I have two dicts:
a = {'a': 1, 'b': 2, 'c': 3}
b = {'a': 2, 'd': 4, 'c': 5}
and i want to get:
{'a': 2, 'b': 2, 'c': 5}
i used {**a, **b} but it return:
{'a': 2, 'b': 2, 'c': 5, 'd': 4}
Help me please exclude keys from b which not in a with the simplest and fastest way.
i have python 3.7
CodePudding user response:
You have to filter the elements of the second dict
first in order to not add any new elements. I got two possible solutions:
a = {'a': 1, 'b': 2, 'c': 3}
b = {'a': 2, 'd': 4, 'c': 5}
for k,v in b.items():
if (k in a.keys()):
a[k] = v
print(a)
a = {'a': 1, 'b': 2, 'c': 3}
b = {'a': 2, 'd': 4, 'c': 5}
a.update([(k,v) for k, v in b.items() if k in a.keys()])
print(a)
Output for both:
{'a': 2, 'b': 2, 'c': 5}
CodePudding user response:
I think a comprehension is easy enough:
{ i : (b[i] if i in b else a[i]) for i in a }