python - If statement with composite condition to validate condition against each item in a list -


this question has answer here:

i found similar question here answers don't seem apply issue.

here code:

y = 3 list1 = [1,2,3,4,5]  if y != 0 or y != list1:     print("y not in range") else:     print(y)

it keeps printing y not in range.

my goal check if y not equal 0 or if y not equal item in list.

i understand above or should and, i'm interested in how check in condition of y being contained in list.

you want check y different 0 and not in list:

if y != 0 , y not in list1:

using or means one of conditions sufficient, since y != 0 returns true without going y != list1 return false because int not list, have use in in case.

if want use or want:

if not (y == 0 or y in list1):     print('y not in range') else:     print(y)

rememer de morgan laws:

not (y == 0 or y in list1) == (not y == 0) , (not y in list1) == y != 0 , y not in list1

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