如何使用相同的键对对象值求和并返回一个对象？

Question

我试图根据数组的每个对象中可用的选定键来对对象值求和。 我的问题是如何处理percentage rate 。 在下面的代码示例中， satisfaction_rate应该有48 。

我的输出应该是：

{ 'students' : 87, 'books': 32, 'satisfaction_rate': 48 }

 const myData = [ { 'students' : 30, 'books': 10, 'satisfaction_rate': "60%", 'returning_students': 14 }, { 'students' : 25, 'books': 8, 'satisfaction_rate': "45%", 'returning_students': 14 }, { 'students' : 32, 'books': 14, 'satisfaction_rate': "39%", 'returning_students': 19 } ]; const keysToConsider = ['students', 'books', 'satisfaction_rate']; function getSumOrAvgValuesOfKeys(data){ const obj = {}; let val = 0; keysToConsider.forEach(el => { data.forEach(element => { if (typeof(element[el] === 'string')) { val += parseInt(element[el], 10); } else { val += element[el]; } }); obj[el] = val; // Reset value val = 0; }); return obj; } console.log(getSumOrAvgValuesOfKeys(myData)); 

Answer 1

您可以使用reduce方法，然后在最后一次迭代中计算每个元素的平均值，其中value是string的类型。

 const myData = [{"students":30,"books":10,"satisfaction_rate":"60%","returning_students":14},{"students":25,"books":8,"satisfaction_rate":"45%","returning_students":14},{"students":32,"books":14,"satisfaction_rate":"39%","returning_students":19}] const keys = ['students', 'books', 'satisfaction_rate']; const result = myData.reduce((r, e, i, a) => { keys.forEach(k => r[k] = (r[k] || 0) + parseInt(e[k])); if(!a[i + 1]) Object.keys(e) .filter(k => typeof e[k] == 'string') .forEach(k => r[k] /= myData.length) return r; }, {}) console.log(result) 

Answer 2

您可能有另一个数组具有要平均的键，并且在准备求和结果后，您可以仅计算这些键的平均值。

 const myData = [{ 'students': 30, 'books': 10, 'satisfaction_rate': "60%", 'returning_students': 14 }, { 'students': 25, 'books': 8, 'satisfaction_rate': "45%", 'returning_students': 14 }, { 'students': 32, 'books': 14, 'satisfaction_rate': "39%", 'returning_students': 19 }]; const keysToConsider = ['students', 'books', 'satisfaction_rate']; let keysWithPercent = ['satisfaction_rate']; const summed = myData.reduce((result, item) => { keysToConsider.forEach(k => { if (!result[k]) { result[k] = 0; } let v = parseInt(item[k], 10); result[k] += v; }); return result; }, {}); keysWithPercent.forEach(k => { summed[k] = summed[k] / myData.length; }) console.log(summed); 

Answer 3

您可以通过采用先前的值或默认值来减少对象。 对于百分比值，需要三分之一进行求和。

 var data = [{ students: 30, books: 10, satisfaction_rate: "60%", returning_students: 14 }, { students: 25, books: 8, satisfaction_rate: "45%", returning_students: 14 }, { students: 32, books: 14, satisfaction_rate: "39%", returning_students: 19 }], result = data.reduce((r, { students, books, satisfaction_rate }, _, a) => ({ students: (students || 0) + students, books: (r.books || 0) + books, satisfaction_rate: (r.satisfaction_rate || 0) + satisfaction_rate.slice(0, -1) / a.length }), {}); console.log(result); 

使用动态密钥列表

 var data = [{ students: 30, books: 10, satisfaction_rate: "60%", returning_students: 14 }, { students: 25, books: 8, satisfaction_rate: "45%", returning_students: 14 }, { students: 32, books: 14, satisfaction_rate: "39%", returning_students: 19 }], keys = ['students', 'books', 'satisfaction_rate'], result = data.reduce((r, o, _, a) => Object.assign( ...keys.map(k => ({ [k]: (r[k] || 0) + (o[k].toString().endsWith('%') ? o[k].slice(0, -1) / a.length : o[k]) })) ), {} ); console.log(result); 

Answer 4

恕我直言，你要做的事情可以使用如下的reduce来完成：

 var myData = [{ students: 30, books: 10, satisfaction_rate: "60%", returning_students: 14 }, { students: 25, books: 8, satisfaction_rate: "45%", returning_students: 14 }, { students: 32, books: 14, satisfaction_rate: "39%", returning_students: 19 }] var result = myData.reduce((accum, { students, books, satisfaction_rate }, idx, arr) => { accum["students"] ? (accum["students"] += students) : (accum["students"] = students); accum["books"] ? (accum["books"] += books) : (accum["books"] = books); accum["satisfaction_rate"] ? (accum["satisfaction_rate"] += parseInt(satisfaction_rate)) : (accum["satisfaction_rate"] = parseInt(satisfaction_rate)); if (idx === arr.length - 1) { accum["satisfaction_rate"] = accum["satisfaction_rate"] / arr.length } return accum; }, {}); console.log(result); 

Answer 5

更通用的解决方案是指定如何减少 keysToConsider对象中的每个键。 你可以使用SUM，AVG，MIN，MAX减少它们，甚至更复杂的标准偏差。

在这种情况下，我会将keysToConsider更改为对象而不是数组：

 const myData = [{ 'students': 30, 'books': 10, 'satisfaction_rate': "60%", 'returning_students': 14 }, { 'students': 25, 'books': 8, 'satisfaction_rate': "45%", 'returning_students': 14 }, { 'students': 32, 'books': 14, 'satisfaction_rate': "39%", 'returning_students': 19 }]; keysToConsider = { 'students': sum, 'books': sum, 'satisfaction_rate': avg } function reduceValuesOfKeys(data) { const obj = {}; let val = 0; for (var p in keysToConsider) { var reduce = keysToConsider[p]; obj[p] = reduce(data, p); } return obj; } function sum(data, property) { var val = 0; data.forEach(element => { if (typeof(element[property] === 'string')) { val += parseInt(element[property], 10); } else { val += element[property]; } }); return val; } function avg(data, property) { var val = 0; data.forEach(element => { if (typeof(element[property] === 'string')) { val += parseInt(element[property], 10); } else { val += element[property]; } }); return val / data.length; } console.log(reduceValuesOfKeys(myData)); 

如果您将来需要添加其他reduce功能，那就更容易了。