sorted()中元素的索引 [英] Index of element in sorted()

问题描述

致电时

sorted(<#source: C#>, <#isOrderedBefore: (C.Generator.Element, C.Generator.Element) -> Bool##(C.Generator.Element, C.Generator.Element) -> Bool#>)

您可以通过$0和$1访问两个源元素进行比较.但是，如何确定从源中获取索引的索引呢?

you can access the two source elements for comparison via $0 and $1. But how can I determine the index from where they were taken from the source?

推荐答案

您可以将sorted传递给sorted，而不是元素的集合，而是元素的索引:

You could pass into sorted not a collection of elements, but the indices of the elements:

let a = ["hello","i","must","be","going"]
let idxs = sorted(indices(a)) { a[$0] < a[$1] }
// produces [3, 4, 0, 1, 2]

或者，如果您不喜欢捕获a并希望将元素本身传递到闭包中，则可以传递一系列的索引和元素对，如下所示:

Or, if you don't like capturing a and wanted the elements themselves passed into the closure, you could pass in a sequence of pairs of the index and the element, like so:

let pairs = sorted(Zip2(indices(a),a)) {
    $0.1 < $1.1
}

请注意，结果将是(index,element)对的数组:[(3, be), (4, going), (0, hello), (1, i), (2, must)].如果您只想将其转换为元素，可以执行map(pairs) { $0.1 }

Note the result would be an array of (index,element) pairs: [(3, be), (4, going), (0, hello), (1, i), (2, must)]. If you wanted to turn that back into just the elements, you can do map(pairs) { $0.1 }

此外，如果您选择了正义指数路线，并希望以后再将其转换为元素，则可以使用PermutationGenerator做到这一点:

Also, if you take the just-indices route and want to turn that back into elements later, you can do it with PermutationGenerator:

let values = PermutationGenerator(elements: a, indices: idxs)
println(" ".join(values)) // prints "be going hello i must"

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