How to combine similar characters in a list?












7















I'm trying to combine similar characters that are next to each other that are in a list. I was wondering if there was a Python way to do it? Here's an example:



test = 'hello###_world###test#test123##'

splitter = re.split("(#)", test)

splitter = filter(None, splitter)


Which returns this in the splitter variable:



['hello', '#', '#', '#', '_world', '#', '#', '#', 'test', '#', 'test123', '#', '#']


I'm trying to combine the hashes so the list turns into this:



['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


Thanks for any help!






python







share|improve this question









New contributor




Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

























    7















    I'm trying to combine similar characters that are next to each other that are in a list. I was wondering if there was a Python way to do it? Here's an example:



    test = 'hello###_world###test#test123##'
    
splitter = re.split("(#)", test)
    
splitter = filter(None, splitter)


    Which returns this in the splitter variable:



    ['hello', '#', '#', '#', '_world', '#', '#', '#', 'test', '#', 'test123', '#', '#']


    I'm trying to combine the hashes so the list turns into this:



    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


    Thanks for any help!






    python







    share|improve this question









    New contributor




    Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      7












      7








      7








      I'm trying to combine similar characters that are next to each other that are in a list. I was wondering if there was a Python way to do it? Here's an example:



      test = 'hello###_world###test#test123##'
      
splitter = re.split("(#)", test)
      
splitter = filter(None, splitter)


      Which returns this in the splitter variable:



      ['hello', '#', '#', '#', '_world', '#', '#', '#', 'test', '#', 'test123', '#', '#']


      I'm trying to combine the hashes so the list turns into this:



      ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


      Thanks for any help!






      python







      share|improve this question









      New contributor




      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      I'm trying to combine similar characters that are next to each other that are in a list. I was wondering if there was a Python way to do it? Here's an example:



      test = 'hello###_world###test#test123##'
      
splitter = re.split("(#)", test)
      
splitter = filter(None, splitter)


      Which returns this in the splitter variable:



      ['hello', '#', '#', '#', '_world', '#', '#', '#', 'test', '#', 'test123', '#', '#']


      I'm trying to combine the hashes so the list turns into this:



      ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


      Thanks for any help!






      python




      python






      share|improve this question









      New contributor




      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 49 mins ago









      Ajax1234

      41.4k42853




      41.4k42853






      New contributor




      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 50 mins ago









      GregGreg

      362




      362




      New contributor




      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Greg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.
























          3 Answers
          3






          active

          oldest

          votes


















          5














          Try:



          splitter = re.split("(#+)", test)





          share|improve this answer



















          • 1





            It gives an empty string at the end.

            – AkshayNevrekar
            26 mins ago






          • 2





            @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

            – DYZ
            23 mins ago











          • This works for me. Thanks for the help!

            – Greg
            17 mins ago



















          1














          Add + at the end of the regular expression and filtering None values will do the trick



          >>> import re
          
>>> test = 'hello###_world###test#test123##'
          
>>> splitter = re.split("(#+)", test)
          
>>> splitter
          
['hello', '###', '_world', '###', 'test', '#', 'test123', '##', '']
          
>>> splitter = list(filter(None, splitter))
          
>>> splitter
          
['hello', '###', '_world', '###', 'test', '#', 'test123', '##']
          
>>>





          share|improve this answer































            1














            You can use itertools.groupby:



            import itertools
            
test = 'hello###_world###test#test123##'
            
new_result = [''.join(b) for _, b in itertools.groupby(test, key=lambda x:x == '#')]


            Output:



            ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


            You can also use re.findall:



            import re
            
result = re.findall('#+|[^#]+', test)


            Output:



            ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']





            share|improve this answer


























            • This one also does the job. Thanks for the help!

              – Greg
              16 mins ago











            Your Answer






            StackExchange.ifUsing("editor", function () {
            StackExchange.using("externalEditor", function () {
            StackExchange.using("snippets", function () {
            StackExchange.snippets.init();
            });
            });
            }, "code-snippets");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "1"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });






            Greg is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54603094%2fhow-to-combine-similar-characters-in-a-list%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            Try:



            splitter = re.split("(#+)", test)





            share|improve this answer



















            • 1





              It gives an empty string at the end.

              – AkshayNevrekar
              26 mins ago






            • 2





              @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

              – DYZ
              23 mins ago











            • This works for me. Thanks for the help!

              – Greg
              17 mins ago
















            5














            Try:



            splitter = re.split("(#+)", test)





            share|improve this answer



















            • 1





              It gives an empty string at the end.

              – AkshayNevrekar
              26 mins ago






            • 2





              @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

              – DYZ
              23 mins ago











            • This works for me. Thanks for the help!

              – Greg
              17 mins ago














            5












            5








            5







            Try:



            splitter = re.split("(#+)", test)





            share|improve this answer













            Try:



            splitter = re.split("(#+)", test)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 31 mins ago









            mingganzmingganz

            1986




            1986








            • 1





              It gives an empty string at the end.

              – AkshayNevrekar
              26 mins ago






            • 2





              @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

              – DYZ
              23 mins ago











            • This works for me. Thanks for the help!

              – Greg
              17 mins ago














            • 1





              It gives an empty string at the end.

              – AkshayNevrekar
              26 mins ago






            • 2





              @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

              – DYZ
              23 mins ago











            • This works for me. Thanks for the help!

