Using regex with list comprehension in python












1















I have following code which will store all the csv filename in a list from a specific folder



import pandas as pd

import re

import os



files = os.listdir('.')

filename=[filename for filename in files if filename.endswith('.csv')]


However, in my folder, I have two types of csv files, one ends with, for example, _20.cvs(or maybe _18.csv,_01.csv), another one ends with _Raw.csv;



However I only need the first type stored in my list. I know regular expression may can help me on that, so I did some google search, and come up with the following code, but it seems doesn't work, can anyone offer a advice?



filename = [re.search(r'^d{2}.csv'),filename).group(0) for filename in files]





regex python-3.x







share|improve this question























  • BTW, do you have _20.cvs or _20.csv?

    – Wiktor Stribiżew
    Nov 22 '18 at 8:10











  • _20.csv, thanks

    – Rowling
    Nov 22 '18 at 8:14
















1















I have following code which will store all the csv filename in a list from a specific folder



import pandas as pd

import re

import os



files = os.listdir('.')

filename=[filename for filename in files if filename.endswith('.csv')]


However, in my folder, I have two types of csv files, one ends with, for example, _20.cvs(or maybe _18.csv,_01.csv), another one ends with _Raw.csv;



However I only need the first type stored in my list. I know regular expression may can help me on that, so I did some google search, and come up with the following code, but it seems doesn't work, can anyone offer a advice?



filename = [re.search(r'^d{2}.csv'),filename).group(0) for filename in files]





regex python-3.x







share|improve this question























  • BTW, do you have _20.cvs or _20.csv?

    – Wiktor Stribiżew
    Nov 22 '18 at 8:10











  • _20.csv, thanks

    – Rowling
    Nov 22 '18 at 8:14














1












1








1








I have following code which will store all the csv filename in a list from a specific folder



import pandas as pd

import re

import os



files = os.listdir('.')

filename=[filename for filename in files if filename.endswith('.csv')]


However, in my folder, I have two types of csv files, one ends with, for example, _20.cvs(or maybe _18.csv,_01.csv), another one ends with _Raw.csv;



However I only need the first type stored in my list. I know regular expression may can help me on that, so I did some google search, and come up with the following code, but it seems doesn't work, can anyone offer a advice?



filename = [re.search(r'^d{2}.csv'),filename).group(0) for filename in files]





regex python-3.x







share|improve this question














I have following code which will store all the csv filename in a list from a specific folder



import pandas as pd

import re

import os



files = os.listdir('.')

filename=[filename for filename in files if filename.endswith('.csv')]


However, in my folder, I have two types of csv files, one ends with, for example, _20.cvs(or maybe _18.csv,_01.csv), another one ends with _Raw.csv;



However I only need the first type stored in my list. I know regular expression may can help me on that, so I did some google search, and come up with the following code, but it seems doesn't work, can anyone offer a advice?



filename = [re.search(r'^d{2}.csv'),filename).group(0) for filename in files]





regex python-3.x




regex python-3.x






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 22 '18 at 8:00









RowlingRowling

8811




8811













  • BTW, do you have _20.cvs or _20.csv?

    – Wiktor Stribiżew
    Nov 22 '18 at 8:10











  • _20.csv, thanks

    – Rowling
    Nov 22 '18 at 8:14



















  • BTW, do you have _20.cvs or _20.csv?

    – Wiktor Stribiżew
    Nov 22 '18 at 8:10











  • _20.csv, thanks

    – Rowling
    Nov 22 '18 at 8:14

















BTW, do you have _20.cvs or _20.csv?

– Wiktor Stribiżew
Nov 22 '18 at 8:10





BTW, do you have _20.cvs or _20.csv?

– Wiktor Stribiżew
Nov 22 '18 at 8:10













_20.csv, thanks

– Rowling
Nov 22 '18 at 8:14





_20.csv, thanks

– Rowling
Nov 22 '18 at 8:14












4 Answers
4






active

oldest

votes


















4














You need to remove ^ (as it matches the start of string location), add $ at the end of the pattern (to make sure the match is at the end of the string) and escape the dot (else, . matches any char but a line break char).



Note you must check if there is a match before accessing .group():



result = [f for f in files if re.search(r'_d{2}.csv$', f)]


Details





  • _ - an underscore


  • d{2} - 2 digits


  • . - a literal dot


  • csv - csv text


  • $ - end of string.


