file_list = os.listdir(os.getcwd)
files = file_list.sort()
If I have a list based on a directory listing as above, why does it sometimes return NoneType if using the files.sort() function but returns intended sort using the sorted(files) function?
list.sort() does an in place sort whereas sorted(list) returns a copy of the sorted list.
This means .sort() will return None as it has no return, thus defaulting in None. Using sorted() is usefull if you want to keep the original list, as .sort() destroys the original order.
>>> my_list = [3, 1, 2]
>>> sorted_list = sorted(my_list)
>>> sorted_list
[1, 2, 3]
>>> my_list
[3, 1, 2]
>>> print my_list.sort()
None
>>> my_list
[1, 2, 3]
sort() will always return None.sort() just call file_list.sort() without assigning it and after the call just operate on file_list as it will be sorted
list.sort() is a method of list and sorts the list that it is called on in-place and doesn't return anything.
sorted(collection) takes any iterable (it can be a list, set, tuple, or even a generator) and returns a list with all the items in sorted order.
Since sorted(collection) works any iterable and returns a list it is very useful when chaining method calls or when you can't use list.sort(), such as when iterating over the items in a dictionary (which is a tuple):
dct = { "c": 1, "b": 3, "a": 7 }
for key, value in sorted(dct.items()):
print("{} = {}".format(key, value))
In your case it doesn't really matter, but I'd go with
file_list = sorted(os.listdir(os.getcwd))
purely out of personal preference (it takes up less space).
