which numpy command could I use to subtract vectors with different dimensions many times?









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i have to write this function:
enter image description here
in which x is a vector with dimensions [150,2] and c is [N,2] (lets suppose N=20). From each component xi (i=1,2) I have to subtract the components of c in this way ([x11-c11,x12-c12])...([x11-cN1, x12-cN2])for all the 150 sample.
I've trasformed them in a way I have the same dimensions and I can subtract them, but the result of the function should be a vector. Maybe How can I write this in numpy?
Thank you
Ok, lets suppose x=(5,2) and c=(3,2)
enter image description here
this is what I have obtained transforming dimensions of the two arrays. the problem is that, I have to do this but with a iteration "for loop" because the exp function should give me as a result a vector. so I have to obtain a sort of matrix divided in N blocks.










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  • 1




    So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
    – iGian
    Nov 11 at 14:31










  • I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
    – xnx
    Nov 11 at 14:45











  • iGian i have added an example :)
    – ggg
    Nov 11 at 14:51










  • For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
    – iGian
    Nov 11 at 14:55










  • You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
    – Andras Deak
    Nov 11 at 15:00















up vote
-1
down vote

favorite
1












i have to write this function:
enter image description here
in which x is a vector with dimensions [150,2] and c is [N,2] (lets suppose N=20). From each component xi (i=1,2) I have to subtract the components of c in this way ([x11-c11,x12-c12])...([x11-cN1, x12-cN2])for all the 150 sample.
I've trasformed them in a way I have the same dimensions and I can subtract them, but the result of the function should be a vector. Maybe How can I write this in numpy?
Thank you
Ok, lets suppose x=(5,2) and c=(3,2)
enter image description here
this is what I have obtained transforming dimensions of the two arrays. the problem is that, I have to do this but with a iteration "for loop" because the exp function should give me as a result a vector. so I have to obtain a sort of matrix divided in N blocks.










share|improve this question



















  • 1




    So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
    – iGian
    Nov 11 at 14:31










  • I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
    – xnx
    Nov 11 at 14:45











  • iGian i have added an example :)
    – ggg
    Nov 11 at 14:51










  • For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
    – iGian
    Nov 11 at 14:55










  • You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
    – Andras Deak
    Nov 11 at 15:00













up vote
-1
down vote

favorite
1









up vote
-1
down vote

favorite
1






1





i have to write this function:
enter image description here
in which x is a vector with dimensions [150,2] and c is [N,2] (lets suppose N=20). From each component xi (i=1,2) I have to subtract the components of c in this way ([x11-c11,x12-c12])...([x11-cN1, x12-cN2])for all the 150 sample.
I've trasformed them in a way I have the same dimensions and I can subtract them, but the result of the function should be a vector. Maybe How can I write this in numpy?
Thank you
Ok, lets suppose x=(5,2) and c=(3,2)
enter image description here
this is what I have obtained transforming dimensions of the two arrays. the problem is that, I have to do this but with a iteration "for loop" because the exp function should give me as a result a vector. so I have to obtain a sort of matrix divided in N blocks.










share|improve this question















i have to write this function:
enter image description here
in which x is a vector with dimensions [150,2] and c is [N,2] (lets suppose N=20). From each component xi (i=1,2) I have to subtract the components of c in this way ([x11-c11,x12-c12])...([x11-cN1, x12-cN2])for all the 150 sample.
I've trasformed them in a way I have the same dimensions and I can subtract them, but the result of the function should be a vector. Maybe How can I write this in numpy?
Thank you
Ok, lets suppose x=(5,2) and c=(3,2)
enter image description here
this is what I have obtained transforming dimensions of the two arrays. the problem is that, I have to do this but with a iteration "for loop" because the exp function should give me as a result a vector. so I have to obtain a sort of matrix divided in N blocks.







python arrays numpy for-loop






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share|improve this question








edited Nov 11 at 15:16

























asked Nov 11 at 14:13









ggg

134




134







  • 1




    So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
    – iGian
    Nov 11 at 14:31










  • I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
    – xnx
    Nov 11 at 14:45











  • iGian i have added an example :)
    – ggg
    Nov 11 at 14:51










  • For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
    – iGian
    Nov 11 at 14:55










  • You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
    – Andras Deak
    Nov 11 at 15:00













  • 1




    So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
    – iGian
    Nov 11 at 14:31










  • I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
    – xnx
    Nov 11 at 14:45











  • iGian i have added an example :)
    – ggg
    Nov 11 at 14:51










  • For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
    – iGian
    Nov 11 at 14:55










  • You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
    – Andras Deak
    Nov 11 at 15:00








1




1




So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
– iGian
Nov 11 at 14:31




So you are subtracting just from the first row of x? Can you add sample data for example for x shape (5,2) and c shape (3,2)?
– iGian
Nov 11 at 14:31












