Python: Importing files and creating a list for each file

Question

I am looping over a list containing 9 file names and trying to create 9 variables (i.e. one for each file that I open).

The code seems to import correctly but I cant get it to create 9 variables, I've only managed to create one.

Code that works and creates one list (the 9th file):

for i in category_list:
 j = category_list.index(i)
 with open(str(path) + category_list[j] + f_ext, 'rb') as f:
 d = pickle.load(f)

Code that gives me error "name 'd_' is not defined

for i in category_list:
 j = category_list.index(i)
 with open(str(path) + category_list[j] + f_ext, 'rb') as f:
 d_[i] = pickle.load(f)

I think I may need to either declare the variable (which doesn't feel right for python) or I'm missing something even more simple.

Appreciate any help.

Thanks.

• Vasyl Kolomiets: hm... it seams Your forgot 4 spaces in second row ))
• PM 2Ring: Python doesn't have declarations, but you still need to define stuff, and I can't see a definition for d_; I assume it's supposed to be a list. And as Vasyl Kolomiets said, your indentation is broken.
• KSQ: The indentation got deleted when I pasted from Spyder, have corrected the above
• PM 2Ring: BTW, you can iterate over any iterable and get an index number at the same time: for idx, item in enumerate(my_sequence):
• KSQ: So do I have to define 9 empty lists before I load the 9 files?

Answer 1

hm... it seams it may works:

d_ = []
for category in category_list:
 with open(str(path) + category + f_ext, 'rb') as f:
 d_.append( pickle.load(f))

Isn't it? If - it is so - try eat more Python-like code )

• KSQ: That works I guess, it creates one list with 9 elements, I can work with that!
• Vasyl Kolomiets: You can write "d_ += pickle.load(f) ," but they say "d_.append(pickle.load(f))" is a bit "profy"