Looking for a way to create a particular matrix in python in a less amount of time

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I want to create a matrix like this in Python:




0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3



0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3



1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1




I have written a code that works, but takes a long time to execute:



row = 3000
col = 4000

row_1 = np.asarray((col)*[0])
for i in range(1,row):
 row_1 = np.append(row_1, np.asarray((col)*[i]))
row_2 = np.asarray(row*[x for x in range(0,col)])
row_3 = np.asarray(col*(row)*[1])

output = np.vstack([row_1,row_2,row_3])


It takes between 55 to 65 seconds to run this code. Is it possible to create the same matrix in a way that is more efficient?






python time-limit-exceeded matrix numpy







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    up vote
    2
    down vote

    favorite












    I want to create a matrix like this in Python:




    0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3



    0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3



    1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1




    I have written a code that works, but takes a long time to execute:



    row = 3000
    col = 4000

    row_1 = np.asarray((col)*[0])
    for i in range(1,row):
     row_1 = np.append(row_1, np.asarray((col)*[i]))
    row_2 = np.asarray(row*[x for x in range(0,col)])
    row_3 = np.asarray(col*(row)*[1])

    output = np.vstack([row_1,row_2,row_3])


    It takes between 55 to 65 seconds to run this code. Is it possible to create the same matrix in a way that is more efficient?






    python time-limit-exceeded matrix numpy







    share|improve this question









    New contributor




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





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I want to create a matrix like this in Python:




      0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3



      0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3



      1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1




      I have written a code that works, but takes a long time to execute:



      row = 3000
      col = 4000

      row_1 = np.asarray((col)*[0])
      for i in range(1,row):
       row_1 = np.append(row_1, np.asarray((col)*[i]))
      row_2 = np.asarray(row*[x for x in range(0,col)])
      row_3 = np.asarray(col*(row)*[1])

      output = np.vstack([row_1,row_2,row_3])


      It takes between 55 to 65 seconds to run this code. Is it possible to create the same matrix in a way that is more efficient?






      python time-limit-exceeded matrix numpy







      share|improve this question









      New contributor




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











      I want to create a matrix like this in Python:




      0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3



      0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3



      1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1




      I have written a code that works, but takes a long time to execute:



      row = 3000
      col = 4000

      row_1 = np.asarray((col)*[0])
      for i in range(1,row):
       row_1 = np.append(row_1, np.asarray((col)*[i]))
      row_2 = np.asarray(row*[x for x in range(0,col)])
      row_3 = np.asarray(col*(row)*[1])

      output = np.vstack([row_1,row_2,row_3])


      It takes between 55 to 65 seconds to run this code. Is it possible to create the same matrix in a way that is more efficient?






      python time-limit-exceeded matrix numpy




      python time-limit-exceeded matrix numpy






      share|improve this question









      New contributor




      stressed out 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




      stressed out 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 1 hour ago









      Gareth Rees

      43.8k397176




      43.8k397176






      New contributor




      stressed out is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 2 hours ago









      stressed out

      1114




      1114




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      New contributor





      stressed out is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          1 Answer
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          The NumPy Reference should be the first place you look when you have a problem like this. The operations you need are nearly always in there somewhere. And functions that you find while browsing the reference are sure to come in useful later in your NumPy career.



          For the first row, you could use numpy.arange and numpy.repeat:



          >>> np.repeat(np.arange(4), 4)
          array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])


          For the second row, you could use numpy.tile:



          >>> np.tile(np.arange(4), 4)
          array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3])


          For the third row, you could use numpy.ones:



          >>> np.ones(4 * 4, dtype=int)
          array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])


          Putting that together (using numpy.ones_like for row3 so that it has the same shape and data type as row2):



          row1 = np.repeat(np.arange(row), col)
          row2 = np.tile(np.arange(col), row)
          row3 = np.ones_like(row2)
          output = np.vstack([row1, row2, row3])


          This is about 100 times as fast as the code in the post.



          But you can do better than that! There is a hint in the documentation for numpy.tile:




          Note: Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.




          Broadcasting is the way that NumPy adapts the inputs to an operation so that they match in shape. Using this we can write the whole function using array indexing and reshaping operations, like this:



          output = np.zeros((3, row, col), dtype=int)
          output[0] = np.arange(row).reshape(-1, 1)
          output[1] = np.arange(col).reshape(1, -1)
          output[2] = 1
          output = output.reshape(3, -1)


          This is about 160 times as fast as that code in the post.



          A slightly clearer way to write this is to use numpy.meshgrid to generate the coordinate arrays:



          output = np.zeros((3, row, col), dtype=int)
          i, j = np.meshgrid(np.arange(row), np.arange(col), indexing='ij', sparse=True)
          output[0] = i
          output[1] = j
          output[2] = 1
          output = output.reshape(3, -1)


          This isn't any faster than the previous version but the use of numpy.meshgrid gives more of a clue to the reader as to what is happening.



