Question

输入类似于

[
  {"name": "person 1", "age": 20, "type": "student"},
  {"name": "person 2", "age": 19, "type": "worker"},
  {"name": "person 3", "age": 30, "type": "student"},
  {"name": "person 4", "age": 25, "type": "worker"},
  {"name": "person 5", "age": 17, "type": "student"}
]

，并且按“类型”字段分组时所需的输出应为

[
  [
    {"name": "person 1", "age": 20, "type": "student"},
    {"name": "person 3", "age": 30, "type": "student"},    
    {"name": "person 5", "age": 17, "type": "student"}
  ],
  [
    {"name": "person 2", "age": 19, "type": "worker"},
    {"name": "person 4", "age": 25, "type": "worker"}
  ]
]

我有以下代码可用于itertools

from itertools import groupby

input = [
  {"name": "person 1", "age": 20, "type": "student"},
  {"name": "person 2", "age": 19, "type": "worker"},
  {"name": "person 3", "age": 30, "type": "student"},
  {"name": "person 4", "age": 25, "type": "worker"},
  {"name": "person 5", "age": 17, "type": "student"}
]

input.sort(key=lambda x: x["type"])
output = [list(v) for k, v in groupby(input, key=lambda x: x["type"])]

可以正确给出结果。但是对于大量数据，我认为使用熊猫应该更有效，但是现在看来我还不太清楚如何使用熊猫来完成上述任务。我现在拥有的代码有些奏效，但我认为它根本没有效率。

import pandas as pd

input = [
  {"name": "person 1", "age": 20, "type": "student"},
  {"name": "person 2", "age": 19, "type": "worker"},
  {"name": "person 3", "age": 30, "type": "student"},
  {"name": "person 4", "age": 25, "type": "worker"},
  {"name": "person 5", "age": 17, "type": "student"}
]

indexes = [list(v) for k, v in pd.DataFrame(input).groupby(["type"]).groups.items()]
output = [[input[y] for y in x] for x in indexes]

我很确定上面的代码是使用pandas groupby功能的一种非常错误的方式，那么如何正确地使用它有帮助吗？谢谢。

Answer 1

您可以使用GroupBy.apply和to_dict进行此操作：

pd.DataFrame(input).groupby('type').apply(lambda x: x.to_dict('r')).to_list()

快一点，

pd.DataFrame(input).groupby('type').apply(
    pd.DataFrame.to_dict, orient='r').tolist()

# [[{'age': 20, 'name': 'person 1', 'type': 'student'},
#   {'age': 30, 'name': 'person 3', 'type': 'student'},
#   {'age': 17, 'name': 'person 5', 'type': 'student'}],
#  [{'age': 19, 'name': 'person 2', 'type': 'worker'},
#   {'age': 25, 'name': 'person 4', 'type': 'worker'}]]

Answer 2

我会做什么

l1=[[y.iloc[0].to_dict() for  z in y.iterrows()] for _ , y in pd.DataFrame(input).groupby('type')]
Out[254]: 
[[{'age': 20, 'name': 'person 1', 'type': 'student'},
  {'age': 20, 'name': 'person 1', 'type': 'student'},
  {'age': 20, 'name': 'person 1', 'type': 'student'}],
 [{'age': 19, 'name': 'person 2', 'type': 'worker'},
  {'age': 19, 'name': 'person 2', 'type': 'worker'}]]

如果只需要与具有该值的键匹配，则可以使用itertuples

进行检查。

l=[list(y.itertuples()) for _ , y in pd.DataFrame(input).groupby('type')]
Out[256]: 
[[Pandas(Index=0, age=20, name='person 1', type='student'),
  Pandas(Index=2, age=30, name='person 3', type='student'),
  Pandas(Index=4, age=17, name='person 5', type='student')],
 [Pandas(Index=1, age=19, name='person 2', type='worker'),
  Pandas(Index=3, age=25, name='person 4', type='worker')]]

比较

l[0][0].age
Out[263]: 20
l1[0][0]['age']
Out[264]: 20

如何使用熊猫将字典列表分组为子列表？

2 个答案: