How to Correctly Sort a String With a Number Inside

How to correctly sort a string with a number inside?

Perhaps you are looking for human sorting (also known as natural sorting):

import re

def atoi(text):
    return int(text) if text.isdigit() else text

def natural_keys(text):
    '''
    alist.sort(key=natural_keys) sorts in human order
    http://nedbatchelder.com/blog/200712/human_sorting.html
    (See Toothy's implementation in the comments)
    '''
    return [ atoi(c) for c in re.split(r'(\d+)', text) ]

alist=[
    "something1",
    "something12",
    "something17",
    "something2",
    "something25",
    "something29"]

alist.sort(key=natural_keys)
print(alist)

yields

['something1', 'something2', 'something12', 'something17', 'something25', 'something29']

PS. I've changed my answer to use Toothy's implementation of natural sorting (posted in the comments here) since it is significantly faster than my original answer.

---

If you wish to sort text with floats, then you'll need to change the regex from one that matches ints (i.e. (\d+)) to a regex that matches floats:

import re

def atof(text):
    try:
        retval = float(text)
    except ValueError:
        retval = text
    return retval

def natural_keys(text):
    '''
    alist.sort(key=natural_keys) sorts in human order
    http://nedbatchelder.com/blog/200712/human_sorting.html
    (See Toothy's implementation in the comments)
    float regex comes from https://stackoverflow.com/a/12643073/190597
    '''
    return [ atof(c) for c in re.split(r'[+-]?([0-9]+(?:[.][0-9]*)?|[.][0-9]+)', text) ]

alist=[
    "something1",
    "something2",
    "something1.0",
    "something1.25",
    "something1.105"]

alist.sort(key=natural_keys)
print(alist)

yields

['something1', 'something1.0', 'something1.105', 'something1.25', 'something2']

How to sort string with numbers in it numerically?

I think

val sortedList = data.sortedWith(compareBy(
    { it.listId },
    { it.name.substring(0, it.name.indexOf(' ')) },
    { it.name.substring(it.name.indexOf(' ') + 1).toInt() }
))

will work but it is not computationally efficient because it will call String.indexOf() many times.

If you have a very long list, you should consider making another list whose each item has String and Int names.

Sort string by multiple separated numbers

Try with sorted, supplying a custom key that uses re to extract all numbers from the path:

import re

>>> sorted(paths, key=lambda x: list(map(int,re.findall("(\d+)", x))))
['apple1/banana1',
 'apple1/banana2',
 'apple2/banana1',
 'apple2/banana2',
 'apple10/banana1/carrot1',
 'apple10/banana1/carrot2',
 'apple10/banana2/carrot1']

Sort list of string based on number in string

If you want to do this in the general case, I would try a natural sorting package like natsort.

from natsort import natsorted
my_list = ['image101.jpg', 'image2.jpg', 'image1.jpg']
natsorted(my_list)

Returns:

['image1.jpg', 'image2.jpg', 'image101.jpg']

You can install it using pip i.e. pip install natsort

How to sort first by letter then by number?

You can use a function to modify the sort key of each list item, without altering the original list contents:

def sort_key(text: str):
    letter = text[0]   # the first char
    digits = text[1:]  # everything after the first char
    return (letter, int(digits))

>>> sws = ['C6', 'A4', 'B8', 'A8', 'B11', 'C3', 'C5']
>>> sorted(sws, key=sort_key)
['A4', 'A8', 'B8', 'B11', 'C3', 'C5', 'C6']

How to sort a list of strings numerically?

You haven't actually converted your strings to ints. Or rather, you did, but then you didn't do anything with the results. What you want is:

list1 = ["1","10","3","22","23","4","2","200"]
list1 = [int(x) for x in list1]
list1.sort()

If for some reason you need to keep strings instead of ints (usually a bad idea, but maybe you need to preserve leading zeros or something), you can use a key function. sort takes a named parameter, key, which is a function that is called on each element before it is compared. The key function's return values are compared instead of comparing the list elements directly:

list1 = ["1","10","3","22","23","4","2","200"]
# call int(x) on each element before comparing it
list1.sort(key=int)

Is there a way to sort string lists by numbers inside of the strings?

You can sort the list like this:

hi.sort();

(because numbers sort before letters in its implementation)

How do I sort strings alphabetically while accounting for value when a string is numeric?

Pass a custom comparer into OrderBy. Enumerable.OrderBy will let you specify any comparer you like.

This is one way to do that:

void Main()
{
    string[] things = new string[] { "paul", "bob", "lauren", "007", "90", "101"};

    foreach (var thing in things.OrderBy(x => x, new SemiNumericComparer()))
    {    
        Console.WriteLine(thing);
    }
}

public class SemiNumericComparer: IComparer<string>
{
    /// <summary>
    /// Method to determine if a string is a number
    /// </summary>
    /// <param name="value">String to test</param>
    /// <returns>True if numeric</returns>
    public static bool IsNumeric(string value)
    {
        return int.TryParse(value, out _);
    }

    /// <inheritdoc />
    public int Compare(string s1, string s2)
    {
        const int S1GreaterThanS2 = 1;
        const int S2GreaterThanS1 = -1;

        var IsNumeric1 = IsNumeric(s1);
        var IsNumeric2 = IsNumeric(s2);

        if (IsNumeric1 && IsNumeric2)
        {
            var i1 = Convert.ToInt32(s1);
            var i2 = Convert.ToInt32(s2);

            if (i1 > i2)
            {
                return S1GreaterThanS2;
            }

            if (i1 < i2)
            {
                return S2GreaterThanS1;
            }

            return 0;
        }

        if (IsNumeric1)
        {
            return S2GreaterThanS1;
        }

        if (IsNumeric2)
        {
            return S1GreaterThanS2;
        }

        return string.Compare(s1, s2, true, CultureInfo.InvariantCulture);
    }
}

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