How can I count the number of matches for a regex?
Let's say I have a string which contains this:
HelloxxxHelloxxxHello
I compile a pattern to look for 'Hello'
Pattern pattern = Pattern.compile("Hello");
Matcher matcher = pattern.matcher("HelloxxxHelloxxxHello");
It should find three matches. How can I get a count of how many matches there were?
I've tried various loops and using the matcher.groupCount()
but it didn't work.
Solution 1:
matcher.find()
does not find all matches, only the next match.
Solution for Java 9+
long matches = matcher.results().count();
Solution for Java 8 and older
You'll have to do the following. (Starting from Java 9, there is a nicer solution)
int count = 0;
while (matcher.find())
count++;
Btw, matcher.groupCount()
is something completely different.
Complete example:
import java.util.regex.*;
class Test {
public static void main(String[] args) {
String hello = "HelloxxxHelloxxxHello";
Pattern pattern = Pattern.compile("Hello");
Matcher matcher = pattern.matcher(hello);
int count = 0;
while (matcher.find())
count++;
System.out.println(count); // prints 3
}
}
Handling overlapping matches
When counting matches of aa
in aaaa
the above snippet will give you 2.
aaaa
aa
aa
To get 3 matches, i.e. this behavior:
aaaa
aa
aa
aa
You have to search for a match at index <start of last match> + 1
as follows:
String hello = "aaaa";
Pattern pattern = Pattern.compile("aa");
Matcher matcher = pattern.matcher(hello);
int count = 0;
int i = 0;
while (matcher.find(i)) {
count++;
i = matcher.start() + 1;
}
System.out.println(count); // prints 3
Solution 2:
This should work for matches that might overlap:
public static void main(String[] args) {
String input = "aaaaaaaa";
String regex = "aa";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
int from = 0;
int count = 0;
while(matcher.find(from)) {
count++;
from = matcher.start() + 1;
}
System.out.println(count);
}
Solution 3:
From Java 9, you can use the stream provided by Matcher.results()
long matches = matcher.results().count();
Solution 4:
If you want to use Java 8 streams and are allergic to while
loops, you could try this:
public static int countPattern(String references, Pattern referencePattern) {
Matcher matcher = referencePattern.matcher(references);
return Stream.iterate(0, i -> i + 1)
.filter(i -> !matcher.find())
.findFirst()
.get();
}
Disclaimer: this only works for disjoint matches.
Example:
public static void main(String[] args) throws ParseException {
Pattern referencePattern = Pattern.compile("PASSENGER:\\d+");
System.out.println(countPattern("[ \"PASSENGER:1\", \"PASSENGER:2\", \"AIR:1\", \"AIR:2\", \"FOP:2\" ]", referencePattern));
System.out.println(countPattern("[ \"AIR:1\", \"AIR:2\", \"FOP:2\" ]", referencePattern));
System.out.println(countPattern("[ \"AIR:1\", \"AIR:2\", \"FOP:2\", \"PASSENGER:1\" ]", referencePattern));
System.out.println(countPattern("[ ]", referencePattern));
}
This prints out:
2
0
1
0
This is a solution for disjoint matches with streams:
public static int countPattern(String references, Pattern referencePattern) {
return StreamSupport.stream(Spliterators.spliteratorUnknownSize(
new Iterator<Integer>() {
Matcher matcher = referencePattern.matcher(references);
int from = 0;
@Override
public boolean hasNext() {
return matcher.find(from);
}
@Override
public Integer next() {
from = matcher.start() + 1;
return 1;
}
},
Spliterator.IMMUTABLE), false).reduce(0, (a, c) -> a + c);
}
Solution 5:
Use the below code to find the count of number of matches that the regex finds in your input
Pattern p = Pattern.compile(regex, Pattern.MULTILINE | Pattern.DOTALL);// "regex" here indicates your predefined regex.
Matcher m = p.matcher(pattern); // "pattern" indicates your string to match the pattern against with
boolean b = m.matches();
if(b)
count++;
while (m.find())
count++;
This is a generalized code not specific one though, tailor it to suit your need
Please feel free to correct me if there is any mistake.