Is there a better way to write this if-statement?

Question

What is the better way of doing this:

if pattern[0] != [0] or pattern[0] != [0 , 0] or ... so on:
 # do something

• paxdiablo: Is there a limit, or do you want to detect up to an infinity of zeros? In any case, since pattern[0] can't be both [0] and [0,0] at the same time, you can change the whole thing to if True: :-) Or did you mean and rather than or?
• b8ckwith: well let's say the limit is 20 of zeros [ 0, 0, ->18 more ]
• ChesterL: what is the else case?
• Brandon Buck: @paxdiablo if pattern is [[0, 0]] could it not fit pattern[0] != [0 , 0]?
• paxdiablo: @Brandon, if it is [[0,0]], then pattern[0] != [0] is true. In fact, the whole if statement condition is true regardless of what pattern is set to. That's my point.

Answer 1

It seems like you are looking for any:

if any(pattern[0]):

This solution tests if any of the items in pattern[0] are not equal to 0. It works because 0 evaluates to False in Python. Of course, it also assumes that pattern[0] is iterable since you were comparing it to lists originally.

Also, the condition of your if-statement is incorrect regardless of what you are trying to do. It will always be True because pattern[0] will always be either not equal to [0] or not equal to [0, 0]. You should be using and instead of or:

if pattern[0] != [0] and pattern[0] != [0 , 0] and ... so on:

• paxdiablo: Until OP clarifies the question re use of and or or, this is technically wrong. It's probably right but you may want to note the discepency in your answer.
• b8ckwith: @paxdiablo well I need it for OR(as i asked above), but will be nice to have with AND for future reference.