python - 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_list 的 item['tag'] 等于 old_list 的 item['tag']， new_list['value'] 等于 old_list['value']，如果 new_list 有新项，
然后 new_list['value'] 初始化 0.

结果:

new_list = [{"tag": "apple","value":4}, {"tag": "orange","value":5}, {"tag": "grape","value":0}]

这是我的工具:

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'] = 0
print(new_list)

因为我使用 2'times 迭代器来实现这个问题，我认为它太慢并且不是 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、方便、高效的实现方式

Answer 1

您可以从 old_list 创建一个 tag 到 value 字典并适本地更新 new_list 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}]