df.loc导致SettingWithCopyWarning警告消息 [英] df.loc causes a SettingWithCopyWarning warning message

问题描述

我的代码的以下行会引发警告:

The following line of my code causes a warning :

import pandas as pd

s = pd.DataFrame(np.random.randint(0,100,size=(100, 4)), columns=list('ABCD'))
s.loc[-1] = [5,np.nan,np.nan,6]
grouped = s.groupby(['A'])
for key_m, group_m in grouped:
    group_m.loc[-1] = [10,np.nan,np.nan,10]

C:\Anaconda3\lib\site-packages\ipykernel\__main__.py:10: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

根据文档这是推荐的方法，那么发生了什么?

According to the documentation this is the recommended way of doing, so what is happening ?

感谢您的帮助.

推荐答案

文档有些混乱.

您的dataframe是另一个dataframe的副本.您可以通过运行bool(df.is_copy)来验证这一点.由于您正在尝试分配给该副本，因此您将收到警告.

Your dataframe is a copy of another dataframe. You can verify this by running bool(df.is_copy) You are getting the warning because you are trying to assign to this copy.

警告/文档首先告诉您应该如何构造df.现在它已经是副本了，不是您应该如何分配它.

The warning/documentation is telling you how you should have constructed df in the first place. Not how you should assign to it now that it is a copy.

df = some_other_df[cols]

将使df成为some_other_df的副本.警告建议改为这样做

will make df a copy of some_other_df. The warning suggests doing this instead

df = some_other_df.loc[:, [cols]]

---

现在已完成，如果您选择忽略此警告，则可以

---

Now that it is done, if you choose to ignore this warning, you could

df = df.copy()

或

df.is_copy = None

这篇关于df.loc导致SettingWithCopyWarning警告消息的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！