在熊猫中按年/月/日分组

Question

假设有以下内容DataFrame:

rng = pd.date_range('1/1/2011', periods=72, freq='H')
np.random.seed(10)
n = 10
df = pd.DataFrame(
    {
        "datetime": np.random.choice(rng,n),
        "cat": np.random.choice(['a','b','b'], n),
        "val": np.random.randint(0,5, size=n)
        }
    )

如果我现在groupby:

gb = df.groupby(['cat','datetime']).sum()

我得到cat每小时的总数:

cat datetime            val
a   2011-01-01 00:00:00 1
    2011-01-01 09:00:00 3
    2011-01-02 16:00:00 1
    2011-01-03 16:00:00 1
b   2011-01-01 08:00:00 4
    2011-01-01 15:00:00 3
    2011-01-01 16:00:00 3
    2011-01-02 04:00:00 4
    2011-01-02 05:00:00 1
    2011-01-02 12:00:00 4

但是,我希望有类似的东西:

cat datetime   val
a   2011-01-01 4
    2011-01-02 1
    2011-01-03 1
b   2011-01-01 10
    2011-01-02 9

我可以通过添加另一个列来获得所需的结果date:

df['date'] = df.datetime.apply(pd.datetime.date)

然后做一个类似的groupby:df.groupby(['cat','date']).sum().但我感兴趣的是,有更多的pythonic方式吗？另外,我可能想看看月份或年级.那么,什么是正确的方法？

Answer 1

您可以尝试set_index然后groupby通过cat和date：

import pandas as pd
import numpy as np

rng = pd.date_range('1/1/2011', periods=72, freq='H')
np.random.seed(10)
n = 10
df = pd.DataFrame(
    {
        "datetime": np.random.choice(rng,n),
        "cat": np.random.choice(['a','b','b'], n),
        "val": np.random.randint(0,5, size=n)
        }
    )
print df
  cat            datetime  val
0   a 2011-01-01 09:00:00    3
1   b 2011-01-01 15:00:00    3
2   a 2011-01-03 16:00:00    1
3   b 2011-01-02 04:00:00    4
4   b 2011-01-02 05:00:00    1
5   b 2011-01-01 08:00:00    4
6   a 2011-01-01 00:00:00    1
7   a 2011-01-02 16:00:00    1
8   b 2011-01-02 12:00:00    4
9   b 2011-01-01 16:00:00    3

df = df.set_index('datetime')
gb = df.groupby(['cat', lambda x: x.date]).sum()
print gb
                val
cat                
a   2011-01-01    4
    2011-01-02    1
    2011-01-03    1
b   2011-01-01   10
    2011-01-02    9