Python 我可以在数据帧中找到替换函数的替代项“.loc”替代项吗

以前，我在问，有3个答案，但其中3个给出了警告，通常我忽略了这一点，但这是为了生产目的，所以我不能。以下是答案和警告：

一,

警告

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

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
  exec(code_obj, self.user_global_ns, self.user_ns)
/home/ubuntu/anaconda3/lib/python3.6/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

二,

警告

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

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
  exec(code_obj, self.user_global_ns, self.user_ns)
/home/ubuntu/anaconda3/lib/python3.6/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

三,

警告

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

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
  exec(code_obj, self.user_global_ns, self.user_ns)
/home/ubuntu/anaconda3/lib/python3.6/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

---

如何减少警告问题显然是在过滤前一行

解决方案

类似于：

说明

：

---

如果您稍后修改

df1

中的值，您会发现修改不会传播回原始数据（

df

），并且

Pandas

会发出警告。

问题出在上面的代码中？代码正在工作，但在我之前的功能工程部署中，jupyter notebook上需要注意，通常jupyter笔记本上的警告会给出生产上的问题。这种错误非常混乱，因为问题是前一行。显然，如果进行一些筛选，则需要

df=df1[df1['col']==10].copy（）

。好的，这是可行的。你可以贴出答案