Split string into repeated characters












20















I want to split the string "aaaabbbccccaaddddcfggghhhh" into "aaaa", "bbb", "cccc". "aa", "dddd", "c", "f" and so on.



I tried this:



String arr = "aaaabbbccccaaddddcfggghhhh".split("(.)(?!\1)");


But this eats away one character, so with the above regular expression I get "aaa" while I want it to be "aaaa" as the first string.



How do I achieve this?






java regex string







share|improve this question




















  • 1





    @Adri1du40: I am open to other options but don't want to use loop.

    – Lokesh
    May 7 '14 at 16:52











  • Check this question : stackoverflow.com/questions/15101577/…

    – Tofandel
    May 7 '14 at 16:57











  • I'm not a Java guy, but wouldn't string.split() be slower than a loop?

    – Amal Murali
    May 7 '14 at 17:49











  • @AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

    – Cruncher
    May 7 '14 at 19:04








  • 1





    Possible duplicate of Split regex to extract Strings of contiguous characters

    – maxxyme
    Jan 12 '17 at 8:50
















20















I want to split the string "aaaabbbccccaaddddcfggghhhh" into "aaaa", "bbb", "cccc". "aa", "dddd", "c", "f" and so on.



I tried this:



String arr = "aaaabbbccccaaddddcfggghhhh".split("(.)(?!\1)");


But this eats away one character, so with the above regular expression I get "aaa" while I want it to be "aaaa" as the first string.



How do I achieve this?






java regex string







share|improve this question




















  • 1





    @Adri1du40: I am open to other options but don't want to use loop.

    – Lokesh
    May 7 '14 at 16:52











  • Check this question : stackoverflow.com/questions/15101577/…

    – Tofandel
    May 7 '14 at 16:57











  • I'm not a Java guy, but wouldn't string.split() be slower than a loop?

    – Amal Murali
    May 7 '14 at 17:49











  • @AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

    – Cruncher
    May 7 '14 at 19:04








  • 1





    Possible duplicate of Split regex to extract Strings of contiguous characters

    – maxxyme
    Jan 12 '17 at 8:50














20












20








20


9






I want to split the string "aaaabbbccccaaddddcfggghhhh" into "aaaa", "bbb", "cccc". "aa", "dddd", "c", "f" and so on.



I tried this:



String arr = "aaaabbbccccaaddddcfggghhhh".split("(.)(?!\1)");


But this eats away one character, so with the above regular expression I get "aaa" while I want it to be "aaaa" as the first string.



How do I achieve this?






java regex string







share|improve this question
















I want to split the string "aaaabbbccccaaddddcfggghhhh" into "aaaa", "bbb", "cccc". "aa", "dddd", "c", "f" and so on.



I tried this:



String arr = "aaaabbbccccaaddddcfggghhhh".split("(.)(?!\1)");


But this eats away one character, so with the above regular expression I get "aaa" while I want it to be "aaaa" as the first string.



How do I achieve this?






java regex string




java regex string






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited May 7 '14 at 22:13









Peter Mortensen

13.6k1986111




13.6k1986111










asked May 7 '14 at 16:44









LokeshLokesh

5,98562762




5,98562762








  • 1





    @Adri1du40: I am open to other options but don't want to use loop.

    – Lokesh
    May 7 '14 at 16:52











  • Check this question : stackoverflow.com/questions/15101577/…

    – Tofandel
    May 7 '14 at 16:57











  • I'm not a Java guy, but wouldn't string.split() be slower than a loop?

    – Amal Murali
    May 7 '14 at 17:49











  • @AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

    – Cruncher
    May 7 '14 at 19:04








  • 1





    Possible duplicate of Split regex to extract Strings of contiguous characters

    – maxxyme
    Jan 12 '17 at 8:50














  • 1





    @Adri1du40: I am open to other options but don't want to use loop.

    – Lokesh
    May 7 '14 at 16:52











  • Check this question : stackoverflow.com/questions/15101577/…

    – Tofandel
    May 7 '14 at 16:57











  • I'm not a Java guy, but wouldn't string.split() be slower than a loop?

