Pandas：按字典中的dict键分组，其中包含字典

Question

我的数据

我有以下pandas数据框：

df = pd.DataFrame({
    'c1': range(5),
    'c2': [
        {'k1': 'x-1', 'k2': 'z'}, 
        {'k1': 'x-2', 'k2': 'z1'},
        {'k1': 'x-3', 'k2': 'z1'},
        {'k1': 'y-1', 'k2': 'z'},
        {'k1': 'y-2', 'k2': 'z1'}
    ]
})

我的目标

现在，我想按'k1'进行分组，这是'c2'列的所有行中的公共密钥，其中包含字典。分组功能为lambda x: x.split('-')[0]以切断破折号后面的数字。

所需的输出是：

'x'     3
'y'     2   

尝试

>>> df.groupby(df['c2']['k1'].str.split('-')[0]).count()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Library/Python/2.7/site-packages/pandas/core/series.py", line 601, in __getitem__
    result = self.index.get_value(self, key)
  File "/Library/Python/2.7/site-packages/pandas/core/indexes/base.py", line 2477, in get_value
    tz=getattr(series.dtype, 'tz', None))
  File "pandas/_libs/index.pyx", line 98, in pandas._libs.index.IndexEngine.get_value (pandas/_libs/index.c:4404)
  File "pandas/_libs/index.pyx", line 106, in pandas._libs.index.IndexEngine.get_value (pandas/_libs/index.c:4087)
  File "pandas/_libs/index.pyx", line 156, in pandas._libs.index.IndexEngine.get_loc (pandas/_libs/index.c:5210)
KeyError: 'k1'

显然，我无法通过k1为行c2的键df['c2']['k1']编制索引。

我怎么能这样做？

Answer 1

您已关闭，只需将dicts的列转换为新的DataFrame：

print (pd.DataFrame(df['c2'].values.tolist()))
    k1  k2
0  x-1   z
1  x-2  z1
2  x-3  z1
3  y-1   z
4  y-2  z1

a = pd.DataFrame(df['c2'].values.tolist())['k1'].str.split('-').str[0]
print (a)
0    x
1    x
2    x
3    y
4    y
Name: k1, dtype: object

df = df.groupby(a).size().reset_index(name='len')
print (df)
  k1  len
0  x    3
1  y    2

另一个解决方案是使用list comprehension用于groupby键：

L = [x['k1'].split('-')[0] for x in df['c2']]
print (L)
['x', 'x', 'x', 'y', 'y']

df = df.groupby(L).size().rename_axis('k1').reset_index(name='len')
print (df)
  k1  len
0  x    3
1  y    2

What is the difference between size and count in pandas?

value_counts的解决方案：

df = a.value_counts().rename_axis('k1').reset_index(name='len')
print (df)
  k1  len
0  x    3
1  y    2

df = pd.Series(L).value_counts().rename_axis('k1').reset_index(name='len')
print (df)
  k1  len
0  x    3
1  y    2