How to efficiently unroll a matrix by value with numpy?





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6















I have a matrix M with values 0 through N within it. I'd like to unroll this matrix to create a new matrix A where each submatrix A[i, :, :] represents whether or not M == i.



The solution below uses a loop.



# Example Setup

import numpy as np



np.random.seed(0)

N = 5

M = np.random.randint(0, N, size=(5,5))



# Solution with Loop

A = np.zeros((N, M.shape[0], M.shape[1]))

for i in range(N):

    A[i, :, :] = M == i


This yields: 



M

array([[4, 0, 3, 3, 3],

       [1, 3, 2, 4, 0],

       [0, 4, 2, 1, 0],

       [1, 1, 0, 1, 4],

       [3, 0, 3, 0, 2]])



M.shape

# (5, 5)





A 

array([[[0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [1, 0, 0, 0, 1],

        [0, 0, 1, 0, 0],

        [0, 1, 0, 1, 0]],

       ...

       [[1, 0, 0, 0, 0],

        [0, 0, 0, 1, 0],

        [0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [0, 0, 0, 0, 0]]])



A.shape

# (5, 5, 5)


Is there a faster way, or a way to do it in a single numpy operation?






python arrays numpy







share|improve this question

























  • It would be better if you explain it in detail.

    – Marios Nikolaou
    6 hours ago






  • 2





    @MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

    – Reedinationer
    6 hours ago













  • @Reedinationer i did it.

    – Marios Nikolaou
    6 hours ago











  • I would not recommend pasting output for this code as the input is randomised without a seed.

    – coldspeed
    6 hours ago






  • 1





    i == M compare int with array 5x5 ? and then save it in A?

    – Marios Nikolaou
    6 hours ago


















6















I have a matrix M with values 0 through N within it. I'd like to unroll this matrix to create a new matrix A where each submatrix A[i, :, :] represents whether or not M == i.



The solution below uses a loop.



# Example Setup

import numpy as np



np.random.seed(0)

N = 5

M = np.random.randint(0, N, size=(5,5))



# Solution with Loop

A = np.zeros((N, M.shape[0], M.shape[1]))

for i in range(N):

    A[i, :, :] = M == i


This yields: 



M

array([[4, 0, 3, 3, 3],

       [1, 3, 2, 4, 0],

       [0, 4, 2, 1, 0],

       [1, 1, 0, 1, 4],

       [3, 0, 3, 0, 2]])



M.shape

# (5, 5)





A 

array([[[0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [1, 0, 0, 0, 1],

        [0, 0, 1, 0, 0],

        [0, 1, 0, 1, 0]],

       ...

       [[1, 0, 0, 0, 0],

        [0, 0, 0, 1, 0],

        [0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [0, 0, 0, 0, 0]]])



A.shape

# (5, 5, 5)


Is there a faster way, or a way to do it in a single numpy operation?






python arrays numpy







share|improve this question

























  • It would be better if you explain it in detail.

    – Marios Nikolaou
    6 hours ago






  • 2





    @MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

    – Reedinationer
    6 hours ago













  • @Reedinationer i did it.

    – Marios Nikolaou
    6 hours ago











  • I would not recommend pasting output for this code as the input is randomised without a seed.

    – coldspeed
    6 hours ago






  • 1





    i == M compare int with array 5x5 ? and then save it in A?

    – Marios Nikolaou
    6 hours ago














6












6








6








I have a matrix M with values 0 through N within it. I'd like to unroll this matrix to create a new matrix A where each submatrix A[i, :, :] represents whether or not M == i.



The solution below uses a loop.



