简体   繁体   English

如何根据列表中的元素从Python字典中删除元素?

[英]How to remove elements from a Python dictionary based on elements in a list?

I have a list of tuples: 我有一个元组列表:

lst=[(6, 'C'), (6, 'H'), (2, 'C'), (2, 'H')]

And a dictionary: 还有一本字典:

dct={'6C': (6, 'C'), '6H': (6, 'H'), '9D': (9, 'D'), '10D': (10, 'D'), '11S': (11, 'S'), '2C': (2, 'C'), '2H': (2, 'H')}

How can I remove the elements from the dictionary that are in the list? 如何从列表中删除列表中的元素? In this example my desired output would be: 在这个例子中,我想要的输出是:

dct2={'9D': (9, 'D'), '10D': (10, 'D'), '11S': (11, 'S')}

我会使用字典理解来映射键与列表中找不到的值:

new_dict = {k: v for k, v in old_dict.items() if v not in the_list} # filter from the list

If you're on Python 2 try this: 如果您使用的是Python 2,请尝试以下操作:

for key, value in dct.items():
    if value in lst:
        del dct[key]

EDIT: 编辑:

A solution that works in both Python 2 and 3: 适用于Python 2和3的解决方案:

dict((key, value) for key, value in dct.items() if value not in lst)

Using the valfilter function from toolz : 使用valfilter函数

from toolz import valfilter
valfilter(lst.__contains__, dct)

I would make the lst set before filtering out elements, since it is data structure which let's you test if element is present more efficiently. 我会在过滤掉元素之前设置第一个lst ,因为它是数据结构,让你测试元素是否更有效。

purge_set = set(lst)
dict(filter(lambda (k, v): v not in purge_set, dct.iteritems()))

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM