如何比较两个＆QUOT;通用＆QUOT;数组中迅速 [英] How to compare two "generic" array in swift

问题描述

我有两个数组：

var tstFrames = [TimeFrame]()
var tstFramesRefresh = [TimeFrame]()

是否有人知道如何将它们进行比较？

Does anybody know how to compare them?

我发现了一些实例添加数组设置，但设置没有初始化，可以接受此数组类型..：（

I found some examples adding array to Set, but Set doesn't have initializer which can accept this arraytype.. :(

两个阵列具有几乎时限对象的同一实例，但tstFramesRefresh具有一个更这是不是在tstFrames阵列我会得到一个时间框架对象。

The two array has almost the same instances of TimeFrame object but tstFramesRefresh has one more which is not in tstFrames array i would get that one TimeFrame object.

推荐答案

当你说比较 - 你的意思是平等的，字典比较或差异？听起来像是你想有一个差异。

When you say "compare" – do you mean equal, lexicographic comparison, or diff? Sounds like you want a diff.

该组解决方案可能是这样的（假设你想在没有其他的一个东西，而不是脱节二）：

The set solution might look like this (assuming you want things in one not in the other, rather than the disjunction of the two):

let a = [1,2,3]
let b = [1,2]

Set(a).subtract(b)  // returns a set of {3}

presumably您的问题是，时间表未哈希，因此它不会里面工作集。

Presumably your issue is that TimeFrame is not Hashable, hence it won’t work inside a set.

你也可以使用包含来检查，如果每个元素 A 为成员乙。不幸的是这将是大型成套相当低效的，因为包含工作于线性时间，所以复杂度将为O（n ^ 2）。但它很容易做到：

You could instead use contains to check if each element of a were a member of b. Unfortunately this will be quite inefficient for large sets, since contains works in linear time so complexity will be O(n^2). But it’s easy to do:

func subtract<S: SequenceType, C: CollectionType, T: Equatable
    where C.Generator.Element == T, S.Generator.Element == T>
    (source: S, toSubtract: C) -> [S.Generator.Element] {
        return filter(source) { !contains(toSubtract, $0) }
}


subtract(a, b)  // returns [3]

更有效的解决方案，如果你的对象是可比可以使用排序或树结构。

More efficient solutions if your object were Comparable could use sorting or a tree structure.

如果你的对象不是 Equatable 要么，你需要一个带一个封闭检查等价版本：

If your objects aren’t Equatable either, you’d need a version that took a closure to check for equivalence:

func subtract<S: SequenceType, C: CollectionType>
    (source: S, toSubtract: C,
     isEquivalent: (S.Generator.Element, C.Generator.Element) -> Bool)
    -> [S.Generator.Element] {
        let x = filter(source) { elem in
            return !contains(toSubtract) { isEquivalent(elem,$0) }
        }
        return x
}

subtract(a, b) { $0 == $1 }

如果你只是要检查它们是否相等，那么 == 将做到这一点 - 除非对象不equatable，在这种情况下使用，它接受一个比较函数等于功能：

If you just want to check if they are equal, then == will do it – unless the objects aren’t equatable, in which case use the equal function that takes a comparator function:

// asuming these are classes and reference equality is reasonable thing to check
equal(tstFrames, tstFramesRefresh) { $0 === $1 } 

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