ios 将 Swift 字符串转换为数组

Convert Swift string to array

提问by Mr.KLD

How can I convert a string "Hello" to an array ["H","e","l","l","o"] in Swift?

如何在 Swift 中将字符串“Hello”转换为数组 ["H","e","l","l","o"]？

In Objective-C I have used this:

在Objective-C中，我使用了这个：

NSMutableArray *characters = [[NSMutableArray alloc] initWithCapacity:[myString length]];
for (int i=0; i < [myString length]; i++) {
    NSString *ichar  = [NSString stringWithFormat:@"%c", [myString characterAtIndex:i]];
    [characters addObject:ichar];
}

回答by Martin R

It is even easier in Swift:

在 Swift 中更简单：

let string : String = "Hello  "
let characters = Array(string)
println(characters)
// [H, e, l, l, o,  , , ,  , ]

This uses the facts that

这使用了以下事实

• an Arraycan be created from a SequenceType, and
• Stringconforms to the SequenceTypeprotocol, and its sequence generator enumerates the characters.

• anArray可以从 a 创建SequenceType，并且
• String符合SequenceType协议，其序列生成器枚举字符。

And since Swift strings have full support for Unicode, this works even with characters outside of the "Basic Multilingual Plane" (such as ) and with extended grapheme clusters (such as , which is actually composed of twoUnicode scalars).

并且由于 Swift 字符串完全支持 Unicode，这甚至适用于“基本多语言平面”之外的字符（例如 ）和扩展的字素簇（例如 ，它实际上由两个Unicode 标量组成）。

---

---

Update: As of Swift 2,Stringdoes no longer conform to SequenceType, but the charactersproperty provides a sequence of the Unicode characters:

更新：从 Swift 2 开始，String不再符合 SequenceType，但该characters属性提供了一系列 Unicode 字符：

let string = "Hello  "
let characters = Array(string.characters)
print(characters)

This works in Swift 3as well.

这也适用于Swift 3。

---

---

Update: As of Swift 4,Stringis (again) a collection of its Characters:

更新：从 Swift 4 开始，String（再次）是其Characters的集合 ：

let string = "Hello  "
let characters = Array(string)
print(characters)
// ["H", "e", "l", "l", "o", " ", "", "", " ", ""]

回答by devxoul

Edit (Swift 4)

编辑（斯威夫特 4）

In Swift 4, you don't have to use charactersto use map(). Just do map()on String.

在 Swift 4 中，您不必characters使用map(). 只需map()在字符串上做。

let letters = "ABC".map { String(
let letters = "ABC".characters.map { String(
$ swift
Welcome to Swift!  Type :help for assistance.
  1> Array("ABC")
$R0: [Character] = 3 values {
  [0] = "A"
  [1] = "B"
  [2] = "C"
}
) }
print(letters) // ["A", "B", "C"]
) }
print(letters) // ["A", "B", "C"]
print(type(of: letters)) // Array<String>

Or if you'd prefer shorter: "ABC".map(String.init)(2-bytes )

或者，如果您更喜欢更短的："ABC".map(String.init)（2 字节）

Edit (Swift 2 & Swift 3)

编辑（Swift 2 和 Swift 3）

In Swift 2 and Swift 3, You can use map()function to charactersproperty.

在 Swift 2 和 Swift 3 中，您可以使用map()函数来设置characters属性。

let str = "ABC"
let arr = map(str) { s -> String in String(s) }

Original (Swift 1.x)

原版 (Swift 1.x)

Accepted answer doesn't seem to be the best, because sequence-converted Stringis not a Stringsequence, but Character:

接受的答案似乎不是最好的，因为序列转换String不是String序列，而是Character：

let string = "1;2;3"
let array = string.components(separatedBy: ";")
print(array) // returns ["1", "2", "3"]

This below works for me:

下面这对我有用：

//array of Characters
let charArr1 = [Character](myString)

//array of String.element
let charArr2 = Array(myString)

for char in myString {
  //char is of type Character
}

Reference for a global function map()is here: http://swifter.natecook.com/func/map/

全局函数的参考在map()这里：http: //swifter.natecook.com/func/map/

回答by Frédéric Adda

There is also this useful function on String: components(separatedBy: String)

String 上也有这个有用的函数：components(separatedBy: String)

//array of String
var strArr = myString.map { String(
let charArr1 = [Character](myString.characters)
let charArr2 = Array(myString.characters)
for char in myString.characters {
  //char is of type Character
}
)}

Works well to deal with strings separated by a character like ";" or even "\n"

可以很好地处理由“;”等字符分隔的字符串 甚至 "\n"

回答by ScottyBlades

Updated for Swift 4

为 Swift 4 更新

Here are 3 ways.

这里有3种方法。

var strArr = myString.characters.map { String(
extension String {
   func letterize() -> [Character] {
     return Array(self.characters)
  }
}
)}

In some cases, what people really want is a way to convert a string into an array of little strings with 1 character length each. Here is a super efficient way to do that:

在某些情况下，人们真正想要的是一种将字符串转换为每个长度为 1 个字符的小字符串数组的方法。这是一个超级有效的方法：

let charArr = "Cat".letterize()

Swift 3

斯威夫特 3

Here are 3 ways.

这里有3种方法。

    let string = "hell0"
    let ar = Array(string.characters)
    print(ar)

In some cases, what people really want is a way to convert a string into an array of little strings with 1 character length each. Here is a super efficient way to do that:

在某些情况下，人们真正想要的是一种将字符串转换为每个长度为 1 个字符的小字符串数组的方法。这是一个超级有效的方法：

let array1 = Array("hello") // Array<Character>
let array2 = Array("hello").map({ "\(
let characters = "Hello"
var charactersArray: [Character] = []

for (index, character) in enumerate(characters) {
    //do something with the character at index
    charactersArray.append(character)
}

println(charactersArray)
)" }) // Array<String>
let array3 = "hello".map(String.init) // Array<String>

Or you can add an extension to String.

或者您可以向 String 添加扩展名。

var strArray = "Hello, playground".Letterize()

extension String {
    func Letterize() -> [String] {
        return map(self) { String(
func letterize() -> [Character] {
    return Array(self.characters)
}
) }
    }
}

Then you can call it like this:

然后你可以这样称呼它：

let someText = "hello"

let array = someText.map({ String(##代码##) }) // [String]

回答by William Hu

##代码##

回答by onmyway133

In Swift 4, as Stringis a collection of Character, you need to use map

在 Swift 4 中，作为String的集合Character，您需要使用map

##代码##

回答by Alberto Barrera

Martin R answer is the best approach, and as he said, because String conforms the SquenceType protocol, you can also enumerate a string, getting each character on each iteration.

Martin R 的答案是最好的方法，正如他所说，因为 String 符合 SquenceType 协议，您还可以枚举一个字符串，在每次迭代中获取每个字符。

##代码##

回答by fabrizioM

You can also create an extension:

您还可以创建扩展：

##代码##

回答by Piotr Pszczó?kowski

##代码##

回答by atolite

An easy way to do this is to mapthe variable and return each Characteras a String:

一个简单的方法是对map变量并将每个Character作为 a返回String：

##代码##

The output should be ["h", "e", "l", "l", "o"].

输出应该是["h", "e", "l", "l", "o"].