Using np.where to find element from sub-arrays












4















I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:



X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])

Y = np.array([1, 2, 3, 4, 4, 5])

a = [2, 3, 4]

matchedX = X[np.where(Y == a)]


I am expecting to get the result like this:



array([[2, 1],

   [1, 3],

   [5, 9],

   [6, 7]])


but I got different results:



array(, shape=(0, 2), dtype=int64)


So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.



matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]





python numpy







share|improve this question



























    4















    I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:



    X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
    
Y = np.array([1, 2, 3, 4, 4, 5])
    
a = [2, 3, 4]
    
matchedX = X[np.where(Y == a)]


    I am expecting to get the result like this:



    array([[2, 1],
    
   [1, 3],
    
   [5, 9],
    
   [6, 7]])


    but I got different results:



    array(, shape=(0, 2), dtype=int64)


    So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.



    matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]





    python numpy







    share|improve this question

























      4












      4








      4








      I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:



      X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
      
Y = np.array([1, 2, 3, 4, 4, 5])
      
a = [2, 3, 4]
      
matchedX = X[np.where(Y == a)]


      I am expecting to get the result like this:



      array([[2, 1],
      
   [1, 3],
      
   [5, 9],
      
   [6, 7]])


      but I got different results:



      array(, shape=(0, 2), dtype=int64)


      So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.



      matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]





      python numpy







      share|improve this question














      I'm trying to find all elements in array X that match element on array Y using np.where() and the condition on where() function is comparing list (a) not one element. Please see the following code:



      X = np.array([[0, 2], [2, 1], [1, 3], [5, 9], [6, 7], [4, 6]])
      
Y = np.array([1, 2, 3, 4, 4, 5])
      
a = [2, 3, 4]
      
matchedX = X[np.where(Y == a)]


      I am expecting to get the result like this:



      array([[2, 1],
      
   [1, 3],
      
   [5, 9],
      
   [6, 7]])


      but I got different results:



      array(, shape=(0, 2), dtype=int64)


      So, I need an alternative solution in a way that I can get the same elements if I do not know about the values of a? This line below give me the exact results that I want, but I do not know in previous the a values.



      matchedX = X[np.where((Y == 2) | (Y==3) | (Y==4))]





      python numpy




      python numpy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 17 '18 at 0:04









      Esraa MohamedEsraa Mohamed

      234




      234
























          2 Answers
          2






          active

          oldest

          votes


















          1














          You can use the set functions of numpy:



          X[np.where(np.isin(Y, a))]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])





          share|improve this answer
























          • It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20



















          1














          You can skip the np.where, which is redundant here, and just index using np.isin:



          X[np.isin(Y,a)]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])


          This is because np.isin gives you a boolean array of where Y is in a:



          array([False,  True,  True,  True,  True, False])


          So by indexing with this array, it only selects those rows where True






          share|improve this answer





















          • 1





            It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20











          Your Answer






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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You can use the set functions of numpy:



          X[np.where(np.isin(Y, a))]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])





          share|improve this answer
























          • It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20
















          1














          You can use the set functions of numpy:



          X[np.where(np.isin(Y, a))]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])





          share|improve this answer
























          • It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20














          1












          1








          1







          You can use the set functions of numpy:



          X[np.where(np.isin(Y, a))]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])





          share|improve this answer













          You can use the set functions of numpy:



          X[np.where(np.isin(Y, a))]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 17 '18 at 0:12









          Matthias OssadnikMatthias Ossadnik

          57427




          57427













          • It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20



















          • It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20

















          It works fine, thank you :)

          – Esraa Mohamed
          Nov 17 '18 at 0:20





          It works fine, thank you :)

          – Esraa Mohamed
          Nov 17 '18 at 0:20













          1














          You can skip the np.where, which is redundant here, and just index using np.isin:



          X[np.isin(Y,a)]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])


          This is because np.isin gives you a boolean array of where Y is in a:



          array([False,  True,  True,  True,  True, False])


          So by indexing with this array, it only selects those rows where True






          share|improve this answer





















          • 1





            It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20
















          1














          You can skip the np.where, which is redundant here, and just index using np.isin:



          X[np.isin(Y,a)]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])


          This is because np.isin gives you a boolean array of where Y is in a:



          array([False,  True,  True,  True,  True, False])


          So by indexing with this array, it only selects those rows where True






          share|improve this answer





















          • 1





            It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20














          1












          1








          1







          You can skip the np.where, which is redundant here, and just index using np.isin:



          X[np.isin(Y,a)]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])


          This is because np.isin gives you a boolean array of where Y is in a:



          array([False,  True,  True,  True,  True, False])


          So by indexing with this array, it only selects those rows where True






          share|improve this answer















          You can skip the np.where, which is redundant here, and just index using np.isin:



          X[np.isin(Y,a)]
          

          
array([[2, 1],
          
       [1, 3],
          
       [5, 9],
          
       [6, 7]])


          This is because np.isin gives you a boolean array of where Y is in a:



          array([False,  True,  True,  True,  True, False])


          So by indexing with this array, it only selects those rows where True







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 17 '18 at 0:20

























          answered Nov 17 '18 at 0:16









          saculsacul

          30k41740




          30k41740








          • 1





            It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20














          • 1





            It works fine, thank you :)

            – Esraa Mohamed
            Nov 17 '18 at 0:20








          1




          1





          It works fine, thank you :)

          – Esraa Mohamed
          Nov 17 '18 at 0:20





          It works fine, thank you :)

          – Esraa Mohamed
          Nov 17 '18 at 0:20


















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