See the regex demo.



Python demo:



import re

files = ["gfrt_32_20.csv", "wertf_18.csv", "12_01.csv", "ith_Raw.csv"]

result = [f for f in files if re.search(r'_d{2}.csv$', f)] 

print(result)

# => ['gfrt_32_20.csv', 'wertf_18.csv', '12_01.csv']





share|improve this answer


























  • @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

    – Wiktor Stribiżew
    Nov 22 '18 at 11:48





















2














re.match would not work because it matches at the beginning. Use re.search instead.
But everything else is fine in the previous solution. 



import os

import re

files = os.listdir('.')

filenames = [f for f in files if re.search(r'(_d+.csv)', f)]

print(filenames)





share|improve this answer































    1














    Try to use re.match method:



    import os
    
import re
    
files = os.listdir('.')
    
filenames = [f for f in files if re.match(r'(_d+.csv)', f)]
    
print(filenames)





    share|improve this answer
























    • it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

      – Rowling
      Nov 22 '18 at 8:13











    • @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

      – Rezvanov Maxim
      Nov 22 '18 at 8:17











    • [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

      – Rowling
      Nov 22 '18 at 8:17











    • Data_100000_11_22.csv

      – Rowling
      Nov 22 '18 at 8:18











    • @Frank try regex here: pythex.org

      – Rezvanov Maxim
      Nov 22 '18 at 8:20



















    1














    You should put the regex operation in the if clause so as to filter out those you don't want.



    You should also escape the . in the regex, since dots have special meaning in regex (match all non-line terminators).



    [filename for filename in files if re.search(r'd{2}.csv$', filename)]


    If you want only the matched bit, you can do a simple substring:



    [filename[-6:] for filename in files if re.search(r'd{2}.csv$', filename)]





    share|improve this answer























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      You need to remove ^ (as it matches the start of string location), add $ at the end of the pattern (to make sure the match is at the end of the string) and escape the dot (else, . matches any char but a line break char).



      Note you must check if there is a match before accessing .group():



      result = [f for f in files if re.search(r'_d{2}.csv$', f)]


      Details





      • _ - an underscore


      • d{2} - 2 digits


      • . - a literal dot


      • csv - csv text


      • $ - end of string.


      See the regex demo.



      Python demo:



      import re
      
files = ["gfrt_32_20.csv", "wertf_18.csv", "12_01.csv", "ith_Raw.csv"]
      
result = [f for f in files if re.search(r'_d{2}.csv$', f)] 
      
print(result)
      
# => ['gfrt_32_20.csv', 'wertf_18.csv', '12_01.csv']





      share|improve this answer


























      • @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

        – Wiktor Stribiżew
        Nov 22 '18 at 11:48


















      4














      You need to remove ^ (as it matches the start of string location), add $ at the end of the pattern (to make sure the match is at the end of the string) and escape the dot (else, . matches any char but a line break char).



      Note you must check if there is a match before accessing .group():



      result = [f for f in files if re.search(r'_d{2}.csv$', f)]


      Details





      • _ - an underscore


      • d{2} - 2 digits


      • . - a literal dot


      • csv - csv text


      • $ - end of string.


      See the regex demo.



      Python demo:



      import re
      
files = ["gfrt_32_20.csv", "wertf_18.csv", "12_01.csv", "ith_Raw.csv"]
      
result = [f for f in files if re.search(r'_d{2}.csv$', f)] 
      
print(result)
      
# => ['gfrt_32_20.csv', 'wertf_18.csv', '12_01.csv']





      share|improve this answer


























      • @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

        – Wiktor Stribiżew
        Nov 22 '18 at 11:48
















      4












      4








      4







      You need to remove ^ (as it matches the start of string location), add $ at the end of the pattern (to make sure the match is at the end of the string) and escape the dot (else, . matches any char but a line break char).



      Note you must check if there is a match before accessing .group():



      result = [f for f in files if re.search(r'_d{2}.csv$', f)]


      Details





      • _ - an underscore


      • d{2} - 2 digits


      • . - a literal dot


      • csv - csv text


      • $ - end of string.


      See the regex demo.



