有条件地在分类列中创建一个“其他”类别

Question

我有一个带有一列的DataFrame df，用下面的代码创建了category：

import pandas as pd
import random as rand
from string import ascii_uppercase

rand.seed(1010)

df = pd.DataFrame()
values = list()
for i in range(0,1000):   
    category = (''.join(rand.choice(ascii_uppercase) for i in range(1)))
    values.append(category)

df['category'] = values

每个值的频率计数是：

df['category'].value_counts()
Out[95]: 
P    54
B    50
T    48
V    46
I    46
R    45
F    43
K    43
U    41
C    40
W    39
E    39
J    39
X    37
M    37
Q    35
Y    35
Z    34
O    33
D    33
H    32
G    32
L    31
N    31
S    29

我想在df['category']列中创建一个称为“其他”的新值，并分配df['category']小于value_count的所有35值。

有人可以帮我这个忙吗？

让我知道您是否还需要我更多

编辑来自@EdChum提出的解决方案

import pandas as pd
import random as rand
from string import ascii_uppercase

rand.seed(1010)

df = pd.DataFrame()
values = list()
for i in range(0,1000):   
    category = (''.join(rand.choice(ascii_uppercase) for i in range(1)))
    values.append(category)

df['category'] = values
df['category'].value_counts()

df.loc[df['category'].isin((df['category'].value_counts([df['category'].value_‌​counts() < 35]).index), 'category'] = 'other'

  File "<stdin>", line 1
    df.loc[df['category'].isin((df['category'].value_counts()[df['category'].value_‌​counts() < 35]).index), 'category'] = 'other'
                                                                                   ^
SyntaxError: invalid syntax

注意，我在Spyder IDE上使用Python 2.7（我在iPython和Python控制台窗口中尝试了建议的解决方案）

Answer 1

您可以使用value_counts生成布尔值掩码以掩盖值，然后使用loc将其设置为“其他”：

In [71]:
df.loc[df['category'].isin((df['category'].value_counts()[df['category'].value_counts() < 35]).index), 'category'] = 'other'
df

Out[71]:
    category
0      other
1      other
2          A
3          V
4          U
5          D
6          T
7          G
8          S
9          H
10     other
11     other
12     other
13     other
14         S
15         D
16         B
17         P
18         B
19     other
20     other
21         F
22         H
23         G
24         P
25     other
26         M
27         V
28         T
29         A
..       ...
970        E
971        D
972    other
973        P
974        V
975        S
976        E
977    other
978        H
979        V
980        O
981    other
982        O
983        Z
984    other
985        P
986        P
987    other
988        O
989    other
990        P
991        X
992        E
993        V
994        B
995        P
996        B
997        P
998        Q
999        X

[1000 rows x 1 columns]

打破以上：

In [74]:
df['category'].value_counts() < 35

Out[74]:
W    False
B    False
C    False
V    False
H    False
P    False
T    False
R    False
U    False
K    False
E    False
Y    False
M    False
F    False
O    False
A    False
D    False
Q    False
N     True
J     True
S     True
G     True
Z     True
I     True
X     True
L     True
Name: category, dtype: bool

In [76]:    
df['category'].value_counts()[df['category'].value_counts() < 35]

Out[76]:
N    34
J    33
S    33
G    33
Z    32
I    31
X    31
L    30
Name: category, dtype: int64

然后我们可以对isin值使用.index并将行设置为'other'

Answer 2

Unit 2 assignment notebook中有一个例子：

# Reduce cardinality for NEIGHBORHOOD feature

# Get a list of the top 10 neighborhoods
top10 = df['NEIGHBORHOOD'].value_counts()[:10].index

# At locations where the neighborhood is NOT in the top 10, 
# replace the neighborhood with 'OTHER'
df.loc[~df['NEIGHBORHOOD'].isin(top10), 'NEIGHBORHOOD'] = 'OTHER'