nested “and/or” if statements

问题

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

回答1:

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):
   ...

回答2:

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):
    ....

回答3:

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)

来源：https://stackoverflow.com/questions/45352597/nested-and-or-if-statements

标签

python

list

boolean-logic

nested-if