从另一列pandas df分配值的有效方法

Question

我正在尝试创建一个更有效的脚本,column在另一列中创建一个新的基于off的值.下面的脚本执行此操作,但我一次只能选择一个string.我想对所有个人价值观做这件事.

对于df下面我目前正在运行的每个单独的脚本string在Location.但是,我想在所有上运行脚本unique strings.

有关如何分配新列的说明:每个人string都Location获取前3个唯一项的值Day.因此,对于每个值Location,将新的字符串分配给前3个唯一值Day.

import pandas as pd
import numpy as np

d = ({
    'Day' : ['Mon','Tues','Wed','Wed','Thurs','Thurs','Fri','Mon','Sat','Fri','Sun'],                 
    'Location' : ['Home','Home','Away','Home','Away','Home','Home','Home','Home','Away','Home'],        
    })

df = pd.DataFrame(data=d)

#Select value
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)

目前我正在选择每个值['Location'],例如mask = df['Location'] == 'Home'.

我想在所有价值观上做到这一点.例如mask = df['Location'] == All unique values

预期产出:

      Day Location Assign
0     Mon     Home     C1
1    Tues     Home     C1
2     Wed     Away     C2
3     Wed     Home     C1
4   Thurs     Away     C2
5   Thurs     Home     C3
6     Fri     Home     C3
7     Mon     Home     C1
8     Sat     Home     C3
9     Fri     Away     C2
10    Sun     Home     C4

Answer 1

您可以使用：

def f(x):
    #get unique days
    u = x['Day'].unique()
    #mapping dictionary
    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)
#add Location column
s = df['new'].astype(str) + df['Location']
#encoding by factorize
df['new'] = pd.Series(pd.factorize(s)[0] + 1).map(str).radd('C')
print (df)
      Day Location new
0     Mon     Home  C1
1    Tues     Home  C1
2     Wed     Away  C2
3     Wed     Home  C1
4   Thurs     Away  C2
5   Thurs     Home  C3
6     Fri     Home  C3
7     Mon     Home  C1
8     Sat     Home  C3
9     Fri     Away  C2
10    Sun     Home  C4