    – Amal Murali
    May 7 '14 at 17:49











  • @AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

    – Cruncher
    May 7 '14 at 19:04








  • 1





    Possible duplicate of Split regex to extract Strings of contiguous characters

    – maxxyme
    Jan 12 '17 at 8:50








1




1





@Adri1du40: I am open to other options but don't want to use loop.

– Lokesh
May 7 '14 at 16:52





@Adri1du40: I am open to other options but don't want to use loop.

– Lokesh
May 7 '14 at 16:52













Check this question : stackoverflow.com/questions/15101577/…

– Tofandel
May 7 '14 at 16:57





Check this question : stackoverflow.com/questions/15101577/…

– Tofandel
May 7 '14 at 16:57













I'm not a Java guy, but wouldn't string.split() be slower than a loop?

– Amal Murali
May 7 '14 at 17:49





I'm not a Java guy, but wouldn't string.split() be slower than a loop?

– Amal Murali
May 7 '14 at 17:49













@AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

– Cruncher
May 7 '14 at 19:04







@AmalMurali would be less readable too. I don't know about you but reading this regex (?<=(.))(?!\1) is going to make me scratch my head.

– Cruncher
May 7 '14 at 19:04






1




1





Possible duplicate of Split regex to extract Strings of contiguous characters

– maxxyme
Jan 12 '17 at 8:50





Possible duplicate of Split regex to extract Strings of contiguous characters

– maxxyme
Jan 12 '17 at 8:50












3 Answers
3






active

oldest

votes


















24














Try this:



String   str = "aaaabbbccccaaddddcfggghhhh";

String out = str.split("(?<=(.))(?!\1)");



System.out.println(Arrays.toString(out));

=> [aaaa, bbb, cccc, aa, dddd, c, f, ggg, hhhh]


Explanation: we want to split the string at groups of same chars, so we need to find out the "boundary" between each group. I'm using Java's syntax for positive look-behind to pick the previous char and then a negative look-ahead with a back reference to verify that the next char is not the same as the previous one. No characters were actually consumed, because only two look-around assertions were used (that is, the regular expresion is zero-width).






share|improve this answer


























  • Your solution works perfectly. Can you please explain this regex. How it works.

    – Lokesh
    May 7 '14 at 17:00











  • @Lokesh you got it

    – Óscar López
    May 7 '14 at 17:05



















5














What about capturing in a lookbehind?



(?<=(.))(?!1|$)


as a Java string:



(?<=(.))(?!\1|$)





share|improve this answer





















  • 1





    @T.J.Crowder seems ok here. Why do you think its not working?

    – Reimeus
    May 7 '14 at 16:58








  • 2





    @Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

    – T.J. Crowder
    May 7 '14 at 17:03



















1














here I am taking each character and Checking two conditions in the if loop i.e String can't exceed the length and if next character is not equaled to the first character continue the for loop else take new line and print it.



for (int i = 0; i < arr.length; i++) {

    char chr= arr[i];

    System.out.print(chr);

    if (i + 1 < arr.length && arr[i + 1] != chr) {

        System.out.print(" n");

    }

}





share|improve this answer


























  • For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

    – pczeus
    Jan 27 '17 at 5:10











  • I was improved the answer@pczeus

    – Shivanandam
    Jan 27 '17 at 14:09











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









24














Try this:



String   str = "aaaabbbccccaaddddcfggghhhh";

String out = str.split("(?<=(.))(?!\1)");



System.out.println(Arrays.toString(out));

=> [aaaa, bbb, cccc, aa, dddd, c, f, ggg, hhhh]


Explanation: we want to split the string at groups of same chars, so we need to find out the "boundary" between each group. I'm using Java's syntax for positive look-behind to pick the previous char and then a negative look-ahead with a back reference to verify that the next char is not the same as the previous one. No characters were actually consumed, because only two look-around assertions were used (that is, the regular expresion is zero-width).






share|improve this answer


























  • Your solution works perfectly. Can you please explain this regex. How it works.

