在 Python 中，如果两个值相同，则从嵌套字典中删除重复条目

发布于 2024-09-30 10:31:09 字数 333 浏览 2 评论 0原文

考虑这种字典格式。

{1:{'name':'chrome', 'author':'google', 'url':'http://www.google.com/' },
 2:{'name':'firefox','author':'mozilla','url':'http://www.mozilla.com/'}}

我想删除所有具有相同名称和作者的项目。

我可以通过将所有键放入一个集合中来轻松删除基于键的重复条目，并且可能将其扩展以处理特定值，但这似乎是一个昂贵的操作，需要多次迭代字典。我不知道如何以有效的方式使用两个值来做到这一点。这是一本包含数千条条目的词典。

原文

Consider this dictionary format.

{1:{'name':'chrome', 'author':'google', 'url':'http://www.google.com/' },
 2:{'name':'firefox','author':'mozilla','url':'http://www.mozilla.com/'}}

I want to remove all items which have the same name and author.

I can easily remove duplicate entries based on keys by putting all keys in a set, and maybe expand this to work on a specific value, but this seems like a costly operation which iterates over a dictionary multiple times. I wouldn't know how to do this with two values in an efficient way. It's a dictionary with thousands of items.

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西瓜 2024-10-07 10:31:09

遍历字典，跟踪遇到的 (name,author) 元组，并删除已经遇到的元组：

def remove_duplicates(d):
    encountered_entries = set()
    for key, entry in d.items():
        if (entry['name'], entry['author']) in encountered_entries:
            del d[key]
        else:
            encountered_entries.add((entry['name'], entry['author']))

Iterate through the dictionary, keeping track of encountered (name, author) tuples as you go and remove those that you have already encountered:

def remove_duplicates(d):
    encountered_entries = set()
    for key, entry in d.items():
        if (entry['name'], entry['author']) in encountered_entries:
            del d[key]
        else:
            encountered_entries.add((entry['name'], entry['author']))

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昔日梦未散 2024-10-07 10:31:09

让我们看看这是否有效...

from itertools import groupby

def entry_key(entry):
    key, value = entry
    return (value['name'], value['author'])

def nub(d):
    items = d.items()
    items.sort(key=entry_key)
    grouped = groupby(items, entry_key)
    return dict([grouper.next() for (key, grouper) in grouped])

Let's see if this works...

from itertools import groupby

def entry_key(entry):
    key, value = entry
    return (value['name'], value['author'])

def nub(d):
    items = d.items()
    items.sort(key=entry_key)
    grouped = groupby(items, entry_key)
    return dict([grouper.next() for (key, grouper) in grouped])

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~没有更多了~