Sort numbered list

Question

I can't seem to find any information on how to correctly sort a list like this. My list looks something like this

["1. Banana", "2. Pear", "11. Apple", "5. Grapes", "4. Orange"]

Using sorted() leaves me with the following

['1. Banana', '11. Apple', '2. Pear', '3. Orange', '4. Grapes']

and I would like it to come out like this

['1. Banana', '2. Pear', '3. Orange', '4. Grapes', '11. Apple']

If it adds any complications I need to use this to sort a list of dictionaries using a specific key's value, so my current code is

list.sort(key=lambda k: k["key"])

Wait, why are you using `list.sort(key=lambda k: k["key"])`? I see no dictionaries in your example. Anyway, what you'll have to do is sort by the *int value of the number* since that is what you seem to want. You can parse out that int. Indeed, if that is something you want, you should have used a different data-structure in the first place, i.e. `(1, 'Banana')` and only construct the final string when you need it. — juanpa.arrivillaga, Feb 17 '18 at 20:22
I think your question's answer is there : https://stackoverflow.com/questions/5967500/how-to-correctly-sort-a-string-with-a-number-inside — syny, Feb 17 '18 at 20:24
@syny that's exactly what I was looking for, thank you . I neglected to mention in the question that some items of the list may not have a leading number but your suggestion is perfect and now I know what it's called! — , Feb 17 '18 at 21:27

score 5 · Accepted Answer · answered Feb 17 '18 at 20:21

5

Sort by . separated first int value as sorting key:

sorted(lst, key=lambda x: int(x.split('.')[0]))

Example:

In [20]: lst = ["1. Banana", "2. Pear", "11. Apple", "5. Grapes", "4. Orange"]

In [21]: sorted(lst, key=lambda x: int(x.split('.')[0]))
Out[21]: ['1. Banana', '2. Pear', '4. Orange', '5. Grapes', '11. Apple']

answered Feb 17 '18 at 20:21

heemayl

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sorry, didn't see your solution, until i submit my answer, which was identical to yours. – marmeladze Feb 17 '18 at 20:26
2

@marmeladze No problem. We were probably writing at the same time; mine came just a bit earlier i guess. It happens :) – heemayl Feb 17 '18 at 20:27
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Yeah @marmeladze, it's no big deal. I happens to the best of us - even to me ;-) I was just trying to give you heads up. – Christian Dean Feb 17 '18 at 20:31
@heemayl hey thank you for the quick answer! I didn't think ahead when posting my question and over-simplified the example list, not all of the strings I need to sort have a leading number. How would you go about sorting that? – Feb 17 '18 at 20:49
@Flameout Hard to say without seeing a concrete example. Also, i think it would be better if you ask it as a new question with all the details. – heemayl Feb 17 '18 at 20:51

Christian Dean · Answer 2 · 2018-02-17T20:52:50.457

4

For a bit more robust solution, grab the digit from the string items using regex, and use that to sort your list:

>>> from re import compile
>>> 
>>> regex = compile('\d+')
>>> lst = ["1. Banana", "2. Pear", "11. Apple", "5. Grapes", "4. Orange"]
>>> sorted(lst, key= lambda el: int(regex.search(el).group()))
['1. Banana', '2. Pear', '4. Orange', '5. Grapes', '11. Apple']
>>>

That way, you code won't fail even if the digit isn't the first part of your strings:

>>> lst = ["Banana 1.", "2. Pear", "11. Apple", "5. Grapes", "4. Orange"]
>>> sorted(lst, key= lambda el: int(regex.search(el).group()))
['Banana 1.', '2. Pear', '4. Orange', '5. Grapes', '11. Apple']
>>>

edited Feb 17 '18 at 20:52

answered Feb 17 '18 at 20:27

Christian Dean

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I fail to see how it's more robust. It will crash if there's no starting digit, just like the `int/split` solution. – Jean-François Fabre Feb 17 '18 at 20:48
Ah, my bad @Jean. I meant to use `search` not `match`. – Christian Dean Feb 17 '18 at 20:52

score 3 · Answer 3 · answered Feb 17 '18 at 20:25

In my opinion, you should be storing this data as a dictionary:

lst = ["1. Banana", "2. Pear", "11. Apple", "5. Grapes", "4. Orange"]

d = dict((int(j[0]), j[1]) for j in [i.split('. ') for i in lst])

# {1: 'Banana', 2: 'Pear', 4: 'Orange', 5: 'Grapes', 11: 'Apple'}

Then sorting is trivial:

sorted(d.items())

# [(1, 'Banana'), (2, 'Pear'), (4, 'Orange'), (5, 'Grapes'), (11, 'Apple')]

And so is formatting:

['. '.join((str(a), b)) for a, b in sorted(d.items())]

# ['1. Banana', '2. Pear', '4. Orange', '5. Grapes', '11. Apple']

It's not that big of an improvement, but it seems using `str.format` over `str.join` is a _tiny bit faster_: https://ideone.com/6iXDXs. Like I said though, in practice the difference probably wouldn't matter. — Christian Dean, Feb 17 '18 at 20:36

Sort numbered list

3 Answers3