[python] Filter dict to contain only certain keys?

I've got a dict that has a whole bunch of entries. I'm only interested in a select few of them. Is there an easy way to prune all the other ones out?

Here's an example in python 2.6:

>>> a = {1:1, 2:2, 3:3}
>>> dict((key,value) for key, value in a.iteritems() if key == 1)
{1: 1}

The filtering part is the if statement.

This method is slower than delnan's answer if you only want to select a few of very many keys.

Given your original dictionary orig and the set of entries that you're interested in keys:

filtered = dict(zip(keys, [orig[k] for k in keys]))

which isn't as nice as delnan's answer, but should work in every Python version of interest. It is, however, fragile to each element of keys existing in your original dictionary.

Slightly more elegant dict comprehension:

foodict = {k: v for k, v in mydict.items() if k.startswith('foo')}

Short form:

[s.pop(k) for k in list(s.keys()) if k not in keep]

As most of the answers suggest in order to maintain the conciseness we have to create a duplicate object be it a list or dict. This one creates a throw-away list but deletes the keys in original dict.

This function will do the trick:

def include_keys(dictionary, keys):
    """Filters a dict by only including certain keys."""
    key_set = set(keys) & set(dictionary.keys())
    return {key: dictionary[key] for key in key_set}

Just like delnan's version, this one uses dictionary comprehension and has stable performance for large dictionaries (dependent only on the number of keys you permit, and not the total number of keys in the dictionary).

And just like MyGGan's version, this one allows your list of keys to include keys that may not exist in the dictionary.

And as a bonus, here's the inverse, where you can create a dictionary by excluding certain keys in the original:

def exclude_keys(dictionary, keys):
    """Filters a dict by excluding certain keys."""
    key_set = set(dictionary.keys()) - set(keys)
    return {key: dictionary[key] for key in key_set}

Note that unlike delnan's version, the operation is not done in place, so the performance is related to the number of keys in the dictionary. However, the advantage of this is that the function will not modify the dictionary provided.

Edit: Added a separate function for excluding certain keys from a dict.

Code 1:

dict = { key: key * 10 for key in range(0, 100) }
d1 = {}
for key, value in dict.items():
    if key % 2 == 0:
        d1[key] = value

Code 2:

dict = { key: key * 10 for key in range(0, 100) }
d2 = {key: value for key, value in dict.items() if key % 2 == 0}

Code 3:

dict = { key: key * 10 for key in range(0, 100) }
d3 = { key: dict[key] for key in dict.keys() if key % 2 == 0}

All pieced of code performance are measured with timeit using number=1000, and collected 1000 times for each piece of code.

For python 3.6 the performance of three ways of filter dict keys almost the same. For python 2.7 code 3 is slightly faster.

You could use python-benedict, it's a dict subclass.

Installation: pip install python-benedict

from benedict import benedict

dict_you_want = benedict(your_dict).subset(keys=['firstname', 'lastname', 'email'])

It's open-source on GitHub: https://github.com/fabiocaccamo/python-benedict

^{Disclaimer: I'm the author of this library.}

This one liner lambda should work:

dictfilt = lambda x, y: dict([ (i,x[i]) for i in x if i in set(y) ])

Here's an example:

my_dict = {"a":1,"b":2,"c":3,"d":4}
wanted_keys = ("c","d")

# run it
In [10]: dictfilt(my_dict, wanted_keys)
Out[10]: {'c': 3, 'd': 4}

It's a basic list comprehension iterating over your dict keys (i in x) and outputs a list of tuple (key,value) pairs if the key lives in your desired key list (y). A dict() wraps the whole thing to output as a dict object.

We can do simply with lambda function like this:

>>> dict_filter = lambda x, y: dict([ (i,x[i]) for i in x if i in set(y) ])
>>> large_dict = {"a":1,"b":2,"c":3,"d":4}
>>> new_dict_keys = ("c","d")
>>> small_dict=dict_filter(large_dict, new_dict_keys)
>>> print(small_dict)
{'c': 3, 'd': 4}
>>> 

Based on the accepted answer by delnan.

What if one of your wanted keys aren't in the old_dict? The delnan solution will throw a KeyError exception that you can catch. If that's not what you need maybe you want to:

1. only include keys that excists both in the old_dict and your set of wanted_keys.

   old_dict = {'name':"Foobar", 'baz':42}
   wanted_keys = ['name', 'age']
   new_dict = {k: old_dict[k] for k in set(wanted_keys) & set(old_dict.keys())}
   
   >>> new_dict
   {'name': 'Foobar'}
   

2. have a default value for keys that's not set in old_dict.

   default = None
   new_dict = {k: old_dict[k] if k in old_dict else default for k in wanted_keys}
   
   >>> new_dict
   {'age': None, 'name': 'Foobar'}
   

Here is another simple method using del in one liner:

for key in e_keys: del your_dict[key]

e_keys is the list of the keys to be excluded. It will update your dict rather than giving you a new one.

If you want a new output dict, then make a copy of the dict before deleting:

new_dict = your_dict.copy()           #Making copy of dict

for key in e_keys: del new_dict[key]

You can do that with project function from my funcy library:

from funcy import project
small_dict = project(big_dict, keys)

Also take a look at select_keys.

Another option:

content = dict(k1='foo', k2='nope', k3='bar')
selection = ['k1', 'k3']
filtered = filter(lambda i: i[0] in selection, content.items())

But you get a list (Python 2) or an iterator (Python 3) returned by filter(), not a dict.

If we want to make a new dictionary with selected keys removed, we can make use of dictionary comprehension
For example:

d = {
'a' : 1,
'b' : 2,
'c' : 3
}
x = {key:d[key] for key in d.keys() - {'c', 'e'}} # Python 3
y = {key:d[key] for key in set(d.keys()) - {'c', 'e'}} # Python 2.*
# x is {'a': 1, 'b': 2}
# y is {'a': 1, 'b': 2}