          (You might be able to do even better than that. Do you really need to construct the whole matrix? Could you combine broadcasting and numpy.meshgrid to get the results you want instead? I can't answer this since you didn't explain what you are going to use this code for. But maybe you can ask another question showing us more of your program.)






          share|improve this answer






















          • This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
            – stressed out
            1 hour ago










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          1 Answer
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          active

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          up vote
          3
          down vote













          The NumPy Reference should be the first place you look when you have a problem like this. The operations you need are nearly always in there somewhere. And functions that you find while browsing the reference are sure to come in useful later in your NumPy career.



          For the first row, you could use numpy.arange and numpy.repeat:



          >>> np.repeat(np.arange(4), 4)
          array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])


          For the second row, you could use numpy.tile:



          >>> np.tile(np.arange(4), 4)
          array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3])


          For the third row, you could use numpy.ones:



          >>> np.ones(4 * 4, dtype=int)
          array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])


          Putting that together (using numpy.ones_like for row3 so that it has the same shape and data type as row2):



          row1 = np.repeat(np.arange(row), col)
          row2 = np.tile(np.arange(col), row)
          row3 = np.ones_like(row2)
          output = np.vstack([row1, row2, row3])


          This is about 100 times as fast as the code in the post.



          But you can do better than that! There is a hint in the documentation for numpy.tile:




          Note: Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.




          Broadcasting is the way that NumPy adapts the inputs to an operation so that they match in shape. Using this we can write the whole function using array indexing and reshaping operations, like this:



          output = np.zeros((3, row, col), dtype=int)
          output[0] = np.arange(row).reshape(-1, 1)
          output[1] = np.arange(col).reshape(1, -1)
          output[2] = 1
          output = output.reshape(3, -1)


          This is about 160 times as fast as that code in the post.



          A slightly clearer way to write this is to use numpy.meshgrid to generate the coordinate arrays:



          output = np.zeros((3, row, col), dtype=int)
          i, j = np.meshgrid(np.arange(row), np.arange(col), indexing='ij', sparse=True)
          output[0] = i
          output[1] = j
          output[2] = 1
          output = output.reshape(3, -1)


          This isn't any faster than the previous version but the use of numpy.meshgrid gives more of a clue to the reader as to what is happening.



          (You might be able to do even better than that. Do you really need to construct the whole matrix? Could you combine broadcasting and numpy.meshgrid to get the results you want instead? I can't answer this since you didn't explain what you are going to use this code for. But maybe you can ask another question showing us more of your program.)






          share|improve this answer






















          • This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
            – stressed out
            1 hour ago














          up vote
          3
          down vote













          The NumPy Reference should be the first place you look when you have a problem like this. The operations you need are nearly always in there somewhere. And functions that you find while browsing the reference are sure to come in useful later in your NumPy career.



          For the first row, you could use numpy.arange and numpy.repeat:



          >>> np.repeat(np.arange(4), 4)
          array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])


          For the second row, you could use numpy.tile:



          >>> np.tile(np.arange(4), 4)
          array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3])


          For the third row, you could use numpy.ones:



          >>> np.ones(4 * 4, dtype=int)
          array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])


          Putting that together (using numpy.ones_like for row3 so that it has the same shape and data type as row2):



          row1 = np.repeat(np.arange(row), col)
          row2 = np.tile(np.arange(col), row)
          row3 = np.ones_like(row2)
          output = np.vstack([row1, row2, row3])


          This is about 100 times as fast as the code in the post.



          But you can do better than that! There is a hint in the documentation for numpy.tile:




          Note: Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.




          Broadcasting is the way that NumPy adapts the inputs to an operation so that they match in shape. Using this we can write the whole function using array indexing and reshaping operations, like this:



          output = np.zeros((3, row, col), dtype=int)
          output[0] = np.arange(row).reshape(-1, 1)
          output[1] = np.arange(col).reshape(1, -1)
          output[2] = 1
          output = output.reshape(3, -1)


          This is about 160 times as fast as that code in the post.



          A slightly clearer way to write this is to use numpy.meshgrid to generate the coordinate arrays:



          output = np.zeros((3, row, col), dtype=int)
          i, j = np.meshgrid(np.arange(row), np.arange(col), indexing='ij', sparse=True)
          output[0] = i
          output[1] = j
          output[2] = 1
          output = output.reshape(3, -1)


          This isn't any faster than the previous version but the use of numpy.meshgrid gives more of a clue to the reader as to what is happening.



