How to translate / shift a numpy array?

Question

I am not sure what key-word to search for so if it has been already asked please link the response and close this thread.

I am trying to shift the non-zero entries of a numpy array by a fixed direction, for instance, imagine I have a 2d array:

Shifting it by (1,1) would produce the following array:

Let's say if the non-zero entries goes out of bound they're simply dropped. How might I do this?

edit: aparently some duplicate from this? Shift elements in a numpy array I don't really see why are they the same question at all because that one talks about looping the things out of bound around, so it's more of a "rolling" action rather than shifting. Also I liked the solution here, it is very simple and readable.

edit again: fixed some formats

Save for the edge behaviour you might be looking for np.roll. You could also just construct a new array from a slice. — Andras Deak -- Слава Україні
– Andras Deak -- Слава Україні, Commented Jul 2, 2017 at 19:41
Please edit you values to be 1,2,3 instead of 2,2,2 so we can see what kind of shift you want, this can be many different things right now — Ofer Sadan
– Ofer Sadan, Commented Jul 2, 2017 at 19:44
Something like shift=(1,1); new=np.zeros_like(old); new[shift[0]:,shift[1]:]=old[:-shift[0],:-shift[1]] give or take a few off-by-one errors. — Andras Deak -- Слава Україні
– Andras Deak -- Слава Україні, Commented Jul 2, 2017 at 19:47

Adrien Pavao · Accepted Answer · 2019-10-27 23:08:24Z

6

Using roll method from numpy.

>>> import numpy as np
>>> m
array([[0, 1, 2, 0],
       [0, 3, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0]])
>>> m = np.roll(m, 1, axis=0) # shift 1 place in horizontal axis
>>> m = np.roll(m, 1, axis=1) # shift 1 place in vertical axis
>>> m
array([[0, 0, 0, 0],
       [0, 0, 1, 2],
       [0, 0, 3, 0],
       [0, 0, 0, 0]])

https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.roll.html

answered Oct 27, 2019 at 23:08

Adrien Pavao

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B. M. · Accepted Answer · 2017-07-02 20:08:08Z

2

To simply manage the edges, you can enlarge your array in a bigger one :

square=\
array([[0, 2, 2, 0],
       [0, 2, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0]], dtype=int64)

n,m=square.shape
bigsquare=np.zeros((3*n,3*m),square.dtype) 
bigsquare[n:2*n,m:2*m]=square

Then shift is just a view :

def shift(dx,dy):
    x=n-dx
    y=m-dy
    return bigsquare[x:x+n,y:y+m]

print(shift(1,1))

#[[0 0 0 0]
# [0 0 2 2]
# [0 0 2 0]
# [0 0 0 0]]

answered Jul 2, 2017 at 20:08

B. M.

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3 Comments

Andras Deak -- Слава Україні Over a year ago

It's a view of a copy which is not particularly informative.

Evan Pu Over a year ago

@AndrasDeak ah i didn't meant to edit in place anyways, a copy is just fine for me as I kept most of my things functional anyways

Andras Deak -- Слава Україні Over a year ago

@EvanPu I know, you can't really do this any other way. I just reflected on B. M.'s "shift is just a view" remark.

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