Slicing a dictionary

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.

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.

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}