将Swift数组过滤器的结果限制为X以提高性能

问题描述

我的数组中有大约2000个元素，当对其进行过滤时，只要过滤的数组中有5个元素，我就想结束过滤.

I have about 2000 elements in my array, and when it is filtered, I would like to end the filtering as soon as I have 5 elements in my filtered array.

当前是:

providerArray.filter({($0.lowercased().range(of:((row.value as? String)?.lowercased())!) != nil)})

最多可以返回2000个结果，这浪费了处理时间和时间.

which can return up to 2000 results which is a waste of processing, and time.

更清楚地说，我需要一个解决方案，它等同于限制过滤器结果，就像我可以使用coreData提取[request setFetchLimit:5];

To be more clear, I need a solution that is equivalent to limiting the filter results like I can with coreData fetches [request setFetchLimit:5];

推荐答案

就执行时间而言，最快的解决方案似乎是显式循环，添加匹配的元素直到达到限制:

The fastest solution in terms of execution time seems to be anexplicit loop which adds matching elements until the limit is reached:

extension Sequence {
    public func filter(where isIncluded: (Iterator.Element) -> Bool, limit: Int) -> [Iterator.Element] {
        var result : [Iterator.Element] = []
        result.reserveCapacity(limit)
        var count = 0
        var it = makeIterator()

        // While limit not reached and there are more elements ...
        while count < limit, let element = it.next() {
            if isIncluded(element) {
                result.append(element)
                count += 1
            }
        }
        return result
    }
}

示例用法:

let numbers = Array(0 ..< 2000)
let result = numbers.filter(where: { $0 % 3 == 0 }, limit: 5)
print(result) // [0, 3, 6, 9, 12]

这篇关于将Swift数组过滤器的结果限制为X以提高性能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！