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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?

It's helpful to say what type of keys (integers? strings? dates? arbitrary objects?) and thus whether there's a simple (string, regex, list membership, or numerical inequality) test to check which keys are in or out. Or else do we need to call an arbitrary function(s) to determine that.
@smci String keys. Don't think it even occurred to me that I could use anything else; I've been coding in JS and PHP for so long...

C
Community

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]

"Uses dictionary comprehension, if you use a version which lacks them" == version <= 2.6
Throws a KeyError if one of the filer keys is not present in old_dict. I would suggest {k:d[k] for k in filter if k in d}
@PeterGibson Yes, if that's part of the requirements, you need to do something about it. Whether it's silently dropping the keys, adding a default value, or something else, depends on what you are doing; there are plenty of use cases where your approach is wrong. There are also many where a key missing in old_dict indicates a bug elsewhere, and in that case I very much prefer an error to silently wrong results.
@PeterGibson It doesn't, dictionary lookup is O(1).
nit: Dictionaries are hash maps, so the normal case is O(1). Worst (highly unlikely) case is O(n), but depends on hash collision likelihood. You'd need an astronomically large dictionary, or a really crude hashing algorithm to start seeing that be an issue. stackoverflow.com/a/1963514/1335793
r
ransford

Slightly more elegant dict comprehension:

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

Upvoted. I was thinking about adding an answer similar to this. Just out of curiosity though, why do {k:v for k,v in dict.items() ...} rather than {k:dict[k] for k in dict ...} Is there a performance difference?
Answered my own question. The {k:dict[k] for k in dict ...} is about 20-25% faster, at least in Python 2.7.6, with a dictionary of 26 items (timeit(..., setup="d = {chr(x+97):x+1 for x in range(26)}")), depending on how many items are being filtered out (filtering out consonant keys is faster than filtering out vowel keys because you're looking up fewer items). The difference in performance may very well become less significant as your dictionary size grows.
Would probably be the same perf if you used mydict.iteritems() instead. .items() creates another list.
j
jnnnnn

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.


except I'd probably use if key in ('x','y','z') I guess.
if you already know which keys you want, use delnan's answer. If you need to test each key with an if statement, use ransford's answer.
This solution has one more advantage. If the dictionary is returned from an expensive function call (i.e. a/old_dict is a function call) this solution calls the function only once. In an imperative environment storing the dictionary returned by the function in a variable is not a big deal but in a functional environment (e.g. in a lambda) this is key observation.
S
Suor

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.


Y
Y.Y

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.

https://i.stack.imgur.com/xqsJC.png

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.


just curious, did you make that plot from Python?
ggplot2 in R - part of tidyverse
better not use dict as it's builtin in python.
J
Jim

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.


Should use a set for wanted_keys, but otherwise looks good.
This gives me a blank dictionary if my original dictionary contains lists in place of values. Any workarounds?
@Francesco, can you provide an example? If I run: dictfilt({'x':['wefwef',52],'y':['iuefiuef','efefij'],'z':['oiejf','iejf']}, ('x','z')), it returns {'x': ['wefwef', 52], 'z': ['oiejf', 'iejf']} as intended.
I tried this with: dict={'0':[1,3], '1':[0,2,4], '2':[1,4]} and the result was {}, which I assumed to be a blank dict.
One thing, "dict" is a reserved word so you shouldn't use it to name a dict. What were the keys you were trying to pull out? If I run: foo = {'0':[1,3], '1':[0,2,4], '2':[1,4]}; dictfilt(foo,('0','2')), I get: {'0': [1, 3], '2': [1, 4]} which is the intended outcome
K
Kai

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.


Well, this is basically an eager version of the "tuple generator version" of my dict comprehension. Very compatible indeed, though generator expressions were introduced in 2.4, spring 2005 - seriously, is anyone still using this?
I don't disagree; 2.3 really shouldn't exist anymore. However, as an outdated survey of 2.3 usage: moinmo.in/PollAboutRequiringPython24 Short version: RHEL4, SLES9, shipped with OS X 10.4
M
MyGGaN

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:

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'} 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'}


You could also do {k: old_dict.get(k, default) for k in ...}
R
Ryan Shea

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.


You should allow keys to by any kind of iterable, like what set accepts.
Ah, good call, thanks for pointing this out. I'll make that update.
I wonder if you are better off with two functions. If you asked 10 people "does invert imply that the keys argument is kept, or that the keys argument is rejected?", how many of them would agree?
Updated. Let me know what you think.
This appears not to be working if the input dict has lists in place of values. In this case you get a void dict. Any workarounds?
m
marsl

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.


