How to move zeros to the end of a list

Question

I have to create a code(function) that moves elements in a list without using temporary list and the function has to return nothing, I've tried the following but It won't work please help

def move_zeros_v2(lst):
    left = []
    right = []
    left_a = left.append
    right_a = right.append
    for x in lst:
        if x:
            left_a(x)
        else:
            right_a(x)
    left.extend(right)
    i = 0
    while i < len(left):
        lst[i] = left[i]
        i = i + 1

x = [1, 0, 3, 0, 0, 5, 7]
z=move_zeros_v2(x)
print(x, z)

Answer 1

Here is one way to do it:

def move_zeros(lst):
  n = len(lst)
  lst[:] = filter(None, lst)
  lst.extend([0] * (n - len(lst)))

Answer 2

The big "gotcha" here is "no temporary list." This means that the usual approach of iterating over a copy of the list while mutating the real list...

for idx, el in enumerate(lst[:]):

...violates the rules of the assignment. Instead you'll have to do some jiggery pokery with indexes and etc.

for idx in range(len(lst)):
    el = lst[idx]
    while el == 0:
        # push the zero to the end of the list
        lst.append(lst.pop(idx))
        el = lst[idx]  # check if the next one is a zero as well!

This is generally a Bad Idea, and definitely code smell in Python. You shouldn't feel bad about using a temporary list if this were Real Code, but I assume it's a homework assignment.

Answer 3

This is very easy to do with normal list methods, although this isn't necessarily the prettiest or most efficient way to accomplish your task.

Take some regular list of integers with some zeroes sprinkled in:

test_list = [1, 2, 4, 0, 3, 5, 0, 9, 0, 3, 4, 7]

Figure out how many zeroes there are. You'll be sticking these on the end later:

zeroes_to_append = test_list._____(0)

You can remove the original zeroes with a simple loop. This method will cause a (catchable) ValueError when you run out of zeroes to get rid of:

>>> while True:
...   try:
...     test_list.______(0)
...   except ValueError:
...     break

Then you put the number of zeroes you removed back on the right end:

for i in range(zeroes_to_append):
    test_list.append(0)

The only thing left for you is to complete your homework by looking up the actual methods I used in order to "fill in the blanks" in this answer. Use this page in the Python docs as a reference to help!

Answer 4

Not a very efficient solution, but I think, that's the type of solution you need.

x = [1, 0, 3, 0, 0, 5, 7]
for _ in range(len(x)-1):
    for index in range(len(x)-1):
        if x[index]==0:
            x[index+1],x[index]=x[index],x[index+1]

print x

Output:

[1, 3, 5, 7, 0, 0, 0]

Answer 5

Updated answer with a solution for same list;

lst = filter(lambda x: x!=0, lst) + lst[:]

#which returns [1, -1, 5, 7, 0, 0, 0] for lst = [1, 0,-1, 0, 0, 5, 7]

You can use a listcomprehension with sorted() and reversed()

x = [1, 0, 3, 0, 0, 5, 7]
sorted_x = [i for i in reversed(sorted(x))]

# will return [7, 5, 3, 1, 0, 0, 0]