check if a pattern is in a list of words












0














I need an output that contains words that are exactly like a pattern - same letters in same spots only (and letters shouldn't show in the word at other places) and the same length
for example:



words = ['hatch','catch','match','chat','mates'] 

pattern = '_atc_


needed output:



['hatch','match']


I have tried to use nested for loops but it didn't work for a pattern that starts and ends with '_'



def filter_words_list(words, pattern):

relevant_words = 

for word in words:

    if len(word) == len(pattern):

        for i in range(len(word)):

            for j in range(len(pattern)):

                if word[i] != pattern[i]:

                    break

                if word[i] == pattern[i]:

                    relevant_words.append(word)


thx !






python list







share|improve this question


















  • 1




    Just replace "_" with ".", then use your pattern as regular expression
    – quant
    Nov 11 at 13:20










  • what do you mean by replacing? the pattern is given
    – user10596917
    Nov 11 at 13:21










  • There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
    – quant
    Nov 11 at 13:24


















0














I need an output that contains words that are exactly like a pattern - same letters in same spots only (and letters shouldn't show in the word at other places) and the same length
for example:



words = ['hatch','catch','match','chat','mates'] 

pattern = '_atc_


needed output:



['hatch','match']


I have tried to use nested for loops but it didn't work for a pattern that starts and ends with '_'



def filter_words_list(words, pattern):

relevant_words = 

for word in words:

    if len(word) == len(pattern):

        for i in range(len(word)):

            for j in range(len(pattern)):

                if word[i] != pattern[i]:

                    break

                if word[i] == pattern[i]:

                    relevant_words.append(word)


thx !






python list







share|improve this question


















  • 1




    Just replace "_" with ".", then use your pattern as regular expression
    – quant
    Nov 11 at 13:20










  • what do you mean by replacing? the pattern is given
    – user10596917
    Nov 11 at 13:21










  • There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
    – quant
    Nov 11 at 13:24
















0












0








0







I need an output that contains words that are exactly like a pattern - same letters in same spots only (and letters shouldn't show in the word at other places) and the same length
for example:



words = ['hatch','catch','match','chat','mates'] 

pattern = '_atc_


needed output:



['hatch','match']


I have tried to use nested for loops but it didn't work for a pattern that starts and ends with '_'



def filter_words_list(words, pattern):

relevant_words = 

for word in words:

    if len(word) == len(pattern):

        for i in range(len(word)):

            for j in range(len(pattern)):

                if word[i] != pattern[i]:

                    break

                if word[i] == pattern[i]:

                    relevant_words.append(word)


thx !






python list







share|improve this question













I need an output that contains words that are exactly like a pattern - same letters in same spots only (and letters shouldn't show in the word at other places) and the same length
for example:



words = ['hatch','catch','match','chat','mates'] 

pattern = '_atc_


needed output:



['hatch','match']


I have tried to use nested for loops but it didn't work for a pattern that starts and ends with '_'



def filter_words_list(words, pattern):

relevant_words = 

for word in words:

    if len(word) == len(pattern):

        for i in range(len(word)):

            for j in range(len(pattern)):

                if word[i] != pattern[i]:

                    break

                if word[i] == pattern[i]:

                    relevant_words.append(word)


thx !






python list




python list






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 11 at 13:18







user10596917















  • 1




    Just replace "_" with ".", then use your pattern as regular expression
    – quant
    Nov 11 at 13:20










  • what do you mean by replacing? the pattern is given
    – user10596917
    Nov 11 at 13:21










  • There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
    – quant
    Nov 11 at 13:24
















  • 1




    Just replace "_" with ".", then use your pattern as regular expression
    – quant
    Nov 11 at 13:20










  • what do you mean by replacing? the pattern is given
    – user10596917
    Nov 11 at 13:21










  • There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
    – quant
    Nov 11 at 13:24










1




1




Just replace "_" with ".", then use your pattern as regular expression
– quant
Nov 11 at 13:20




Just replace "_" with ".", then use your pattern as regular expression
– quant
Nov 11 at 13:20












what do you mean by replacing? the pattern is given
– user10596917
Nov 11 at 13:21




what do you mean by replacing? the pattern is given
– user10596917
Nov 11 at 13:21












There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
– quant
Nov 11 at 13:24






