如何在熊猫中用组名替换多行值？

Question

我正在尝试将日名转换为“工作日”或“周末”。

0         Saturday
1         Saturday
2         Saturday
3         Saturday
4         Saturday
            ...   
209526      Friday
209527      Friday
209528      Friday
209529      Friday
209530      Friday

我想要这个

0         Weekend
1         Weekend
2         Weekend
3         Weekend
4         Weekend            
             ...   
209526      Workday
209527      Workday
209528      Workday
209529      Workday
209530      Workday

我的代码什么也不返回：

df['Weekday_Or_Workday'] = df.replace(["Monday", "Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"],["Workday","Workday","Workday","Workday","Workday","Weekend","Weekend"])

我怎样才能使这个转换？

Answer 1

尝尝这个

df['Weekday_Or_Workday'] = df['Weekday_Or_Workday'].map({'Monday':'Workday','Tuesday':'Workday','Wednesday':'Workday','Thursday':'Workday','Friday':'Workday','Saturday':'Weekend','Sunday':'Weekend'})

Answer 2

我抄袭了你的代码对我有用。

因为这一行代码设置一个对象等于另一个对象，所以它不会返回任何内容。

如果您现在运行df (或print(df))，您应该看到已经添加了该列。

编辑:下面是我开始使用的数据格式：

df = pd.DataFrame(['Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday'])

Answer 3

如果将dict(zip(..包装到代码中，它可能会更易读：

import calendar

days = calendar.day_name
to_replace = *["Workday"]*5, *["Weekend"]*2

df["Weekday_Or_Workday"] = df.replace(dict(zip(days, to_replace)))