Why Do I Still Get a SettingWithCopyWarning Even When Using .loc[row_indexer,col_indexer] = value?

SettingWithCopyWarning Despite Using .loc[row_indexer,col_indexer] = value

When using .loc[row_indexer,col_indexer] = value to assign values to a DataFrame slice, it's possible to encounter the SettingWithCopyWarning. This warning arises when a slice view of the DataFrame is being modified, potentially leading to confusion regarding whether the changes are made on the original or the copy.

To resolve this warning and ensure that changes are applied directly to the original DataFrame, it's crucial to use .copy() when creating a new DataFrame based on a subset of the original. This ensures that the new DataFrame is a true copy and not a slice view.

For instance, consider the following code:

import pandas as pd

df = pd.DataFrame({'col1': [1, 2, 3, 4], 'col2': [3, 4, 5, 6]})

new_df = df[df['col1'] > 2]  # Create a new DataFrame with a filter

new_df['new_column'] = None  # Create a new column in the new DataFrame

new_df.loc[2, 'new_column'] = 100  # Assign a value using .loc

Without using .copy(), this code will result in the SettingWithCopyWarning because new_df is a slice view of df. To fix it:

import pandas as pd

df = pd.DataFrame({'col1': [1, 2, 3, 4], 'col2': [3, 4, 5, 6]})

new_df = df[df['col1'] > 2].copy()  # Use .copy() to create a true copy

new_df['new_column'] = None

new_df.loc[2, 'new_column'] = 100

By using .copy(), you can avoid the warning and ensure that the changes made to the new DataFrame are directly reflected in the original DataFrame.

Similarly, using .astype() or .convert_objects() on a copy of a DataFrame slice using .loc may also trigger the warning. To resolve it, apply the functions beforehand or create a true copy before making the changes.

The above is the detailed content of Why Do I Still Get a SettingWithCopyWarning Even When Using .loc[row_indexer,col_indexer] = value?. For more information, please follow other related articles on the PHP Chinese website!