blocks|key|1966847|text|var+testString+=+"abcdefghijklmnopqrstu"
var+startingPoint:+Int+=+0
var+substringLength:+Int+=+1
var+substringArray+=+[AnyHashable]()
for+i+in+0..<(testString.count+??+0)+/+substringLength+{
++++var+substring:+String+=+(testString+as+NSString).substring(with:+NSRange(location:+startingPoint,+length:+substringLength))
++++substringArray.append(substring)
++++startingPoint+%2B=+substringLength
}
print("\(substringArray)")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1966848|OutPut：(+a，b，c，d，e，f，g，h，i，j，k，l，m，n，o，p，q，r，s，t，u)|unstyled|1966849|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>var testString = "abcdefghijklmnopqrstu"
var startingPoint: Int = 0
var substringLength: Int = 1
var substringArray = [AnyHashable]()
for i in 0..&lt;(testString.count ?? 0) / substringLength {
 var substring: String = (testString as NSString).substring(with: NSRange(location: startingPoint, length: substringLength))
 substringArray.append(substring)
 startingPoint += substringLength
}
print("\(substringArray)")
</code></pre>

OutPut :
(
 a,
 b,
 c,
 d,
 e,
 f,
 g,
 h,
 i,
 j,
 k,
 l,
 m,
 n,
 o,
 p,
 q,
 r,
 s,
 t,
 u
 )

blocks|key|550514|text|可以将集合元素(在本例中为字符)分组为n个元素，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550515|extension+Collection+{
++++func+unfoldSubSequences(limitedTo+maxLength:+Int)+->+UnfoldSequence<SubSequence,Index>+{
++++++++sequence(state:+startIndex)+{+start+in
++++++++++++guard+start+<+self.endIndex+else+{+return+nil+}
++++++++++++let+end+=+self.index(start,+offsetBy:+maxLength,+limitedBy:+self.endIndex)+??+self.endIndex
++++++++++++defer+{+start+=+end+}
++++++++++++return+self[start..<end]
++++++++}
++++}
++++func+subSequences(of+n:+Int)+->+[SubSequence]+{
++++++++.init(unfoldSubSequences(limitedTo:+n))
++++}
}|code-block|syntax|javascript|550516|let+numbers+=+"1234567"
let+subSequences+=+numbers.subSequences(of:+2)
print(subSequences)++++//+["12",+"34",+"56",+"7"]|550517|编辑/更新|offset|length|style|BOLD|550518|如果要在最后一组中附加超出的字符：|550519|extension+Collection+{
++++func+unfoldSubSequencesWithTail(lenght:+Int)+->+UnfoldSequence<SubSequence,Index>+{
++++++++let+n+=+count+/+lenght
++++++++var+counter+=+0
++++++++return+sequence(state:+startIndex)+{+start+in
++++++++++++guard+start+<+endIndex+else+{+return+nil+}
++++++++++++let+end+=+index(start,+offsetBy:+lenght,+limitedBy:+endIndex)+??+endIndex
++++++++++++counter+%2B=+1
++++++++++++if+counter+==+n+{
++++++++++++++++defer+{+start+=+endIndex+}
++++++++++++++++return+self[start...]
++++++++++++}+else+{
++++++++++++++++defer+{+start+=+end+}
++++++++++++++++return+self[start..<end]
++++++++++++}
++++++++}
++++}
++++func+subSequencesWithTail(n:+Int)+->+[SubSequence]+{
++++++++.init(unfoldSubSequencesWithTail(lenght:+n))
++++}
}|550520|let+numbers+=+"1234567"
let+subSequencesWithTail+=+numbers.subSequencesWithTail(n:+2)
print(subSequencesWithTail)++++//+["12",+"34",+"567"]|550521|entityMap^0|0|0|0|0|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

