find out type/value of x in lambda x while using apply/pipe-CodePudding

I'm still running in errors when trying to write code where I use pipe/lambda on Series/Dataframes so I wondered if there is a way to debug my code to get better understanding. Just a simple example: I have a Series:

s = pd.Series([10,20,30,40,50])

If I want to add some number to each element I could use:

s.apply(lambda x : x 5)
#or
s.pipe(lambda x : x 5)

both works fine. But if I want to use conditional lambda:

s.apply(lambda x : x 5 if x>20 else 0) # works

s.pipe(lambda x : x 5 if x>20 else 0) # doesn't work

I get ValueError, the truth of a Series is ambiguous. If I wouldn't know why I get the error here I would need to know what x is... This is just an example, it's NOT about how to get the conditional adding done. I just want to know what is happening there and therefore I need to know what is x while executing my code.

Here is my question. When working on more complex matters and running into errors, is there a way I can debug my code to get to know the type or value of the x at any moment in the lambda while executing the code? For example a print statement? It would help to understand why I fail sometimes.

CodePudding user response：

This case:

The difference is that the x in apply is a single element of the Series, since the lambda function is applied to all elements one after the other. From the docs:

Parameters:
   func, function: Function to apply to each column or row.

In the case of pipe, the x is the series itself, which is why the conditional lambda doesn't work (if x>20 doesn't make sense). This is evident from the pipe docs:

Parameters:
    func, function: Function to apply to the Series/DataFrame.

General debugging:

Define a function instead of using a lambda expression, then you can add print statements as you desire:

def my_func(x):
    print(x)
    return x 5 if x>20 else 0


s.pipe(my_func)