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?

edited yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

It would be better if you explain it in detail.

– Marios Nikolaou
yesterday

2

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

– Reedinationer
yesterday

@Reedinationer i did it.

– Marios Nikolaou
yesterday

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

– coldspeed
yesterday

1

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

– Marios Nikolaou
yesterday

|
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 yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

It would be better if you explain it in detail.

– Marios Nikolaou
yesterday

2

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

– Reedinationer
yesterday

@Reedinationer i did it.

– Marios Nikolaou
yesterday

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

– coldspeed
yesterday

1

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

– Marios Nikolaou
yesterday

|
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 yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

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 yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

edited yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

edited yesterday

coldspeed

140k25156241

edited yesterday

coldspeed

140k25156241

edited yesterday

coldspeed

140k25156241

asked yesterday

seveibar

1,29711225

asked yesterday

seveibar

1,29711225

asked yesterday

seveibar

1,29711225

It would be better if you explain it in detail.

– Marios Nikolaou
yesterday

2

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

– Reedinationer
yesterday

@Reedinationer i did it.

– Marios Nikolaou
yesterday

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

– coldspeed
yesterday

1

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

– Marios Nikolaou
yesterday

|
show 2 more comments

It would be better if you explain it in detail.

– Marios Nikolaou
yesterday

2

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

– Reedinationer
yesterday

@Reedinationer i did it.

– Marios Nikolaou
yesterday

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

– coldspeed
yesterday

1

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

– Marios Nikolaou
yesterday

It would be better if you explain it in detail.

– Marios Nikolaou
yesterday

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

– Reedinationer
yesterday

@Reedinationer i did it.

– Marios Nikolaou
yesterday

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

– coldspeed
yesterday

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

– Marios Nikolaou
yesterday

|
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)

edited yesterday

answered yesterday

user3483203

31.8k82857

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

– seveibar
yesterday

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 yesterday

answered yesterday

coldspeed

140k25156241

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
yesterday

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

– coldspeed
yesterday

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 yesterday

answered yesterday

Paul Panzer

31.5k21845

1

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

1

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

– Paul Panzer
yesterday

add a comment |

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3 Answers
3

active

oldest

votes

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)

edited yesterday

answered yesterday

user3483203

31.8k82857

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

– seveibar
yesterday

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 yesterday

answered yesterday

user3483203

31.8k82857

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

– seveibar
yesterday

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 yesterday

answered yesterday

user3483203

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)

edited yesterday

answered yesterday

user3483203

31.8k82857

edited yesterday

answered yesterday

user3483203

31.8k82857

answered yesterday

user3483203

31.8k82857

answered yesterday

user3483203

31.8k82857

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

– seveibar
yesterday

add a comment |

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

– seveibar
yesterday

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

– seveibar
yesterday

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 yesterday

answered yesterday

coldspeed

140k25156241

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
yesterday

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

– coldspeed
yesterday

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 yesterday

answered yesterday

coldspeed

140k25156241

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
yesterday

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

– coldspeed
yesterday

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 yesterday

answered yesterday

coldspeed

140k25156241

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 yesterday

answered yesterday

coldspeed

140k25156241

edited yesterday

answered yesterday

coldspeed

140k25156241

answered yesterday

coldspeed

140k25156241

answered yesterday

coldspeed

140k25156241

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
yesterday

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

– coldspeed
yesterday

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
yesterday

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

– coldspeed
yesterday

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
yesterday

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

– coldspeed
yesterday

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 yesterday

answered yesterday

Paul Panzer

31.5k21845

1

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

1

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

– Paul Panzer
yesterday

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 yesterday

answered yesterday

Paul Panzer

31.5k21845

1

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

1

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

– Paul Panzer
yesterday

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 yesterday

answered yesterday

Paul 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))

edited yesterday

answered yesterday

Paul Panzer

31.5k21845

edited yesterday

answered yesterday

Paul Panzer

31.5k21845

answered yesterday

Paul Panzer

31.5k21845

answered yesterday

Paul Panzer

31.5k21845

1

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

1

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

– Paul Panzer
yesterday

add a comment |

1

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

1

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

– Paul Panzer
yesterday

This is great, really interesting answer.

– user3483203
yesterday

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
yesterday

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

– Paul Panzer
yesterday

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