There is a standard way of writing patterns in python. Your pattern however is using "_" - which is another syntax. So you simply need to convert your syntax to the standard syntax, then you can use the standard pattern matching library see docs.python.org/2/library/re.html ... as shown in the way of the answer from Daniel Masejo
– quant
Nov 11 at 13:24














2 Answers
2






active

oldest

votes


















1














So you should use regex. and replace the underscore with '.' which means any single character.
so the input looks like:



words = ['hatch','catch','match','chat','mates'] 

pattern = '.atc.'


and the code is:



import re

def filter_words_list(words, pattern):

ret = 

for word in words:

    if(re.match(pattern,word)):ret.append(word)

return ret


Hopes tha helped






share|improve this answer





























    1














    You could use a regular expression:



    import re
    

    
words = ['hatch','catch','match','chat','mates']
    
pattern = re.compile('[^atc]atc[^atc]')
    

    
result = list(filter(pattern.fullmatch, words))
    
print(result)


    Output



    ['hatch', 'match']


    The pattern '[^atc]atc[^atc]' matches everything that is not a or t or c ([^atc]) followed by 'atc' and again everything that is not a or t or c.



    As an alternative you could write your own matching function that will work with any given pattern:



    from collections import Counter
    

    

    
def full_match(word, pattern='_atc_'):
    
    if len(pattern) != len(word):
    
        return False
    

    
    pattern_letter_counts = Counter(e for e in pattern if e != '_')  # count characters that are not wild card
    
    word_letter_counts = Counter(word) # count letters
    

    
    if any(count != word_letter_counts.get(ch, 0) for ch, count in pattern_letter_counts.items()):
    
        return False
    

    
    return all(p == w for p, w in zip(pattern, word) if p != '_')  # the word must match in all characters that are not wild card
    

    

    
words = ['hatch', 'catch', 'match', 'chat', 'mates']
    

    

    
result = list(filter(full_match, words))
    
print(result)


    Output



    ['hatch', 'match']


    Further




    1. See the documentation on the built-in functions any and all.

    2. See the documentation on Counter.






    share|improve this answer























      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
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53249133%2fcheck-if-a-pattern-is-in-a-list-of-words%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown
























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      So you should use regex. and replace the underscore with '.' which means any single character.
      so the input looks like:



      words = ['hatch','catch','match','chat','mates'] 
      
pattern = '.atc.'


      and the code is:



      import re
      
def filter_words_list(words, pattern):
      
ret = 
      
for word in words:
      
    if(re.match(pattern,word)):ret.append(word)
      
return ret


      Hopes tha helped






      share|improve this answer


























        1














        So you should use regex. and replace the underscore with '.' which means any single character.
        so the input looks like:



        words = ['hatch','catch','match','chat','mates'] 
        
pattern = '.atc.'


        and the code is:



        import re
        
def filter_words_list(words, pattern):
        
ret = 
        
for word in words:
        
    if(re.match(pattern,word)):ret.append(word)
        
return ret


        Hopes tha helped






        share|improve this answer
























          1












          1








          1






          So you should use regex. and replace the underscore with '.' which means any single character.
          so the input looks like:



          words = ['hatch','catch','match','chat','mates'] 
          
pattern = '.atc.'


          and the code is:



          import re
          
def filter_words_list(words, pattern):
          
ret = 
          
for word in words:
          
    if(re.match(pattern,word)):ret.append(word)
          
return ret


          Hopes tha helped






          share|improve this answer












          So you should use regex. and replace the underscore with '.' which means any single character.
          so the input looks like:



          words = ['hatch','catch','match','chat','mates'] 
          
pattern = '.atc.'


          and the code is:



          import re
          
def filter_words_list(words, pattern):
          
ret = 
          
for word in words:
          
    if(re.match(pattern,word)):ret.append(word)
          
return ret


          Hopes tha helped







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 11 at 13:38









          Mbjahnoon

          212




          212

























              1














              You could use a regular expression:



              import re
              

              
words = ['hatch','catch','match','chat','mates']
              
pattern = re.compile('[^atc]atc[^atc]')
              

              
result = list(filter(pattern.fullmatch, words))
              
print(result)


              Output



              ['hatch', 'match']


              The pattern '[^atc]atc[^atc]' matches everything that is not a or t or c ([^atc]) followed by 'atc' and again everything that is not a or t or c.



