Python:将元组列表重组为字典的聚合列表。

Question

我有一些data，它包含在Python的元组列表中，如下所示。

data = [('Test-1', 1, 0.203194), ('Test-1', 2, 0.0143804), ('Test-1', 3, 0.0769853), ('Test-2', 1, 0.00173769), ('Test-3', 1, 0.00842112), ('Test-3', 2, 0.128969), ('Test-4', 1, 0.0481806)]

每个元组都包含test_name、session数和percentile分数的值(按这个顺序)。我需要将这个data重新组合成一个字典列表，其中每个唯一的test_name按如下方式分组：

[
    {
        "test_name": "Test-1",
        "session": [
            {"submission": 1, "percentile": 0.203194},
            {"submission": 2, "percentile": 0.0143804},
            {"submission": 3, "percentile": 0.0769853}
        ]
    },
    {
        "test_name": "Test-2",
        "session": [
            {"submission": 1, "percentile": 0.0}
        ]
    },
    {
        "test_name": "Test-3",
        "session": [
            {"submission": 1, "percentile": 0.0},
            {"submission": 2, "percentile": 0.0}
        ]
    },
    {
        "test_name": "Test-4",
        "session": [
            {"submission": 1, "percentile": 0.0}
        ]
    }
]

我怎么能用Python做这个呢？

Answer 1

在这里，我假设data中相同的data是连续的，所以我可以将它写成一行：

>>> from itertools import groupby
>>> from operator import itemgetter
>>> [{'test_name': test_name,
...   'session': [{'submission': submission, 'percentile': percentile}
...               for _, submission, percentile in group]}
...  for test_name, group in groupby(data, key=itemgetter(0))]
[{'test_name': 'Test-1',
  'session': [{'submission': 1, 'percentile': 0.203194},
              {'submission': 2, 'percentile': 0.0143804},
              {'submission': 3, 'percentile': 0.0769853}]},
 {'test_name': 'Test-2',
  'session': [{'submission': 1, 'percentile': 0.00173769}]},
 {'test_name': 'Test-3',
  'session': [{'submission': 1, 'percentile': 0.00842112},
              {'submission': 2, 'percentile': 0.128969}]},
 {'test_name': 'Test-4',
  'session': [{'submission': 1, 'percentile': 0.0481806}]}]

如果假设不成立：

>>> mp = {}
>>> for test_name, submission, percentile in data:
...    mp.setdefault(test_name, []).append({'submission': submission, 'percentile': percentile})
...
>>> [{'test_name': test_name, 'session': session}
...  for test_name, session in mp.items()]
[{'test_name': 'Test-1',
  'session': [{'submission': 1, 'percentile': 0.203194},
              {'submission': 2, 'percentile': 0.0143804},
              {'submission': 3, 'percentile': 0.0769853}]},
 {'test_name': 'Test-2',
  'session': [{'submission': 1, 'percentile': 0.00173769}]},
 {'test_name': 'Test-3',
  'session': [{'submission': 1, 'percentile': 0.00842112},
              {'submission': 2, 'percentile': 0.128969}]},
 {'test_name': 'Test-4',
  'session': [{'submission': 1, 'percentile': 0.0481806}]}]

Answer 2

让我们把这个问题分成几个步骤

这就是我将如何使用简单的原语(循环，列表，字典)。

为每个元组标识“key”dictionary

• collect

• ，为以前未见过的密钥准备一个二叉键，如果您以前见过该密钥，则将该“键”更新为“

”列表

intermediate_dict = {} # dictionaries of dictionaries

# 1. For each tuple identify the "key"
for tup in data:
    key, submission, percentile = tup
    if key in intermediate_dict:
        # 3. If you have seen the key before, update the dictionary
        intermediate_dict[key]["session"].append({"submission": submission, "percentile": percentile})

    else:
        # 2. Prepare a dictonary for a key that you haven't seen before
        intermediate_dict[key] = {"session": [{"submission": submission, "percentile": percentile}]}

你的中间白痴可能会像这样

{'Test-1': {'session': [{'submission': 1, 'percentile': 0.203194},
   {'submission': 1, 'percentile': 0.203194},
   {'submission': 2, 'percentile': 0.0143804},
   {'submission': 3, 'percentile': 0.0769853}]},
 'Test-2': {'session': [{'submission': 1, 'percentile': 0.00173769}]},
 'Test-3': {'session': [{'submission': 1, 'percentile': 0.00842112},
   {'submission': 2, 'percentile': 0.128969}]},
 'Test-4': {'session': [{'submission': 1, 'percentile': 0.0481806}]}}

现在，下一步是简单地将字典转换为列表。

return [{"test_name": key, "session": value["session"]} for key, value in intermediate_dict.items()]