# Example Setup

import numpy as np



np.random.seed(0)

N = 5

M = np.random.randint(0, N, size=(5,5))



# Solution with Loop

A = np.zeros((N, M.shape[0], M.shape[1]))

for i in range(N):

    A[i, :, :] = M == i


This yields: 



M

array([[4, 0, 3, 3, 3],

       [1, 3, 2, 4, 0],

       [0, 4, 2, 1, 0],

       [1, 1, 0, 1, 4],

       [3, 0, 3, 0, 2]])



M.shape

# (5, 5)





A 

array([[[0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [1, 0, 0, 0, 1],

        [0, 0, 1, 0, 0],

        [0, 1, 0, 1, 0]],

       ...

       [[1, 0, 0, 0, 0],

        [0, 0, 0, 1, 0],

        [0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [0, 0, 0, 0, 0]]])



A.shape

# (5, 5, 5)


Is there a faster way, or a way to do it in a single numpy operation?






python arrays numpy







share|improve this question
















I have a matrix M with values 0 through N within it. I'd like to unroll this matrix to create a new matrix A where each submatrix A[i, :, :] represents whether or not M == i.



The solution below uses a loop.



# Example Setup

import numpy as np



np.random.seed(0)

N = 5

M = np.random.randint(0, N, size=(5,5))



# Solution with Loop

A = np.zeros((N, M.shape[0], M.shape[1]))

for i in range(N):

    A[i, :, :] = M == i


This yields: 



M

array([[4, 0, 3, 3, 3],

       [1, 3, 2, 4, 0],

       [0, 4, 2, 1, 0],

       [1, 1, 0, 1, 4],

       [3, 0, 3, 0, 2]])



M.shape

# (5, 5)





A 

array([[[0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [1, 0, 0, 0, 1],

        [0, 0, 1, 0, 0],

        [0, 1, 0, 1, 0]],

       ...

       [[1, 0, 0, 0, 0],

        [0, 0, 0, 1, 0],

        [0, 1, 0, 0, 0],

        [0, 0, 0, 0, 1],

        [0, 0, 0, 0, 0]]])



A.shape

# (5, 5, 5)


Is there a faster way, or a way to do it in a single numpy operation?






python arrays numpy




python arrays numpy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 6 hours ago









coldspeed

140k24156241




140k24156241










asked 6 hours ago









seveibarseveibar

1,29211225




1,29211225













  • It would be better if you explain it in detail.

    – Marios Nikolaou
    6 hours ago






  • 2





    @MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

    – Reedinationer
    6 hours ago













  • @Reedinationer i did it.

    – Marios Nikolaou
    6 hours ago











  • I would not recommend pasting output for this code as the input is randomised without a seed.

    – coldspeed
    6 hours ago






  • 1





    i == M compare int with array 5x5 ? and then save it in A?

    – Marios Nikolaou
    6 hours ago



















  • It would be better if you explain it in detail.

    – Marios Nikolaou
    6 hours ago






  • 2





    @MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

    – Reedinationer
    6 hours ago













  • @Reedinationer i did it.

    – Marios Nikolaou
    6 hours ago











  • I would not recommend pasting output for this code as the input is randomised without a seed.

    – coldspeed
    6 hours ago






  • 1





    i == M compare int with array 5x5 ? and then save it in A?

    – Marios Nikolaou
    6 hours ago

















It would be better if you explain it in detail.

– Marios Nikolaou
6 hours ago





It would be better if you explain it in detail.

– Marios Nikolaou
6 hours ago




2




2





@MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

– Reedinationer
6 hours ago







@MariosNikolaou just copy/paste his code and print(M);print(A)...I edited it for you though

– Reedinationer
6 hours ago















@Reedinationer i did it.

– Marios Nikolaou
6 hours ago





@Reedinationer i did it.

– Marios Nikolaou
6 hours ago













I would not recommend pasting output for this code as the input is randomised without a seed.

– coldspeed
6 hours ago





I would not recommend pasting output for this code as the input is randomised without a seed.

– coldspeed
6 hours ago




1




1





i == M compare int with array 5x5 ? and then save it in A?

– Marios Nikolaou
6 hours ago





i == M compare int with array 5x5 ? and then save it in A?

