Is There a Built in Function For String Natural Sort

Is there a built in function for string natural sort?

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.

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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']

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Update November 2020

Given that a popular request/question is "how to sort like Windows Explorer?" (or whatever is your operating system's file system browser), as of natsort version 7.1.0 there is a function called os_sorted to do exactly this. On Windows, it will sort in the same order as Windows Explorer, and on other operating systems it should sort like whatever is the local file system browser.

>>> from natsort import os_sorted
>>> os_sorted(list_of_paths)
# your paths sorted like your file system browser

For those needing a sort key, you can use os_sort_keygen (or os_sort_key if you just need the defaults).

Caveat - Please read the API documentation for this function before you use to understand the limitations and how to get best results.

Natural sorting

Google: Python natural sorting.

Result 1: The page you linked to.

But don't stop there!

Result 2: Jeff Atwood's blog that explains how to do it properly.

Result 3: An answer I posted based on Jeff Atwood's blog.

Here's the code from that answer:

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)

Results for your data:

PresserInc-1.jpg
PresserInc-1_10.jpg
PresserInc-1_11.jpg
PresserInc-2.jpg
PresserInc-3.jpg
etc...

See it working online: ideone

Natural sorting of a list in Python3

Your code is correct, but your data look incorrect: all the entries have a leading whitespace, which implies they are "before" the one you identify as least, that actually have no leading whitespace.

If the data is fine as they are I suggest you to revise the code to ignore leading whitespaces (check this: How do I remove leading whitespace in Python?).

How to natural sort string with multiple numbers in it

This should do the trick:

z = sorted(list_list, key=lambda s: (s[-5:-3], s[-2:]))

Sort list of string in Python

Solved.

from natsort import natsorted, ns

natsorted(listElements, alg=ns.IGNORECASE)

Python analog of PHP's natsort function (sort a list using a natural order algorithm)

From my answer to Natural Sorting algorithm:

import re
def natural_key(string_):
    """See https://blog.codinghorror.com/sorting-for-humans-natural-sort-order/"""
    return [int(s) if s.isdigit() else s for s in re.split(r'(\d+)', string_)]

Example:

>>> L = ['image1.jpg', 'image15.jpg', 'image12.jpg', 'image3.jpg']
>>> sorted(L)
['image1.jpg', 'image12.jpg', 'image15.jpg', 'image3.jpg']
>>> sorted(L, key=natural_key)
['image1.jpg', 'image3.jpg', 'image12.jpg', 'image15.jpg']

To support Unicode strings, .isdecimal() should be used instead of .isdigit(). See example in @phihag's comment. Related: How to reveal Unicodes numeric value property.

.isdigit() may also fail (return value that is not accepted by int()) for a bytestring on Python 2 in some locales e.g., '\xb2' ('²') in cp1252 locale on Windows.

Natural sort order string comparison in Java - is one built in?

In java the "natural" order meaning is "lexicographical" order, so there is no implementation in the core like the one you're looking for.

There are open source implementations.

Here's one:

NaturalOrderComparator.java

Make sure you read the:

Cougaar Open Source License

I hope this helps!

Sorting a list of strings numerically

Yes:

flist.sort(key=lambda fname: int(fname.split('.')[0]))

Explanation: strings are lexically sorted so "10" comes before "3" (because "1" < "3", so whatever comes after "1" in the first string is ignored). So we use list.sort()'s key argument which is a callback function that takes a list item and return the value to be used for ordering for this item - in your case, an integer built from the first part of the filename. This way the list is properly sorted on the numerical values.

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