You can group your collection elements (in this case Characters) every n elements as follow:
<pre><code>extension Collection {
 func unfoldSubSequences(limitedTo maxLength: Int) -&gt; UnfoldSequence&lt;SubSequence,Index&gt; {
 sequence(state: startIndex) { start in
 guard start &lt; self.endIndex else { return nil }
 let end = self.index(start, offsetBy: maxLength, limitedBy: self.endIndex) ?? self.endIndex
 defer { start = end }
 return self[start..&lt;end]
 }
 }
 func subSequences(of n: Int) -&gt; [SubSequence] {
 .init(unfoldSubSequences(limitedTo: n))
 }
}
</code></pre>
<hr />
<pre><code>let numbers = &quot;1234567&quot;
let subSequences = numbers.subSequences(of: 2)
print(subSequences) // [&quot;12&quot;, &quot;34&quot;, &quot;56&quot;, &quot;7&quot;]
</code></pre>
edit/update:
If you would like to append the exceeding characters to the last group:
<pre><code>extension Collection {
 func unfoldSubSequencesWithTail(lenght: Int) -&gt; UnfoldSequence&lt;SubSequence,Index&gt; {
 let n = count / lenght
 var counter = 0
 return sequence(state: startIndex) { start in
 guard start &lt; endIndex else { return nil }
 let end = index(start, offsetBy: lenght, limitedBy: endIndex) ?? endIndex
 counter += 1
 if counter == n {
 defer { start = endIndex }
 return self[start...]
 } else {
 defer { start = end }
 return self[start..&lt;end]
 }
 }
 }
 func subSequencesWithTail(n: Int) -&gt; [SubSequence] {
 .init(unfoldSubSequencesWithTail(lenght: n))
 }
}
</code></pre>
<hr />
<pre><code>let numbers = &quot;1234567&quot;
let subSequencesWithTail = numbers.subSequencesWithTail(n: 2)
print(subSequencesWithTail) // [&quot;12&quot;, &quot;34&quot;, &quot;567&quot;]
</code></pre>

blocks|key|4003875|text|extension+String+{
++++func+split(len:+Int)+->+[String]+{
++++++++var+currentIndex+=+0
++++++++var+array+=+[String]()
++++++++let+length+=+self.characters.count
++++++++while+currentIndex+<+length+{
++++++++++++let+startIndex+=+self.startIndex.advancedBy(currentIndex)
++++++++++++let+endIndex+=+startIndex.advancedBy(len,+limit:+self.endIndex)
++++++++++++let+substr+=+self.substringWithRange(Range(start:+startIndex,+end:+endIndex))
++++++++++++array.append(substr)
++++++++++++currentIndex+%2B=+len
++++++++}
++++++++return+array
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4003876|"123456789".split(2)
//产出："12“、"34”、"56“、"78”、"9“|blockquote|4003877|unstyled|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|K|8|@]|9|@]|A|$]]|$1|G|3|-4|5|H|7|L|8|@]|9|@]|A|$]]]|I|$]]

<pre><code>extension String {
 func split(len: Int) -&gt; [String] {
 var currentIndex = 0
 var array = [String]()
 let length = self.characters.count
 while currentIndex &lt; length {
 let startIndex = self.startIndex.advancedBy(currentIndex)
 let endIndex = startIndex.advancedBy(len, limit: self.endIndex)
 let substr = self.substringWithRange(Range(start: startIndex, end: endIndex))
 array.append(substr)
 currentIndex += len
 }
 return array
 }
}
</code></pre>

<blockquote>
 "123456789".split(2)
 
 //output: ["12", "34", "56", "78", "9"]
</blockquote>

blocks|key|4003879|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4003880|func+SplitString(stringToBeSplitted:String,+By:Int)+->+[String]
++++{
++++++++var+newArray+=+[String]()
++++++++var+newStr+=+String()
++++++++for+char+in+stringToBeSplitted
++++++++{
++++++++++++newStr+%2B=+String(char)
++++++++++++if+newStr.count+==+By
++++++++++++{
++++++++++++++++newArray.append(newStr)
++++++++++++++++newStr+=+""
++++++++++++}

++++++++}
++++++++return+newArray
++++}|code-block|syntax|javascript|4003881|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this

<pre><code>func SplitString(stringToBeSplitted:String, By:Int) -&gt; [String]
 {
 var newArray = [String]()
 var newStr = String()
 for char in stringToBeSplitted
 {
 newStr += String(char)
 if newStr.count == By
 {
 newArray.append(newStr)
 newStr = ""
 }

 }
 return newArray
 }
</code></pre>

blocks|key|550614|text|我在目标c中写了一个方法如下，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550615|-(NSMutableArray*)splitString+:+(NSString*)str+withRange+:+(int)range{


NSMutableArray+*arr+=+[[NSMutableArray+alloc]init];

NSMutableString+*mutableStr+=+[[NSMutableString+alloc]initWithString:str];

int+j+=+0;
int+counter+=+0;

for+(int+i+=+0;+i+<+str.length;+i%2B%2B)+{

++++j%2B%2B;

++++if+(range+==+j)+{

++++++++j+=+0;


++++++++if+(!(i+==+str.length+-+1))+{

+++++++++++++[mutableStr+insertString:@"$"+atIndex:i%2B1%2Bcounter];
++++++++}



++++++++counter%2B%2B;
++++}
}



arr+=+(NSMutableArray*)[mutableStr+componentsSeparatedByString:@"$"];

NSLog(@"%25@",arr);

return+arr;
}|code-block|syntax|javascript|550616|你可以把这个方法叫做，|550617|+[self+splitString:@"123456"+withRange:2];|550618|结果会是，|550619|(
12,
34,
56
)|550620|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I have write one method in objective c as below,

<pre><code>-(NSMutableArray*)splitString : (NSString*)str withRange : (int)range{


NSMutableArray *arr = [[NSMutableArray alloc]init];

NSMutableString *mutableStr = [[NSMutableString alloc]initWithString:str];

int j = 0;
int counter = 0;

for (int i = 0; i &lt; str.length; i++) {

 j++;

 if (range == j) {

 j = 0;


 if (!(i == str.length - 1)) {

 [mutableStr insertString:@"$" atIndex:i+1+counter];
 }



 counter++;
 }
}



arr = (NSMutableArray*)[mutableStr componentsSeparatedByString:@"$"];

NSLog(@"%@",arr);

return arr;
}
</code></pre>

You can call this method like,

<pre><code> [self splitString:@"123456" withRange:2];
</code></pre>

and result will be,

<pre><code>(
12,
34,
56
)
</code></pre>

blocks|key|550640|text|您还可以尝试下面的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550641|var+arrStr:+[Substring]+=+[]
++++let+str+=+"1234567"
++++var+i+=+0
++++while+i+<+str.count+-+1+{
++++++++let+index+=+str.index(str.startIndex,+offsetBy:+i)
++++++++//Below+line+gets+current+index+and+advances+by+2
++++++++let+substring+=+str[index..<str.index(index,+offsetBy:+2)]
++++++++arrStr.append(substring)
++++++++i+%2B=+2
++++}
++++if+str.count+%25+2+==+1+{
++++++++arrStr.append(str.suffix(1))
++++}
++++print(arrStr)|code-block|syntax|javascript|550642|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also try below code:

<pre><code>var arrStr: [Substring] = []
 let str = "1234567"
 var i = 0
 while i &lt; str.count - 1 {
 let index = str.index(str.startIndex, offsetBy: i)
 //Below line gets current index and advances by 2
 let substring = str[index..&lt;str.index(index, offsetBy: 2)]
 arrStr.append(substring)
 i += 2
 }
 if str.count % 2 == 1 {
 arrStr.append(str.suffix(1))
 }
 print(arrStr)
</code></pre>

blocks|key|4003959|text|有一种愚蠢的方法，你可以考虑数据模型的规则。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4003960|++++var+strOld+=+"123456"
++++print("The+original+string:\(strOld)")

++++strOld.insert("、",+at:+strOld.index(before:+strOld.index(strOld.startIndex,+offsetBy:+3)))

++++strOld.insert("、",+at:+strOld.index(before:+strOld.index(strOld.startIndex,+offsetBy:+6)))

++++print("After+inserting:\(strOld)")

++++let+str+=+strOld

++++let+splitedArray+=+str.components(separatedBy:+"、")
++++print("After+the+split+of+the+array:\(splitedArray)")

++++let+splitedArrayOther+=+str.split{$0+==+"、"}.map(String.init)
++++print("After+break+up+the+array+(method+2):\(splitedArrayOther)")|code-block|syntax|javascript|4003961|结果：|4003962|The+original+string:123456
After+inserting:12、34、56
After+the+split+of+the+array:["12",+"34",+"56"]
After+break+up+the+array+(method+2):["12",+"34",+"56"]|4003963|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

There's a stupid way, you can think about the rules of the data model.

<pre><code> var strOld = "123456"
 print("The original string:\(strOld)")

 strOld.insert("、", at: strOld.index(before: strOld.index(strOld.startIndex, offsetBy: 3)))

 strOld.insert("、", at: strOld.index(before: strOld.index(strOld.startIndex, offsetBy: 6)))

 print("After inserting:\(strOld)")

 let str = strOld

 let splitedArray = str.components(separatedBy: "、")
 print("After the split of the array:\(splitedArray)")

 let splitedArrayOther = str.split{$0 == "、"}.map(String.init)
 print("After break up the array (method 2):\(splitedArrayOther)")
</code></pre>

The results:

<pre><code>The original string:123456
After inserting:12、34、56
After the split of the array:["12", "34", "56"]
After break up the array (method 2):["12", "34", "56"]
</code></pre>

blocks|key|1967045|text|下面是一个简短的(干净的)解决方案，这要归功于递归：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1967046|extension+Collection+{
++++func+chunks(ofSize+size:+Int)+->+[SubSequence]+{
++++++++//+replace+this+by+`guard+count+>=+size+else+{+return+[]+}`
++++++++//+if+you+want+to+omit+incomplete+chunks
++++++++guard+!isEmpty+else+{+return+[]+}
++++++++return+[prefix(size)]+%2B+dropFirst(size).chunks(ofSize:+size)
++++}
}|code-block|syntax|javascript|1967047|递归不应造成性能问题，因为Swift支持尾调用优化。|1967048|另外，如果Swift数组在预置或追加元素(像Objective元素那样)时非常快，那么数组操作应该是快速的。|1967049|因此，您可以获得快速和可读的代码(假设我的数组假设是正确的)。|1967050|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's a short (and clean) solution, thanks to recursion:
<pre class="lang-swift prettyprint-override"><code>extension Collection {
 func chunks(ofSize size: Int) -&gt; [SubSequence] {
 // replace this by `guard count &gt;= size else { return [] }`
 // if you want to omit incomplete chunks
 guard !isEmpty else { return [] }
 return [prefix(size)] + dropFirst(size).chunks(ofSize: size)
 }
}
</code></pre>
The recursion should not pose a performance problem, as Swift has support for tail call optimization.
Also if Swift arrays are really fast when it comes to prepending or appending elements (like the Objective-C ones are), then the array operations should be fast.
Thus you get both fast and readable code (assuming my array assumptions are true).

blocks|key|4004053|text|Swift+5|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4004054|extension+Array+{
++++func+chunks(size:+Int)+->+[[Element]]+{
++++++++return+stride(from:+0,+to:+count,+by:+size).map+{
++++++++++++Array(self[$0+..<+Swift.min($0+%2B+size,+count)])
++++++++}
++++}
}

extension+String+{
++++func+chunks(size:+Int)+->+[String]+{
++++++++map+{+$0+}.chunks(size:+size).compactMap+{+String($0)+}
++++}
}

let+s+=+"1234567"
print(s.chunks(size:+2))+//+["12",+"34",+"56",+"7"]|code-block|syntax|javascript|4004055|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Swift 5

<pre><code>extension Array {
 func chunks(size: Int) -&gt; [[Element]] {
 return stride(from: 0, to: count, by: size).map {
 Array(self[$0 ..&lt; Swift.min($0 + size, count)])
 }
 }
}

extension String {
 func chunks(size: Int) -&gt; [String] {
 map { $0 }.chunks(size: size).compactMap { String($0) }
 }
}

let s = "1234567"
print(s.chunks(size: 2)) // ["12", "34", "56", "7"]
</code></pre>

I want to split string by chunks of given size <code>2</code>

Example :

String <code>"1234567"</code> and output should be <code>["12", "34", "56","7"]</code>

How can split from string to array by chunks of given size

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我想将字符串按给定大小的块( 2 )分割例子：字符串"1234567"和输出应该是["12", "34", "56","7"]

问如何根据给定大小的块从字符串拆分到数组
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何根据给定大小的块从字符串拆分到数组EN