– Marios Nikolaou
6 hours ago












3 Answers
3






active

oldest

votes


















6














You can make use of some broadcasting here:



P = np.arange(N)

Y = np.broadcast_to(P[:, None], M.shape)

T = np.equal(M, Y[:, None]).astype(int)



Alternative using indices:



X, Y = np.indices(M.shape)

Z = np.equal(M, X[:, None]).astype(int)





share|improve this answer


























  • This answer was really helpful towards my understanding of broadcasting, thank you!

    – seveibar
    4 hours ago



















6














Broadcasted comparison is your friend:



B = (M[None, :] == np.arange(N)[:, None, None]).view(np.int8)



 np.array_equal(A, B)

# True


The idea is to expand the dimensions in such a way that the comparison can be broadcasted in the manner desired.




As pointed out by @Alex Riley in the comments, you can use np.equal.outer to avoid having to do the indexing stuff yourself,



B = np.equal.outer(np.arange(N), M).view(np.int8)



np.array_equal(A, B)

# True





share|improve this answer





















  • 1





    Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

    – Alex Riley
    6 hours ago













  • @AlexRiley Thanks for that! And the outer solution is quite neat.

    – coldspeed
    6 hours ago



















3














You can index into the identity matrix like so



 A = np.identity(N, int)[:, M]


or so



 A = np.identity(N, int)[M.T].T


Or use the new (v1.15.0) put_along_axis



A = np.zeros((N,5,5), int)

np.put_along_axis(A, M[None], 1, 0)


Note if N is much larger than 5 then creating an NxN identity matrix may be considered wasteful. We can mitigate this using stride tricks:



def read_only_identity(N, dtype=float):

    z = np.zeros(2*N-1, dtype)

    s, = z.strides

    z[N-1] = 1

    return np.lib.stride_tricks.as_strided(z[N-1:], (N, N), (-s, s))





share|improve this answer





















  • 1





    This is great, really interesting answer.

    – user3483203
    5 hours ago






  • 1





    Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

    – seveibar
    4 hours ago






  • 1





    @seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

    – Paul Panzer
    3 hours ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














You can make use of some broadcasting here:



P = np.arange(N)

Y = np.broadcast_to(P[:, None], M.shape)

T = np.equal(M, Y[:, None]).astype(int)



Alternative using indices:



X, Y = np.indices(M.shape)

Z = np.equal(M, X[:, None]).astype(int)





share|improve this answer


























  • This answer was really helpful towards my understanding of broadcasting, thank you!

    – seveibar
    4 hours ago
















6














You can make use of some broadcasting here:



P = np.arange(N)

Y = np.broadcast_to(P[:, None], M.shape)

T = np.equal(M, Y[:, None]).astype(int)



Alternative using indices:



X, Y = np.indices(M.shape)

Z = np.equal(M, X[:, None]).astype(int)





share|improve this answer


























  • This answer was really helpful towards my understanding of broadcasting, thank you!

    – seveibar
    4 hours ago














6












6








6







You can make use of some broadcasting here:



P = np.arange(N)

Y = np.broadcast_to(P[:, None], M.shape)

T = np.equal(M, Y[:, None]).astype(int)



Alternative using indices:



X, Y = np.indices(M.shape)

Z = np.equal(M, X[:, None]).astype(int)





share|improve this answer















You can make use of some broadcasting here:



P = np.arange(N)

Y = np.broadcast_to(P[:, None], M.shape)

T = np.equal(M, Y[:, None]).astype(int)



Alternative using indices:



X, Y = np.indices(M.shape)

Z = np.equal(M, X[:, None]).astype(int)






share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 6 hours ago









user3483203user3483203

31.8k82857




31.8k82857













  • This answer was really helpful towards my understanding of broadcasting, thank you!

    – seveibar
    4 hours ago



















  • This answer was really helpful towards my understanding of broadcasting, thank you!

