I want to insert a key-value pair into dict if key not in dict.keys(). Basically I could do it with:
if key not in d.keys():
d[key] = value
But is there a better way? Or what's the pythonic solution to this problem?
This question is related to
python
dictionary
The answer is
You do not need to call d.keys(), so
if key not in d:
d[key] = value
is enough. There is no clearer, more readable method.
You could update again with dict.get(), which would return an existing value if the key is already present:
d[key] = d.get(key, value)
but I strongly recommend against this; this is code golfing, hindering maintenance and readability.
Use dict.setdefault():
>>> d = {1: 'one'}
>>> d.setdefault(1, '1')
'one'
>>> d # d has not changed because the key already existed
{1: 'one'}
>>> d.setdefault(2, 'two')
'two'
>>> d
{1: 'one', 2: 'two'}
According to the above answers setdefault() method worked for me.
old_attr_name = mydict.setdefault(key, attr_name)
if attr_name != old_attr_name:
raise RuntimeError(f"Key '{key}' duplication: "
f"'{old_attr_name}' and '{attr_name}'.")
Though this solution is not generic. Just suited me in this certain case. The exact solution would be checking for the key first (as was already advised), but with setdefault() we avoid one extra lookup on the dictionary, that is, though small, but still a performance gain.
With the following you can insert multiple values and also have default values but you're creating a new dictionary.
d = {**{ key: value }, **default_values}
I've tested it with the most voted answer and on average this is faster as it can be seen in the following example, .
Speed test comparing a for loop based method with a dict comprehension with unpack operator method.
if no copy (d = default_vals.copy()) is made on the first case then the most voted answer would be faster once we reach orders of magnitude of 10**5 and greater. Memory footprint of both methods are the same.
Since Python 3.9 you can use the merge operator | to merge two dictionaries. The dict on the right takes precedence:
new_dict = old_dict | { key: val }
For example:
new_dict = { 'a': 1, 'b': 2 } | { 'b': 42 }
print(new_dict} # {'a': 1, 'b': 42}
Note: this creates a new dictionary with the updated values.
Source: Stackoverflow.com