How do you sequentially flip each dimension in a NumPy array?

Question

I have encountered the following function in MATLAB that sequentially flips all of the dimensions in a matrix:

function X=flipall(X)
    for i=1:ndims(X)
        X = flipdim(X,i);
    end
end

Where X has dimensions (M,N,P) = (24,24,100). How can I do this in Python, given that X is a NumPy array?

What is the format of X in Python? Is it a list or perhaps a NumPy array? There is too little information to answer your question. — rayryeng
– rayryeng, Commented Sep 15, 2016 at 5:39

rayryeng · Accepted Answer · 2016-09-15 06:34:53Z

6

The equivalent to flipdim in MATLAB is flip in numpy. Be advised that this is only available in version 1.12.0.

Therefore, it's simply:

import numpy as np

def flipall(X):
    Xcopy = X.copy()
    for i in range(X.ndim):
        Xcopy = np.flip(Xcopy, i)
     return Xcopy

As such, you'd simply call it like so:

Xflip = flipall(X)

However, if you know a priori that you have only three dimensions, you can hard code the operation by simply doing:

def flipall(X):
    return X[::-1,::-1,::-1]

This flips each dimension one right after the other.

If you don't have version 1.12.0 (thanks to user hpaulj), you can use slice to do the same operation:

import numpy as np

def flipall(X):
    return X[[slice(None,None,-1) for _ in X.shape]]

answered Sep 15, 2016 at 5:56

rayryeng

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12 Comments

How about this one, X[::-1,::-1,::-1] Will it be the same?

@Wajih Correct. That's assuming that you have three dimensions. The code will work for any amount of dimensions of your numpy array.

Thanks. Python syntax is so confusing at times. Yes the dimension will be 3, so it will work just fine :)

@Wajih haha! Don't I know it!

x[[slice(None,None,-1) for _ in x.shape]] for x of unknown dimension.

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