Merge a dict in Python using 1 dict as base
This question already has an answer here:
"Pythonicness" is a hard measure to assess, but here is my take on it:
def merge_dicts(base_dict, other_dict):
""" Merge two dicts
Ensure that the base_dict remains as is and overwrite info from other_dict
"""
if other_dict is None:
return base_dict
t = type(base_dict)
if type(other_dict) != t:
raise TypeError("Mismatching types: {} and {}."
.format(t, type(other_dict)))
if not issubclass(t, dict):
return other_dict
return {k: merge_dicts(v, other_dict.get(k)) for k, v in base_dict.items()}
Example:
merge_dicts({"a":2, "b":{"b1": 5, "b2": 7}}, {"b": {"b1": 9}})
>>> {'a': 2, 'b': {'b1': 9, 'b2': 7}}
If you're using python 3.5 or later you can simply do:
merged_dict = {**base_dict, **other_dict}
In case you're using any prior version you can do it with the update method:
merged_dict = {}
merged_dict.update(base_dict)
merged_dict.update(other_dict)
For more information about it you can check The Idiomatic Way to Merge Dictionaries in Python
You can use the dict.update method, along with a generator expression:
base_dict.update((k, v) for k, v in other_dict.items() if k in base_dict)
Explanation
base_dict.update(other_dict) will overwrite the values in base_dict with those in other_dict . If a key exists in other_dict but not in base_dict , it will be added to base_dict , which is not what you want. Therefore, you need to test if each key in other_dict is in base_dict .
dict.update can take an iterable as first parameter. If the latter contains 2-tuples (k, v) , then base_dict[k] will be set to v .
Abstract from help(dict.update) :
update(...) method of builtins.dict instance
If E is present and lacks a .keys() method, then does: for k, v in E: D[k] = v
Therefore, it's convenient to pass a generator expression to it. If you're not familiar with generator expressions, the stuff inside of the parentheses is more or less equivalent to the following:
l = []
for k, v in other.dict_items():
if k in base_dict:
l.append((k, v))
Then l is passed to update , like base_dict.update(l) .
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