英文:
Deepdiff Printing all keys if any mismatched
问题
我正在使用deepdiff来比较两个数据库的数据。以下是示例：
from deepdiff import DeepDiff
users1 = [{'id': 1, 'name': 'John', 'age': 30}, {'id': 2, 'name': 'Jane', 'age': 25}]
users2 = [{'id': 1, 'name': 'John', 'age': 30}, {'id': 2, 'name': 'Bob', 'age': 35}]
diff = DeepDiff(users1, users2)
print(diff)
它给我输出如下：
{'values_changed': {"root[0]['age']": {'new_value': 20, 'old_value': 30}, "root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[1]['age']": {'new_value': 35, 'old_value': 25}}}
但我想要的是像ID一样的键也被打印出来，这样我就能知道哪个ID不匹配。
示例输出可以是：
{'values_changed': {"root[ID_VALUE]['age']": {'new_value': 20, 'old_value': 30}, "root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[ID_VALUE]['age']": {'new_value': 35, 'old_value': 25}}}
或者
{'values_changed': {1: {"root[0]['age']": {'new_value': 20, 'old_value': 30}}, {2: "root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[1]['age']": {'new_value': 35, 'old_value': 25}}}
有没有办法做到这一点。
谢谢。
英文:
I am using a deepdiff to compare data of two databases. here is example
from deepdiff import DeepDiff
users1 = [{'id': 1, 'name': 'John', 'age': 30}, {'id': 2, 'name': 'Jane', 'age': 25}]
users2 = [{'id': 1, 'name': 'John', 'age': 30}, {'id': 2, 'name': 'Bob', 'age': 35}]
diff = DeepDiff(users1, users2)
print(diff)
It is giving me output like below
{'values_changed': {"root[0]['age']": {'new_value': 20, 'old_value': 30}, "root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[1]['age']": {'new_value': 35, 'old_value': 25}}}
But I wanted the keys like ids should also be printed so that i will be able to know which id has mismatched.
The sample example output could be
{'values_changed': {"root[ID_VALUE]['age']": {'new_value': 20, 'old_value': 30}, "root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[ID_VALUE]['age']": {'new_value': 35, 'old_value': 25}}}
or
{'values_changed': {1:{"root[0]['age']": {'new_value': 20, 'old_value': 30}}, "{2:root[1]['name']": {'new_value': 'Bob', 'old_value': 'Jane'}, "root[1]['age']": {'new_value': 35, 'old_value': 25}}}}
Is there any way to do this.
Thanks
答案1
得分: 1
# 创建一个稍微大一些的示例
users1 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Jane', 'age': 25}, {'id': 123, 'name': 'Janet', 'age': 32}, {'id': 42, 'name': 'Jack', 'age': 22}]
users2 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Bob', 'age': 35}, {'id': 123, 'name': 'Janet', 'age': 69}, {'id': 42, 'name': 'Jack', 'age': 22}]
dd = DeepDiff(users1, users2)
k = [users1[i]['id'] for i in dd.affected_root_keys] # [420, 123]
v = dd["values_changed"].values()
dict(zip(k, v))
英文:
# making a slightly larger example
users1 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Jane', 'age': 25}, {'id': 123, 'name': 'Janet', 'age': 32}, {'id': 42, 'name': 'Jack', 'age': 22}]
users2 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Bob', 'age': 35}, {'id': 123, 'name': 'Janet', 'age': 69}, {'id': 42, 'name': 'Jack', 'age': 22}]
dd = DeepDiff(users1, users2)
k = [users1[i]['id'] for i in dd.affected_root_keys] # [420, 123]
v = dd["values_changed"].values()
dict(zip(k, v))
This assumes that the IDs don't change between the two lists
答案2
得分: 1
以下是翻译好的部分：
"我尝试过，并且成功地获得了期望的结果，但我想知道是否有更有效的方法来实现这个。"
"这给我提供了以下结果，对我来说是可以接受的："
"{420: {'new_value': 35, 'old_value': 25}, 123: {'new_value': 69, 'old_value': 32}}"
英文:
I tried and was able to get the desired result but wanted to know If there is any efficient way to this
users1 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Jane', 'age': 25}, {'id': 123, 'name': 'Janet', 'age': 32}, {'id': 42, 'name': 'Jack', 'age': 22}]
df = pd.DataFrame(users1) # I making df here for example but in real i am doing from df to dict
users2 = [{'id': 69, 'name': 'John', 'age': 30}, {'id': 420, 'name': 'Bob', 'age': 35}, {'id': 123, 'name': 'Janet', 'age': 69}, {'id': 42, 'name': 'Jack', 'age': 22}]
diff = DeepDiff(users1, users2)
# print(diff)
diff = diff.get('values_changed')
if diff is not None:
for key in list(diff.keys()):
find_index = re.search(r'\b\d+\b', key)
index = int(find_index.group())
primary_key_from_df = df['id'].iloc[index]
diff[primary_key_from_df] = diff.pop(key)
print(diff)
This is giving me below result which is fine for me
{420: {'new_value': 35, 'old_value': 25}, 123: {'new_value': 69, 'old_value': 32}}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论