如何使用默认字典更新python字典：另一个字典中给出的值对

Question

Suppose a dictionary is expected to have certain keys present. Is there a simple way of adding these specific keys with default values if they are missing?

For example:

default_dict = {'name': '', 'height': 100, 'age': 20} 

d = {'name': 'James', 'age': 65}
d.set_defaults(default_dict)

would update the dictionary d to

{'name': 'James', 'age': 65, 'height': 100}

where the original values of d are kept and only the missing keys are added.

The default_dict should not be destroyed in the process.

Answer 1

Create a copy of the defaults, and update it with d ; if all keys in d are strings, you can do so with one dict() call:

d = dict(default_dict, **d)

For dictionaries with non-string keys, you'd use two steps:

d = default_dict.copy()
d.update({'name': 'James', 'age': 65})

or you could use a loop to update d with any keys not present using dictionary views; this is not as fast however:

d = {'name': 'James', 'age': 65}
d.update((k, default_dict[v]) for k in default_dict.viewkeys() - d)

Replace viewkeys with keys in Python 3.

If you are using Python 3.5 or newer, you can use similar syntax to create a new dictionary:

d = {**default_dict, 'name': 'James', 'age': 65}

The key-value pairs of default_dict are applied first, followed by whatever new keys you set; these will override the old. See PEP 448 - Additional Unpacking Generalizations in the 3.5 What's New documentation.

Any of the methods creating a new dictionary can update an existing dictionary simply by wrapping in a dict.update() call. So the first could update d in-place with:

d.update(dict(default_dict, **d))

Answer 2

To update an existing dictionary, I might use the dictionary's setdefault method:

for key, value in dict_of_defaults.items():
    dict_maybe_without_defaults.setdefault(key, value)

If I was creating a new dictionary where I had a small number of keys, I'd probably do something more along the lines of the solution that was posted by Martijn.

Answer 3

You may consider using a collections.ChainMap to associate a fallback:

import collections

default_dict = {'name': '', 'height': 100, 'age': 20} 

d = collections.ChainMap({'name': 'James', 'age': 65},default_dict)

>>> d
ChainMap({'name': 'James', 'age': 65}, {'name': '', 'height': 100, 'age': 20})
>>> dict(d) #this flattens the map but isn't necessary for it to work
{'name': 'James', 'height': 100, 'age': 65}

Also note that since all mutating methods will only modify the first mapping default_dict is safe from pop or other methods.