Sfdwhf

Do infinite dimensional systems make sense?

tikz convert color string to hex value

I'm flying to France today and my passport expires in less than 2 months

What's the output of a record needle playing an out-of-speed record

What typically incentivizes a professor to change jobs to a lower ranking university?

Meaning of に in 本当に

How does quantile regression compare to logistic regression with the variable split at the quantile?

Why are electrically insulating heatsinks so rare? Is it just cost?

Why can't we play rap on piano?

What does "Puller Prush Person" mean?

How do I draw and define two right triangles next to each other?

If human space travel is limited by the G force vulnerability, is there a way to counter G forces?

How to efficiently unroll a matrix by value with numpy?

Watching something be written to a file live with tail

Are the number of citations and number of published articles the most important criteria for a tenure promotion?

Why does Kotter return in Welcome Back Kotter?

Why is 150k or 200k jobs considered good when there's 300k+ births a month?

High voltage LED indicator 40-1000 VDC without additional power supply

What are these boxed doors outside store fronts in New York?

Is it unprofessional to ask if a job posting on GlassDoor is real?

Theorems that impeded progress

Intersection point of 2 lines defined by 2 points each

Important Resources for Dark Age Civilizations?

Why can't I see bouncing of a switch on an oscilloscope?

How to efficiently unroll a matrix by value with numpy?

How to merge two dictionaries in a single expression?How do I check if a list is empty?How do I check whether a file exists without exceptions?How do I check if an array includes an object in JavaScript?How to append something to an array?How do I sort a dictionary by value?How do I determine whether an array contains a particular value in Java?How do I list all files of a directory?How do I remove a particular element from an array in JavaScript?How to use foreach with array in JavaScript?

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

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

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

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

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

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

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?

python arrays numpy

share|improve this question

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

1,29211225

# 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

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)

share|improve this question

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

1,29211225

share|improve this question

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

1,29211225

edited 7 hours ago

coldspeed

140k24156241

edited 7 hours ago

coldspeed

140k24156241

asked 7 hours ago

seveibarseveibar

1,29211225

asked 7 hours ago

seveibarseveibar

1,29211225

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

edited 7 hours ago

answered 7 hours ago

user3483203user3483203

31.8k82857

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This answer was really helpful towards my understanding of broadcasting, thank you!
  
  – seveibar
  4 hours ago
  

  
  
  
  

add a comment | 

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

edited 6 hours ago

answered 7 hours ago

coldspeedcoldspeed

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 | 

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

edited 4 hours ago

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
  4 hours ago
  

  
  
  
  

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');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

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

Email

Required, but never shown

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

edited 7 hours ago

answered 7 hours ago

user3483203user3483203

31.8k82857

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This answer was really helpful towards my understanding of broadcasting, thank you!
  
  – seveibar
  4 hours ago
  

  
  
  
  

add a comment | 

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

edited 7 hours ago

answered 7 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)

share|improve this answer

edited 7 hours ago

answered 7 hours ago

user3483203user3483203

31.8k82857

P = np.arange(N)

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

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

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

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

user3483203user3483203

31.8k82857

answered 7 hours ago

user3483203user3483203

31.8k82857

answered 7 hours ago

user3483203user3483203

31.8k82857

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

edited 6 hours ago

answered 7 hours ago

coldspeedcoldspeed

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 | 

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

edited 6 hours ago

answered 7 hours ago

coldspeedcoldspeed

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

share|improve this answer

edited 6 hours ago

answered 7 hours ago

coldspeedcoldspeed

140k24156241

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



 np.array_equal(A, B)

# True

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



np.array_equal(A, B)

# True

share|improve this answer

edited 6 hours ago

answered 7 hours ago

coldspeedcoldspeed

140k24156241

answered 7 hours ago

coldspeedcoldspeed

140k24156241

answered 7 hours ago

coldspeedcoldspeed

140k24156241

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

edited 4 hours ago

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
  4 hours ago
  

  
  
  
  

add a comment | 

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

edited 4 hours ago

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
  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))

share|improve this answer

edited 4 hours ago

answered 5 hours ago

Paul PanzerPaul Panzer

31.5k21845

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

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

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

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

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

edited 4 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