Python 如何比较数据帧中的多个布尔值

我有这样一个df：

         customer_id                       offer_id                   viewed_withintime completed_withintime
0   389bc3fa690240e798340f5a15918d5c    f19421c1d4aa40978ebb69ca19b0e20d    True           True
1   389bc3fa690240e798340f5a15918d5c    f19421c1d4aa40978ebb69ca19b0e20d    True           False
2   389bc3fa690240e798340f5a15918d5c    f19421c1d4aa40978ebb69ca19b0e20d    True           False

我想创建一个名为

offer\u succusful

的新列，如果

viewsed\u withintime

和

completed\u withintime

都是

True

，那么

offer\u succusful

的值是

1

，否则是

0

这是我的密码：

# label offer_successful to 1 if an offer is viewed and completed within offer_endtime, if not, set it to 0
offer_success_merge['offer_successful'] = (offer_success_merge['completed_withintime'] == True) and (offer_success_merge['viewed_withintime'] == True)

这将返回错误：

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

---

有人能帮我看一下吗，不知道如何修改这段代码…提前谢谢。

您可以使用numpy模块来解决此问题：

import numpy as np
offer_success_merge['offer_successful'] = np.logical_and((offer_success_merge['completed_withintime'] == True), (offer_success_merge['viewed_withintime'] == True))

---

您可以使用numpy模块解决此问题：

import numpy as np
offer_success_merge['offer_successful'] = np.logical_and((offer_success_merge['completed_withintime'] == True), (offer_success_merge['viewed_withintime'] == True))

您还可以使用：

df['offer_successful'] = (df['viewed_withintime']== True) & (df['completed_withintime']== True)

您还可以使用：

df['offer_successful'] = (df['viewed_withintime']== True) & (df['completed_withintime']== True)