python - Filter dict to contain only certain keys?

Question

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?

Answer 1

Constructing a new dict:

dict_you_want = { your_key: old_dict[your_key] for your_key in your_keys }

Uses dictionary comprehension.

If you use a version which lacks them (ie Python 2.6 and earlier), make it dict((your_key, old_dict[your_key]) for ...). It's the same, though uglier.

Note that this, unlike jnnnnn's version, has stable performance (depends only on number of your_keys) for old_dicts of any size. Both in terms of speed and memory. Since this is a generator expression, it processes one item at a time, and it doesn't looks through all items of old_dict.

Removing everything in-place:

unwanted = set(keys) - set(your_dict)
for unwanted_key in unwanted: del your_dict[unwanted_key]

Answer 2

Slightly more elegant dict comprehension:

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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'}
   

Answer 9

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.

Answer 10

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.

Answer 11

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}

Answer 12

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.}

Answer 13

This seems to me the easiest way:

d1 = {'a':1, 'b':2, 'c':3}
d2 = {k:v for k,v in d1.items() if k in ['a','c']}

I like doing this to unpack the values too:

a, c = {k:v for k,v in d1.items() if k in ['a','c']}.values()

Answer 14

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.

Answer 15

This is my approach, supports nested fields like mongo query.

How to use:

>>> obj = { "a":1, "b":{"c":2,"d":3}}
>>> only(obj,["a","b.c"])
{'a': 1, 'b': {'c': 2}}

only function:

def only(object,keys):
    obj = {}
    for path in keys:
        paths = path.split(".")
        rec=''
        origin = object
        target = obj
        for key in paths:
            rec += key
            if key in target:
                target = target[key]
                origin = origin[key]
                rec += '.'
                continue
            if key in origin:
                if rec == path:
                    target[key] = origin[key]
                else:
                    target[key] = {}
                target = target[key]
                origin = origin[key]
                rec += '.'
            else:
                target[key] = None
                break
    return obj

Answer 16

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]

Answer 17

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}
>>>