基于最后一个字符修改Pandas DataFrame中stores变量的值

针对需求，我们可以用两种高效的方法实现stores列的规则替换：

方法一：使用np.where多层条件判断 #

通过提取字符串最后一个字符进行条件匹配，逻辑直观清晰：

import pandas as pd
import numpy as np

data = {
         'stores': ['Lexinton1','ROYAl2','Mall1','Mall2','Levis1','Levis2','Shark1','Shark','Lexinton'],
         'quantity':[1,1,1,1,1,1,1,1,1]
        }

df = pd.DataFrame(data, columns = ['stores', 'quantity'])

# 按规则替换
df['stores'] = np.where(
    df['stores'].str[-1] == '1',
    'open',
    np.where(
        df['stores'].str[-1] == '2',
        'closed',
        df['stores']
    )
)

print(df)

执行结果：

stores  quantity
0      open         1
1    closed         1
2      open         1
3    closed         1
4      open         1
5    closed         1
6      open         1
7     Shark         1
8  Lexinton         1

方法二：使用正则表达式替换 #

利用正则匹配以1或2结尾的字符串，代码更简洁：

import pandas as pd
import numpy as np

data = {
         'stores': ['Lexinton1','ROYAl2','Mall1','Mall2','Levis1','Levis2','Shark1','Shark','Lexinton'],
         'quantity':[1,1,1,1,1,1,1,1,1]
        }

df = pd.DataFrame(data, columns = ['stores', 'quantity'])

# 替换以1结尾的字符串为open
df['stores'] = df['stores'].replace(r'.*1$', 'open', regex=True)
# 替换以2结尾的字符串为closed
df['stores'] = df['stores'].replace(r'.*2$', 'closed', regex=True)

print(df)

执行结果与方法一完全一致。

内容的提问来源于stack exchange，提问作者silent_hunter