Specific sort a list of numbers separated by dots

If you need to sort by all digits, produce a sequence of integers in the key function:

sorted(L, key=lambda v: [int(p) for p in v.split('.') if p.isdigit()])

This method is robust in the face of non-digit values between the dots.

Demo:

>>> L = ['1.1.1.', '1.1.10.', '1.1.11.', '1.1.12.', '1.1.13.', '1.1.2.', '1.1.3.', '1.1.4.']
>>> sorted(L, key=lambda v: [int(p) for p in v.split('.') if p.isdigit()])
['1.1.1.', '1.1.2.', '1.1.3.', '1.1.4.', '1.1.10.', '1.1.11.', '1.1.12.', '1.1.13.']

Your specific attempt only returns the second number in the list, which for your sample data is always 1.

Just get the last part, convert that to an int and return it as the key for comparison

print(sorted(L, key=lambda x: int(x.split(".")[2])))

If you want all the parts to be considered, you can do like this

print(sorted(L, key=lambda x: [int(i) for i in x.rstrip(".").split(".")]))

It removes . at the end of the strings, splits them based on . and then converts each and every part of it to an int. The returned list will be used for comparison.

You can read more about how various sequences will be compared by Python, here

Output

['1.1.1.','1.1.2.','1.1.3.','1.1.4.','1.1.10.','1.1.11.','1.1.12.','1.1.13.']