Python 3.6 vs Python 3.8.2 if statement with 'is' - shows different results

Question

I have this question where I need to put every 0 or 0.0 at the back of the list while saving the order of the other elements, for example:

move_zero([1,3,4,[],False,None,0,0,3,0.0,4])

Will output:

[1,3,4,[],False,None,3,4,0,0,0]

Note that the function turned especially 0.0 to 0 and it's fine, this is what it should do

Now, this is my go at this question:

def move_zeros(array):
   new_arr = []
   count_zero = 0
   for x in array:
       if (x is 0) or (x is 0.0):
           count_zero += 1
       else:
           new_arr.append(x)

   return new_arr + count_zero * [0]

But for some reason which I cannot call why, it does not enter the first if statement where my input is:

[9,0.0,0,9,1,2,0,1,0,1,0.0,3,0,1,9,0,0,0,0,9]

The output is:

[9, 0.0, 9, 1, 2, 1, 1, 0.0, 3, 1, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0]

It outputs the wrong list on Python 3.6 While on python 3.8.2 it works fine.. (However I need to use 3.6 for this)
Where am I wrong? the if statement seems ok!

Thank you!

Answer 1

But for some reason which I cannot call why, it does not enter the first if statement where my input is:

You need to use == for comparing values. is is used to compare identity in Python. See this for information on == vs is in Python.

Try this:

def move_zeros(a):
    r = [i for i in a if not isinstance(i, (float, int)) or str(i) not in ('0.0', '0')]
    r += [0] * (len(a) - len(r))
    return r

print(move_zeros([9,0.0,0,9,1,2,0,1,0,1,0.0,3,0,1,9,0,0,0,0,9]))

Outputs:

[9, 9, 1, 2, 1, 1, 3, 1, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]