    – seveibar
    4 hours ago

















This answer was really helpful towards my understanding of broadcasting, thank you!

– seveibar
4 hours ago





This answer was really helpful towards my understanding of broadcasting, thank you!

– seveibar
4 hours ago













6














Broadcasted comparison is your friend:



B = (M[None, :] == np.arange(N)[:, None, None]).view(np.int8)



 np.array_equal(A, B)

# True


The idea is to expand the dimensions in such a way that the comparison can be broadcasted in the manner desired.




As pointed out by @Alex Riley in the comments, you can use np.equal.outer to avoid having to do the indexing stuff yourself,



B = np.equal.outer(np.arange(N), M).view(np.int8)



np.array_equal(A, B)

# True





share|improve this answer





















  • 1





    Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

    – Alex Riley
    6 hours ago













  • @AlexRiley Thanks for that! And the outer solution is quite neat.

    – coldspeed
    6 hours ago
















6














Broadcasted comparison is your friend:



B = (M[None, :] == np.arange(N)[:, None, None]).view(np.int8)



 np.array_equal(A, B)

# True


The idea is to expand the dimensions in such a way that the comparison can be broadcasted in the manner desired.




As pointed out by @Alex Riley in the comments, you can use np.equal.outer to avoid having to do the indexing stuff yourself,



B = np.equal.outer(np.arange(N), M).view(np.int8)



np.array_equal(A, B)

# True





share|improve this answer





















  • 1





    Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

    – Alex Riley
    6 hours ago













  • @AlexRiley Thanks for that! And the outer solution is quite neat.

    – coldspeed
    6 hours ago














6












6








6







Broadcasted comparison is your friend:



B = (M[None, :] == np.arange(N)[:, None, None]).view(np.int8)



 np.array_equal(A, B)

# True


The idea is to expand the dimensions in such a way that the comparison can be broadcasted in the manner desired.




As pointed out by @Alex Riley in the comments, you can use np.equal.outer to avoid having to do the indexing stuff yourself,



B = np.equal.outer(np.arange(N), M).view(np.int8)



np.array_equal(A, B)

# True





share|improve this answer















Broadcasted comparison is your friend:



B = (M[None, :] == np.arange(N)[:, None, None]).view(np.int8)



 np.array_equal(A, B)

# True


The idea is to expand the dimensions in such a way that the comparison can be broadcasted in the manner desired.




As pointed out by @Alex Riley in the comments, you can use np.equal.outer to avoid having to do the indexing stuff yourself,



B = np.equal.outer(np.arange(N), M).view(np.int8)



np.array_equal(A, B)

# True






share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 6 hours ago









coldspeedcoldspeed

140k24156241




140k24156241








  • 1





    Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

    – Alex Riley
    6 hours ago













  • @AlexRiley Thanks for that! And the outer solution is quite neat.

    – coldspeed
    6 hours ago














  • 1





    Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

    – Alex Riley
    6 hours ago













  • @AlexRiley Thanks for that! And the outer solution is quite neat.

    – coldspeed
    6 hours ago








1




1





Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

– Alex Riley
6 hours ago







Good answer - just to point out there's a superfluous newaxis in your indexing for M (resulting in a 4D array). You could use M[None, :] instead to get the 3D array. An alternative to avoid fiddly indexing is to use np.equal.outer(np.arange(N), M).view(np.int8).

– Alex Riley
6 hours ago















@AlexRiley Thanks for that! And the outer solution is quite neat.

– coldspeed
6 hours ago





@AlexRiley Thanks for that! And the outer solution is quite neat.

– coldspeed
6 hours ago











3














You can index into the identity matrix like so



 A = np.identity(N, int)[:, M]


or so



 A = np.identity(N, int)[M.T].T


Or use the new (v1.15.0) put_along_axis



A = np.zeros((N,5,5), int)

np.put_along_axis(A, M[None], 1, 0)


Note if N is much larger than 5 then creating an NxN identity matrix may be considered wasteful. We can mitigate this using stride tricks:



def read_only_identity(N, dtype=float):

    z = np.zeros(2*N-1, dtype)

    s, = z.strides

    z[N-1] = 1

    return np.lib.stride_tricks.as_strided(z[N-1:], (N, N), (-s, s))





share|improve this answer





















  • 1





    This is great, really interesting answer.

