👀 Views: 74 💬 Answers: 1 📅 Created: 2025-06-06

pandas dataframe apply python

After trying multiple solutions online, I still can't figure this out... I'm having trouble with `DataFrame.apply()` in Pandas 1.3.5 when trying to apply a lambda function that calls a custom function. The intention is to transform one column based on the values of another column, but I'm encountering unexpected results. Here's a simplified version of my DataFrame: ```python import pandas as pd data = { 'A': [1, 2, 3, 4], 'B': ['x', 'y', 'z', 'x'] } df = pd.DataFrame(data) ``` I want to create a new column `C` where each value is the result of calling `my_function()` on the corresponding value in column `A`, but only if the value in column `B` is 'x'. My function looks like this: ```python def my_function(x): return x * 10 ``` I used the following code to apply this logic: ```python df['C'] = df.apply(lambda row: my_function(row['A']) if row['B'] == 'x' else None, axis=1) ``` After running this, I expected the new column `C` to contain `[10, None, None, 40]`. However, I'm getting this result instead: ```python A B C 0 1 x 10.0 1 2 y NaN 2 3 z NaN 3 4 x 40.0 ``` The output is correct, but I'm puzzled why the `None` values are being represented as `NaN` instead. I was expecting to see `None` explicitly in the DataFrame. Is there a way to get `None` instead of `NaN`? Also, is there a more efficient approach for this operation to improve performance with larger datasets? I also tried using `.loc` like this: ```python df.loc[df['B'] == 'x', 'C'] = df['A'].apply(my_function) ``` ``` This didn't work as expected either because it still resulted in `NaN` for the other entries. Any insights would be greatly appreciated. I'm working on a web app that needs to handle this.