blocks|key|2024137|text|Pattern+p+=+Pattern.compile("-?\\d%2B");
Matcher+m+=+p.matcher("There+are+more+than+-2+and+less+than+12+numbers+here");
while+(m.find())+{
++System.out.println(m.group());
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2024138|..。打印-2和12。|unstyled|offset|length|style|CODE|2024139|2024140|-?匹配前导负号--可选。\d匹配一个数字，但是我们需要在Java+String中将\写成\\。因此，\d%2B匹配一个或多个数字。|2024141|entityMap^0|0|5|2|8|2|0|0|16|1|19|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Q|8|@$G|R|H|S|I|J]|$G|T|H|U|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|V|8|@]|9|@]|A|$]]|$1|L|3|M|5|F|7|W|8|@$G|X|H|Y|I|J]|$G|Z|H|10|I|J]]|9|@]|A|$]]|$1|N|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|O|$]]

<pre><code>Pattern p = Pattern.compile("-?\\d+");
Matcher m = p.matcher("There are more than -2 and less than 12 numbers here");
while (m.find()) {
 System.out.println(m.group());
}
</code></pre>

... prints <code>-2</code> and <code>12</code>.

<hr>

-? matches a leading negative sign -- optionally. \d matches a digit, and we need to write <code>\</code> as <code>\\</code> in a Java String though. So, \d+ matches 1 or more digits.

blocks|key|1810942|text|Pattern+p+=+Pattern.compile("[0-9]%2B");
Matcher+m+=+p.matcher(myString);
while+(m.find())+{
++++int+n+=+Integer.parseInt(m.group());
++++//+append+n+to+list
}
//+convert+list+to+array,+etc|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1810943|您实际上可以将0-9替换为\d，但这涉及到双反斜杠转义，这会使其更难阅读。|unstyled|1810944|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>Pattern p = Pattern.compile("[0-9]+");
Matcher m = p.matcher(myString);
while (m.find()) {
 int n = Integer.parseInt(m.group());
 // append n to list
}
// convert list to array, etc
</code></pre>

You can actually replace [0-9] with \d, but that involves double backslash escaping, which makes it harder to read.

blocks|key|2024213|text|++StringBuffer+sBuffer+=+new+StringBuffer();
++Pattern+p+=+Pattern.compile("[0-9]%2B.[0-9]*%7C[0-9]*.[0-9]%2B%7C[0-9]%2B");
++Matcher+m+=+p.matcher(str);
++while+(m.find())+{
++++sBuffer.append(m.group());
++}
++return+sBuffer.toString();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2024214|这是为了提取保留小数的数字。|unstyled|2024215|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> StringBuffer sBuffer = new StringBuffer();
 Pattern p = Pattern.compile("[0-9]+.[0-9]*|[0-9]*.[0-9]+|[0-9]+");
 Matcher m = p.matcher(str);
 while (m.find()) {
 sBuffer.append(m.group());
 }
 return sBuffer.toString();
</code></pre>

This is for extracting numbers retaining the decimal

blocks|key|1811052|text|接受的答案检测数字，但不检测格式化数字，例如2000，也不检测小数，例如4.8。对于这种用法，请使用-?\\d%2B(,\\d%2B)*?\\.?\\d%2B?|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1811053|Pattern+p+=+Pattern.compile("-?\\d%2B(,\\d%2B)*?\\.?\\d%2B?");
List<String>+numbers+=+new+ArrayList<String>();
Matcher+m+=+p.matcher("Government+has+distributed+4.8+million+textbooks+to+2,000+schools");
while+(m.find())+{++
++++numbers.add(m.group());
}+++
System.out.println(numbers);|code-block|syntax|javascript|1811054|输出：[4.8,+2,000]|1811055|entityMap^0|1E|O|0|0|3|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

The accepted answer detects digits but does not detect formated numbers, e.g. 2,000, nor decimals, e.g. 4.8. For such use <code>-?\\d+(,\\d+)*?\\.?\\d+?</code>:
<pre><code>Pattern p = Pattern.compile(&quot;-?\\d+(,\\d+)*?\\.?\\d+?&quot;);
List&lt;String&gt; numbers = new ArrayList&lt;String&gt;();
Matcher m = p.matcher(&quot;Government has distributed 4.8 million textbooks to 2,000 schools&quot;);
while (m.find()) { 
 numbers.add(m.group());
} 
System.out.println(numbers);
</code></pre>
Output:
<code>[4.8, 2,000]</code>

blocks|key|1810844|text|对于有理数，请使用这个：(([0-9]%2B.[0-9]*)%7C([0-9]*.[0-9]%2B)%7C([0-9]%2B))|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1810845|entityMap^0|C|16|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

for rational numbers use this one: <code>(([0-9]+.[0-9]*)|([0-9]*.[0-9]+)|([0-9]+))</code>