      Python demo:



      import re
      
files = ["gfrt_32_20.csv", "wertf_18.csv", "12_01.csv", "ith_Raw.csv"]
      
result = [f for f in files if re.search(r'_d{2}.csv$', f)] 
      
print(result)
      
# => ['gfrt_32_20.csv', 'wertf_18.csv', '12_01.csv']





      share|improve this answer















      You need to remove ^ (as it matches the start of string location), add $ at the end of the pattern (to make sure the match is at the end of the string) and escape the dot (else, . matches any char but a line break char).



      Note you must check if there is a match before accessing .group():



      result = [f for f in files if re.search(r'_d{2}.csv$', f)]


      Details





      • _ - an underscore


      • d{2} - 2 digits


      • . - a literal dot


      • csv - csv text


      • $ - end of string.


      See the regex demo.



      Python demo:



      import re
      
files = ["gfrt_32_20.csv", "wertf_18.csv", "12_01.csv", "ith_Raw.csv"]
      
result = [f for f in files if re.search(r'_d{2}.csv$', f)] 
      
print(result)
      
# => ['gfrt_32_20.csv', 'wertf_18.csv', '12_01.csv']






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 22 '18 at 8:23

























      answered Nov 22 '18 at 8:17









      Wiktor StribiżewWiktor Stribiżew

      313k16133207




      313k16133207













      • @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

        – Wiktor Stribiżew
        Nov 22 '18 at 11:48





















      • @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

        – Wiktor Stribiżew
        Nov 22 '18 at 11:48



















      @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

      – Wiktor Stribiżew
      Nov 22 '18 at 11:48







      @Borisu There is no need adding the details about re.match and re.search difference into my answer as OP problem is not related to it. Here is a good thread on that.

      – Wiktor Stribiżew
      Nov 22 '18 at 11:48















      2














      re.match would not work because it matches at the beginning. Use re.search instead.
      But everything else is fine in the previous solution. 



      import os
      
import re
      
files = os.listdir('.')
      
filenames = [f for f in files if re.search(r'(_d+.csv)', f)]
      
print(filenames)





      share|improve this answer




























        2














        re.match would not work because it matches at the beginning. Use re.search instead.
        But everything else is fine in the previous solution. 



        import os
        
import re
        
files = os.listdir('.')
        
filenames = [f for f in files if re.search(r'(_d+.csv)', f)]
        
print(filenames)





        share|improve this answer


























          2












          2








          2







          re.match would not work because it matches at the beginning. Use re.search instead.
          But everything else is fine in the previous solution. 



          import os
          
import re
          
files = os.listdir('.')
          
filenames = [f for f in files if re.search(r'(_d+.csv)', f)]
          
print(filenames)





          share|improve this answer













          re.match would not work because it matches at the beginning. Use re.search instead.
          But everything else is fine in the previous solution. 



          import os
          
import re
          
files = os.listdir('.')
          
filenames = [f for f in files if re.search(r'(_d+.csv)', f)]
          
print(filenames)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 22 '18 at 8:18









          AResemAResem

          1114




          1114























              1














              Try to use re.match method:



              import os
              
import re
              
files = os.listdir('.')
              
filenames = [f for f in files if re.match(r'(_d+.csv)', f)]
              
print(filenames)





              share|improve this answer
























              • it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

                – Rowling
                Nov 22 '18 at 8:13











              • @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

                – Rezvanov Maxim
                Nov 22 '18 at 8:17











              • [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

                – Rowling
                Nov 22 '18 at 8:17











              • Data_100000_11_22.csv

                – Rowling
                Nov 22 '18 at 8:18











              • @Frank try regex here: pythex.org

                – Rezvanov Maxim
                Nov 22 '18 at 8:20
















              1














              Try to use re.match method:



              import os
              
import re
              
files = os.listdir('.')
              
filenames = [f for f in files if re.match(r'(_d+.csv)', f)]
              
print(filenames)





              share|improve this answer
























              • it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

                – Rowling
                Nov 22 '18 at 8:13











              • @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

                – Rezvanov Maxim
                Nov 22 '18 at 8:17











              • [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

                – Rowling
                Nov 22 '18 at 8:17











              • Data_100000_11_22.csv

                – Rowling
                Nov 22 '18 at 8:18











              • @Frank try regex here: pythex.org

                – Rezvanov Maxim
                Nov 22 '18 at 8:20














              1












              1








              1







              Try to use re.match method:



              import os
              
import re
              
files = os.listdir('.')
              
