如何在Python中对带有结尾零的数字字符串进行排序？

Question

I want to sort a list of numeric strings (w/ trailing zeroes) like how you normally sort number. I'd also like to return a True bool if a provided numeric string is in that list.

Example:

old_list = [1.1, 1.8, 1.50, 1.5, 1.9, 2.1]

to

sorted = [1.1, 1.5, 1.8, 1.9, 1.50, 2.1]

So, obviously the decimals are treated like numbers and sorted like numbers. 1.50 should not be simplified as 1.5, etc. It is used for product id for my company and has been bothering me for a year already.

I have tried converting the numeric strings into str then sort. I also tried using the Decimal module but it turns 1.1 into sth like 1.10000000000000008881...

Here is the code I used:

object_num = list(map(str, ['1.1', '1.8', '1.50', '1.5', '1.9', '2.1']))
object_num.sort()
print(object_num)

Result:

['1.1', '1.5', '1.50', '1.8', '1.9', '2.1']

I appreciate that Python is smart enough to sort the list numerically, but I hope it could also get me this result:

sorted = [1.1, 1.5, 1.8, 1.9, 1.50, 2.1]

And when I typed

print(1.50 in sorted)

It could return True.

Edit1: sorry for not making myself clear. Let me explain the numbering system my company is using:

ID for first product from series A: 1.1

ID for second product from series A: 1.2

...

ID for fiftieth product from series A: 1.50

ID for first product from series B: 2.1

All I want is to sort products from series A first, then series B

Thus, 1.5 is indeed different from 1.50 in my case

Answer 1

One solution using re module:

import re

lst = ['1.1', '1.8', '1.50', '1.5', '1.9', '2.1']

l = sorted(lst, key=lambda k: [*map(int, re.findall(r'\d+', k))])
print(l)

Prints:

['1.1', '1.5', '1.8', '1.9', '1.50', '2.1']

Or without re :

l = sorted(lst, key=lambda k: [*map(int, k.split('.'))])
print(l)

Answer 2

For version numbers you could use the (unfortunately not so well documented) distutils.version module:

from distutils.version import LooseVersion  # or StrictVersion

object_num = ['1.1', '1.8', '1.50', '1.5', '1.9', '2.1']
object_num.sort(key=LooseVersion)
print(object_num)

Result:

['1.1', '1.5', '1.8', '1.9', '1.50', '2.1']

It will also handle version numbers like "1.5.2b2" (eg for beta versions or release candidates). Depending on your needs you can choose between LooseVersion or StrictVersion (see the docstrings for the differences between them).