blocks|key|2024360|text|使用Java+8，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2024361|String+str+=+"There+0+are+1+some+-2-34+-numbers+567+here+890+.";
int[]+ints+=+Arrays.stream(str.replaceAll("-",+"+-").split("[%5E-\\d]%2B"))
+++++++++++++++++.filter(s+->+!s.matches("-?"))
+++++++++++++++++.mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(ints));+//+prints+[0,+1,+-2,+-34,+567,+890]|code-block|syntax|javascript|2024362|如果没有负数，可以去掉replaceAll+(在filter中使用!s.isEmpty()+)，因为这只是为了正确拆分像2-34这样的东西(在split中，这也可以完全使用正则表达式来处理，但它相当复杂)。|offset|length|style|CODE|2024363|Arrays.stream把我们的String[]变成了Stream。|2024364|filter去掉了前导和尾随的空字符串以及任何不是数字一部分的-。|2024365|mapToInt(Integer::parseInt).toArray()在每个String上调用parseInt来给我们一个int[]。|2024366|2024367|或者，Java9有一个Matcher.results方法，它应该允许这样的东西：|2024368|Pattern+p+=+Pattern.compile("-?\\d%2B");
Matcher+m+=+p.matcher("There+0+are+1+some+-2-34+-numbers+567+here+890+.");
int[]+ints+=+m.results().map(MatchResults::group).mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(ints));+//+prints+[0,+1,+-2,+-34,+567,+890]|2024369|2024370|就目前而言，与其他答案中所示的仅使用Pattern+/+Matcher循环结果相比，这两种方法都没有大的改进，但如果您想要使用streams显著简化的更复杂的操作，那么它应该会更简单。|2024371|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html|1|https://docs.oracle.com/javase/9/docs/api/java/util/regex/Matcher.html#results--^0|0|0|B|A|O|6|X|C|1O|4|1Z|5|0|0|D|H|8|S|6|S|6|0|0|0|6|V|1|0|0|11|14|6|1D|8|1R|5|0|0|B|F|1|0|0|0|I|7|S|7|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1C|8|@$I|1D|J|1E|K|L]|$I|1F|J|1G|K|L]|$I|1H|J|1I|K|L]|$I|1J|J|1K|K|L]|$I|1L|J|1M|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1N|8|@$I|1O|J|1P|K|L]|$I|1Q|J|1R|K|L]|$I|1S|J|1T|K|L]]|9|@$I|1U|J|1V|1|1W]]|A|$]]|$1|O|3|P|5|6|7|1X|8|@$I|1Y|J|1Z|K|L]|$I|20|J|21|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|22|8|@$I|23|J|24|K|L]|$I|25|J|26|K|L]|$I|27|J|28|K|L]|$I|29|J|2A|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|2B|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|2C|8|@]|9|@$I|2D|J|2E|1|2F]]|A|$]]|$1|V|3|W|5|D|7|2G|8|@]|9|@]|A|$E|F]]|$1|X|3|-4|5|6|7|2H|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|2I|8|@$I|2J|J|2K|K|L]|$I|2L|J|2M|K|L]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|2N|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]|18|$5|13|14|15|A|$16|19]]]]

Using Java 8, you can do:

<pre><code>String str = "There 0 are 1 some -2-34 -numbers 567 here 890 .";
int[] ints = Arrays.stream(str.replaceAll("-", " -").split("[^-\\d]+"))
 .filter(s -&gt; !s.matches("-?"))
 .mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(ints)); // prints [0, 1, -2, -34, 567, 890]
</code></pre>

If you don't have negative numbers, you can get rid of the <code>replaceAll</code> (and use <code>!s.isEmpty()</code> in <code>filter</code>), as that's only to properly split something like <code>2-34</code> (this can also be handled purely with regex in <code>split</code>, but it's fairly complicated).

<code>Arrays.stream</code> turns our <code>String[]</code> into a <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html" rel="nofollow noreferrer"><code>Stream&lt;String&gt;</code></a>.

<code>filter</code> gets rid of the leading and trailing empty strings as well as any <code>-</code> that isn't part of a number.

<code>mapToInt(Integer::parseInt).toArray()</code> calls <code>parseInt</code> on each <code>String</code> to give us an <code>int[]</code>.

<hr>

Alternatively, Java 9 has a <a href="https://docs.oracle.com/javase/9/docs/api/java/util/regex/Matcher.html#results--" rel="nofollow noreferrer">Matcher.results</a> method, which should allow for something like:

<pre><code>Pattern p = Pattern.compile("-?\\d+");
Matcher m = p.matcher("There 0 are 1 some -2-34 -numbers 567 here 890 .");
int[] ints = m.results().map(MatchResults::group).mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(ints)); // prints [0, 1, -2, -34, 567, 890]
</code></pre>

<hr>

As it stands, neither of these is a big improvement over just looping over the results with <code>Pattern</code> / <code>Matcher</code> as shown in the other answers, but it should be simpler if you want to follow this up with more complex operations which are significantly simplified with the use of streams.

blocks|key|1811179|text|用这个来提取所有的实数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1811180|public+static+ArrayList<Double>+extractNumbersInOrder(String+str){

++++str%2B='a';
++++double[]+returnArray+=+new+double[]{};

++++ArrayList<Double>+list+=+new+ArrayList<Double>();
++++String+singleNum="";
++++Boolean+numStarted;
++++for(char+c:str.toCharArray()){

++++++++if(isNumber(c)){
++++++++++++singleNum%2B=c;

++++++++}+else+{
++++++++++++if(!singleNum.equals("")){++//number+ended
++++++++++++++++list.add(Double.valueOf(singleNum));
++++++++++++++++System.out.println(singleNum);
++++++++++++++++singleNum="";
++++++++++++}
++++++++}
++++}

++++return+list;
}


public+static+boolean+isNumber(char+c){
++++if(Character.isDigit(c)%7C%7Cc=='-'%7C%7Cc=='%2B'%7C%7Cc=='.'){
++++++++return+true;
++++}+else+{
++++++++return+false;
++++}
}|code-block|syntax|javascript|1811181|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|$]]

Extract all real numbers using this.

<pre><code>public static ArrayList&lt;Double&gt; extractNumbersInOrder(String str){

 str+='a';
 double[] returnArray = new double[]{};

 ArrayList&lt;Double&gt; list = new ArrayList&lt;Double&gt;();
 String singleNum="";
 Boolean numStarted;
 for(char c:str.toCharArray()){

 if(isNumber(c)){
 singleNum+=c;

 } else {
 if(!singleNum.equals("")){ //number ended
 list.add(Double.valueOf(singleNum));
 System.out.println(singleNum);
 singleNum="";
 }
 }
 }

 return list;
}


public static boolean isNumber(char c){
 if(Character.isDigit(c)||c=='-'||c=='+'||c=='.'){
 return true;
 } else {
 return false;
 }
}
</code></pre>

blocks|key|2024692|text|如果您想排除单词中包含的数字，比如bar1或aa1bb，那么可以将单词边界\b添加到任何基于正则表达式的答案中。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2024693|Pattern+p+=+Pattern.compile("\\b-?\\d%2B\\b");
Matcher+m+=+p.matcher("9There+9are+more9+th9an+-2+and+less+than+12+numbers+here9");
while+(m.find())+{
++System.out.println(m.group());
}|code-block|syntax|javascript|2024694|显示：|2024695|2
12|2024696|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|$]]

If you want to exclude numbers that are contained within words, such as bar1 or aa1bb, then add word boundaries \b to any of the regex based answers. For example:

<pre><code>Pattern p = Pattern.compile("\\b-?\\d+\\b");
Matcher m = p.matcher("9There 9are more9 th9an -2 and less than 12 numbers here9");
while (m.find()) {
 System.out.println(m.group());
}
</code></pre>

displays:

<pre><code>2
12
</code></pre>

blocks|key|1811154|text|我发现这个表达式最简单|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1811155|String[]+extractednums+=+msg.split("\\\\D%2B%2B");|code-block|syntax|javascript|1811156|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|$]]

I found this expression simplest

<pre><code>String[] extractednums = msg.split("\\\\D++");
</code></pre>

blocks|key|2024735|text|public+static+String+extractNumberFromString(String+number)+{
++++String+num+=+number.replaceAll("[%5E0-9]%2B",+"+");
++++return+num.replaceAll("+",+"");
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2024736|仅从字符串中提取数字|unstyled|2024737|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>public static String extractNumberFromString(String number) {
 String num = number.replaceAll("[^0-9]+", " ");
 return num.replaceAll(" ", "");
}
</code></pre>

extracts only numbers from string

I have a String variable (basically an English sentence with an unspecified number of numbers) and I'd like to extract all the numbers into an array of integers. I was wondering whether there was a quick solution with regular expressions?

<hr>

I used Sean's solution and changed it slightly:

<pre><code>LinkedList&lt;String&gt; numbers = new LinkedList&lt;String&gt;();

Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(line); 
while (m.find()) {
 numbers.add(m.group());
}
</code></pre>

How to extract numbers from a string and get an array of ints?

Java

我有一个字符串变量(基本上是一个包含未指定数字的英语句子)，我想将所有数字提取到一个整数数组中。我想知道是否有正则表达式的快速解决方案？我使用了Sean的解决方案，并对其进行了细微的修改：LinkedList<String> numbers = new LinkedList<String>();Pattern p = Pattern.compile("\\d+");Matcher m = p.ma

问如何从字符串中提取数字并获取整型数组？
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