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 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

1,29211225

It would be better if you explain it in detail.

– Marios Nikolaou
7 hours ago

2

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

– Reedinationer
7 hours ago

@Reedinationer i did it.

– Marios Nikolaou
7 hours ago

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

– coldspeed
7 hours ago

1

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

– Marios Nikolaou
7 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?

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

1,29211225

It would be better if you explain it in detail.

– Marios Nikolaou
7 hours ago

2

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

– Reedinationer
7 hours ago

@Reedinationer i did it.

– Marios Nikolaou
7 hours ago

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

– coldspeed
7 hours ago

1

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

– Marios Nikolaou
7 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?

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

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

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

1,29211225

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

1,29211225

edited 7 hours ago

coldspeed

140k24156241

edited 7 hours ago

coldspeed

140k24156241

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibar

1,29211225

asked 7 hours ago

seveibar

1,29211225

asked 7 hours ago

seveibar

1,29211225

It would be better if you explain it in detail.

– Marios Nikolaou
7 hours ago

2

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

– Reedinationer
7 hours ago

@Reedinationer i did it.

– Marios Nikolaou
7 hours ago

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

– coldspeed
7 hours ago

1

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

– Marios Nikolaou
7 hours ago

|
show 2 more comments

It would be better if you explain it in detail.

– Marios Nikolaou
7 hours ago

2

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

– Reedinationer
7 hours ago

@Reedinationer i did it.

– Marios Nikolaou
7 hours ago

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

– coldspeed
7 hours ago

1

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

– Marios Nikolaou
7 hours ago

It would be better if you explain it in detail.

– Marios Nikolaou
7 hours ago

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

– Reedinationer
7 hours ago

@Reedinationer i did it.

– Marios Nikolaou
7 hours ago

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

– coldspeed
7 hours ago

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

– Marios Nikolaou
7 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)

edited 6 hours ago

answered 7 hours ago

user3483203

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

edited 6 hours ago

answered 7 hours ago

coldspeed

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
7 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))

edited 4 hours ago

answered 5 hours ago

Paul 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
4 hours ago

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 6 hours ago

answered 7 hours ago

user3483203

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)

edited 6 hours ago

answered 7 hours ago

user3483203

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)

edited 6 hours ago

answered 7 hours ago

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 6 hours ago

answered 7 hours ago

user3483203

31.8k82857

edited 6 hours ago

answered 7 hours ago

user3483203

31.8k82857

answered 7 hours ago

user3483203

31.8k82857

answered 7 hours ago

user3483203

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

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 6 hours ago

answered 7 hours ago

coldspeed

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
7 hours ago

@AlexRiley Thanks for that! And the outer solution 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

edited 6 hours ago

answered 7 hours ago

coldspeed

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
7 hours ago

@AlexRiley Thanks for that! And the outer solution 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

edited 6 hours ago

answered 7 hours ago

coldspeed

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

edited 6 hours ago

answered 7 hours ago

coldspeed

140k24156241

edited 6 hours ago

answered 7 hours ago

coldspeed

140k24156241

answered 7 hours ago

coldspeed

140k24156241

answered 7 hours ago

coldspeed

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
7 hours ago

@AlexRiley Thanks for that! And the outer solution 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 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
7 hours ago

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

– coldspeed
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
7 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))

edited 4 hours ago

answered 5 hours ago

Paul 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
4 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))

edited 4 hours ago

answered 5 hours ago

Paul 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
4 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))

edited 4 hours ago

answered 5 hours ago

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 4 hours ago

answered 5 hours ago

Paul Panzer

31.5k21845

edited 4 hours ago

answered 5 hours ago

Paul Panzer

31.5k21845

answered 5 hours ago

Paul Panzer

31.5k21845

answered 5 hours ago

Paul 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
4 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
4 hours ago

This is great, really interesting answer.

– user3483203
5 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

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

– Paul Panzer
4 hours ago

add a comment |

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