              As an alternative you could write your own matching function that will work with any given pattern:



              from collections import Counter
              

              

              
def full_match(word, pattern='_atc_'):
              
    if len(pattern) != len(word):
              
        return False
              

              
    pattern_letter_counts = Counter(e for e in pattern if e != '_')  # count characters that are not wild card
              
    word_letter_counts = Counter(word) # count letters
              

              
    if any(count != word_letter_counts.get(ch, 0) for ch, count in pattern_letter_counts.items()):
              
        return False
              

              
    return all(p == w for p, w in zip(pattern, word) if p != '_')  # the word must match in all characters that are not wild card
              

              

              
words = ['hatch', 'catch', 'match', 'chat', 'mates']
              

              

              
result = list(filter(full_match, words))
              
print(result)


              Output



              ['hatch', 'match']


              Further




              1. See the documentation on the built-in functions any and all.

              2. See the documentation on Counter.






              share|improve this answer




























                1














                You could use a regular expression:



                import re
                

                
words = ['hatch','catch','match','chat','mates']
                
pattern = re.compile('[^atc]atc[^atc]')
                

                
result = list(filter(pattern.fullmatch, words))
                
print(result)


                Output



                ['hatch', 'match']


                The pattern '[^atc]atc[^atc]' matches everything that is not a or t or c ([^atc]) followed by 'atc' and again everything that is not a or t or c.



                As an alternative you could write your own matching function that will work with any given pattern:



                from collections import Counter
                

                

                
def full_match(word, pattern='_atc_'):
                
    if len(pattern) != len(word):
                
        return False
                

                
    pattern_letter_counts = Counter(e for e in pattern if e != '_')  # count characters that are not wild card
                
    word_letter_counts = Counter(word) # count letters
                

                
    if any(count != word_letter_counts.get(ch, 0) for ch, count in pattern_letter_counts.items()):
                
        return False
                

                
    return all(p == w for p, w in zip(pattern, word) if p != '_')  # the word must match in all characters that are not wild card
                

                

                
words = ['hatch', 'catch', 'match', 'chat', 'mates']
                

                

                
result = list(filter(full_match, words))
                
print(result)


                Output



                ['hatch', 'match']


                Further




                1. See the documentation on the built-in functions any and all.

                2. See the documentation on Counter.






                share|improve this answer


























                  1












                  1








                  1






                  You could use a regular expression:



                  import re
                  

                  
words = ['hatch','catch','match','chat','mates']
                  
pattern = re.compile('[^atc]atc[^atc]')
                  

                  
result = list(filter(pattern.fullmatch, words))
                  
print(result)


                  Output



                  ['hatch', 'match']


                  The pattern '[^atc]atc[^atc]' matches everything that is not a or t or c ([^atc]) followed by 'atc' and again everything that is not a or t or c.



                  As an alternative you could write your own matching function that will work with any given pattern:



                  from collections import Counter
                  

                  

                  
def full_match(word, pattern='_atc_'):
                  
    if len(pattern) != len(word):
                  
        return False
                  

                  
    pattern_letter_counts = Counter(e for e in pattern if e != '_')  # count characters that are not wild card
                  
    word_letter_counts = Counter(word) # count letters
                  

                  
    if any(count != word_letter_counts.get(ch, 0) for ch, count in pattern_letter_counts.items()):
                  
        return False
                  

                  
    return all(p == w for p, w in zip(pattern, word) if p != '_')  # the word must match in all characters that are not wild card
                  

                  

                  
words = ['hatch', 'catch', 'match', 'chat', 'mates']
                  

                  

                  
result = list(filter(full_match, words))
                  
print(result)


                  Output



                  ['hatch', 'match']


                  Further




                  1. See the documentation on the built-in functions any and all.

                  2. See the documentation on Counter.






                  share|improve this answer














                  You could use a regular expression:



                  import re
                  

                  
words = ['hatch','catch','match','chat','mates']
                  
pattern = re.compile('[^atc]atc[^atc]')
                  

                  
result = list(filter(pattern.fullmatch, words))
                  
print(result)


                  Output



                  ['hatch', 'match']


                  The pattern '[^atc]atc[^atc]' matches everything that is not a or t or c ([^atc]) followed by 'atc' and again everything that is not a or t or c.



