Bypassing and printing out the name of an Error in a Loop - Python [duplicate]

Question

Although I know it is not good practice to bypass an error, I'm trying to figure out in which of the data sets that I am working one, I get an error.

I am looping through all of the Data Sets, and everythime I get an error I flag it by using:

    try:
                # block raising an exception
    except:
       pass     # doing nothing on exception

I managed to flag the ones where the errors appeared by printing their names on the except block, but I would also like to know if anyone knows a way of also printing out the error I am getting (just to make sure it is all the same Error as I expect)

for j in list:
  for i in range(len(j)):
        try:
            run_path = j.iloc[i]
            name = run_path.processed_file
            name1 = name.split('/')
            name_final = name1[2]
            print(name_final)

            time, angle, couple, rpm, Fx, Fy, Fz, Mx, My , Mz , U , V, H = load_data(path_data + run_path.processed_file)
            dt = time[1] - time[0]

            minrose, maxrose, minwave, maxwave = minmaxrose_minmmaxH(Fx, angle, H)

            recap['Cxmin_rose'] = minrose
            recap['Cymmax_rose'] = maxrose
            recap['Cxmin_houle'] = minwave
            recap['Cxmax_houle'] = maxwave

            recap.to_excel(os.path.join(path_data,'recap_essai - JRS.xlsx'))  

            print('')
        except:
            run_path = j.iloc[i]
            name = run_path.processed_file
            name1 = name.split('/')
            name_final = name1[2]
            print('')
            print(name_final + ' ERROR ERROR ERROR ERROR ERROR ERROR ERROR ERROR ERROR')
            print('')
            pass

In this case, list is a list of "blocks of data sets" and i is a data set inside each block.

My Goal here would be to print, alongside the name, the type of error that I get.

Thank you!!

Answer 1

To print the type of error, you can use the following code

try:
# Your code
except Exception as e:
    print(type(e).__name__)

Example:

try:
    1/0
except Exception as e:
    print(type(e).__name__)
# Output:
# ZeroDivisionError

Also see How to print an exception in Python?