如何将python列表或numpy数组中的所有非零元素移到一侧？

Question

I'm going to do the following operation of a list or numpy array:

[0, 0, 0, 1, 0, 0, 4, 2, 0, 7, 0, 0, 0]

move all non-zeros to the right side:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7]

How can I do this efficiently?

Thanks

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Sorry I didn't make it clear, I need the order of non-zeros elements remains.

Answer 1

You could sort the list by their boolean value. All falsy values (just zero for numbers) will get pushed to the front of the list. Python's builtin sort appears stable, so other values will keep their relative position.

Example:

>>> a = [0, 0, 0, 1, 0, 0, 5, 2, 0, 7, 0, 0, 0]
>>> sorted(a, key=bool)
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 5, 2, 7]

Answer 2

Using NumPy:

>>> a = np.array([0, 0, 0, 1, 0, 0, 4, 2, 0, 7, 0, 0, 0])
>>> np.concatenate((a[a==0], a[a!=0]))
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7])

You can do this in O(N) time in Python as well by using a simple for-loop. But will take some extra memory which we can prevent in @grc's solution by using a.sort(key=bool) :

>>> from collections import deque
#Using a deque
>>> def solve_deque(lst):
    d = deque()
    append_l = d.appendleft
    append_r = d.append
    for x in lst:
        if x:
            append_r(x)
        else:
            append_l(x)
    return list(d) #Convert to list if you want O(1) indexing.
...
#Using simple list
>>> def solve_list(lst):                           
    left = []                                    
    right = []
    left_a = left.append
    right_a = right.append
    for x in lst:
        if x:
            right_a(x)
        else:
            left_a(x)
    left.extend(right)
    return left

>>> solve_list([0, 0, 0, 1, 0, 0, 4, 2, 0, 7, 0, 0, 0])
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7]
>>> solve_deque([0, 0, 0, 1, 0, 0, 4, 2, 0, 7, 0, 0, 0])
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7]