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How to efficiently unroll a matrix by value with numpy?
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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
edited 6 hours ago
coldspeed
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
|
show 2 more comments
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
edited 6 hours ago
coldspeed
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
It would be better if you explain it in detail.
– Marios Nikolaou
6 hours ago
2
@MariosNikolaou just copy/paste his code andprint(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 == Mcompare int with array 5x5 ? and then save it in A?
– Marios Nikolaou
6 hours ago
|
show 2 more comments
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
edited 6 hours ago
coldspeed
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
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
edited 6 hours ago
coldspeed
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
edited 6 hours ago
coldspeed
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
edited 6 hours ago
coldspeed
140k24156241
edited 6 hours ago
coldspeed
140k24156241
edited 6 hours ago
coldspeed
140k24156241
140k24156241
asked 6 hours ago
seveibarseveibar
1,29211225
asked 6 hours ago
seveibarseveibar
1,29211225
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 andprint(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 == Mcompare int with array 5x5 ? and then save it in A?
– Marios Nikolaou
6 hours ago
|
show 2 more comments
It would be better if you explain it in detail.
– Marios Nikolaou
6 hours ago
2
@MariosNikolaou just copy/paste his code andprint(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 == Mcompare 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
|
show 2 more comments
3 Answers
3
active
oldest
votes
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)
answered 6 hours ago
user3483203user3483203
31.8k82857
This answer was really helpful towards my understanding of broadcasting, thank you!
– seveibar
4 hours ago
add a comment |
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
answered 6 hours ago
coldspeedcoldspeed
140k24156241
1
Good answer - just to point out there's a superfluous newaxis in your indexing forM(resulting in a 4D array). You could useM[None, :]instead to get the 3D array. An alternative to avoid fiddly indexing is to usenp.equal.outer(np.arange(N), M).view(np.int8).
– Alex Riley
6 hours ago
@AlexRiley Thanks for that! And theoutersolution is quite neat.
– coldspeed
6 hours ago
add a comment |
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))
answered 5 hours ago
Paul PanzerPaul Panzer
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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)
answered 6 hours ago
user3483203user3483203
31.8k82857
This answer was really helpful towards my understanding of broadcasting, thank you!
– seveibar
4 hours ago
add a comment |
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)
answered 6 hours ago
user3483203user3483203
31.8k82857
This answer was really helpful towards my understanding of broadcasting, thank you!
– seveibar
4 hours ago
add a comment |
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)
answered 6 hours ago
user3483203user3483203
31.8k82857
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)
answered 6 hours ago
user3483203user3483203
31.8k82857
edited 6 hours ago
answered 6 hours ago
user3483203user3483203
31.8k82857
answered 6 hours ago
user3483203user3483203
31.8k82857
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
add a comment |
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
add a comment |
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
answered 6 hours ago
coldspeedcoldspeed
140k24156241
1
Good answer - just to point out there's a superfluous newaxis in your indexing forM(resulting in a 4D array). You could useM[None, :]instead to get the 3D array. An alternative to avoid fiddly indexing is to usenp.equal.outer(np.arange(N), M).view(np.int8).
– Alex Riley
6 hours ago
@AlexRiley Thanks for that! And theoutersolution is quite neat.
– coldspeed
6 hours ago
add a comment |
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
answered 6 hours ago
coldspeedcoldspeed
140k24156241
1
Good answer - just to point out there's a superfluous newaxis in your indexing forM(resulting in a 4D array). You could useM[None, :]instead to get the 3D array. An alternative to avoid fiddly indexing is to usenp.equal.outer(np.arange(N), M).view(np.int8).
– Alex Riley
6 hours ago
@AlexRiley Thanks for that! And theoutersolution is quite neat.
– coldspeed
6 hours ago
add a comment |
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
answered 6 hours ago
coldspeedcoldspeed
140k24156241
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
answered 6 hours ago
coldspeedcoldspeed
140k24156241
edited 6 hours ago
answered 6 hours ago
coldspeedcoldspeed
140k24156241
answered 6 hours ago
coldspeedcoldspeed
140k24156241
answered 6 hours ago
coldspeedcoldspeed
140k24156241
140k24156241
1
Good answer - just to point out there's a superfluous newaxis in your indexing forM(resulting in a 4D array). You could useM[None, :]instead to get the 3D array. An alternative to avoid fiddly indexing is to usenp.equal.outer(np.arange(N), M).view(np.int8).
– Alex Riley
6 hours ago
@AlexRiley Thanks for that! And theoutersolution is quite neat.
– coldspeed
6 hours ago
add a comment |
1
Good answer - just to point out there's a superfluous newaxis in your indexing forM(resulting in a 4D array). You could useM[None, :]instead to get the 3D array. An alternative to avoid fiddly indexing is to usenp.equal.outer(np.arange(N), M).view(np.int8).
– Alex Riley
6 hours ago
@AlexRiley Thanks for that! And theoutersolution 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
add a comment |
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))
answered 5 hours ago
Paul PanzerPaul Panzer
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
add a comment |
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))
answered 5 hours ago
Paul PanzerPaul Panzer
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
add a comment |
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))
answered 5 hours ago
Paul PanzerPaul Panzer
31.5k21845
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))
answered 5 hours ago
Paul PanzerPaul Panzer
31.5k21845
edited 3 hours ago
answered 5 hours ago
Paul PanzerPaul Panzer
31.5k21845
answered 5 hours ago
Paul PanzerPaul Panzer
31.5k21845
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
add a comment |
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
add a comment |
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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 == Mcompare int with array 5x5 ? and then save it in A?– Marios Nikolaou
6 hours ago