How to efficiently unroll a matrix by value with numpy?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

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?

edited Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

It would be better if you explain it in detail.

– Marios Nikolaou
Apr 5 at 22:16

2

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

– Reedinationer
Apr 5 at 22:18

@Reedinationer i did it.

– Marios Nikolaou
Apr 5 at 22:21

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

– coldspeed
Apr 5 at 22:21

1

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

– Marios Nikolaou
Apr 5 at 22:26

|
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?

edited Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

It would be better if you explain it in detail.

– Marios Nikolaou
Apr 5 at 22:16

2

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

– Reedinationer
Apr 5 at 22:18

@Reedinationer i did it.

– Marios Nikolaou
Apr 5 at 22:21

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

– coldspeed
Apr 5 at 22:21

1

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

– Marios Nikolaou
Apr 5 at 22:26

|
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?

edited Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

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 Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

edited Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

edited Apr 5 at 22:24

coldspeed

141k25157242

edited Apr 5 at 22:24

coldspeed

141k25157242

edited Apr 5 at 22:24

coldspeed

141k25157242

asked Apr 5 at 22:06

seveibar

1,29911225

asked Apr 5 at 22:06

seveibar

1,29911225

asked Apr 5 at 22:06

seveibar

1,29911225

It would be better if you explain it in detail.

– Marios Nikolaou
Apr 5 at 22:16

2

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

– Reedinationer
Apr 5 at 22:18

@Reedinationer i did it.

– Marios Nikolaou
Apr 5 at 22:21

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

– coldspeed
Apr 5 at 22:21

1

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

– Marios Nikolaou
Apr 5 at 22:26

|
show 2 more comments

It would be better if you explain it in detail.

– Marios Nikolaou
Apr 5 at 22:16

2

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

– Reedinationer
Apr 5 at 22:18

@Reedinationer i did it.

– Marios Nikolaou
Apr 5 at 22:21

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

– coldspeed
Apr 5 at 22:21

1

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

– Marios Nikolaou
Apr 5 at 22:26

It would be better if you explain it in detail.

– Marios Nikolaou
Apr 5 at 22:16

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

– Reedinationer
Apr 5 at 22:18

@Reedinationer i did it.

– Marios Nikolaou
Apr 5 at 22:21

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

– coldspeed
Apr 5 at 22:21

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

– Marios Nikolaou
Apr 5 at 22:26

|
show 2 more comments

3 Answers
3

active

oldest

votes

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

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

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)

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

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

– seveibar
Apr 6 at 0:58

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))

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

1

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

1

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

– Paul Panzer
Apr 6 at 1:13

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55543949%2fhow-to-efficiently-unroll-a-matrix-by-value-with-numpy%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

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

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

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

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

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

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

edited Apr 5 at 22:45

answered Apr 5 at 22:19

coldspeed

141k25157242

answered Apr 5 at 22:19

coldspeed

141k25157242

answered Apr 5 at 22:19

coldspeed

141k25157242

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

add a comment |

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

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
Apr 5 at 22:34

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

– coldspeed
Apr 5 at 22:45

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)

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

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

– seveibar
Apr 6 at 0:58

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)

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

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

– seveibar
Apr 6 at 0:58

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)

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

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)

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

edited Apr 5 at 22:39

answered Apr 5 at 22:19

user3483203

31.9k82857

answered Apr 5 at 22:19

user3483203

31.9k82857

answered Apr 5 at 22:19

user3483203

31.9k82857

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

– seveibar
Apr 6 at 0:58

add a comment |

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

– seveibar
Apr 6 at 0:58

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

– seveibar
Apr 6 at 0:58

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))

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

1

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

1

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

– Paul Panzer
Apr 6 at 1:13

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))

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

1

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

1

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

– Paul Panzer
Apr 6 at 1:13

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))

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

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))

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

edited Apr 6 at 1:08

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

answered Apr 5 at 23:45

Paul Panzer

31.6k21845

1

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

1

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

– Paul Panzer
Apr 6 at 1:13

add a comment |

1

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

1

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

– Paul Panzer
Apr 6 at 1:13

This is great, really interesting answer.

– user3483203
Apr 5 at 23:47

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
Apr 6 at 0:55

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

– Paul Panzer
Apr 6 at 1:13

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Cfxtrjtrk