Find most elegant way to return key if condition is set to true-CodePudding

I have a python dictionary

slot_a = 'a'
slot_b = 'b'

# dict which lists all possible conditions
con_dict = {"branch_1": slot_a == 'a' and slot_b == 'b',
            "branch_2": slot_a == 'a' and slot_b == 'c'}

Now I want to return the key for the first true condition. In this case it's branch_1.

My solution is:

# Pick only the condition which is set to True
true_branch = [k for k, v in con_dict.items() if v == True]

true_branch
>>> branch_1

Since the number of branches can be very long, I was wondering, if there might be a more elegant way to get the same result?! Maybe if / elif / else and then return key? Or even something completely different? All I need at the end is the name of the true condition. Therefore working with a dict may not even be necessary.

Just asking for inspiration!

CodePudding user response：

You could try to use an iterator. It will stop as soon it gets the first match without going through the whole "object".

ks, vs = zip(*con_dict.items()) # decoupling the list of pairs
i = 0
vs = iter(vs)     # terms are all booleans
while not next(vs):
    i  = 1

del vs # "free" the iterator

print(ks[i])

true_branch = next((k for k, condition in con_dict.items() if condition), None)
print(true_branch)