在Python中减去两个字典列表

huangapple
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英文:

subtract two list of dict in python

问题

以下是翻译好的部分：

第一个列表

我的第一个列表看起来像这样：

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2023.01.04",
  6.                 "platform": "web"
  7.             },
  8.             {
  9.                 "date": "2022.12.13",
  10.                 "platform": "web"
  11.             }
  12.         ]
  13.     },
  14.     {
  15.         "another_app_name": [
  16.             {
  17.                 "date": "2022.12.25",
  18.                 "platform": "windows"
  19.             },
  20.             {
  21.                 "date": "2022.12.31",
  22.                 "platform": "ios"
  23.             }
  24.         ]
  25.     }
  26. ]

第二个列表

我的第二个列表看起来像这样：

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2023.01.04",
  6.                 "platform": "web"
  7.             }
  8.         ]
  9.     },
  10.     {
  11.         "another_app_name": [
  12.             {
  13.                 "date": "2022.12.31",
  14.                 "platform": "ios"
  15.             }
  16.         ]
  17.     }
  18. ]

结果应该如下所示：

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2022.12.13",
  6.                 "platform": "web"
  7.             }
  8.         ]
  9.     },
  10.     {
  11.         "another_app_name": [
  12.             {
  13.                 "date": "2022.12.25",
  14.                 "platform": "windows"
  15.             }
  16.         ]
  17.     }
  18. ]

这是我的解决方案：

  1. for f, b in zip(all_pcap, failed_pcap):
  2.         res = {key: f[key] - b.get(key, 0) for key in f}
英文:

I have two list like below and i want to subtract second one from first one, but i failed with solutions in the other questions.

First list

My first list is somthing look like below:

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2023.01.04",
  6.                 "platform": "web"
  7.             },
  8.             {
  9.                 "date": "2022.12.13",
  10.                 "platform": "web"
  11.             }
  12.         ]
  13.     },
  14.     {
  15.         "another_app_name": [
  16.             {
  17.                 "date": "2022.12.25",
  18.                 "platform": "windows"
  19.             },
  20.             {
  21.                 "date": "2022.12.31",
  22.                 "platform": "ios"
  23.             }
  24.         ]
  25.     }
  26. ]

Second list

My Second list is somthing look like below:

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2023.01.04",
  6.                 "platform": "web"
  7.             }
  8.         ]
  9.     },
  10.     {
  11.         "another_app_name": [
  12.             {
  13.                 "date": "2022.12.31",
  14.                 "platform": "ios"
  15.             }
  16.         ]
  17.     }
  18. ]

Resulte must be like:

  1. [
  2.     {
  3.         "app_name": [
  4.             {
  5.                 "date": "2022.12.13",
  6.                 "platform": "web"
  7.             }
  8.         ]
  9.     },
  10.     {
  11.         "another_app_name": [
  12.             {
  13.                 "date": "2022.12.25",
  14.                 "platform": "windows"
  15.             }
  16.         ]
  17.     }
  18. ]

there is my solution:

  1. for f, b in zip(all_pcap, failed_pcap):
  2.         res = {key: f[key] - b.get(key, 0) for key in f}

答案1

得分: 0

如果您的列表中的字典始终只包含单个{键: 值}对，那么更好的数据结构可能是这样的：

  1. apps = {
  2.     "app_name": [{"date": "2023.01.04", "platform": "web"}, {"date": "2022.12.13", "platform": "web"}],
  3.     "another_app_name": [{"date": "2022.12.25", "platform": "windows"}, {"date": "2022.12.31", "platform": "ios"}],
  4. }

假设是这种情况，并且每个字典中的应用程序名称是唯一的，那么问题就会变得简单一些。

使用您提供的值：

  1. a = [
  2.     {"app_name": [{"date": "2023.01.04", "platform": "web"}, {"date": "2022.12.13", "platform": "web"}]},
  3.     {"another_app_name": [{"date": "2022.12.25", "platform": "windows"}, {"date": "2022.12.31", "platform": "ios"}]}
  4. ]
  6. b = [
  7.     {"app_name": [{"date": "2023.01.04", "platform": "web"}]},
  8.     {"another_app_name": [{"date": "2022.12.31", "platform": "ios"}]}
  9. ]
  11. expected = [
  12.     {"app_name": [{"date": "2022.12.13", "platform": "web"}]},
  13.     {"another_app_name": [{"date": "2022.12.25", "platform": "windows"}]}
  14. ]

我们可以定义一个函数，将数据转换为更容易处理的格式，执行移除操作，然后将其转换回当前预期的格式：

  1. def remove_app_entries(a, b):
  2.     a_transformed = {k: v for row in a for k, v in row.items()}
  3.     b_transformed = {k: v for row in b for k, v in row.items()}
  5.     results = {}
  7.     for app, data in a_transformed.items():
  8.         to_remove = b_transformed.get(app)
  9.         if not to_remove:
  10.             results[app] = data
  11.         else:
  12.             results[app] = [row for row in data if row not in to_remove]
  14.     return [{app: data} for app, data in results.items()]

这将生成与指定的预期数据相匹配的数据：

  1. >>> remove_app_entries(a, b) == expected
  2. True

如果数据已经按照我建议的格式，则可以跳过a_transformed / b_transformed步骤，直接使用a和b。如果对返回值也适用，那么results可以在最后简单地返回，而无需将其转换为列表。

英文:

If the dictionaries in your lists only ever contain a single {key: value} pair, then a better data structure for the data would be something like this:

  1. apps = {
  2.     "app_name": [{"date": "2023.01.04", "platform": "web"}, {"date": "2022.12.13", "platform": "web"}],
  3.     "another_app_name": [{"date": "2022.12.25", "platform": "windows"}, {"date": "2022.12.31", "platform": "ios"}],
  4. }

Assuming that that is the case, and the app names are unique in each dictionary, then it makes the problem a bit easier.

Using the values you provided:

  1. a = [
  2.     {"app_name": [{"date": "2023.01.04", "platform": "web"}, {"date": "2022.12.13", "platform": "web"}]},
  3.     {"another_app_name": [{"date": "2022.12.25", "platform": "windows"}, {"date": "2022.12.31", "platform": "ios"}]}
  4. ]
  6. b = [
  7.     {"app_name": [{"date": "2023.01.04", "platform": "web"}]},
  8.     {"another_app_name": [{"date": "2022.12.31", "platform": "ios"}]}
  9. ]
  11. expected = [
  12.     {"app_name": [{"date": "2022.12.13", "platform": "web"}]},
  13.     {"another_app_name": [{"date": "2022.12.25", "platform": "windows"}]}
  14. ]

We can define a function that transforms the data into a format that is easier to work with, perform the removals, and then transform it back into the currently expected format:

  1. def remove_app_entries(a, b):
  2.     a_transformed = {k: v for row in a for k, v in row.items()}
  3.     b_transformed = {k: v for row in b for k, v in row.items()}
  5.     results = {}
  7.     for app, data in a_transformed.items():
  8.         to_remove = b_transformed.get(app)
  9.         if not to_remove:
  10.             results[app] = data
  11.         else:
  12.             results[app] = [row for row in data if row not in to_remove]
  14.     return [{app: data} for app, data in results.items()]

This results in data that matches the specified expected data:

  1. >>> remove_app_entries(a, b) == expected
  2. True

If the data was already in the format I suggested, then you would skip the a_transformed / b_transformed steps and use a and b directly. If it also works for the return value, then results could be simply returned at the end instead of transforming it into a list.

huangapple
  • 本文由 发表于 2023年1月9日 04:51:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/75051192.html
  • dictionary
  • list
  • python
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