    – Lokesh
    May 7 '14 at 17:00











  • @Lokesh you got it

    – Óscar López
    May 7 '14 at 17:05
















24














Try this:



String   str = "aaaabbbccccaaddddcfggghhhh";

String out = str.split("(?<=(.))(?!\1)");



System.out.println(Arrays.toString(out));

=> [aaaa, bbb, cccc, aa, dddd, c, f, ggg, hhhh]


Explanation: we want to split the string at groups of same chars, so we need to find out the "boundary" between each group. I'm using Java's syntax for positive look-behind to pick the previous char and then a negative look-ahead with a back reference to verify that the next char is not the same as the previous one. No characters were actually consumed, because only two look-around assertions were used (that is, the regular expresion is zero-width).






share|improve this answer


























  • Your solution works perfectly. Can you please explain this regex. How it works.

    – Lokesh
    May 7 '14 at 17:00











  • @Lokesh you got it

    – Óscar López
    May 7 '14 at 17:05














24












24








24







Try this:



String   str = "aaaabbbccccaaddddcfggghhhh";

String out = str.split("(?<=(.))(?!\1)");



System.out.println(Arrays.toString(out));

=> [aaaa, bbb, cccc, aa, dddd, c, f, ggg, hhhh]


Explanation: we want to split the string at groups of same chars, so we need to find out the "boundary" between each group. I'm using Java's syntax for positive look-behind to pick the previous char and then a negative look-ahead with a back reference to verify that the next char is not the same as the previous one. No characters were actually consumed, because only two look-around assertions were used (that is, the regular expresion is zero-width).






share|improve this answer















Try this:



String   str = "aaaabbbccccaaddddcfggghhhh";

String out = str.split("(?<=(.))(?!\1)");



System.out.println(Arrays.toString(out));

=> [aaaa, bbb, cccc, aa, dddd, c, f, ggg, hhhh]


Explanation: we want to split the string at groups of same chars, so we need to find out the "boundary" between each group. I'm using Java's syntax for positive look-behind to pick the previous char and then a negative look-ahead with a back reference to verify that the next char is not the same as the previous one. No characters were actually consumed, because only two look-around assertions were used (that is, the regular expresion is zero-width).







share|improve this answer














share|improve this answer



share|improve this answer








edited May 7 '14 at 17:05

























answered May 7 '14 at 16:58









Óscar LópezÓscar López

178k25227322




178k25227322













  • Your solution works perfectly. Can you please explain this regex. How it works.

    – Lokesh
    May 7 '14 at 17:00











  • @Lokesh you got it

    – Óscar López
    May 7 '14 at 17:05



















  • Your solution works perfectly. Can you please explain this regex. How it works.

    – Lokesh
    May 7 '14 at 17:00











  • @Lokesh you got it

    – Óscar López
    May 7 '14 at 17:05

















Your solution works perfectly. Can you please explain this regex. How it works.

– Lokesh
May 7 '14 at 17:00





Your solution works perfectly. Can you please explain this regex. How it works.

– Lokesh
May 7 '14 at 17:00













@Lokesh you got it

– Óscar López
May 7 '14 at 17:05





@Lokesh you got it

– Óscar López
May 7 '14 at 17:05













5














What about capturing in a lookbehind?



(?<=(.))(?!1|$)


as a Java string:



(?<=(.))(?!\1|$)





share|improve this answer





















  • 1





    @T.J.Crowder seems ok here. Why do you think its not working?

    – Reimeus
    May 7 '14 at 16:58








  • 2





    @Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

    – T.J. Crowder
    May 7 '14 at 17:03
















5














What about capturing in a lookbehind?



(?<=(.))(?!1|$)


as a Java string:



(?<=(.))(?!\1|$)





share|improve this answer





















  • 1





    @T.J.Crowder seems ok here. Why do you think its not working?

    – Reimeus
    May 7 '14 at 16:58








  • 2





    @Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

    – T.J. Crowder
    May 7 '14 at 17:03














5












5








5







What about capturing in a lookbehind?



(?<=(.))(?!1|$)


as a Java string:



(?<=(.))(?!\1|$)





share|improve this answer















What about capturing in a lookbehind?



