Question

如何将此字符串转换为忽略前2个字符和换行符（\ n）或空格的字符数组？所以我有这个：

var leds = """
1 1
XXX
000
00X
"""

将其退还给我：

['x','x','x','0','0','0','0','0','x']

我有这个，但它不会忽略空格，换行符或第一行的编号：

let characters = Array(leds)
print(characters)

非常感谢您

Answer 1

将字符串拆分为行数组，然后放置第一行。然后使用flatMap将每一行映射到一个字符数组并连接结果。

let array = leds.components(separatedBy: .newlines)
    .dropFirst()
    .flatMap(Array.init)

print(array) // ["X", "X", "X", "0", "0", "0", "0", "0", "X"]

使用map代替flatMap，您将获得一个与输入字符串中的行和列相对应的“嵌套数组”：

let board = leds.components(separatedBy: .newlines)
    .dropFirst()
    .map(Array.init)

print(board) // [["X", "X", "X"], ["0", "0", "0"], ["0", "0", "X"]]

Answer 2

这也可能有用-

if let firstLine = leds.components(separatedBy: CharacterSet.newlines).first {
        let removeFirstLineStr = leds.replacingOccurrences(of: firstLine, with: "")
        let characters = String(removeFirstLineStr.filter { !" \n\t\r".contains($0) })
        print(Array(characters))
    }

OR

func getCharacters() {
    guard let firstLineStr = leds.components(separatedBy: "\n").first else {
        return
    }
    let removeFirstLineStr = leds.replacingOccurrences(of: firstLineStr, with: "")
    let characters = String(removeFirstLineStr.filter { !" \n\t\r".contains($0) })
    print(Array(characters))
}

快速将字符串转换并格式化为字符数组

2 个答案: