I have a list of strings for which I would like to perform a natural alphabetical sort.

For instance, the following list is naturally sorted (what I want):

['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']

And here’s the “sorted” version of the above list (what I get using sorted()):

['Elm11', 'Elm12', 'Elm2', 'elm0', 'elm1', 'elm10', 'elm13', 'elm9']

I’m looking for a sort function which behaves like the first one.

There is a third party library for this on PyPI called natsort (full disclosure, I am the package’s author). For your case, you can do either of the following:

>>> from natsort import natsorted, ns
>>> x = ['Elm11', 'Elm12', 'Elm2', 'elm0', 'elm1', 'elm10', 'elm13', 'elm9']
>>> natsorted(x, key=lambda y: y.lower())
['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']
>>> natsorted(x, alg=ns.IGNORECASE)  # or alg=ns.IC
['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']

You should note that natsort uses a general algorithm so it should work for just about any input that you throw at it. If you want more details on why you might choose a library to do this rather than rolling your own function, check out the natsort documentation’s How It Works page, in particular the Special Cases Everywhere! section.

If you need a sorting key instead of a sorting function, use either of the below formulas.

>>> from natsort import natsort_keygen, ns
>>> l1 = ['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']
>>> l2 = l1[:]
>>> natsort_key1 = natsort_keygen(key=lambda y: y.lower())
>>> l1.sort(key=natsort_key1)
>>> l1
['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']
>>> natsort_key2 = natsort_keygen(alg=ns.IGNORECASE)
>>> l2.sort(key=natsort_key2)
>>> l2
['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']

Try this:

import re

def natural_sort(l): 
    convert = lambda text: int(text) if text.isdigit() else text.lower() 
    alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ] 
    return sorted(l, key = alphanum_key)

Output:

['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']

Code adapted from here: Sorting for Humans : Natural Sort Order.

Here’s a much more pythonic version of Mark Byer’s answer:

import re

def natural_sort_key(s, _nsre=re.compile('([0-9]+)')):
    return [int(text) if text.isdigit() else text.lower()
            for text in _nsre.split(s)]    

Now this function can be used as a key in any function that uses it, like list.sort, sorted, max, etc.

As a lambda:

lambda s: [int(t) if t.isdigit() else t.lower() for t in re.split('(\d+)', s)]

I wrote a function based on http://www.codinghorror.com/blog/2007/12/sorting-for-humans-natural-sort-order.html which adds the ability to still pass in your own ‘key’ parameter. I need this in order to perform a natural sort of lists that contain more complex objects (not just strings).

import re

def natural_sort(list, key=lambda s:s):
    """
    Sort the list into natural alphanumeric order.
    """
    def get_alphanum_key_func(key):
        convert = lambda text: int(text) if text.isdigit() else text 
        return lambda s: [convert(c) for c in re.split('([0-9]+)', key(s))]
    sort_key = get_alphanum_key_func(key)
    list.sort(key=sort_key)

For example:

my_list = [{'name':'b'}, {'name':'10'}, {'name':'a'}, {'name':'1'}, {'name':'9'}]
natural_sort(my_list, key=lambda x: x['name'])
print my_list
[{'name': '1'}, {'name': '9'}, {'name': '10'}, {'name': 'a'}, {'name': 'b'}]

data = ['elm13', 'elm9', 'elm0', 'elm1', 'Elm11', 'Elm2', 'elm10']

Let’s analyse the data. The digit capacity of all elements is 2. And there are 3 letters in common literal part 'elm'.

So, the maximal length of element is 5. We can increase this value to make sure (for example, to 8).

Bearing that in mind, we’ve got a one-line solution:

data.sort(key=lambda x: '{0:0>8}'.format(x).lower())

without regular expressions and external libraries!

print(data)

>>> ['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'elm13']

Explanation:

for elm in data:
    print('{0:0>8}'.format(elm).lower())

>>>
0000elm0
0000elm1
0000elm2
0000elm9
000elm10
000elm11
000elm13

Given:

data=['Elm11', 'Elm12', 'Elm2', 'elm0', 'elm1', 'elm10', 'elm13', 'elm9']

Similar to SergO’s solution, a 1-liner without external libraries would be:

data.sort(key=lambda x : int(x[3:]))

or

sorted_data=sorted(data, key=lambda x : int(x[3:]))

Explanation:

This solution uses the key feature of sort to define a function that will be employed for the sorting. Because we know that every data entry is preceded by ‘elm’ the sorting function converts to integer the portion of the string after the 3rd character (i.e. int(x[3:])). If the numerical part of the data is in a different location, then this part of the function would have to change.

Cheers

There are many implementations out there, and while some have come close, none quite captured the elegance modern python affords.

• Tested using python(3.5.1)
• Included an additional list to demonstrate that it works when the numbers are mid string
• Didn’t test, however, I am assuming that if your list was sizable it would be more efficient to compile the regex beforehand
  • I’m sure someone will correct me if this is an erroneous assumption

from re import compile, split    
dre = compile(r'(\d+)')
mylist.sort(key=lambda l: [int(s) if s.isdigit() else s.lower() for s in split(dre, l)])

#!/usr/bin/python3
# coding=utf-8
"""
Natural-Sort Test
"""

from re import compile, split

dre = compile(r'(\d+)')
mylist = ['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13', 'elm']
mylist2 = ['e0lm', 'e1lm', 'E2lm', 'e9lm', 'e10lm', 'E12lm', 'e13lm', 'elm', 'e01lm']

mylist.sort(key=lambda l: [int(s) if s.isdigit() else s.lower() for s in split(dre, l)])
mylist2.sort(key=lambda l: [int(s) if s.isdigit() else s.lower() for s in split(dre, l)])

print(mylist)  
  # ['elm', 'elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']
print(mylist2)  
  # ['e0lm', 'e1lm', 'e01lm', 'E2lm', 'e9lm', 'e10lm', 'E12lm', 'e13lm', 'elm']

Caution when using

• from os.path import split
  • you will need to differentiate the imports

Inspiration from

• Python Documentation- Sorting HOW TO
• Sorting for Humans : Natural Sort Order
• Human Sorting
• Contributors/Commentators to this and referenced posts

Value Of This Post

My point is to offer a non regex solution that can be applied generally.
I’ll create three functions:

1. find_first_digit which I borrowed from @AnuragUniyal. It will find the position of the first digit or non-digit in a string.
2. split_digits which is a generator that picks apart a string into digit and non digit chunks. It will also yield integers when it is a digit.
3. natural_key just wraps split_digits into a tuple. This is what we use as a key for sorted, max, min.

Functions

def find_first_digit(s, non=False):
    for i, x in enumerate(s):
        if x.isdigit() ^ non:
            return i
    return -1

def split_digits(s, case=False):
    non = True
    while s:
        i = find_first_digit(s, non)
        if i == 0:
            non = not non
        elif i == -1:
            yield int(s) if s.isdigit() else s if case else s.lower()
            s = ''
        else:
            x, s = s[:i], s[i:]
            yield int(x) if x.isdigit() else x if case else x.lower()

def natural_key(s, *args, **kwargs):
    return tuple(split_digits(s, *args, **kwargs))

We can see that it is general in that we can have multiple digit chunks:

# Note that the key has lower case letters
natural_key('asl;dkfDFKJ:sdlkfjdf809lkasdjfa_543_hh')

('asl;dkfdfkj:sdlkfjdf', 809, 'lkasdjfa_', 543, '_hh')

Or leave as case sensitive:

natural_key('asl;dkfDFKJ:sdlkfjdf809lkasdjfa_543_hh', True)

('asl;dkfDFKJ:sdlkfjdf', 809, 'lkasdjfa_', 543, '_hh')

We can see that it sorts the OP’s list in the appropriate order

sorted(
    ['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13'],
    key=natural_key
)

['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']

But it can handle more complicated lists as well:

sorted(
    ['f_1', 'e_1', 'a_2', 'g_0', 'd_0_12:2', 'd_0_1_:2'],
    key=natural_key
)

['a_2', 'd_0_1_:2', 'd_0_12:2', 'e_1', 'f_1', 'g_0']

My regex equivalent would be

def int_maybe(x):
    return int(x) if str(x).isdigit() else x

def split_digits_re(s, case=False):
    parts = re.findall('\d+|\D+', s)
    if not case:
        return map(int_maybe, (x.lower() for x in parts))
    else:
        return map(int_maybe, parts)
    
def natural_key_re(s, *args, **kwargs):
    return tuple(split_digits_re(s, *args, **kwargs))

One option is to turn the string into a tuple and replace digits using expanded form http://wiki.answers.com/Q/What_does_expanded_form_mean

that way a90 would become (“a”,90,0) and a1 would become (“a”,1)

below is some sample code (which isn’t very efficient due to the way It removes leading 0’s from numbers)

alist=["something1",
    "something12",
    "something17",
    "something2",
    "something25and_then_33",
    "something25and_then_34",
    "something29",
    "beta1.1",
    "beta2.3.0",
    "beta2.33.1",
    "a001",
    "a2",
    "z002",
    "z1"]

def key(k):
    nums=set(list("0123456789"))
        chars=set(list(k))
    chars=chars-nums
    for i in range(len(k)):
        for c in chars:
            k=k.replace(c+"0",c)
    l=list(k)
    base=10
    j=0
    for i in range(len(l)-1,-1,-1):
        try:
            l[i]=int(l[i])*base**j
            j+=1
        except:
            j=0
    l=tuple(l)
    print l
    return l

print sorted(alist,key=key)

output:

('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 1)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 10, 2)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 10, 7)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 2)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 20, 5, 'a', 'n', 'd', '_', 't', 'h', 'e', 'n', '_', 30, 3)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 20, 5, 'a', 'n', 'd', '_', 't', 'h', 'e', 'n', '_', 30, 4)
('s', 'o', 'm', 'e', 't', 'h', 'i', 'n', 'g', 20, 9)
('b', 'e', 't', 'a', 1, '.', 1)
('b', 'e', 't', 'a', 2, '.', 3, '.')
('b', 'e', 't', 'a', 2, '.', 30, 3, '.', 1)
('a', 1)
('a', 2)
('z', 2)
('z', 1)
['a001', 'a2', 'beta1.1', 'beta2.3.0', 'beta2.33.1', 'something1', 'something2', 'something12', 'something17', 'something25and_then_33', 'something25and_then_34', 'something29', 'z1', 'z002']

Based on the answers here, I wrote a natural_sorted function that behaves like the built-in function sorted:

# Copyright (C) 2018, Benjamin Drung <bdrung@posteo.de>
#
# Permission to use, copy, modify, and/or distribute this software for any
# purpose with or without fee is hereby granted, provided that the above
# copyright notice and this permission notice appear in all copies.
#
# THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
# WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
# MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
# ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
# WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
# ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
# OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.

import re

def natural_sorted(iterable, key=None, reverse=False):
    """Return a new naturally sorted list from the items in *iterable*.

    The returned list is in natural sort order. The string is ordered
    lexicographically (using the Unicode code point number to order individual
    characters), except that multi-digit numbers are ordered as a single
    character.

    Has two optional arguments which must be specified as keyword arguments.

    *key* specifies a function of one argument that is used to extract a
    comparison key from each list element: ``key=str.lower``.  The default value
    is ``None`` (compare the elements directly).

    *reverse* is a boolean value.  If set to ``True``, then the list elements are
    sorted as if each comparison were reversed.

    The :func:`natural_sorted` function is guaranteed to be stable. A sort is
    stable if it guarantees not to change the relative order of elements that
    compare equal --- this is helpful for sorting in multiple passes (for
    example, sort by department, then by salary grade).
    """
    prog = re.compile(r"(\d+)")

    def alphanum_key(element):
        """Split given key in list of strings and digits"""
        return [int(c) if c.isdigit() else c for c in prog.split(key(element)
                if key else element)]

    return sorted(iterable, key=alphanum_key, reverse=reverse)

The source code is also available in my GitHub snippets repository: https://github.com/bdrung/snippets/blob/master/natural_sorted.py

The above answers are good for the specific example that was shown, but miss several useful cases for the more general question of natural sort. I just got bit by one of those cases, so created a more thorough solution:

def natural_sort_key(string_or_number):
    """
    by Scott S. Lawton <scott@ProductArchitect.com> 2014-12-11; public domain and/or CC0 license

    handles cases where simple 'int' approach fails, e.g.
        ['0.501', '0.55'] floating point with different number of significant digits
        [0.01, 0.1, 1]    already numeric so regex and other string functions won't work (and aren't required)
        ['elm1', 'Elm2']  ASCII vs. letters (not case sensitive)
    """

    def try_float(astring):
        try:
            return float(astring)
        except:
            return astring

    if isinstance(string_or_number, basestring):
        string_or_number = string_or_number.lower()

        if len(re.findall('[.]\d', string_or_number)) <= 1:
            # assume a floating point value, e.g. to correctly sort ['0.501', '0.55']
            # '.' for decimal is locale-specific, e.g. correct for the Anglosphere and Asia but not continental Europe
            return [try_float(s) for s in re.split(r'([\d.]+)', string_or_number)]
        else:
            # assume distinct fields, e.g. IP address, phone number with '.', etc.
            # caveat: might want to first split by whitespace
            # TBD: for unicode, replace isdigit with isdecimal
            return [int(s) if s.isdigit() else s for s in re.split(r'(\d+)', string_or_number)]
    else:
        # consider: add code to recurse for lists/tuples and perhaps other iterables
        return string_or_number

Test code and several links (on and off of StackOverflow) are here: http://productarchitect.com/code/better-natural-sort.py

Feedback welcome. That’s not meant to be a definitive solution; just a step forward.

Most likely functools.cmp_to_key() is closely tied to the underlying implementation of python’s sort. Besides, the cmp parameter is legacy. The modern way is to transform the input items into objects that support the desired rich comparison operations.

Under CPython 2.x, objects of disparate types can be ordered even if the respective rich comparison operators haven’t been implemented. Under CPython 3.x, objects of different types must explicitly support the comparison. See How does Python compare string and int? which links to the official documentation. Most of the answers depend on this implicit ordering. Switching to Python 3.x will require a new type to implement and unify comparisons between numbers and strings.

Python 2.7.12 (default, Sep 29 2016, 13:30:34) 
>>> (0,"foo") < ("foo",0)
True  

Python 3.5.2 (default, Oct 14 2016, 12:54:53) 
>>> (0,"foo") < ("foo",0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  TypeError: unorderable types: int() < str()

There are three different approaches. The first uses nested classes to take advantage of Python’s Iterable comparison algorithm. The second unrolls this nesting into a single class. The third foregoes subclassing str to focus on performance. All are timed; the second is twice as fast while the third almost six times faster. Subclassing str isn’t required, and was probably a bad idea in the first place, but it does come with certain conveniences.

The sort characters are duplicated to force ordering by case, and case-swapped to force lower case letter to sort first; this is the typical definition of “natural sort”. I couldn’t decide on the type of grouping; some might prefer the following, which also brings significant performance benefits:

d = lambda s: s.lower()+s.swapcase()

Where utilized, the comparison operators are set to that of object so they won’t be ignored by functools.total_ordering.

import functools
import itertools


@functools.total_ordering
class NaturalStringA(str):
    def __repr__(self):
        return "{}({})".format\
            ( type(self).__name__
            , super().__repr__()
            )
    d = lambda c, s: [ c.NaturalStringPart("".join(v))
                        for k,v in
                       itertools.groupby(s, c.isdigit)
                     ]
    d = classmethod(d)
    @functools.total_ordering
    class NaturalStringPart(str):
        d = lambda s: "".join(c.lower()+c.swapcase() for c in s)
        d = staticmethod(d)
        def __lt__(self, other):
            if not isinstance(self, type(other)):
                return NotImplemented
            try:
                return int(self) < int(other)
            except ValueError:
                if self.isdigit():
                    return True
                elif other.isdigit():
                    return False
                else:
                    return self.d(self) < self.d(other)
        def __eq__(self, other):
            if not isinstance(self, type(other)):
                return NotImplemented
            try:
                return int(self) == int(other)
            except ValueError:
                if self.isdigit() or other.isdigit():
                    return False
                else:
                    return self.d(self) == self.d(other)
        __le__ = object.__le__
        __ne__ = object.__ne__
        __gt__ = object.__gt__
        __ge__ = object.__ge__
    def __lt__(self, other):
        return self.d(self) < self.d(other)
    def __eq__(self, other):
        return self.d(self) == self.d(other)
    __le__ = object.__le__
    __ne__ = object.__ne__
    __gt__ = object.__gt__
    __ge__ = object.__ge__

import functools
import itertools


@functools.total_ordering
class NaturalStringB(str):
    def __repr__(self):
        return "{}({})".format\
            ( type(self).__name__
            , super().__repr__()
            )
    d = lambda s: "".join(c.lower()+c.swapcase() for c in s)
    d = staticmethod(d)
    def __lt__(self, other):
        if not isinstance(self, type(other)):
            return NotImplemented
        groups = map(lambda i: itertools.groupby(i, type(self).isdigit), (self, other))
        zipped = itertools.zip_longest(*groups)
        for s,o in zipped:
            if s is None:
                return True
            if o is None:
                return False
            s_k, s_v = s[0], "".join(s[1])
            o_k, o_v = o[0], "".join(o[1])
            if s_k and o_k:
                s_v, o_v = int(s_v), int(o_v)
                if s_v == o_v:
                    continue
                return s_v < o_v
            elif s_k:
                return True
            elif o_k:
                return False
            else:
                s_v, o_v = self.d(s_v), self.d(o_v)
                if s_v == o_v:
                    continue
                return s_v < o_v
        return False
    def __eq__(self, other):
        if not isinstance(self, type(other)):
            return NotImplemented
        groups = map(lambda i: itertools.groupby(i, type(self).isdigit), (self, other))
        zipped = itertools.zip_longest(*groups)
        for s,o in zipped:
            if s is None or o is None:
                return False
            s_k, s_v = s[0], "".join(s[1])
            o_k, o_v = o[0], "".join(o[1])
            if s_k and o_k:
                s_v, o_v = int(s_v), int(o_v)
                if s_v == o_v:
                    continue
                return False
            elif s_k or o_k:
                return False
            else:
                s_v, o_v = self.d(s_v), self.d(o_v)
                if s_v == o_v:
                    continue
                return False
        return True
    __le__ = object.__le__
    __ne__ = object.__ne__
    __gt__ = object.__gt__
    __ge__ = object.__ge__

import functools
import itertools
import enum


class OrderingType(enum.Enum):
    PerWordSwapCase         = lambda s: s.lower()+s.swapcase()
    PerCharacterSwapCase    = lambda s: "".join(c.lower()+c.swapcase() for c in s)


class NaturalOrdering:
    @classmethod
    def by(cls, ordering):
        def wrapper(string):
            return cls(string, ordering)
        return wrapper
    def __init__(self, string, ordering=OrderingType.PerCharacterSwapCase):
        self.string = string
        self.groups = [ (k,int("".join(v)))
                            if k else
                        (k,ordering("".join(v)))
                            for k,v in
                        itertools.groupby(string, str.isdigit)
                      ]
    def __repr__(self):
        return "{}({})".format\
            ( type(self).__name__
            , self.string
            )
    def __lesser(self, other, default):
        if not isinstance(self, type(other)):
            return NotImplemented
        for s,o in itertools.zip_longest(self.groups, other.groups):
            if s is None:
                return True
            if o is None:
                return False
            s_k, s_v = s
            o_k, o_v = o
            if s_k and o_k:
                if s_v == o_v:
                    continue
                return s_v < o_v
            elif s_k:
                return True
            elif o_k:
                return False
            else:
                if s_v == o_v:
                    continue
                return s_v < o_v
        return default
    def __lt__(self, other):
        return self.__lesser(other, default=False)
    def __le__(self, other):
        return self.__lesser(other, default=True)
    def __eq__(self, other):
        if not isinstance(self, type(other)):
            return NotImplemented
        for s,o in itertools.zip_longest(self.groups, other.groups):
            if s is None or o is None:
                return False
            s_k, s_v = s
            o_k, o_v = o
            if s_k and o_k:
                if s_v == o_v:
                    continue
                return False
            elif s_k or o_k:
                return False
            else:
                if s_v == o_v:
                    continue
                return False
        return True
    # functools.total_ordering doesn't create single-call wrappers if both
    # __le__ and __lt__ exist, so do it manually.
    def __gt__(self, other):
        op_result = self.__le__(other)
        if op_result is NotImplemented:
            return op_result
        return not op_result
    def __ge__(self, other):
        op_result = self.__lt__(other)
        if op_result is NotImplemented:
            return op_result
        return not op_result
    # __ne__ is the only implied ordering relationship, it automatically
    # delegates to __eq__

>>> import natsort
>>> import timeit
>>> l1 = ['Apple', 'corn', 'apPlE', 'arbour', 'Corn', 'Banana', 'apple', 'banana']
>>> l2 = list(map(str, range(30)))
>>> l3 = ["{} {}".format(x,y) for x in l1 for y in l2]
>>> print(timeit.timeit('sorted(l3+["0"], key=NaturalStringA)', number=10000, globals=globals()))
362.4729259099986
>>> print(timeit.timeit('sorted(l3+["0"], key=NaturalStringB)', number=10000, globals=globals()))
189.7340817489967
>>> print(timeit.timeit('sorted(l3+["0"], key=NaturalOrdering.by(OrderingType.PerCharacterSwapCase))', number=10000, globals=globals()))
69.34636392899847
>>> print(timeit.timeit('natsort.natsorted(l3+["0"], alg=natsort.ns.GROUPLETTERS | natsort.ns.LOWERCASEFIRST)', number=10000, globals=globals()))
98.2531585780016

Natural sorting is both pretty complicated and vaguely defined as a problem. Don’t forget to run unicodedata.normalize(...) beforehand, and consider use str.casefold() rather than str.lower(). There are probably subtle encoding issues I haven’t considered. So I tentatively recommend the natsort library. I took a quick glance at the github repository; the code maintenance has been stellar.

All the algorithms I’ve seen depend on tricks such as duplicating and lowering characters, and swapping case. While this doubles the running time, an alternative would require a total natural ordering on the input character set. I don’t think this is part of the unicode specification, and since there are many more unicode digits than [0-9], creating such a sorting would be equally daunting. If you want locale-aware comparisons, prepare your strings with locale.strxfrm per Python’s Sorting HOW TO.

Let me submit my own take on this need:

from typing import Tuple, Union, Optional, Generator


StrOrInt = Union[str, int]


# On Python 3.6, string concatenation is REALLY fast
# Tested myself, and this fella also tested:
# https://blog.ganssle.io/articles/2019/11/string-concat.html
def griter(s: str) -> Generator[StrOrInt, None, None]:
    last_was_digit: Optional[bool] = None
    cluster: str = ""
    for c in s:
        if last_was_digit is None:
            last_was_digit = c.isdigit()
            cluster += c
            continue
        if c.isdigit() != last_was_digit:
            if last_was_digit:
                yield int(cluster)
            else:
                yield cluster
            last_was_digit = c.isdigit()
            cluster = ""
        cluster += c
    if last_was_digit:
        yield int(cluster)
    else:
        yield cluster
    return


def grouper(s: str) -> Tuple[StrOrInt, ...]:
    return tuple(griter(s))

Now if we have the list like such:

filelist = [
    'File3', 'File007', 'File3a', 'File10', 'File11', 'File1', 'File4', 'File5',
    'File9', 'File8', 'File8b1', 'File8b2', 'File8b11', 'File6'
]

We can simply use the key= kwarg to do a natural sort:

>>> sorted(filelist, key=grouper)
['File1', 'File3', 'File3a', 'File4', 'File5', 'File6', 'File007', 'File8', 
'File8b1', 'File8b2', 'File8b11', 'File9', 'File10', 'File11']

The drawback here is of course, as it is now, the function will sort uppercase letters before lowercase letters.

I’ll leave the implementation of a case-insenstive grouper to the reader :-)

I suggest you simply use the key keyword argument of sorted to achieve your desired list
For example:

to_order= [e2,E1,e5,E4,e3]
ordered= sorted(to_order, key= lambda x: x.lower())
    # ordered should be [E1,e2,e3,E4,e5]

Following @Mark Byers answer, here is an adaptation which accepts the key parameter, and is more PEP8-compliant.

def natsorted(seq, key=None):
    def convert(text):
        return int(text) if text.isdigit() else text

    def alphanum(obj):
        if key is not None:
            return [convert(c) for c in re.split(r'([0-9]+)', key(obj))]
        return [convert(c) for c in re.split(r'([0-9]+)', obj)]

    return sorted(seq, key=alphanum)

I also made a Gist

An improvement on Claudiu’s improvement on Mark Byers’ answer ;-)

import re

def natural_sort_key(s, _re=re.compile(r'(\d+)')):
    return [int(t) if i & 1 else t.lower() for i, t in enumerate(_re.split(s))]

...
my_naturally_sorted_list = sorted(my_list, key=natural_sort_key)

BTW, maybe not everyone remembers that function argument defaults are evaluated at def time

a = ['H1', 'H100', 'H10', 'H3', 'H2', 'H6', 'H11', 'H50', 'H5', 'H99', 'H8']
b = ''
c = []

def bubble(bad_list):#bubble sort method
        length = len(bad_list) - 1
        sorted = False

        while not sorted:
                sorted = True
                for i in range(length):
                        if bad_list[i] > bad_list[i+1]:
                                sorted = False
                                bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i] #sort the integer list 
                                a[i], a[i+1] = a[i+1], a[i] #sort the main list based on the integer list index value

for a_string in a: #extract the number in the string character by character
        for letter in a_string:
                if letter.isdigit():
                        #print letter
                        b += letter
        c.append(b)
        b = ''

print 'Before sorting....'
print a
c = map(int, c) #converting string list into number list
print c
bubble(c)

print 'After sorting....'
print c
print a

Acknowledgments:

Bubble Sort Homework

How to read a string one letter at a time in python

>>> import re
>>> sorted(lst, key=lambda x: int(re.findall(r'\d+$', x)[0]))
['elm0', 'elm1', 'Elm2', 'elm9', 'elm10', 'Elm11', 'Elm12', 'elm13']