Wrap filtered in dict and you get back the dictionary!
t
this.srivastava

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}

Neat. Only works in Python 3. Python 2 says "TypeError: unsupported operand type(s) for -: 'list' and 'set'"
Added set(d.keys()) for Python 2. This is working when I run.
J
Joey

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()


This method is the easiest one to parse and understand for newbies like me.
For efficiency, I'd recommend filtering keys using a set: if k in {'a','c'} rather than if k in ['a','c'].
S
Sachin

We can also achieve this by slightly more elegant dict comprehension:

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

filtdict = {k: v for k, v in my_dict.items() if k.startswith('a')}
print(filtdict)

M
Mr_and_Mrs_D

According to the title of the question, one would expect to filter the dictionary in place - a couple of answers suggest methods for doing that - still it's not obvious what is the one obvious way - I added some timings:

import random
import timeit
import collections

repeat = 3
numbers = 10000

setup = ''
def timer(statement, msg='', _setup=None):
    print(msg, min(
        timeit.Timer(statement, setup=_setup or setup).repeat(
            repeat, numbers)))

timer('pass', 'Empty statement')

dsize = 1000
d = dict.fromkeys(range(dsize))
keep_keys = set(random.sample(range(dsize), 500))
drop_keys = set(random.sample(range(dsize), 500))

def _time_filter_dict():
    """filter a dict"""
    global setup
    setup = r"""from __main__ import dsize, collections, drop_keys, \
keep_keys, random"""
    timer('d = dict.fromkeys(range(dsize));'
          'collections.deque((d.pop(k) for k in drop_keys), maxlen=0)',
          "pop inplace - exhaust iterator")
    timer('d = dict.fromkeys(range(dsize));'
          'drop_keys = [k for k in d if k not in keep_keys];'
          'collections.deque('
              '(d.pop(k) for k in list(d) if k not in keep_keys), maxlen=0)',
          "pop inplace - exhaust iterator (drop_keys)")
    timer('d = dict.fromkeys(range(dsize));'
          'list(d.pop(k) for k in drop_keys)',
          "pop inplace - create list")
    timer('d = dict.fromkeys(range(dsize));'
          'drop_keys = [k for k in d if k not in keep_keys];'
          'list(d.pop(k) for k in drop_keys)',
          "pop inplace - create list (drop_keys)")
    timer('d = dict.fromkeys(range(dsize))\n'
          'for k in drop_keys: del d[k]', "del inplace")
    timer('d = dict.fromkeys(range(dsize));'
          'drop_keys = [k for k in d if k not in keep_keys]\n'
          'for k in drop_keys: del d[k]', "del inplace (drop_keys)")
    timer("""d = dict.fromkeys(range(dsize))
{k:v for k,v in d.items() if k in keep_keys}""", "copy dict comprehension")
    timer("""keep_keys=random.sample(range(dsize), 5)
d = dict.fromkeys(range(dsize))
{k:v for k,v in d.items() if k in keep_keys}""",
          "copy dict comprehension - small keep_keys")

if __name__ == '__main__':
    _time_filter_dict()

results:

Empty statement 8.375600000000427e-05
pop inplace - exhaust iterator 1.046749841
pop inplace - exhaust iterator (drop_keys) 1.830537424
pop inplace - create list 1.1531293939999987
pop inplace - create list (drop_keys) 1.4512304149999995
del inplace 0.8008298079999996
del inplace (drop_keys) 1.1573763689999979
copy dict comprehension 1.1982901489999982
copy dict comprehension - small keep_keys 1.4407784069999998

So seems del is the winner if we want to update in place - the dict comprehension solution depends on the size of the dict being created of course and deleting half the keys is already too slow - so avoid creating a new dict if you can filter in place.

Edited to address a comment by @mpen - I calculated drop keys from keep_keys (given we do not have drop keys) - I assumed keep_keys/drop_keys are sets for this iteration or would take ages. With these assumptions del is still faster - but to be sure the moral is: if you have a (set, list, tuple) of drop keys, go for del


drop_keys isn't a fair comparison. Question is more akin to keep_keys. We know which keys we want, not which ones we don't want.
Thanks @mpen - indeed if we try to calculate drop_keys this slows down a lot pop/del methods. Will post some timings for that
There @mpen - seems del beats the dict comprehension even if I calculate drop_keys (I assumed keep keys are sets for O(1) k in keep_keys). Probably this means that creating a dict with 500 entries is a bit slower than creating a list with 500 elements :P
n
nehem

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.


Can you profile what is faster? Note you don't need to create the list: stackoverflow.com/a/36763172/281545
G
Georgy

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.


S
Steve Fan

If you know the negation set (aka not keys) in advance:

v = {'a': 'foo', 'b': 'bar', 'command': 'fizz', 'host': 'buzz'  }
args = {k: v[k] for k in v if k not in ["a", "b"]}
args # {'command': 'fizz', 'host': 'buzz'}

B
Black Thunder

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]

Can you profile what is faster in case you don't need to create a copy (so creating the dict vs for key in e_keys: del your_dict[key])?
@Mr_and_Mrs_D Not creating a copy would be definitely faster. About 15% faster
Turns out dict comprehension is indeed slower but greatly depends on size of filtered keys - see: stackoverflow.com/a/69973383/281545
S
Sachin

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}
>>> 

A
Adán Escobar

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