問題描述

我有一個字典列表:

[{"id":"1", "name":"Alice", "age":"25", "languages":"German"},
 {"id":"1", "name":"Alice", "age":"25", "languages":"French"},
 {"id":"2", "name":"John", "age":"30", "languages":"English"},
 {"id":"2", "name":"John", "age":"30", "languages":"Spanish"}]

我希望最終結果是(我在檢查重復項時只考慮 id):

I'd like the end result to be (I am only considering the id when checking for duplicates):

[{"id":"1", "name":"Alice", "age":"25", "languages":"German, French"},
 {"id":"2", "name":"John", "age":"30", "languages":"English, Spanish"}]

看著類似的問題，我認為使用集合可能是答案，但一直無法正確實現.

looking at similar questions, I thought that using a set might be the answer, but haven't been able to implement it correctly.

提前感謝您的回答.

推薦答案

在這里有點冗長以幫助查看結構.絕對可以做一些很酷的 lambda 東西來解決這個問題，并使列表理解更加pythonic".但這里有一個快速的解決方案！

Being a little verbose here to help see the structure. Definitely some cool lambda stuff you can do to solve this and list comprehension to be more "pythonic". But here is a quick solution!

# Set up initial data
unmerged = [
    {"id":"1", "name":"Alice", "age":"25", "languages":"German"},
    {"id":"1", "name":"Alice", "age":"25", "languages":"French"},
    {"id":"2", "name":"John", "age":"30", "languages":"English"},
    {"id":"2", "name":"John", "age":"30", "languages":"Spanish"}]

# merge the data by your composite key of id-name-age
merged = {}
for entry in unmerged:
    entry_id = entry['id']
    entry_name = entry['name']
    entry_age = entry['age']
    entry_languages = entry['languages']
    composite_key = entry_id + entry_name + entry_age
    if composite_key in merged:
        merged[composite_key]['languages'].append(entry_languages)
    else:
        merged[composite_key] = {
            'id': entry_id,
            'name': entry_name,
            'age': entry_age,
            'languages': [entry_languages]
        }

# reconstruct your list with just your unique entries
cleaned = []
for key, value in merged.items():
    print(key, value)
    cleaned.append({
        'id': value['id'],
        'name': value['name'],
        'age': value['age'],
        'languages': ', '.join(value['languages']) # string join langauges by ", "
    })

for clean in cleaned:
    print(clean)

然后給你你的最終輸出，其中清理的是你的合并條目列表:

And than gives you your final output where cleaned is your list of merged entries:

{'id': '1', 'name': 'Alice', 'age': '25', 'languages': 'German, French'}
{'id': '2', 'name': 'John', 'age': '30', 'languages': 'English, Spanish'}

謝謝，如果這有幫助，請告訴我！

Thank, and let me know if this helps!

這篇關于合并 id 重復的字典列表 - python3的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！