nested "and/or" if statements

Question

I am working on code which creates a list and then applies both the "or" and "and" conditions to do further action:

a= ["john", "carlos", "22", "70"]

if (("qjohn" or "carlos") in a) and (("272" or "70") in a):
    print "true"
else:
    print "not true"

output:

not true

when I do this:

a= ["john", "carlos", "22", "70"]

if ("qjohn" or "cdarlos" in a) and ("272" or "d70" in a):
    print "true"
else:
    print "not true"

output is "true"

What I am not getting is **carlos and 70** should be equal to true but it's printing "not true". What is the cause of this error? Thanks

Not quite a dupe, but deals with the same basic issue: [How do I test one variable against multiple values?](https://stackoverflow.com/questions/15112125/how-do-i-test-one-variable-against-multiple-values) — glibdud, Jul 27 '17 at 13:53

Moses Koledoye · Accepted Answer · 2017-07-27T13:53:15.883

Both approaches are incorrect. Keep in mind or is a short-circuit operator so it's not doing what you think it does:

it only evaluates the second argument if the first one is false.

However, non-empty strings are always True, so that the first case only checks for the containment of the first non-empty string while the second never performs the containment check with in at all, therefore, it is always True.

What you want is:

if ("qjohn" in a or "carlos" in a) and ("272"  in a or "70" in a):
   ...

If the items to test were longer, you could avoid repeating the or by using any which like or also short-circuits once one of the items test True:

if any(x in a for x in case1) and any(x in a for x in case2):
   ...

Could be useful : https://stackoverflow.com/questions/16679272/priority-of-the-logical-statements-not-and-or-in-python — Darkaird, Jul 27 '17 at 13:46

cs95 · Answer 2 · 2017-07-27T13:55:23.817

1

b = set(a)
if {"qjohn", "carlos"} & b and {"272", "70"} & b:
    ....

The conditional is True if the intersection of the sets results in a non-empty set (membership test) - testing the truthiness of a non-empty set is quite pythonic in this regard.

Alternatively, using set.intersection:

if {"qjohn", "carlos"}.insersection(a) and {"272", "70"}.insersection(a):
    ....

edited Jul 27 '17 at 13:55

answered Jul 27 '17 at 13:46

cs95

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If you use the `intersection` method, there's no need to precompute the set `b`; the method can take any iterable as its argument. `{"qjohn", "carlos"}.intersection(a)`, etc. (Whether you would want to do this depends on how big `a` is in the first place.) – chepner Jul 27 '17 at 13:54
@chepner Thanks. I'm of the opinion that this is an underrated solution. – cs95 Jul 27 '17 at 13:55

score 0 · Answer 3 · edited Jul 27 '17 at 13:56

0

The both are incorrect. You understood the logic false.

("qjohn" or "carlos") in a is equivalent to "qjohn" in a

and

"qjohn" or "cdarlos" in a is equivalent to "qjohn" or ("cdarlos" in a)

edited Jul 27 '17 at 13:56

chepner

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answered Jul 27 '17 at 13:48

qvpham

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nested "and/or" if statements

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