If you use .loc[row, column]
and still get the same error, it’s probably because of copying another dataframe. You have to use .copy()
.
This is a step-by-step error reproduction:
import pandas as pd
d = {'col1': [1, 2, 3, 4], 'col2': [3, 4, 5, 6]}
df = pd.DataFrame(data=d)
df
# col1 col2
#0 1 3
#1 2 4
#2 3 5
#3 4 6
Creating a new column and updating its value:
df['new_column'] = None
df.loc[0, 'new_column'] = 100
df
# col1 col2 new_column
#0 1 3 100
#1 2 4 None
#2 3 5 None
#3 4 6 None
No error I receive. But, let’s create another dataframe given the previous one:
new_df = df.loc[df.col1>2]
new_df
#col1 col2 new_column
#2 3 5 None
#3 4 6 None
Now, using .loc
, I will try to replace some values in the same manner:
new_df.loc[2, 'new_column'] = 100
However, I got this hateful warning again:
A value is trying to be set on a copy of a slice from a DataFrame. Try
using .loc[row_indexer,col_indexer] = value insteadSee the caveats in the documentation:
https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
SOLUTION
use .copy()
while creating the new dataframe will solve the warning:
new_df_copy = df.loc[df.col1>2].copy()
new_df_copy.loc[2, 'new_column'] = 100
Now, you won’t receive any warnings!
If your dataframe is created using a filter on top of another dataframe, always use .copy()
.