Use a set to intersect on the dict.viewkeys() dictionary view:

l = {1, 5}
{key: d[key] for key in d.viewkeys() & l}

This is Python 2 syntax, in Python 3 use d.keys().

This still uses a loop, but at least the dictionary comprehension is a lot more readable. Using set intersections is very efficient, even if d or l is large.

Demo:

>>> d = {1:2, 3:4, 5:6, 7:8}
>>> l = {1, 5}
>>> {key: d[key] for key in d.viewkeys() & l}
{1: 2, 5: 6}

You should be iterating over the tuple and checking if the key is in the dict not the other way around, if you don't check if the key exists and it is not in the dict you are going to get a key error:

print({k:d[k] for k in l if k in d})

Some timings:

 {k:d[k] for k in set(d).intersection(l)}


In [22]: %%timeit
l = xrange(100000)
{k:d[k] for k in l}
....:
100 loops, best of 3: 11.5 ms per loop


In [23]: %%timeit
l = xrange(100000)
{k:d[k] for k in set(d).intersection(l)}
....:
10 loops, best of 3: 20.4 ms per loop


In [24]: %%timeit
l = xrange(100000)
l = set(l)
{key: d[key] for key in d.viewkeys() & l}
....:
10 loops, best of 3: 24.7 ms per


In [25]: %%timeit


l = xrange(100000)
{k:d[k] for k in l if k in d}
....:
100 loops, best of 3: 17.9 ms per loop

I don't see how {k:d[k] for k in l} is not readable or elegant and if all elements are in d then it is pretty efficient.

Write a dict subclass that accepts a list of keys as an "item" and returns a "slice" of the dictionary:

class SliceableDict(dict):
default = None
def __getitem__(self, key):
if isinstance(key, list):   # use one return statement below
# uses default value if a key does not exist
return {k: self.get(k, self.default) for k in key}
# raises KeyError if a key does not exist
return {k: self[k] for k in key}
# omits key if it does not exist
return {k: self[k] for k in key if k in self}
return dict.get(self, key)

Usage:

d = SliceableDict({1:2, 3:4, 5:6, 7:8})
d[[1, 5]]   # {1: 2, 5: 6}

Or if you want to use a separate method for this type of access, you can use * to accept any number of arguments:

class SliceableDict(dict):
def slice(self, *keys):
return {k: self[k] for k in keys}
# or one of the others from the first example


d = SliceableDict({1:2, 3:4, 5:6, 7:8})
d.slice(1, 5)     # {1: 2, 5: 6}
keys = 1, 5
d.slice(*keys)    # same

set intersection and dict comprehension can be used here

# the dictionary
d = {1:2, 3:4, 5:6, 7:8}


# the subset of keys I'm interested in
l = (1,5)


>>>{key:d[key] for key in set(l) & set(d)}
{1: 2, 5: 6}

On Python 3 you can use the itertools islice to slice the dict.items() iterator

import itertools


d = {1: 2, 3: 4, 5: 6}


dict(itertools.islice(d.items(), 2))


{1: 2, 3: 4}

Note: this solution does not take into account specific keys. It slices by internal ordering of d, which in Python 3.7+ is guaranteed to be insertion-ordered.

the dictionary

d = {1:2, 3:4, 5:6, 7:8}

the subset of keys I'm interested in

l = (1,5)

answer

{key: d[key] for key in l}

To slice a dictionary, Convert it to a list of tuples using d.items(), slice the list and create a dictionary out of it.

Here.

d = {1:2, 3:4, 5:6, 7:8}

To get the first 2 items

first_two = dict(list(d.items())[:2])

first_two

{1: 2, 3: 4}

Another option is to convert the dictionary into a pandas Series object and then locating the specified indexes:

>>> d = {1:2, 3:4, 5:6, 7:8}
>>> l = [1,5]


>>> import pandas as pd
>>> pd.Series(d).loc[l].to_dict()
{1: 2, 5: 6}

My case is probably relatively uncommon, but, I'm posting it here nonetheless in case it helps someone (though not OP directly).

I came across this question searching how to slice a dictionary that had item counts. Basically I had a dictionary where the keys were letters, and the values were the number of times the letter appeared (i.e. abababc --> {'a': 3, 'b': 3, 'c': 1} I wanted to 'slice' the dictionary so that I could return the most common n keys.

It turns out that this is exactly what a Collections Counter object is for, and instead of needing to 'slice' my dictionary, I could easily just convert it to a collections.Counter and then call most_common(n): https://docs.python.org/3/library/collections.html#collections.Counter.most_common