find a substring in a selected portion of a list elements











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I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.



I tried the below code but it finds all the elements where str b is found.



list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
b=str(1)
print([s for s in list1 if b in s])


it returns the following:



['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']


However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?










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    up vote
    1
    down vote

    favorite












    I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.



    I tried the below code but it finds all the elements where str b is found.



    list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
    b=str(1)
    print([s for s in list1 if b in s])


    it returns the following:



    ['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']


    However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?










    share|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.



      I tried the below code but it finds all the elements where str b is found.



      list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
      b=str(1)
      print([s for s in list1 if b in s])


      it returns the following:



      ['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']


      However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?










      share|improve this question















      I have the below string list (list1) and I want to find if str b is present anywhere in left hand side portion of an element before the decimal in list1.



      I tried the below code but it finds all the elements where str b is found.



      list1= ['4.39', '5.25', '2.29', '3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16', '4.4']
      b=str(1)
      print([s for s in list1 if b in s])


      it returns the following:



      ['3.16', '4.19', '1.5', '4.17', '2.18', '5.18', '4.18', '5.16']


      However, I want to get only 1.5 because this is the only element where string b matches the left hand side part before decimal. Remember the elements are in string format. Any fast way of checking this thing?







      python string list list-comprehension python-3.5






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 22 at 15:37









      jpp

      87.8k195099




      87.8k195099










      asked Nov 22 at 15:33









      HT121

      817




      817
























          1 Answer
          1






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          oldest

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          up vote
          2
          down vote



          accepted










          You need to split each string by . and extract the first split:



          print([s for s in list1 if '1' in s.split('.')[0]])

          ['1.5']


          For a precise match, use ==:



          print([s for s in list1 if s.split('.')[0] == '1'])

          ['1.5']





          share|improve this answer























          • And do you know how could i find the actual index of matched element in list1?
            – HT121
            Nov 22 at 15:47






          • 1




            [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
            – jpp
            Nov 22 at 15:59












          • Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
            – HT121
            Nov 22 at 16:08










          • @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
            – jpp
            Nov 22 at 16:10













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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          You need to split each string by . and extract the first split:



          print([s for s in list1 if '1' in s.split('.')[0]])

          ['1.5']


          For a precise match, use ==:



          print([s for s in list1 if s.split('.')[0] == '1'])

          ['1.5']





          share|improve this answer























          • And do you know how could i find the actual index of matched element in list1?
            – HT121
            Nov 22 at 15:47






          • 1




            [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
            – jpp
            Nov 22 at 15:59












          • Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
            – HT121
            Nov 22 at 16:08










          • @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
            – jpp
            Nov 22 at 16:10

















          up vote
          2
          down vote



          accepted










          You need to split each string by . and extract the first split:



          print([s for s in list1 if '1' in s.split('.')[0]])

          ['1.5']


          For a precise match, use ==:



          print([s for s in list1 if s.split('.')[0] == '1'])

          ['1.5']





          share|improve this answer























          • And do you know how could i find the actual index of matched element in list1?
            – HT121
            Nov 22 at 15:47






          • 1




            [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
            – jpp
            Nov 22 at 15:59












          • Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
            – HT121
            Nov 22 at 16:08










          • @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
            – jpp
            Nov 22 at 16:10















          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          You need to split each string by . and extract the first split:



          print([s for s in list1 if '1' in s.split('.')[0]])

          ['1.5']


          For a precise match, use ==:



          print([s for s in list1 if s.split('.')[0] == '1'])

          ['1.5']





          share|improve this answer














          You need to split each string by . and extract the first split:



          print([s for s in list1 if '1' in s.split('.')[0]])

          ['1.5']


          For a precise match, use ==:



          print([s for s in list1 if s.split('.')[0] == '1'])

          ['1.5']






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 22 at 15:37

























          answered Nov 22 at 15:35









          jpp

          87.8k195099




          87.8k195099












          • And do you know how could i find the actual index of matched element in list1?
            – HT121
            Nov 22 at 15:47






          • 1




            [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
            – jpp
            Nov 22 at 15:59












          • Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
            – HT121
            Nov 22 at 16:08










          • @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
            – jpp
            Nov 22 at 16:10




















          • And do you know how could i find the actual index of matched element in list1?
            – HT121
            Nov 22 at 15:47






          • 1




            [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
            – jpp
            Nov 22 at 15:59












          • Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
            – HT121
            Nov 22 at 16:08










          • @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
            – jpp
            Nov 22 at 16:10


















          And do you know how could i find the actual index of matched element in list1?
          – HT121
          Nov 22 at 15:47




          And do you know how could i find the actual index of matched element in list1?
          – HT121
          Nov 22 at 15:47




          1




          1




          [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
          – jpp
          Nov 22 at 15:59






          [idx for idx, s in enumerate(list1) if '1' in s.split('.')[0]] should do it.
          – jpp
          Nov 22 at 15:59














          Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
          – HT121
          Nov 22 at 16:08




          Thank you jpp. it works fine. Just a small comment, there seems to be a small typo in the code. There is one additional bracket by mistake after [0] in the split s.split('.')[0])
          – HT121
          Nov 22 at 16:08












          @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
          – jpp
          Nov 22 at 16:10






          @HT121, I think it should be fine, there are 2 square open brackets and 2 closing ones? I edited it to make a list comprehension earlier (you may wish to refresh).
          – jpp
          Nov 22 at 16:10




















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