这不是副本。请看我下面的评论!
有没有人知道比ES6中循环更有效的解决方案?
我写了以下内容,性能欠佳。有什么改进的想法吗?非常感谢。
基本上,我有一个关于汽车的对象和一个关于用户偏好的数组。预期的行为是将所有相关的汽车名称放入一个数组中。
用户可以提供任意数量的首选项。只有在首选项中提到了所有规范时,才应该推送汽车名称。因此,一些偏好将是“剩余物”。
出于这个原因,在下面的示例中出现了本田,但不是宝马,这是预期的(但非常慢的行为)。
// Car objects
const cars = [{
name: "Honda",
category: "eco",
specs: {
0: "green",
1: "fast",
2: "automatic"
}
},
{
name: "BMW",
category: "sport",
specs: {
0: "blue",
1: "fast",
2: "automatic"
}
}
]
// User preferences
const preferences = ["green", "fast", "4x4", "automatic", "panorama"]
// function to get length/amount of car specifications
function objsize(Myobj) {
var osize = 0,
key;
for (key in Myobj) {
if (Myobj.hasOwnProperty(key)) osize++;
}
return Object(osize);
};
//function to check if ALL specifications are included in the user preferences
function checkSpecs(spec_item) {
return preferences.includes(spec_item)
}
// main function
function filter_func() {
//final results
let matched_cars = []
for (i = 0; i < objsize(cars); i++) {
let specs_collector = []
for (j = 0; j < objsize(cars[i].specs); j++) {
specs_collector.push(cars[i].specs[j])
}
if (specs_collector.every(checkSpecs) === true) {
matched_cars.push(cars[i].name)
specs_collector = []
}
}
console.log(matched_cars)
}
filter_func()
发布于 2018-10-17 07:58:30
你不能不看每一辆车,也不能不看每一辆车的规格,因为你想测试每一辆车。通过使用Set,您可以避免每次循环遍历首选项。
因此,这可能会更快,也可能不会更快,但它要简单得多,也更容易理解,因为代码读起来几乎像英语: filter cars,其中每个规范都在首选项中:
// Car objects
const cars = [{
name: "Honda",
category: "eco",
specs: ["green", "fast","automatic"]
},
{
name: "BMW",
category: "sport",
specs: ["blue", "fast","automatic"]
}
]
const preferences = new Set(["green", "fast", "4x4", "automatic", "panorama"])
let filtered = cars.filter(car => car.specs.every(spec => preferences.has(spec)))
console.log(filtered)
发布于 2018-10-17 07:58:41
-编辑--
使用操作中的数据:
const array_intersect = (a, b) => a.filter( i => (b.indexOf(i) >= 0) )
const a_contains_b = (a, b) => array_intersect(a, b).length == b.length
var cars = [{
name: "Honda",
category: "eco",
specs: ["green", "fast", "automatic"]
},
{
name: "BMW",
category: "sport",
specs: ["blue", "fast", "automatic"]
}
]
const preferences = ["green", "fast", "4x4", "automatic", "panorama"]
let filtered = cars.filter(car => a_contains_b(preferences, car.specs))
console.log(filtered);
发布于 2018-10-17 09:58:17
至少有一个循环是无法逃脱的。你总是要在所有的车里循环,不管是for...或者使用另一个构造,如array.filter()。但是,还有另一种方法可以获得性能。您可以使用位掩码。这将需要更改car对象的数据结构,以便每辆汽车已经包含与其规格相对应的位掩码,并且当用户选择所需的规格时,同样应该添加规格代码。(然而,我怀疑这可能会带来很大的麻烦,但收获很少。)
// Let's pretend there are preset binary digits corresponding
// to each one of the available preferences:
//
// "blue" => 1
// "green" => 2
// "red" => 4
// "fast" => 8
// "slow" => 16
// "automatic" => 32
// "4x4" => 64
// "panorama" => 128
//
// You would encode this into the data before processing
var cars = [{
name: "Honda",
category: "eco",
specs: ["green", "fast", "automatic"],
bin_specs: 42 // 2 + 8 + 32
},
{
name: "BMW",
category: "sport",
specs: ["blue", "fast", "automatic"],
bin_specs: 41 // 1 + 8 + 32
}
]
const preferences = ["green", "fast", "4x4", "automatic", "panorama"]
const bin_preferences = 234 // 2 + 8 + 64 + 32 + 128]
let filtered = cars.filter(car => (car.bin_specs & bin_preferences) === car.bin_specs)
console.log(filtered);
https://stackoverflow.com/questions/52845297
复制相似问题