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'>

`code[0]` and `'A'` may be the same type, but that doesn't mean the operation you're applying is the right operation. — user2357112, Feb 01 '23 at 08:23
if variable = ('N' or 'A' or 'C' ), then variable == "N". That's a property of the 'or' operator. Remember that some code in parenthesis is executed first. — Kerighan, Feb 01 '23 at 08:23
@Kerighan why does ('N' or 'A' or 'C') take in only "N"? Could you advise as i think i have misunderstood a concept here somewhere — stella, Feb 01 '23 at 08:26
@KlausD. either 'N' or 'A' or 'C'. I'm not sure what went wrong. — stella, Feb 01 '23 at 08:30

score 2 · Answer 1 · answered Feb 01 '23 at 08:35

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 Feb 01 '23 at 08:35

Karen Petrosyan

372
2
7

I tried the method if code[0] == 'N' or code[0] == 'A' or code[0] == 'C': and it gave me the same results. But if I use in it works – stella Feb 01 '23 at 08:39

score 0 · Answer 2 · answered Feb 01 '23 at 08:33

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

score -3 · Accepted Answer · answered Feb 01 '23 at 08:44

-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 Feb 01 '23 at 08:44

coffee_cat

10
1

_"what's happening from the above code is, comparing one by one."_: That isn't correct, imho. What's happening is that `('N' or 'A' or 'C')` evaluates to `'N'` since `'N'` is truthy, and the whole expression is reduced to `code[0] != 'N'`. – Timus Feb 01 '23 at 09:04
This answer is completely wrong. – Klaus D. Feb 01 '23 at 10:26

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

3 Answers3