[英]ES6: How do I reduce this array of objects to unique values?
我一直在旋转我的轮子太久试图解决这个问题,我准备把我的电脑扔出窗外。 我该如何减少这个数组:
const array = [
{'location': "Plovdiv", 'department': "Finance"},
{'location': "Plovdiv", 'department': "Client & Employee Support"},
{'location': "Plovdiv", 'department': "Client & Employee Support"},
{'location': "London", 'department': "Engineering"},
{'location': "London", 'department': "Engineering"},
{'location': "Plovdiv", 'department': "Engineering"}
];
对此:
{'location': "Plovdiv", 'department': ["Finance", "Client & Employee Support", "Engineering"]},
{'location': "London", 'department': ["Engineering"]},
我的目标是删除具有位置的对象中的重复数组,并将它们合并为一个键。 其中每个键都会有对应的部门作为一个列表。
编辑:我终于设法用普通的 ol' JS 解决了这个问题,但它很笨重,并且有很多循环,我可能可以用更现代的方法摆脱它们。
let locArray = [];
let newGrouping = [];
// For each location
for (let i = 0; i < locations.length; i += 1) {
// Check to see if the current locations is in the new locArray
if (locArray.indexOf(locations[i].location) === -1) {
// Get in there!
locArray.push(locations[i].location);
}
}
// Loop through the new set of unique locations
for (let i = 0; i < locArray.length; i += 1) {
let depArray = [];
// Loop through our original locations array
for (let j = 0; j < locations.length; j += 1) {
// Check to see if the current unique location matches the current location
// AND make sure that it's not already in depArray
if (locArray[i] === locations[j].location && depArray.indexOf(locations[j].department) === -1) {
// Get in there!
depArray.push(locations[j].department);
}
}
// Push our current unique location and its unique departments into a new object
newGrouping.push({
'location': locArray[i],
'departments': depArray
});
}
使用Set
和map
我们想要唯一地键入location
来做到这一点,我们使用Set
来确保唯一性。
// Notice we use the map function to pull just location.
new Set(array.map(({ location }) => location))
但是现在我们需要迭代这些唯一键并重建我们的数组。
所以我们将Set
加载到一个Array
const unique = new Array(...new Set(array.map(({ location }) => location)))
现在我们有一个唯一location
array
,从这里我们可以使用 map 函数来构建我们想要的array
输出。
请注意,当我们构建最终的object
array
时, department
参数是如何使用原始数组的filter
和map
重新水化的。
[Unique Location Array].map(location => ({
location, // ES6 the property name it is inferred
department: array.filter(({ location: l}) => location === l)
.map(({ department }) => department)
}));
const array = [ {'location': "Plovdiv", 'department': "Finance"}, {'location': "Plovdiv", 'department': "Client & Employee Support"}, {'location': "Plovdiv", 'department': "Client & Employee Support"}, {'location': "London", 'department': "Engineering"}, {'location': "London", 'department': "Engineering"}, {'location': "Plovdiv", 'department': "Engineering"} ]; const unique = new Array(...new Set(array.map(({ location }) => location))) .map(location => ({ location, department: array.filter(({ location: l}) => location === l) .map(({ department }) => department) })); console.log(unique);
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