              – Greg
              17 mins ago








            1




            1





            It gives an empty string at the end.

            – AkshayNevrekar
            26 mins ago





            It gives an empty string at the end.

            – AkshayNevrekar
            26 mins ago




            2




            2





            @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

            – DYZ
            23 mins ago





            @AkshayNevrekar With splitter = filter(None, splitter) (as in the original post), it does not.

            – DYZ
            23 mins ago













            This works for me. Thanks for the help!

            – Greg
            17 mins ago





            This works for me. Thanks for the help!

            – Greg
            17 mins ago













            1














            Add + at the end of the regular expression and filtering None values will do the trick



            >>> import re
            
>>> test = 'hello###_world###test#test123##'
            
>>> splitter = re.split("(#+)", test)
            
>>> splitter
            
['hello', '###', '_world', '###', 'test', '#', 'test123', '##', '']
            
>>> splitter = list(filter(None, splitter))
            
>>> splitter
            
['hello', '###', '_world', '###', 'test', '#', 'test123', '##']
            
>>>





            share|improve this answer




























              1














              Add + at the end of the regular expression and filtering None values will do the trick



              >>> import re
              
>>> test = 'hello###_world###test#test123##'
              
>>> splitter = re.split("(#+)", test)
              
>>> splitter
              
['hello', '###', '_world', '###', 'test', '#', 'test123', '##', '']
              
>>> splitter = list(filter(None, splitter))
              
>>> splitter
              
['hello', '###', '_world', '###', 'test', '#', 'test123', '##']
              
>>>





              share|improve this answer


























                1












                1








                1







                Add + at the end of the regular expression and filtering None values will do the trick



                >>> import re
                
>>> test = 'hello###_world###test#test123##'
                
>>> splitter = re.split("(#+)", test)
                
>>> splitter
                
['hello', '###', '_world', '###', 'test', '#', 'test123', '##', '']
                
>>> splitter = list(filter(None, splitter))
                
>>> splitter
                
['hello', '###', '_world', '###', 'test', '#', 'test123', '##']
                
>>>





                share|improve this answer













                Add + at the end of the regular expression and filtering None values will do the trick



                >>> import re
                
>>> test = 'hello###_world###test#test123##'
                
>>> splitter = re.split("(#+)", test)
                
>>> splitter
                
['hello', '###', '_world', '###', 'test', '#', 'test123', '##', '']
                
>>> splitter = list(filter(None, splitter))
                
>>> splitter
                
['hello', '###', '_world', '###', 'test', '#', 'test123', '##']
                
>>>






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 20 mins ago









                Bodhi94Bodhi94

                901520




                901520























                    1














                    You can use itertools.groupby:



                    import itertools
                    
test = 'hello###_world###test#test123##'
                    
new_result = [''.join(b) for _, b in itertools.groupby(test, key=lambda x:x == '#')]


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


                    You can also use re.findall:



                    import re
                    
result = re.findall('#+|[^#]+', test)


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']





                    share|improve this answer


























                    • This one also does the job. Thanks for the help!

                      – Greg
                      16 mins ago
















                    1














                    You can use itertools.groupby:



                    import itertools
                    
test = 'hello###_world###test#test123##'
                    
new_result = [''.join(b) for _, b in itertools.groupby(test, key=lambda x:x == '#')]


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


                    You can also use re.findall:



                    import re
                    
result = re.findall('#+|[^#]+', test)


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']





                    share|improve this answer


























                    • This one also does the job. Thanks for the help!

                      – Greg
                      16 mins ago














                    1












                    1








                    1







                    You can use itertools.groupby:



                    import itertools
                    
test = 'hello###_world###test#test123##'
                    
new_result = [''.join(b) for _, b in itertools.groupby(test, key=lambda x:x == '#')]


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


                    You can also use re.findall:



                    import re
                    
result = re.findall('#+|[^#]+', test)


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']





                    share|improve this answer















                    You can use itertools.groupby:



                    import itertools
                    
test = 'hello###_world###test#test123##'
                    
new_result = [''.join(b) for _, b in itertools.groupby(test, key=lambda x:x == '#')]


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']


                    You can also use re.findall:



                    import re
                    
result = re.findall('#+|[^#]+', test)


                    Output:



                    ['hello', '###', '_world', '###', 'test', '#', 'test123', '##']






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 13 mins ago

























                    answered 48 mins ago









                    Ajax1234Ajax1234

                    41.4k42853




                    41.4k42853













                    • This one also does the job. Thanks for the help!

                      – Greg
                      16 mins ago



















                    • This one also does the job. Thanks for the help!

                      – Greg
                      16 mins ago

















                    This one also does the job. Thanks for the help!

                    – Greg
                    16 mins ago





                    This one also does the job. Thanks for the help!

                    – Greg
                    16 mins ago










                    Greg is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    Greg is a new contributor. Be nice, and check out our Code of Conduct.













                    Greg is a new contributor. Be nice, and check out our Code of Conduct.












                    Greg is a new contributor. Be nice, and check out our Code of Conduct.
















                    Thanks for contributing an answer to Stack Overflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54603094%2fhow-to-combine-similar-characters-in-a-list%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Bundesstraße 106

                    Image
                    Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                    Read more

                    Verónica Boquete

                    Image
                    Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                    Read more

                    Ida-Boy-Ed-Garten

                    Image
                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                    Read more