如何根据列表有条件地更新 Pandas 中的 DataFrame 列

Question

假设我有一个包含一列的数据框：

df = pd.DataFrame(np.random.randint(0,9,size=(100, 1)), columns=['number'])

我有两个列表，一个包含偶数，另一个包含奇数。

odd_numbers = [1,3,5,7,9]
even_numbers = [0,2,4,6,8]

我想根据 df['number'] 中的值在数据框上创建另一个系列，显示“偶数”或“奇数”

类似：

df['odd_or_even'] = 'even' if df[number].isin(even_numbers)
df['odd_or_even'] = 'odd' if df[number].isin(odd_numbers)

Answer 1

map 版本：

首先，创建一个字典：

d = {**{o: "odd" for o in odd_numbers}, **{e: "even" for e in even_numbers}}

然后在系列上使用地图：

df['odd_or_even'] = df['number'].map(d)

Answer 2

我觉得你可以用numpy.where:

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(0,9,size=(100, 1)), columns=['number'])

df['odd_or_even'] = np.where(df.number % 2, 'odd', 'even')
print (df)

    number odd_or_even
0        1         odd
1        0        even
2        4        even
3        5         odd
4        0        even
5        0        even
6        1         odd
7        0        even
8        7         odd
9        8        even

通过评论编辑

使用loc:

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randint(0,9,size=(20, 1)), columns=['number'])

odd_numbers = [1,3,5,7,9]
even_numbers = [0,2,4,6,8]

df.loc[df.number.isin(odd_numbers), 'odd_or_even'] = 'odd'
df.loc[df.number.isin(even_numbers), 'odd_or_even'] = 'even'

print (df)
    number odd_or_even
0        5         odd
1        1         odd
2        2        even
3        3         odd
4        5         odd
5        6        even
6        3         odd
7        4        even
8        2        even
9        8        even
10       8        even
11       1         odd
12       2        even
13       1         odd
14       3         odd
15       3         odd
16       5         odd
17       4        even
18       2        even
19       5         odd