Python:如何根据其键值对字典进行切片? [英] Python: how to slice a dictionary based on the values of its keys?

问题描述

说我有一本这样的字典:

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}

，我想通过切片d来创建子词典d1，使d1包含以下键:0, 1, 2, 100, 101, 102.最终输出应为:

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}

鉴于我的真实词典包含超过2,000,000个项目，是否有一种有效的Pythonic方法?

我认为这个问题适用于所有键都是整数的情况，当切片需要遵循某些不等式规则，并且最终结果需要将一堆切片放在同一块中时字典.

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.

推荐答案

您可以通过以下方式使用字典理解:

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}

在python 2.7中，您还可以使用(在python 3.x中将it.ifilter(...)替换为filter(...))来计算密钥:

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())}

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