Why doesn't != work for string index comparison?

Question

I tried to run this code with input "AC039"

 code = input("Enter code: ").upper()
 if code[0] != ('N' or 'A' or 'C' ):
     print("The first character must be N, A or C")
else:
    print("Pass!")

It gave me the output error result:

The first character must be N, A or C

However, if I input "AC039" into the below code using ‘not in’,

code = input("Enter code: ").upper()
if code[0] not in ["N", "A", "C"]:
    print("The first character must be N, A or C")
else:
    print("Pass!")

The resulting output is:

print("Pass!")

Why doesn’t "!=" work for the first set of code, since both code[0] and ‘A’ are string types?

I ran a check using type function on code[0] and it returned string type.

code = input("Enter code: ").upper()
print(type(code[0]))
print(type('A'))

returns:

<class 'str'>
<class 'str'>

Asked By: stella

||

Source

Answer 1

I think this is best explained with a rough implementation of what a or b or c or ... does.

import typing as t


T = t.TypeVar("T")


def Or(*args: T) -> T:
    arg: T
    for arg in args[:-1]:
        if bool(arg) is True:
            return arg
    return args[-1]

>>> Or("N", "A", "T")  # equivalent to `"N" or "A" or "T"`
'N'
>>> Or("", None, 0)  # equivalent to `"" or None or 0`
0

Answered By: dROOOze

Answer 2

These two lines are identical.

if code[0] != ('N' or 'A' or 'C' ):

if code[0] != 'N':

Because the operator ‘or’ checks if the first operand is true, then it takes it; otherwise, it checks another one.

Examples:

>>> 1 or 2
1
>>> 0 or 3
3
>>> 0 or 2 or 3
2
>>> False or 2
2
>>> False or 0
0

As a result, you can apply this pattern to solve your problem.

if code[0] in ('N', 'A', 'C'):

or

if code[0] == 'N' or code[0] == 'A' or code[0] == 'C':

Answered By: Karen Petrosyan

Answer 3

It’s because the "or" operator compares sequentially.
from the given input "AC039",

code[0] != ('N' or 'A' or 'C' ):

what’s happening from the above code is, comparing one by one.

(code[0] != 'N') -> True
(code[0] != 'A') -> False
(code[0] != 'C') -> True

since the first ‘N’ doesn’t match with ‘A’, it stops going through other comparisons, ‘A’ and ‘C’.

print('N' != ('N' or 'A' or 'C' )) -> False
print('A' != ('N' or 'A' or 'C' )) -> True
print('C' != ('N' or 'A' or 'C' )) -> True

However, if code[0] not in ["N", "A", "C"]: works because "not in" checks whole elements in the array.

I hope it helps.

Answered By: coffee_cat

Why doesn't != work for string index comparison?

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