我有兩個列表是old_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}]new_list = [{"tag": "apple","value":1}, {"tag": "orange","value":2}, {"tag": "grape","value":3}]我想要一個new_list，如果 new_listitem['tag']等于 old_list item['tag']，則new_list['value']等于old_list['value']，如果 new_list 有新項目，則new_list['value']初始化 0。結果：new_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}, {"tag": "grape","value":0}]這是我的實現：old_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}]new_list = [{"tag": "apple","value":1}, {"tag": "orange","value":2}, {"tag": "grape","value":3}]update_list = []for new in new_list: flag_new_item = True for old in old_list: if new["tag"] == old['tag']: new["value"] = old['value'] flag_new_item = False break if flag_new_item: new['value'] = 0print(new_list)因為我使用 2' 次迭代器來實現這個問題，我認為這太慢了，而且是非 Python 風格，而且我引入了看起來不太好的新標志。有沒有更直接、更方便、更有效的方式來實現它
2 回答
繁星coding
TA貢獻1797條經驗 獲得超4個贊
您可以創建一個tagtovalue字典old_list并new_list相應地更新項目：
old_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}]
new_list = [{"tag": "apple","value":1}, {"tag": "orange","value":2}, {"tag": "grape","value":3}]
# create a dictionary from old_list
old_tags = {i['tag']: i['value'] for i in old_list}
# update new_list with old_list values if present
for item in new_list:
item['value'] = old_tags.get(item['tag'], 0)
print(new_list)
輸出
[{'tag': 'apple', 'value': 4}, {'tag': 'orange', 'value': 5}, {'tag': 'grape', 'value': 0}]
慕尼黑8549860
TA貢獻1818條經驗 獲得超11個贊
使用包含字典理解的列表理解，并使用條件：
print([{k:(next(iter(old_list[idx:idx+1]),{}).get(k,0) if k=='value' else v) for k,v in i.items()} for idx,i in enumerate(new_list)])enumerate首先使用then 循環，然后使用帶有值條件的字典理解，并在可能的情況下獲取 的值old_list。
添加回答
舉報
0/150
提交
取消