How can I reset the _is_copy flag on a Pandas dataframe to avoid a SettingWithCopyWarning, without copying the df?

Question

I have this code:

import pandas as pd
df = pd.DataFrame({'a': [1, 2, 3, 4]})
df2 = df[df['a'] > 2]
df2['b'] = df2['a'] * 2

This code raises a SettingWithCopyWarning. The warning is a false positive because I've assigned the result of the slice operation to another dataframe, and I intentionally only want to modify df2 and not df.

To avoid this warning, I usually call .copy():

import pandas as pd
df = pd.DataFrame({'a': [1, 2, 3, 4]})
df2 = df[df['a'] > 2].copy()
df2['b'] = df2['a'] * 2

However, this is inefficient; df2 is already a copy and not a view, so there's no point in creating another copy.

A more efficient way to do this is to set the _is_copy attribute:

import pandas as pd
df = pd.DataFrame({'a': [1, 2, 3, 4]})
df2 = df[df['a'] > 2]
df2._is_copy = None
df2['b'] = df2['a'] * 2

However, this relies on the private attributes of the dataframe and isn't future-proof.

How can I reset the _is_copy attribute without making doing a heavyweight copy operation?

I'm using Pandas 1.5.3 and cannot easily switch to Pandas 2.0 or above.

Answer 1

You can use method chaining.

Code:

import pandas as pd


df = pd.DataFrame({"a": [1, 2, 3, 4]})

df2 = (df
       .query(expr="a.gt(2)")
       .assign(b=df.a * 2)
       )

print(df2)

Output:

   a  b
2  3  6
3  4  8