(?<=(.))(?!1|$)


as a Java string:



(?<=(.))(?!\1|$)






share|improve this answer














share|improve this answer



share|improve this answer








edited May 7 '14 at 17:08

























answered May 7 '14 at 16:51









Jonny 5Jonny 5

8,95621535




8,95621535








  • 1





    @T.J.Crowder seems ok here. Why do you think its not working?

    – Reimeus
    May 7 '14 at 16:58








  • 2





    @Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

    – T.J. Crowder
    May 7 '14 at 17:03














  • 1





    @T.J.Crowder seems ok here. Why do you think its not working?

    – Reimeus
    May 7 '14 at 16:58








  • 2





    @Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

    – T.J. Crowder
    May 7 '14 at 17:03








1




1





@T.J.Crowder seems ok here. Why do you think its not working?

– Reimeus
May 7 '14 at 16:58







@T.J.Crowder seems ok here. Why do you think its not working?

– Reimeus
May 7 '14 at 16:58






2




2





@Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

– T.J. Crowder
May 7 '14 at 17:03





@Reimeus: Because I copied and pasted it without doing the escape. I really wish Java had regex literals. :-)

– T.J. Crowder
May 7 '14 at 17:03











1














here I am taking each character and Checking two conditions in the if loop i.e String can't exceed the length and if next character is not equaled to the first character continue the for loop else take new line and print it.



for (int i = 0; i < arr.length; i++) {

    char chr= arr[i];

    System.out.print(chr);

    if (i + 1 < arr.length && arr[i + 1] != chr) {

        System.out.print(" n");

    }

}





share|improve this answer


























  • For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

    – pczeus
    Jan 27 '17 at 5:10











  • I was improved the answer@pczeus

    – Shivanandam
    Jan 27 '17 at 14:09
















1














here I am taking each character and Checking two conditions in the if loop i.e String can't exceed the length and if next character is not equaled to the first character continue the for loop else take new line and print it.



for (int i = 0; i < arr.length; i++) {

    char chr= arr[i];

    System.out.print(chr);

    if (i + 1 < arr.length && arr[i + 1] != chr) {

        System.out.print(" n");

    }

}





share|improve this answer


























  • For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

    – pczeus
    Jan 27 '17 at 5:10











  • I was improved the answer@pczeus

    – Shivanandam
    Jan 27 '17 at 14:09














1












1








1







here I am taking each character and Checking two conditions in the if loop i.e String can't exceed the length and if next character is not equaled to the first character continue the for loop else take new line and print it.



for (int i = 0; i < arr.length; i++) {

    char chr= arr[i];

    System.out.print(chr);

    if (i + 1 < arr.length && arr[i + 1] != chr) {

        System.out.print(" n");

    }

}





share|improve this answer















here I am taking each character and Checking two conditions in the if loop i.e String can't exceed the length and if next character is not equaled to the first character continue the for loop else take new line and print it.



for (int i = 0; i < arr.length; i++) {

    char chr= arr[i];

    System.out.print(chr);

    if (i + 1 < arr.length && arr[i + 1] != chr) {

        System.out.print(" n");

    }

}






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 27 '17 at 14:07

























answered Nov 22 '16 at 7:18









ShivanandamShivanandam

8531317




8531317













  • For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

    – pczeus
    Jan 27 '17 at 5:10











  • I was improved the answer@pczeus

    – Shivanandam
    Jan 27 '17 at 14:09



















  • For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

    – pczeus
    Jan 27 '17 at 5:10











  • I was improved the answer@pczeus

    – Shivanandam
    Jan 27 '17 at 14:09

















For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

– pczeus
Jan 27 '17 at 5:10





For a quality answer, @Shiva can you add some explanation to your answer on how the code accomplish what the author is trying to achieve?

– pczeus
Jan 27 '17 at 5:10













I was improved the answer@pczeus

– Shivanandam
Jan 27 '17 at 14:09





I was improved the answer@pczeus

– Shivanandam
Jan 27 '17 at 14:09


















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