Pandas 将索引复制到新列

Question

I am trying to copy the index of a dataframe ( newdf ) to a new column ( temp_index ) using the answer here but am receiving the infamous SettingWithCopyWarning . I tried the answer here , which says to add .loc[:, colname] but it throws even more warnings. All errors are thrown on the last line of code; no errors come up if the code stops when newdf is created.

What's the correct way to copy the index? Would prefer not to reset the index, I'd like the indices from df and newdf to be agreeable. I just need the copy column for something else.

Example Reproducible Code

col1 = [0,1,1,0,0,0,1,1,1]
col2 = [1,5,9,2,4,2,5,6,1]
df = pd.DataFrame(list(zip(col1, col2)), columns =['col1', 'col2'])
newdf = df[df.col2 >= 3]
display(df, newdf)
newdf.loc[:, 'temp_index'] = newdf.index

Errors

C:\Users\...\lib\site-packages\pandas\core\indexing.py:845: 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: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  self.obj[key] = _infer_fill_value(value)
C:\Users\...\lib\site-packages\pandas\core\indexing.py:966: 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: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  self.obj[item] = s

Answer 1

There's nothing wrong with how you are setting the temp_index column in the last line. The issue is as it says in the warning. What are you actually trying to achieve? To avoid this warning do newdf = df[df.col2 >= 3].copy() . Note you are indexing with a Boolean key which, AFAIK, creates a copy anyway so the above will not increase your memory footprint. If you actually want to insert the index to df but only to a subset of the rows try

key = df.col2 >= 3
df = df.loc[key, 'temp_index'] = df.index[key]