Slicing a dictionary
PythonDictionaryPython Problem Overview
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 matching the tuple (1,5)
, the same as d[1,5]
.
d{1,5}
isn't even valid Python (as far as I can tell ...), though it might be handy: The curly braces suggest an unordered set or a dictionary, so returning a dictionary containing the specified keys would look very plausible to me.
d[{1,5}]
would also make sense ("here's a set of keys, give me the matching items"), and {1, 5}
is an unhashable set, so there can't be a key that matches it -- but of course it throws an error, too.
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)
Python Solutions
Solution 1 - Python
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.
Solution 2 - Python
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.
Solution 3 - Python
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}
Solution 4 - Python
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}
Solution 5 - Python
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
Solution 6 - Python
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}
Solution 7 - Python
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}
Solution 8 - Python
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}