filenames = [f for f in files if re.match(r'(_d+.csv)', f)]
              
print(filenames)





              share|improve this answer













              Try to use re.match method:



              import os
              
import re
              
files = os.listdir('.')
              
filenames = [f for f in files if re.match(r'(_d+.csv)', f)]
              
print(filenames)






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 22 '18 at 8:09









              Rezvanov MaximRezvanov Maxim

              1216




              1216













              • it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

                – Rowling
                Nov 22 '18 at 8:13











              • @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

                – Rezvanov Maxim
                Nov 22 '18 at 8:17











              • [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

                – Rowling
                Nov 22 '18 at 8:17











              • Data_100000_11_22.csv

                – Rowling
                Nov 22 '18 at 8:18











              • @Frank try regex here: pythex.org

                – Rezvanov Maxim
                Nov 22 '18 at 8:20



















              • it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

                – Rowling
                Nov 22 '18 at 8:13











              • @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

                – Rezvanov Maxim
                Nov 22 '18 at 8:17











              • [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

                – Rowling
                Nov 22 '18 at 8:17











              • Data_100000_11_22.csv

                – Rowling
                Nov 22 '18 at 8:18











              • @Frank try regex here: pythex.org

                – Rezvanov Maxim
                Nov 22 '18 at 8:20

















              it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

              – Rowling
              Nov 22 '18 at 8:13





              it gives me empty; the filename is like :Mydate_2018_11_22.csv; but we only care about 22.csv? right? for this instance, i think match and search are the same?

              – Rowling
              Nov 22 '18 at 8:13













              @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

              – Rezvanov Maxim
              Nov 22 '18 at 8:17





              @Frank could you provide exmples with full names of files, which you need to collect? Yes, they are same. In this case re.search and re.match can replace each other.

              – Rezvanov Maxim
              Nov 22 '18 at 8:17













              [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

              – Rowling
              Nov 22 '18 at 8:17





              [f for f in files if re.search(r'(_d+.csv)', f)] works; [f for f in files if re.match(r'(_d+.csv)', f)] doesn't

              – Rowling
              Nov 22 '18 at 8:17













              Data_100000_11_22.csv

              – Rowling
              Nov 22 '18 at 8:18





              Data_100000_11_22.csv

              – Rowling
              Nov 22 '18 at 8:18













              @Frank try regex here: pythex.org

              – Rezvanov Maxim
              Nov 22 '18 at 8:20





              @Frank try regex here: pythex.org

              – Rezvanov Maxim
              Nov 22 '18 at 8:20











              1














              You should put the regex operation in the if clause so as to filter out those you don't want.



              You should also escape the . in the regex, since dots have special meaning in regex (match all non-line terminators).



              [filename for filename in files if re.search(r'd{2}.csv$', filename)]


              If you want only the matched bit, you can do a simple substring:



              [filename[-6:] for filename in files if re.search(r'd{2}.csv$', filename)]





              share|improve this answer




























                1














                You should put the regex operation in the if clause so as to filter out those you don't want.



                You should also escape the . in the regex, since dots have special meaning in regex (match all non-line terminators).



                [filename for filename in files if re.search(r'd{2}.csv$', filename)]


                If you want only the matched bit, you can do a simple substring:



                [filename[-6:] for filename in files if re.search(r'd{2}.csv$', filename)]





                share|improve this answer


























                  1












                  1








                  1







                  You should put the regex operation in the if clause so as to filter out those you don't want.



                  You should also escape the . in the regex, since dots have special meaning in regex (match all non-line terminators).



                  [filename for filename in files if re.search(r'd{2}.csv$', filename)]


                  If you want only the matched bit, you can do a simple substring:



                  [filename[-6:] for filename in files if re.search(r'd{2}.csv$', filename)]





                  share|improve this answer













                  You should put the regex operation in the if clause so as to filter out those you don't want.



                  You should also escape the . in the regex, since dots have special meaning in regex (match all non-line terminators).



                  [filename for filename in files if re.search(r'd{2}.csv$', filename)]


                  If you want only the matched bit, you can do a simple substring:



                  [filename[-6:] for filename in files if re.search(r'd{2}.csv$', filename)]






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 22 '18 at 8:16









                  SweeperSweeper

                  66k1073139




                  66k1073139






























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