          (You might be able to do even better than that. Do you really need to construct the whole matrix? Could you combine broadcasting and numpy.meshgrid to get the results you want instead? I can't answer this since you didn't explain what you are going to use this code for. But maybe you can ask another question showing us more of your program.)






          share|improve this answer






















          • This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
            – stressed out
            1 hour ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          The NumPy Reference should be the first place you look when you have a problem like this. The operations you need are nearly always in there somewhere. And functions that you find while browsing the reference are sure to come in useful later in your NumPy career.



          For the first row, you could use numpy.arange and numpy.repeat:



          >>> np.repeat(np.arange(4), 4)
          array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])


          For the second row, you could use numpy.tile:



          >>> np.tile(np.arange(4), 4)
          array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3])


          For the third row, you could use numpy.ones:



          >>> np.ones(4 * 4, dtype=int)
          array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])


          Putting that together (using numpy.ones_like for row3 so that it has the same shape and data type as row2):



          row1 = np.repeat(np.arange(row), col)
          row2 = np.tile(np.arange(col), row)
          row3 = np.ones_like(row2)
          output = np.vstack([row1, row2, row3])


          This is about 100 times as fast as the code in the post.



          But you can do better than that! There is a hint in the documentation for numpy.tile:




          Note: Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.




          Broadcasting is the way that NumPy adapts the inputs to an operation so that they match in shape. Using this we can write the whole function using array indexing and reshaping operations, like this:



          output = np.zeros((3, row, col), dtype=int)
          output[0] = np.arange(row).reshape(-1, 1)
          output[1] = np.arange(col).reshape(1, -1)
          output[2] = 1
          output = output.reshape(3, -1)


          This is about 160 times as fast as that code in the post.



          A slightly clearer way to write this is to use numpy.meshgrid to generate the coordinate arrays:



          output = np.zeros((3, row, col), dtype=int)
          i, j = np.meshgrid(np.arange(row), np.arange(col), indexing='ij', sparse=True)
          output[0] = i
          output[1] = j
          output[2] = 1
          output = output.reshape(3, -1)


          This isn't any faster than the previous version but the use of numpy.meshgrid gives more of a clue to the reader as to what is happening.



          (You might be able to do even better than that. Do you really need to construct the whole matrix? Could you combine broadcasting and numpy.meshgrid to get the results you want instead? I can't answer this since you didn't explain what you are going to use this code for. But maybe you can ask another question showing us more of your program.)






          share|improve this answer














          The NumPy Reference should be the first place you look when you have a problem like this. The operations you need are nearly always in there somewhere. And functions that you find while browsing the reference are sure to come in useful later in your NumPy career.



          For the first row, you could use numpy.arange and numpy.repeat:



          >>> np.repeat(np.arange(4), 4)
          array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])


          For the second row, you could use numpy.tile:



          >>> np.tile(np.arange(4), 4)
          array([0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3])


          For the third row, you could use numpy.ones:



          >>> np.ones(4 * 4, dtype=int)
          array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])


          Putting that together (using numpy.ones_like for row3 so that it has the same shape and data type as row2):



          row1 = np.repeat(np.arange(row), col)
          row2 = np.tile(np.arange(col), row)
          row3 = np.ones_like(row2)
          output = np.vstack([row1, row2, row3])


          This is about 100 times as fast as the code in the post.



          But you can do better than that! There is a hint in the documentation for numpy.tile:




          Note: Although tile may be used for broadcasting, it is strongly recommended to use numpy’s broadcasting operations and functions.




          Broadcasting is the way that NumPy adapts the inputs to an operation so that they match in shape. Using this we can write the whole function using array indexing and reshaping operations, like this:



          output = np.zeros((3, row, col), dtype=int)
          output[0] = np.arange(row).reshape(-1, 1)
          output[1] = np.arange(col).reshape(1, -1)
          output[2] = 1
          output = output.reshape(3, -1)


          This is about 160 times as fast as that code in the post.



          A slightly clearer way to write this is to use numpy.meshgrid to generate the coordinate arrays:



          output = np.zeros((3, row, col), dtype=int)
          i, j = np.meshgrid(np.arange(row), np.arange(col), indexing='ij', sparse=True)
          output[0] = i
          output[1] = j
          output[2] = 1
          output = output.reshape(3, -1)


          This isn't any faster than the previous version but the use of numpy.meshgrid gives more of a clue to the reader as to what is happening.



          (You might be able to do even better than that. Do you really need to construct the whole matrix? Could you combine broadcasting and numpy.meshgrid to get the results you want instead? I can't answer this since you didn't explain what you are going to use this code for. But maybe you can ask another question showing us more of your program.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Gareth Rees

          43.8k397176




          43.8k397176











          • This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
            – stressed out
            1 hour ago
















          • This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
            – stressed out
            1 hour ago















          This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
          – stressed out
          1 hour ago




          This is an amazing post, Gareth. Thank you for all the explanations. I think I need some more time to read the documents you linked to grasp what is really going on with your second and third versions.
          – stressed out
          1 hour ago










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









           

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












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