I am currently doing this:
if x in a and y in a and z in a and q in a and r in a and s in a:
print(b)
Is there a more Pythonic way to express this if
statement?
I am currently doing this:
if x in a and y in a and z in a and q in a and r in a and s in a:
print(b)
Is there a more Pythonic way to express this if
statement?
Using the all function allows to write this in a nice and compact way:
if all(i in a for i in (x, y, z, q, r, s)):
print b
This code should do almost exactly the same as your example, even if the objects are not hashable or if the a
object has some funny __contains__
method. The all
function also has similar short-circuit behavior as the chain of and
in the original problem. Collecting all objects to be tested in a tuple (or a list) will guarantee the same order of execution of the tests as in the original problem. If you use a set, the order might be random.
Another way to do this is to use subsets:
if {x, y, z, q, r, s}.issubset(a):
print(b)
REPL example:
>>> {0, 1, 2}.issubset([0, 1, 2, 3])
True
>>> {0, 1, 2}.issubset([1, 2, 3])
False
One caveat with this approach is that all of x
, y
, z
, etc. must be hashable.
if all(v in a for v in {x, y, z, q, r, s}):
print(b)
Converting to set, either:
if len({x, y, z, q, r, s} - set(a)) == 0:
print b
or
t = {x, y, z, q, r, s}
if t & set(a) == t:
print b