I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
– xnx
Nov 11 at 14:45





I think you need another axis in x to account for the 20 values of the vector c: i.e. x[:, None] - c to give an array of shape (150,20,2)
– xnx
Nov 11 at 14:45













iGian i have added an example :)
– ggg
Nov 11 at 14:51




iGian i have added an example :)
– ggg
Nov 11 at 14:51












For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
– iGian
Nov 11 at 14:55




For the element 1,1 of the matrix you mean (8-1)+(8-5)+(8-5)? Anyway, it seems that the result is still a matrix having the same shape of x...
– iGian
Nov 11 at 14:55












You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
– Andras Deak
Nov 11 at 15:00





You probably need something like (x[:,None,:] - c).reshape(-1,2). Without the reshape you have 150 blocks of (N,2) arrays, like res[0,...].
– Andras Deak
Nov 11 at 15:00













2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










From what I understand of the issue, the problem seems to be in the way you are calculating the vector norm, not in the subtraction. Using your example, but calculating exp(-||x-c||), try:



x = np.linspace(8,17,10).reshape((5,2))
c = np.linspace(1,6,6).reshape((3,2))
sub = np.linalg.norm(x[:,None] - c, axis=-1)
np.exp(-sub)

array([[ 5.02000299e-05, 8.49325705e-04, 1.43695961e-02],
[ 2.96711024e-06, 5.02000299e-05, 8.49325705e-04],
[ 1.75373266e-07, 2.96711024e-06, 5.02000299e-05],
[ 1.03655678e-08, 1.75373266e-07, 2.96711024e-06],
[ 6.12664624e-10, 1.03655678e-08, 1.75373266e-07]])

np.exp(-sub).shape
(5, 3)


numpy.linalg.norm will try to return some kind of matrix norm across all the dimensions of its input unless you tell it explicitly which axis represents the vector components.






share|improve this answer





























    up vote
    0
    down vote













    I I understand, try if this give the expected result, but there is still the problem that the result has the same shape of x:



    import numpy as np

    x = np.arange(10).reshape(5,2)
    c = np.arange(6).reshape(3,2)

    c_col_sum = np.sum(c, axis=0)

    for (h,k), value in np.ndenumerate(x):
    x[h,k] = c.shape[0] * x[h,k] - c_col_sum[k]


    Initially x is:



    [[0 1]
    [2 3]
    [4 5]
    [6 7]
    [8 9]]


    And c is:



    [[0 1]
    [2 3]
    [4 5]]


    After the function x becomes:



    [[-6 -6]
    [ 0 0]
    [ 6 6]
    [12 12]
    [18 18]]





    share|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote



      accepted










      From what I understand of the issue, the problem seems to be in the way you are calculating the vector norm, not in the subtraction. Using your example, but calculating exp(-||x-c||), try:



      x = np.linspace(8,17,10).reshape((5,2))
      c = np.linspace(1,6,6).reshape((3,2))
      sub = np.linalg.norm(x[:,None] - c, axis=-1)
      np.exp(-sub)

      array([[ 5.02000299e-05, 8.49325705e-04, 1.43695961e-02],
      [ 2.96711024e-06, 5.02000299e-05, 8.49325705e-04],
      [ 1.75373266e-07, 2.96711024e-06, 5.02000299e-05],
      [ 1.03655678e-08, 1.75373266e-07, 2.96711024e-06],
      [ 6.12664624e-10, 1.03655678e-08, 1.75373266e-07]])

      np.exp(-sub).shape
      (5, 3)


      numpy.linalg.norm will try to return some kind of matrix norm across all the dimensions of its input unless you tell it explicitly which axis represents the vector components.






      share|improve this answer


























        up vote
        0
        down vote



        accepted










        From what I understand of the issue, the problem seems to be in the way you are calculating the vector norm, not in the subtraction. Using your example, but calculating exp(-||x-c||), try:



        x = np.linspace(8,17,10).reshape((5,2))
        c = np.linspace(1,6,6).reshape((3,2))
        sub = np.linalg.norm(x[:,None] - c, axis=-1)
        np.exp(-sub)

        array([[ 5.02000299e-05, 8.49325705e-04, 1.43695961e-02],
        [ 2.96711024e-06, 5.02000299e-05, 8.49325705e-04],
        [ 1.75373266e-07, 2.96711024e-06, 5.02000299e-05],
        [ 1.03655678e-08, 1.75373266e-07, 2.96711024e-06],
        [ 6.12664624e-10, 1.03655678e-08, 1.75373266e-07]])

        np.exp(-sub).shape
        (5, 3)


        numpy.linalg.norm will try to return some kind of matrix norm across all the dimensions of its input unless you tell it explicitly which axis represents the vector components.






