Test 0 and false since false is 0

Thu Jul 6 22:46:26 EDT 2017

On Thursday, July 6, 2017 at 9:29:29 PM UTC-5, Sayth Renshaw wrote:
> I was trying to solve a problem and cannot determine how to filter 0's but not false.
>> Given a list like this
> ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]
>> I want to be able to return this list
> ["a","b",None,"c","d",1,False,1,3,[],1,9,{},9,0,0,0,0,0,0,0,0,0,0]
>> However if I filter like this 
>> def move_zeros(array):
> l1 = [v for v in array if v != 0]
> l2 = [v for v in array if v == 0]
> return l1 + l2
>> I get this 
> ['a', 'b', None, 'c', 'd', 1, 1, 3, [], 1, 9, {}, 9, 0, 0, 0, False, 0, 0, 0, 0, 0, 0, 0] 
>> I have tried or conditions of v == False etc but then the 0's being false also aren't moved. How can you check this at once?

Yep. This is a common pitfall for noobs, as no logic can
explain to them why integer 0 should bool False, and integer
1 should bool True. But what's really going to cook your
noodle is when you find out that any integer greater than 1
bools True. Go figure! They'll say it's for consistency
sake. But i say it's just a foolish consistency.
You need to learn the subtle difference between `==` and
`is`.
 ## PYTHON 2.x
 >>> 1 == True
 True
 >>> 1 is True
 False
 >>> 0 == False
 True
 >>> 0 is False
 False