Python, comparing identical strings return False -

i'm aware many questions of kind posted here, couldn't find 1 matches case.

i have list made of dictionaries, each dictionary contains single key, , list value. example: keylist = [{'key1': [1,2,3]}, {'key2': [3, 4, 5]}, ...]

now, want create simple function receives 2 arguments: aforementioned list, , key, , returns matching dictionary given list.

the function is:

def foo(somekey, somelist):     in somelist:         if str(i.keys()).lower() == str(somekey).lower():             return 

when called: foo('key1', keylist), function returned none object (instead of {'key1': [1,2,3]}.

the 2 compared values have the same length , of same type (<type 'str'>), yet comparison yields false value.

thanks advance assistance or/and suggestions nature of problem.

dict.keys() returns list in python 2 , view object in python 3, you're comparing string representation string passed here. instead of can use in operator check if dictionary contains key somekey, want case insensitive search you'll have apply str.lower each key first:

def foo(somekey, somelist):     in somelist:         if somekey.lower() in (k.lower() k in i):             return 

also if dicts contains single key can key name using iter() , next():

>>> d = {'key1': [1,2,3]} >>> next(iter(d)) 'key1' 

so, foo be:

def foo(somekey, somelist):     in somelist:         if somekey.lower() == next(iter(i)).lower():             return