Python: how to slice a dictionary based on the values of its keys?

问题

Say I have a dictionary built like this:

d={0:1, 1:2, 2:3, 10:4, 11:5, 12:6, 100:7, 101:8, 102:9, 200:10, 201:11, 202:12}

and I want to create a subdictionary d1 by slicing d in such a way that d1 contains the following keys: 0, 1, 2, 100, 101, 102. The final output should be:

d1={0:1, 1:2, 2:3, 100:7, 101:8, 102:9}

Is there an efficient Pythonic way of doing this, given that my real dictionary contains over 2,000,000 items?

I think this question applies to all cases where keys are integers, when the slicing needs to follow certain inequality rules, and when the final result needs to be a bunch of slices put together in the same dictionary.

回答1:

You could use dictionary comprehension with:

d = {0:1, 1:2, 2:3, 10:4, 11:5, 12:6, 100:7, 101:8, 102:9, 200:10, 201:11, 202:12}
keys = (0, 1, 2, 100, 101, 102)
d1 = {k: d[k] for k in keys}

In python 2.7 you can also compute keys with (in python 3.x replace it.ifilter(...) by filter(...)):

import itertools as it

d = {0:1, 1:2, 2:3, 10:4, 11:5, 12:6, 100:7, 101:8, 102:9, 200:10, 201:11, 202:12}
d1 = {k: d[k] for k in it.ifilter(lambda x: 1 < x <= 11, d.keys())}

回答2:

One succinct way of creating the sub-dictionary is to use operator.itemgetter. This function takes multiple arguments and returns a new function to return a tuple containing the corresponding elements of a given iterable.

from operator import itemgetter as ig

k = [0, 1, 2, 100, 101, 102]
# ig(0,1,2,100,101,102) == lambda d : (d[0], d[1], d[2], d[100], d[101], d[102])
d1 = dict(zip(k, ig(*k)(d)))

来源：https://stackoverflow.com/questions/40440373/python-how-to-slice-a-dictionary-based-on-the-values-of-its-keys