        share|improve this answer
























          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          From what I understand of the issue, the problem seems to be in the way you are calculating the vector norm, not in the subtraction. Using your example, but calculating exp(-||x-c||), try:



          x = np.linspace(8,17,10).reshape((5,2))
          c = np.linspace(1,6,6).reshape((3,2))
          sub = np.linalg.norm(x[:,None] - c, axis=-1)
          np.exp(-sub)

          array([[ 5.02000299e-05, 8.49325705e-04, 1.43695961e-02],
          [ 2.96711024e-06, 5.02000299e-05, 8.49325705e-04],
          [ 1.75373266e-07, 2.96711024e-06, 5.02000299e-05],
          [ 1.03655678e-08, 1.75373266e-07, 2.96711024e-06],
          [ 6.12664624e-10, 1.03655678e-08, 1.75373266e-07]])

          np.exp(-sub).shape
          (5, 3)


          numpy.linalg.norm will try to return some kind of matrix norm across all the dimensions of its input unless you tell it explicitly which axis represents the vector components.






          share|improve this answer














          From what I understand of the issue, the problem seems to be in the way you are calculating the vector norm, not in the subtraction. Using your example, but calculating exp(-||x-c||), try:



          x = np.linspace(8,17,10).reshape((5,2))
          c = np.linspace(1,6,6).reshape((3,2))
          sub = np.linalg.norm(x[:,None] - c, axis=-1)
          np.exp(-sub)

          array([[ 5.02000299e-05, 8.49325705e-04, 1.43695961e-02],
          [ 2.96711024e-06, 5.02000299e-05, 8.49325705e-04],
          [ 1.75373266e-07, 2.96711024e-06, 5.02000299e-05],
          [ 1.03655678e-08, 1.75373266e-07, 2.96711024e-06],
          [ 6.12664624e-10, 1.03655678e-08, 1.75373266e-07]])

          np.exp(-sub).shape
          (5, 3)


          numpy.linalg.norm will try to return some kind of matrix norm across all the dimensions of its input unless you tell it explicitly which axis represents the vector components.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 11 at 15:53

























          answered Nov 11 at 15:33









          xnx

          15.1k43671




          15.1k43671






















              up vote
              0
              down vote













              I I understand, try if this give the expected result, but there is still the problem that the result has the same shape of x:



              import numpy as np

              x = np.arange(10).reshape(5,2)
              c = np.arange(6).reshape(3,2)

              c_col_sum = np.sum(c, axis=0)

              for (h,k), value in np.ndenumerate(x):
              x[h,k] = c.shape[0] * x[h,k] - c_col_sum[k]


              Initially x is:



              [[0 1]
              [2 3]
              [4 5]
              [6 7]
              [8 9]]


              And c is:



              [[0 1]
              [2 3]
              [4 5]]


              After the function x becomes:



              [[-6 -6]
              [ 0 0]
              [ 6 6]
              [12 12]
              [18 18]]





              share|improve this answer
























                up vote
                0
                down vote













                I I understand, try if this give the expected result, but there is still the problem that the result has the same shape of x:



                import numpy as np

                x = np.arange(10).reshape(5,2)
                c = np.arange(6).reshape(3,2)

                c_col_sum = np.sum(c, axis=0)

                for (h,k), value in np.ndenumerate(x):
                x[h,k] = c.shape[0] * x[h,k] - c_col_sum[k]


                Initially x is:



                [[0 1]
                [2 3]
                [4 5]
                [6 7]
                [8 9]]


                And c is:



                [[0 1]
                [2 3]
                [4 5]]


                After the function x becomes:



                [[-6 -6]
                [ 0 0]
                [ 6 6]
                [12 12]
                [18 18]]





                share|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  I I understand, try if this give the expected result, but there is still the problem that the result has the same shape of x:



                  import numpy as np

                  x = np.arange(10).reshape(5,2)
                  c = np.arange(6).reshape(3,2)

                  c_col_sum = np.sum(c, axis=0)

                  for (h,k), value in np.ndenumerate(x):
                  x[h,k] = c.shape[0] * x[h,k] - c_col_sum[k]


                  Initially x is:



                  [[0 1]
                  [2 3]
                  [4 5]
                  [6 7]
                  [8 9]]


                  And c is:



                  [[0 1]
                  [2 3]
                  [4 5]]


                  After the function x becomes:



                  [[-6 -6]
                  [ 0 0]
                  [ 6 6]
                  [12 12]
                  [18 18]]





                  share|improve this answer












                  I I understand, try if this give the expected result, but there is still the problem that the result has the same shape of x:



                  import numpy as np

                  x = np.arange(10).reshape(5,2)
                  c = np.arange(6).reshape(3,2)

                  c_col_sum = np.sum(c, axis=0)

                  for (h,k), value in np.ndenumerate(x):
                  x[h,k] = c.shape[0] * x[h,k] - c_col_sum[k]


                  Initially x is:



                  [[0 1]
                  [2 3]
                  [4 5]
                  [6 7]
                  [8 9]]


                  And c is:



                  [[0 1]
                  [2 3]
                  [4 5]]


                  After the function x becomes:



                  [[-6 -6]
                  [ 0 0]
                  [ 6 6]
                  [12 12]
                  [18 18]]






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 11 at 15:10









                  iGian

                  2,6642622




                  2,6642622



























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