Swift 中的条件删除集合的最后一个元素

Question

我试图从字符串数组的后面删除""& ，直到最后一项包含一些文本，但我的实现没有拾取." "" "

到目前为止我的实现：

var array = ["A", "B", "", "C", "D", " ", " ", ""]

while true {
    if (array.last == " " || array.last == "") {
        array.removeLast()
    } else {
        break
    }
}

我想要的输出是

["A", "B", "", "C", "D"]

，但我当前的输出是

["A", "B", "", "C", "D", " ", " "]

，其中在遇到后while立即循环breaks" "

有什么建议为什么它不接听吗" "？

Answer 1

我不知道为什么他们有drop(while:)但没有实施dropLast(while:)。下面的实现适用于任何集合：

extension Collection {
    func dropLast(while predicate: (Element) throws -> Bool) rethrows -> SubSequence {
        guard let index = try indices.reversed().first(where: { try !predicate(self[$0]) }) else {
            return self[startIndex..<startIndex]
        }
        return self[...index]
    }
}

"123".dropLast(while: \.isWholeNumber)    // ""
"abc123".dropLast(while: \.isWholeNumber) // "abc"
"123abc".dropLast(while: \.isWholeNumber) // "123abc"

并扩展 RangeReplaceableCollection 我们也可以remove(while:)实现：removeLast(while:)

extension RangeReplaceableCollection {
     mutating func remove(while predicate: (Element) throws -> Bool) rethrows {
        guard let index = try indices.first(where: { try !predicate(self[$0]) }) else {
            removeAll()
            return
        }
        removeSubrange(..<index)
    }
    mutating func removeLast(while predicate: (Element) throws -> Bool) rethrows {
        guard let index = try indices.reversed().first(where: { try !predicate(self[$0]) }) else {
            removeAll()
            return
        }
        removeSubrange(self.index(after: index)...)
    }
}

var string = "abc123"
string.removeLast(while: \.isWholeNumber)
string  // "abc"

var string2 = "abc123"
string2.remove(while: \.isLetter)
string2 // "123"

var array = ["A", "B", "", "C", "D", " ", " ", ""]
array.removeLast { $0 == "" || $0 == " " }
array  // ["A", "B", "", "C", "D"]

Answer 2

解决此问题的一种方法是反转集合（这是延迟完成的）并删除不需要的项目，直到遇到想要的项目。之后，再次反转集合。

let array = ["A", "B", "", "C", "D", " ", " ", ""]

let filtered = array.reversed().drop(while: {
    $0.trimmingCharacters(in: .whitespacesAndNewlines).isEmpty
}).reversed() as [String]

print(filtered) // "["A", "B", "", "C", "D"]\n"

" "请注意，如果不是普通空格，例如不间断空格（Unicode 检查点 U+00A0），则检查可能会失败。这可能是您首先遇到的问题。因此，修剪字符串（仅删除开头和结尾的字符）并检查结果是否为空字符串。