    – user3483203
    5 hours ago






  • 1





    Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

    – seveibar
    4 hours ago






  • 1





    @seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

    – Paul Panzer
    3 hours ago
















3














You can index into the identity matrix like so



 A = np.identity(N, int)[:, M]


or so



 A = np.identity(N, int)[M.T].T


Or use the new (v1.15.0) put_along_axis



A = np.zeros((N,5,5), int)

np.put_along_axis(A, M[None], 1, 0)


Note if N is much larger than 5 then creating an NxN identity matrix may be considered wasteful. We can mitigate this using stride tricks:



def read_only_identity(N, dtype=float):

    z = np.zeros(2*N-1, dtype)

    s, = z.strides

    z[N-1] = 1

    return np.lib.stride_tricks.as_strided(z[N-1:], (N, N), (-s, s))





share|improve this answer





















  • 1





    This is great, really interesting answer.

    – user3483203
    5 hours ago






  • 1





    Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

    – seveibar
    4 hours ago






  • 1





    @seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

    – Paul Panzer
    3 hours ago














3












3








3







You can index into the identity matrix like so



 A = np.identity(N, int)[:, M]


or so



 A = np.identity(N, int)[M.T].T


Or use the new (v1.15.0) put_along_axis



A = np.zeros((N,5,5), int)

np.put_along_axis(A, M[None], 1, 0)


Note if N is much larger than 5 then creating an NxN identity matrix may be considered wasteful. We can mitigate this using stride tricks:



def read_only_identity(N, dtype=float):

    z = np.zeros(2*N-1, dtype)

    s, = z.strides

    z[N-1] = 1

    return np.lib.stride_tricks.as_strided(z[N-1:], (N, N), (-s, s))





share|improve this answer















You can index into the identity matrix like so



 A = np.identity(N, int)[:, M]


or so



 A = np.identity(N, int)[M.T].T


Or use the new (v1.15.0) put_along_axis



A = np.zeros((N,5,5), int)

np.put_along_axis(A, M[None], 1, 0)


Note if N is much larger than 5 then creating an NxN identity matrix may be considered wasteful. We can mitigate this using stride tricks:



def read_only_identity(N, dtype=float):

    z = np.zeros(2*N-1, dtype)

    s, = z.strides

    z[N-1] = 1

    return np.lib.stride_tricks.as_strided(z[N-1:], (N, N), (-s, s))






share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 5 hours ago









Paul PanzerPaul Panzer

31.5k21845




31.5k21845








  • 1





    This is great, really interesting answer.

    – user3483203
    5 hours ago






  • 1





    Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

    – seveibar
    4 hours ago






  • 1





    @seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

    – Paul Panzer
    3 hours ago














  • 1





    This is great, really interesting answer.

    – user3483203
    5 hours ago






  • 1





    Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

    – seveibar
    4 hours ago






  • 1





    @seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

    – Paul Panzer
    3 hours ago








1




1





This is great, really interesting answer.

– user3483203
5 hours ago





This is great, really interesting answer.

– user3483203
5 hours ago




1




1





Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

– seveibar
4 hours ago





Hi Paul, this answer is really elegant, but the identity answers seem specific to the case where the N=M.shape[0]=M.shape[1]. Is the solution similarly elegant for N=/=M.shape[0]=/=M.shape[1]? Thanks for the answer, learning a lot!

– seveibar
4 hours ago




1




1





@seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

– Paul Panzer
3 hours ago





@seveibar I've updated the answer. It is really just a matter of replacing the correct 5s with Ns.

– Paul Panzer
3 hours ago


















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