                  As an alternative you could write your own matching function that will work with any given pattern:



                  from collections import Counter
                  

                  

                  
def full_match(word, pattern='_atc_'):
                  
    if len(pattern) != len(word):
                  
        return False
                  

                  
    pattern_letter_counts = Counter(e for e in pattern if e != '_')  # count characters that are not wild card
                  
    word_letter_counts = Counter(word) # count letters
                  

                  
    if any(count != word_letter_counts.get(ch, 0) for ch, count in pattern_letter_counts.items()):
                  
        return False
                  

                  
    return all(p == w for p, w in zip(pattern, word) if p != '_')  # the word must match in all characters that are not wild card
                  

                  

                  
words = ['hatch', 'catch', 'match', 'chat', 'mates']
                  

                  

                  
result = list(filter(full_match, words))
                  
print(result)


                  Output



                  ['hatch', 'match']


                  Further




                  1. See the documentation on the built-in functions any and all.

                  2. See the documentation on Counter.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 11 at 13:49

























                  answered Nov 11 at 13:23









                  Daniel Mesejo

                  12.5k1924




                  12.5k1924






























                      draft saved

                      draft discarded




















































                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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%2f53249133%2fcheck-if-a-pattern-is-in-a-list-of-words%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







                      -

                      這個網誌中的熱門文章

                      Academy of Television Arts & Sciences

                      圖片
                      American television organization Academy of Television Arts & Sciences Founded 1946 ; 73 years ago  ( 1946 ) Location North Hollywood, California, United States Area served Television industry Product Emmy Awards Key people Frank Scherma ( Chairman and CEO ) Website televisionacademy.com [redirects to emmys.com , official website of the Emmys and the Television Academy The Academy of Television Arts & Sciences ( ATAS ), also colloquially known as the Television Academy , is a professional honorary organization dedicated to the advancement of the television industry in the United States. Founded in 1946, the organization presents the Primetime Emmy Awards, an annual ceremony honoring achievement in U.S. primetime television. Contents 1 History 2 Emmy Award 3 Publications and programs 4 Current governance 4.1 Board of Governors 5 Television Academy honors 5.1 1st Annual (2008) 5.2 2nd Annual (2009)...
                      閱讀完整內容

                      L'Équipe

                      圖片
                      For other uses, see Équipe (disambiguation). L'Équipe The front page of L'Équipe on 4 July 2011 Type Daily newspaper Format Tabloid Owner(s) Éditions Philippe Amaury Editor François Morinière Editor-in-chief Fabrice Jouhaud Founded 1946 Language French Headquarters Boulogne-Billancourt ISSN 0153-1069 Website www.lequipe.fr L'Équipe ( pronounced  [lekip] , French for "the team") is a French nationwide daily newspaper devoted to sport, owned by Éditions Philippe Amaury. The paper is noted for coverage of association football , rugby, motorsport and cycling. Its predecessor was L'Auto , a general sports paper whose name reflected not any narrow interest but the excitement of the time in car racing. L'Auto originated the Tour de France cycling stage race in 1903 as a circulation booster. The race leader's yellow jersey (French: maillot jaune ) was instituted in 1919, probably to reflect the dist...
                      閱讀完整內容

                      1995 France bombings

                      圖片
                      This article is a rough translation from French . It may have been generated by a computer or by a translator without dual proficiency. Please help to enhance the translation. The original article is under "français" in the "languages" sidebar. See this article's entry on Pages needing translation into English for discussion. 1995 France bombings Part of Algerian Civil War (spillover) Plaque commemorating the victims of the Saint-Michel station bombing on 25 July 1995 Location Paris and Lyon, France Date 25 July 1995  ( 1995-07-25 ) – 17 October 1995  ( 1995-10-17 ) Weapons Improvised explosive devices, school bombing Deaths 8 Non-fatal injuries 157 Perpetrator Armed Islamic Group Motive To induce the French government to withdraw support from the Algerian government during the Algerian Civil War The 1995 bombings in France were carried out by the Armed Islamic Group (GIA), who were broadening the Alger...
                      閱讀完整內容