[python] Slicing a dictionary

I have a dictionary, and would like to pass a part of it to a function, that part being given by a list (or tuple) of keys. Like so:

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

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

Now, ideally I'd like to be able to do this:

>>> d[l]
{1:2, 5:6}

... but that's not working, since it will look for a key named (1,5). And d[1,5] isn't even valid Python (though it seems it would be handy).

I know I can do this:

>>> dict([(key, value) for key,value in d.iteritems() if key in l])
{1: 2, 5: 6}

or this:

>>> dict([(key, d[key]) for key in l])

which is more compact ... but I feel there must be a "better" way of doing this. Am I missing a more elegant solution?

(I'm using Python 2.7)

This question is related to python dictionary

The answer is


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.


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}

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}

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}

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.


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