How to sort image sequences (with different extensions) in python

Question

I have this:

list = ["0001.exr", "0003.exr", "0002.dpx", "0001.dpx", "0002.exr", "0003.dpx"]

list.sort()

for line in list:
    print (line)

And it's returning this:

0001.dpx

0001.exr

0002.dpx

0002.exr

0003.dpx

0003.exr

How can I return this result? (in a nice simple way)

0001.dpx

0002.dpx

0003.dpx

0001.exr

0002.exr

0003.exr

Answer 1

Just use a key for .sort() that puts what's after "." as more important than what is before it:

list.sort(key=lambda l: "".join(reversed(l.split(".")))

Answer 2

You sort according to name, then use the OS library to get the extension and sort according to it (as a key):

list.sort()
list.sort(key=lambda list_item: os.path.splitext(list_item)[-1])

Answer 3

You could use the built in sorted() function with a custom cmp function called sort :

l = ["0001.exr", "0003.exr", "0002.dpx", "0001.dpx", "0002.exr", "0003.dpx"]

def sort(a, b):
    if 'dpx' in a and 'exr' in b:
        return -1
    if float(a.split('.')[0]) < float(b.split('.')[0]):
        return -1
    return 1

answer = sorted(l, cmp=sort)
print(answer)

Output

['0001.dpx', '0002.dpx', '0003.dpx', '0001.exr', '0002.exr', '0003.exr']

As an aside, you shouldn't mask the built-in list() function by using list as a variable name.

Answer 4

list = ["0001.exr", "0003.exr", "0002.dpx", "0001.dpx", "0002.exr", "0003.dpx"]

lst_tuple=[]
for item in list:
  lst_tuple.append((item.split('.')[0],item.split('.')[1]))


print sorted(lst_tuple, key=lambda element: (element[1], element[0]))