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}

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

You can do slicing of the dictionary with the help of the module dictionarify
This is the link for documentation-https://github.com/suryavenom/Flexi/blob/main/README.md.
Installation –
pip install dictionarify

Slicing a dict using numpy

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

{*np.array([*d])[[1,-2]]}   # {'b', 'c'}

Various slice types are supported, see https://numpy.org/doc/stable/user/basics.indexing.html.

With operator.itemgetter:

dict(zip(l, itemgetter(*l)(d)))

Try it online!