Question

我正在尝试根据另外2个值在column中分配新的pandas df。

在下面的df中，对于Location (Home, Away etc)中的每个单独的值，我想为其中的第一个integer对应的3值分配一个递增的unique Day。

import pandas as pd
import numpy as np

d = ({
    'Time' : ['7:00:00','8:00:00','9:00:00','11:00:00','12:00:00','1:00:00','2:00:00','3:00:00'],                         
    'Day' : ['Mon','Tues','Wed','Thurs','Fri','Thurs','Fri','Sat'],                 
    'Location' : ['Home','Home','Home','Away','Away','Home','Home','Home'],              
    })

df = pd.DataFrame(data=d)

#Assign values from Home
mask = df['Location'] == 'Home'
df1 = df[mask].drop_duplicates('Day')
d = dict(zip(df1['Day'], np.arange(len(df1)) // 3 + 1))

df.loc[mask, 'Assign'] = df.loc[mask, 'Day'].map(d)

#Assign values from Away
mask = df['Location'] == 'Away'
df1 = df[mask].drop_duplicates('Day')
d = dict(zip(df1['Day'], np.arange(len(df1)) // 3 + 1))

df.loc[mask, 'Assign'] = df.loc[mask, 'Day'].map(d)

出局：

       Time    Day Location  Assign
0   7:00:00    Mon     Home     1.0
1   8:00:00   Tues     Home     1.0
2   9:00:00    Wed     Home     1.0
3  11:00:00  Thurs     Away     1.0
4  12:00:00    Fri     Away     1.0
5   1:00:00  Thurs     Home     2.0
6   2:00:00    Fri     Home     2.0
7   3:00:00    Sat     Home     2.0

预期输出：

       Time    Day Location Assign
0   7:00:00    Mon     Home    1.0
1   8:00:00   Tues     Home    1.0
2   9:00:00    Wed     Home    1.0
3  11:00:00  Thurs     Away    2.0
4  12:00:00    Fri     Away    2.0
5   1:00:00  Thurs     Home    3.0
6   2:00:00    Fri     Home    3.0
7   3:00:00    Sat     Home    3.0

Answer 1

我认为需要使用GroupBy.apply的自定义函数，然后通过factorize将值转换为数字值：

def f(x):
    x1 = x.drop_duplicates('Day')
    d = dict(zip(x1['Day'], np.arange(len(x1)) // 3 + 1))
    x['new'] = x['Day'].map(d)
    return x

df = df.groupby('Location', sort=False, group_keys=False).apply(f)
df['new'] = pd.factorize(df['new'].astype(str) + df['Location'])[0] + 1
print (df)
       Time    Day Location  new
0   7:00:00    Mon     Home    1
1   8:00:00   Tues     Home    1
2   9:00:00    Wed     Home    1
3  11:00:00  Thurs     Away    2
4  12:00:00    Fri     Away    2
5   1:00:00  Thurs     Home    3
6   2:00:00    Fri     Home    3
7   3:00:00    Sat     Home    3

使用unique代替drop_duplicates的另一种类似解决方案：

def f(x):
    u = x['Day'].unique()
    d = dict(zip(u, np.arange(len(u)) // 3 + 1))
    x['new'] = x['Day'].map(d)
    return x

df = df.groupby('Location', sort=False).apply(f)

s = df['new'].astype(str) + df['Location']
df['new'] = pd.factorize(s)[0] + 1
print (df)
      Day Location  new
0     Mon     Home    1
1    Tues     Home    1
2     Wed     Away    2
3     Wed     Home    1
4   Thurs     Away    2
5   Thurs     Home    3
6     Fri     Home    3
7     Mon     Home    1
8     Sat     Home    3
9     Fri     Away    2
10    Sun     Home    4

将运行计数分配给以3个熊猫为一组的新列

1 个答案: