Groupby熊猫等级

Question

与我之前的问题类似，我想按groupby拆分数据帧并应用计算。

现在，我想引入一个新列，以将计算拆分到数据框上。 这是代码：

import pandas as pd
import numpy as np

d = {'year' : [2000, 2000, 2000, 2000, 2001, 2001, 2001],
 'home': ['A', 'B', 'B', 'A', 'B', 'A', 'A'],
 'away': ['B', 'A', 'A', 'B', 'A', 'B', 'B'],
 'aw': [1, 0, 0, 0, 1, 0, np.nan],
 'hw': [0, 1, 0, 1, 0, 1, np.nan]}

df = pd.DataFrame(d, columns=['home', 'away', 'hw', 'aw'])
df.index = range(1, len(df) + 1)
df.index.name = 'game'

df = df.set_index(['hw', 'aw'], append=True).stack().reset_index().rename(columns={'level_3': 'role', 0: 'team'}).loc[:,
 ['game', 'team', 'role', 'hw', 'aw']]

def wins(row):
    if row['role'] == 'home':
        return row['hw']
    else:
        return row['aw']
df['wins'] = df.apply(wins, axis=1)

df['expanding_mean'] = df.groupby('team')['wins'].apply(lambda x: pd.expanding_mean(x).shift())

print df

运行上面的操作将给出整个数据帧的扩展平均值。 但是，如何重新开始每个year的计算？

我尝试在df声明中将year添加到column =中，但是它包含在不需要的role中。 我在理解上的差距在于水平，因此任何启发都值得赞赏。

编辑：下面的期望结果

    game team  role  hw  aw  wins  expanding_mean    year
0      1    A  home   0   1     0             NaN    2000
1      1    B  away   0   1     1             NaN    2000
2      2    B  home   1   0     1        1.000000    2000
3      2    A  away   1   0     0        0.000000    2000
4      3    B  home   0   0     0        1.000000    2000
5      3    A  away   0   0     0        0.000000    2000
6      4    A  home   1   0     1        0.000000    2000
7      4    B  away   1   0     0        0.666667    2000
8      5    B  home   0   1     0             NaN    2001
9      5    A  away   0   1     1             NaN    2001
10     6    A  home   1   0     1        0.000000    2001
11     6    B  away   1   0     0        1.000000    2001
12     7    A  home NaN NaN   NaN        0.500000    2001
13     7    B  away NaN NaN   NaN        0.500000    2001

Answer 1

您可以添加year来df.groupby(['team', 'year'])并添加列year在上面的代码groupby不断变化level_3到level_4在功能上rename ，因为列year加入指数：

import pandas as pd
import numpy as np

d = {'year' : [2000, 2000, 2000, 2000, 2001, 2001, 2001],
 'home': ['A', 'B', 'B', 'A', 'B', 'A', 'A'],
 'away': ['B', 'A', 'A', 'B', 'A', 'B', 'B'],
 'aw': [1, 0, 0, 0, 1, 0, np.nan],
 'hw': [0, 1, 0, 1, 0, 1, np.nan]}

df = pd.DataFrame(d, columns=['home', 'away', 'hw', 'aw', 'year'])
df.index = range(1, len(df) + 1)
df.index.name = 'game'

df = df.set_index(['hw', 'aw', 'year'], append=True).stack().reset_index().rename(columns={'level_4': 'role', 0: 'team'}).loc[:,
 ['game', 'team', 'role', 'hw', 'aw', 'year']]

def wins(row):
    if row['role'] == 'home':
        return row['hw']
    else:
        return row['aw']
df['wins'] = df.apply(wins, axis=1)

df['expanding_mean'] = df.groupby(['team', 'year'])['wins'].apply(lambda x: pd.expanding_mean(x).shift())

print df

    game team  role  hw  aw  year  wins  expanding_mean
0      1    A  home   0   1  2000     0             NaN
1      1    B  away   0   1  2000     1             NaN
2      2    B  home   1   0  2000     1        1.000000
3      2    A  away   1   0  2000     0        0.000000
4      3    B  home   0   0  2000     0        1.000000
5      3    A  away   0   0  2000     0        0.000000
6      4    A  home   1   0  2000     1        0.000000
7      4    B  away   1   0  2000     0        0.666667
8      5    B  home   0   1  2001     0             NaN
9      5    A  away   0   1  2001     1             NaN
10     6    A  home   1   0  2001     1        1.000000
11     6    B  away   1   0  2001     0        0.000000
12     7    A  home NaN NaN  2001   NaN        1.000000
13     7    B  away NaN NaN  2001   NaN        0.000000

Answer 2

groupby year和team并使用transform ：

import pandas as pd
import numpy as np


d = {
    'year': [2000, 2000, 2000, 2000, 2001, 2001, 2001],
    'team': ['A', 'B', 'B', 'A', 'B', 'A', 'A'],
    'value': [1, 0, 0, 1, 2, 3, 3],
}

df = pd.DataFrame(d)

df['mean_per_team_and_year'] = df.groupby(['team', 'year']).transform('mean')
print(df)