Swap zeros in numpy matrix

Question

I have a numpy matrix like so:

array([[2,  1, 23, 32],
       [34, 3, 3, 0],
       [3, 33, 0, 0],
       [32, 0, 0, 0]], dtype=int32)

Now

Answer 1

Here's a vectorized approach with masking -

valid_mask = a!=0
flipped_mask = valid_mask.sum(1,keepdims=1) > np.arange(a.shape[1]-1,-1,-1)
a[flipped_mask] = a[valid_mask]
a[~flipped_mask] = 0

Sample run -

In [90]: a
Out[90]: 
array([[ 2,  1, 23, 32],
       [34,  0,  3,  0],  # <== Added a zero in between for variety
       [ 3, 33,  0,  0],
       [32,  0,  0,  0]])

# After code run -

In [92]: a
Out[92]: 
array([[ 2,  1, 23, 32],
       [ 0,  0, 34,  3],
       [ 0,  0,  3, 33],
       [ 0,  0,  0, 32]])

One more generic sample run -

In [94]: a
Out[94]: 
array([[1, 1, 2, 3, 1, 0, 3, 0, 2, 1],
       [2, 1, 0, 1, 2, 0, 1, 3, 1, 1],
       [1, 2, 0, 3, 0, 3, 2, 0, 2, 2]])

# After code run -

In [96]: a
Out[96]: 
array([[0, 0, 1, 1, 2, 3, 1, 3, 2, 1],
       [0, 0, 2, 1, 1, 2, 1, 3, 1, 1],
       [0, 0, 0, 1, 2, 3, 3, 2, 2, 2]])

Runtime test

Approaches that work on generic cases -

# Proposed in this post
def masking_based(a):
    valid_mask = a!=0
    flipped_mask = valid_mask.sum(1,keepdims=1) > np.arange(a.shape[1]-1,-1,-1)
    a[flipped_mask] = a[valid_mask]
    a[~flipped_mask] = 0
    return a

# @Psidom's soln            
def sort_based(a):
    return a[np.arange(a.shape[0])[:, None], (a != 0).argsort(1, kind="mergesort")]

Timings -

In [205]: a = np.random.randint(0,4,(1000,1000))

In [206]: %timeit sort_based(a)
10 loops, best of 3: 30.8 ms per loop

In [207]: %timeit masking_based(a)
100 loops, best of 3: 6.46 ms per loop

In [208]: a = np.random.randint(0,4,(5000,5000))

In [209]: %timeit sort_based(a)
1 loops, best of 3: 961 ms per loop

In [210]: %timeit masking_based(a)
1 loops, best of 3: 151 ms per loop

Answer 2

Trivial attempt in non-numpy based python -

>>> arr = [[2,  1, 23, 32],
...        [34, 3, 3, 0],
...        [3, 33, 0, 0],
...        [32, 0, 0, 0]]
... 
>>> t_arr = [[0 for _ in range(cur_list.count(0))]\
            + [i for i in cur_list if i!=0]\
            for cur_list in arr]
>>> t_arr
[[2, 1, 23, 32], [0, 34, 3, 3], [0, 0, 3, 33], [0, 0, 0, 32]]

Answer 3

You can also perform sorting on the masked array with the help of numpy.ma.sort() that sorts the array in-place along the last axis, axis=-1 as shown:

np.ma.array(a, mask=a!=0).sort()

Now a becomes:

array([[ 2,  1, 23, 32],
       [ 0, 34,  3,  3],
       [ 0,  0,  3, 33],
       [ 0,  0,  0, 32]])

The only downside is that it is not as fast as some of the approaches mentioned above but nevertheless a short one-liner to have.

Answer 4

A row roll based solution, in the spirit of @EDChum's pandas version:

def rowroll(arr):
    for row in arr:
        row[:] = np.roll(row,-np.count_nonzero(row))
    return arr
In [221]: rowroll(arr.copy())
Out[221]: 
array([[ 2,  1, 23, 32],
       [ 0, 34,  3,  3],
       [ 0,  0,  3, 33],
       [ 0,  0,  0, 32]])

np.count_nonzero is a fast compiled way of finding the number of nonzeros. It is used by np.where to find its return size.

But looking at the np.roll code, I think it's overly complicated for the task, since it can work with several axes.

This looks messier, but I suspect it's as fast, if not faster than roll:

def rowroll(arr):
    for row in arr:
        n = np.count_nonzero(row)
        temp = np.zeros_like(row)
        temp[-n:] = row[:n]
        row[:] = temp
    return arr

The roll solutions require trailing 0s in the original, not scattered 0s.

Answer 5

pandas method:

In [181]:
# construct df from array
df = pd.DataFrame(a)
# call apply and call np.roll rowise and roll by the number of zeroes
df.apply(lambda x: np.roll(x, (x == 0).sum()), axis=1).values

Out[181]:
array([[ 2,  1, 23, 32],
       [ 0, 34,  3,  3],
       [ 0,  0,  3, 33],
       [ 0,  0,  0, 32]])

This uses apply so we can call np.roll on each row by the number of zeroes in each row

Answer 6

You can also use numpy.argsort with advanced indexing:

arr[np.arange(arr.shape[0])[:, None], (arr != 0).argsort(1, kind="mergesort")]

#array([[ 2,  1, 23, 32],
#       [ 0, 34,  3,  3],
#       [ 0,  0,  3, 33],
#       [ 0,  0,  0, 32]], dtype=int32)