我有一個(gè)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(jié)果是（我在檢查重復(fù)項(xiàng)時(shí)只考慮 id）：[{"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(rèn)為使用集合可能是答案，但無法正確實(shí)現(xiàn)。提前感謝您的回答。
1 回答
海綿寶寶撒
TA貢獻(xiàn)1809條經(jīng)驗(yàn) 獲得超8個(gè)贊
在這里有點(diǎn)冗長(zhǎng)以幫助查看結(jié)構(gòu)。絕對(duì)可以做一些很酷的 lambda 東西來解決這個(gè)問題，并使列表理解更加“pythonic”。但這里有一個(gè)快速的解決方案！
# 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)
然后為您提供最終輸出，其中清理的是您的合并條目列表：
{'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(qǐng)告訴我！
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