Sort list of strings by integer suffix

I have a list of strings:

[song_1, song_3, song_15, song_16, song_4, song_8]

I would like to sort them by the # at the end, unfortunately since the lower numbers aren't "08" and are "8", they are treated as larger than 15 in lexicographical order.

I know I have to pass a key to the sort function, I saw this somewhere on this site to sort decimal numbers that are strings:

sorted(the_list, key=lambda a:map(int,a.split('.'))

But that was for "1.2, 2.5, 2.3" but I don't have that case. I thought of replacing '开发者_JS百科.' with '_' but from what I understand it converts both sides to ints, which will fail since the left side of the _ is a string.

EDIT: I forgot to mention that all the prefixes are the same (song in this example)

---

You're close.

sorted(the_list, key = lambda x: int(x.split("_")[1]))

should do it. This splits on the underscore, takes the second part (i.e. the one after the first underscore), and converts it to integer to use as a key.

---

Well, you want to sort by the filename first, then on the int part:

def splitter( fn ):
    try:
        name, num = fn.rsplit('_',1)  # split at the rightmost `_`
        return name, int(num)
    except ValueError: # no _ in there
        return fn, None

sorted(the_list, key=splitter)

---

sorted(the_list, key = lambda k: int(k.split('_')[1]))

---

def number_key(name):
   parts = re.findall('[^0-9]+|[0-9]+', name)
   L = []
   for part in parts:
       try:
          L.append(int(part))
       except ValueError:
          L.append(part)
   return L
sorted(your_list, key=number_key)