如何更自然地获取更新列表而不使用标志

Question

我有两个列表是old_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}]new_list = [{"tag": "apple","value":1}, {"tag": "orange","value":2}, {"tag": "grape","value":3}]我想要一个new_list，如果 new_listitem['tag']等于 old_list item['tag']，则new_list['value']等于old_list['value']，如果 new_list 有新项目，则new_list['value']初始化 0。结果：new_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}, {"tag": "grape","value":0}]这是我的实现：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)因为我使用 2' 次迭代器来实现这个问题，我认为这太慢了，而且是非 Python 风格，而且我引入了看起来不太好的新标志。有没有更直接、更方便、更有效的方式来实现它

Answer 1

您可以创建一个tagtovalue字典old_list并new_list相应地更新项目：

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)

输出

[{'tag': 'apple', 'value': 4}, {'tag': 'orange', 'value': 5}, {'tag': 'grape', 'value': 0}]

Answer 2

使用包含字典理解的列表理解，并使用条件：

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 循环，然后使用带有值条件的字典理解，并在可能的情况下获取 的值old_list。