge two arrays including some fields from each

Answer 1

您可以通过“键”将Map中的这些元素分组。在您的情况下，键可以是日期和部门。所以

从费用数组创建映射
遍历出席数组，从上面创建的分组中获取匹配键的费用，并将费用附加到该数组

const attendance = [
  ["09-30-2019", "00_Infants", 22],
  ["09-30-2019", "01_Ones", 38],
  ["09-30-2019", "02_Twos", 40],
  ["09-30-2019", "03_Threes", 42],
  ["09-30-2019", "04_Fours", 19],
  ["10-01-2019", "00_Infants", 27],
  ["10-01-2019", "01_Ones", 42],
  ["10-01-2019", "02_Twos", 51]
];
const expenses = [
  ["09-30-2019", "00_Infants", "584.56"],
  ["09-30-2019", "01_Ones", "701.51"],
  ["09-30-2019", "02_Twos", "614.02"],
  ["09-30-2019", "03_Threes", "442.50"],
  ["09-30-2019", "04_Fours", "166.65"],
  ["09-30-2019", "06_Floater", "141.37"],
  ["09-30-2019", "07_Office", "246.91"],
  ["09-30-2019", "08_Administration", "0.00"],
  ["09-30-2019", "09_Director", "0.00"],
  ["09-30-2019", "12_Kitchen", "0.00"],
  ["10-01-2019", "00_Infants", "760.37"],
  ["10-01-2019", "01_Ones", "756.48"],
  ["10-01-2019", "02_Twos", "640.23"],
  ["10-01-2019", "03_Threes", "552.66"]
];

const key = (date, department) => `${date};${department}`;
const expenseMap = new Map(expenses.map(([date, dept, expense]) => [key(date, dept), expense]));

attendance.forEach(a => a.push(expenseMap.get(key(a[0], a[1]))));
console.log(attendance)

Answer 2

遍历Attendance，并用Array.prototype.find()检查Departments中是否有匹配项。如果找到匹配的部门，我们会将（解析的）金额推送到出勤条目中。

const attendance = [["09-30-2019", "00_Infants", 22],["09-30-2019", "01_Ones", 38],["09-30-2019", "02_Twos", 40],["09-30-2019", "03_Threes", 42],["09-30-2019", "04_Fours", 19],["10-01-2019", "00_Infants", 27],["10-01-2019", "01_Ones", 42],["10-01-2019", "02_Twos", 51]];
const departments = [["09-30-2019", "00_Infants", "584.56"],["09-30-2019", "01_Ones", "701.51"],["09-30-2019", "02_Twos", "614.02"],["09-30-2019", "03_Threes", "442.50"],["09-30-2019", "04_Fours", "166.65"],["09-30-2019", "06_Floater", "141.37"],["09-30-2019", "07_Office", "246.91"],["09-30-2019", "08_Administration", "0.00"],["09-30-2019", "09_Director", "0.00"],["09-30-2019", "12_Kitchen", "0.00"],["10-01-2019", "00_Infants", "760.37"],["10-01-2019", "01_Ones", "756.48"],["10-01-2019", "02_Twos", "640.23"],["10-01-2019", "03_Threes", "552.66"]];


attendance.forEach(a => {
  const department = departments.find(d => a[0] === d[0] && a[1] === d[1]);
  
  if (department) {
    a.push(parseFloat(department[2]));
  }
});

console.log(attendance);

Answer 3

您可以在出勤阵列上使用map()，并使用find()查找该日期和部门的费用。

如果存在费用，则追加费用，否则返回相同的项目。

var attendance = [ ["09-30-2019", "00_Infants", 22], ["09-30-2019", "01_Ones", 38], ["09-30-2019", "02_Twos", 40], ["09-30-2019", "03_Threes", 42], ["09-30-2019", "04_Fours", 19], ["10-01-2019", "00_Infants", 27], ["10-01-2019", "01_Ones", 42], ["10-01-2019", "02_Twos", 51] ];

var expenses = [ ["09-30-2019", "00_Infants", "584.56"], ["09-30-2019", "01_Ones", "701.51"], ["09-30-2019", "02_Twos", "614.02"], ["09-30-2019", "03_Threes", "442.50"], ["09-30-2019", "04_Fours", "166.65"], ["09-30-2019", "06_Floater", "141.37"], ["09-30-2019", "07_Office", "246.91"], ["09-30-2019", "08_Administration", "0.00"], ["09-30-2019", "09_Director", "0.00"], ["09-30-2019", "12_Kitchen", "0.00"], ["10-01-2019", "00_Infants", "760.37"], ["10-01-2019", "01_Ones", "756.48"], ["10-01-2019", "02_Twos", "640.23"], ["10-01-2019", "03_Threes", "552.66"] ];

var result = attendance.map(item => {
  let expense = expenses.find(expense => expense[0] === item[0] && expense[1] === item[1]);
  if (expense) {
    return [...item, expense[2]];
  } else {
    return item;
  }
});

console.log(result);

过滤并合并两个数组，匹配两个字段/列：Javascript

实际结果：除了一切之外的一切。

我的浏览器搜索记录显示昨天19:00至现在08:30之间的115条相关查询。

Applying filter within a filter in JavaScript

3 个答案: