警告:尝试使用.loc[row_indexer，col_indexer] = value

Question

我将数据文件的一部分切片，只保留两列。

description_category = titles[['listed_in','description']]

摘录是这样的

description_category.head()

    Listed_in                                           description
0   International TV Shows, TV Dramas, TV Sci-Fi &...   In a future where the elite inhabit an island ...
1   Dramas, International Movies                        After a devastating earthquake hits Mexico Cit...
2   Horror Movies, International Movies                 When an army recruit is found dead, his fellow...
3   Action & Adventure, Independent Movies, Sci-Fi...   In a postapocalyptic world, rag-doll robots hi...
4   Dramas                                              A brilliant group of students become card-coun...

我想要做的是在"Listed_in“专栏中插入每个主题，所以看起来是这样的：

    listed_in                                           description
0   [International TV Shows, TV Dramas, TV Sci-Fi ...   In a future where the elite inhabit an island ...
1   [Dramas, International Movies]                      After a devastating earthquake hits Mexico Cit...
2   [Horror Movies, International Movies]               When an army recruit is found dead, his fellow...
3   [Action & Adventure, Independent Movies, Sci-F...   In a postapocalyptic world, rag-doll robots hi...
4   [Dramas]                                            A brilliant group of students become card-coun...

我试过了，但它给了我一个警告：

description_category['listed_in'] = description_category['listed_in'].apply(lambda x: x.split(', '))

警告：

C:\Anaconda\envs\nlp_course\lib\site-packages\ipykernel_launcher.py:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  """Entry point for launching an IPython kernel.

我在这个问题上检查了几个线程，但我仍然无法修复它。

你建议我怎么做？

如果你需要更多关于我问题的背景资料，请告诉我。

Answer 1

如果您想在保留titles的同时创建一个新的数据格式，那么

• 要么用.loc[]切片：

'description'] = description_category = titles.loc[:，'listed_in'，

• 或创建.copy()：

description_category =标题[‘listed_in’，description_category

另外，使用.str.split()而不是apply()更快

description_category['listed_in'] = description_category['listed_in'].str.split(', ')

Answer 2

试试这个..。会成功的！！用DataFrame.loc‘结束整个任务

description_category.loc[description_category['listed_in'] = description_category['listed_in'].apply(lambda x: x.split(', '))]

它不会显示任何警告