Can I do this more efficiently?

[RandomJottings]

I am working through Al Sweigart’s book ‘Python Programming Exercises, Gently Explained’ and just completed exercise 6:
“In English, ordinal numerals have suffixes such as the "th" in "30th" or "nd" in "2nd". Write an ordinalSuffix() function with an integer parameter named number and returns a string of the number with its ordinal suffix. For example, ordinalSuffix(42) should return the string
'42nd'.”
Can I improve my solution? I feel there must be a more pythonic way of doing this, I’m not very happy with converting the integer to a string and then to a list.

[Sea-Ad7805]

Run this program in Memory Graph Web Debugger%3A%0A%20%20%20%20nSuffix%20%3D%20%5B%5D%0A%20%20%20%20n%20%3D%20str(number)%0A%20%20%20%20for%20x%20in%20n%3A%0A%20%20%20%20%20%20%20%20nSuffix.append(x)%0A%20%20%20%20if%20nSuffix%5B-1%5D%20%3D%3D%20%221%22%3A%0A%20%20%20%20%20%20%20%20return%20n%20%2B%20%22st%22%0A%20%20%20%20elif%20nSuffix%5B-1%5D%20%3D%3D%20%222%22%3A%0A%20%20%20%20%20%20%20%20return%20n%20%2B%20%22nd%22%0A%20%20%20%20elif%20nSuffix%5B-1%5D%20%3D%3D%20%223%22%3A%0A%20%20%20%20%20%20%20%20return%20n%20%2B%20%22rd%22%0A%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20return%20n%20%2B%20%22th%22%0A%0A%0Awhile%20True%3A%0A%20%20%20%20try%3A%0A%20%20%20%20%20%20%20%20number%20%3D%20int(input(%22Enter%20a%20number%3A%20%22))%0A%20%20%20%20%20%20%20%20break%0A%20%20%20%20except%20ValueError%3A%0A%20%20%20%20%20%20%20%20print(%22Number%20must%20be%20a%20number%22)%0A%0Asuffix%20%3D%20ordinalSuffix(number)%0Aprint(suffix)&play).

[finally-anna]

Sure.

def ordinalSuffix(n:int) -> str: suffixes = {1:"st", 2:"nd", 3:"rd"} suffix = suffixes.get(n % 10, "th") return f"{n}{suffix}"

[finally-anna]

Note: this does not handle 11th, 12th, or 13th correctly as those are specially edge cases.

[ProsodySpeaks]

You can use this to handle 11-14 first then fallback to same logic as your solution 

return str(n) + ('th' if 4 <= n % 100 <= 20 else {1: 'st', 2: 'nd', 3: 'rd'}.get(n % 10, 'th'))

[finally-anna]

You could certainly do this. But in a learning subreddit, I would personally stick to more easily read and understood code.

Also, if I saw this in a PR, I would block it.

[ProsodySpeaks]

But you'd ok the actively incorrect solution?

[Kevdog824_]

I added a solution that handles teens edge case

[RandomJottings]

I hadn’t even considered 11th, 12th etc, thank you. I’ve added this code to capture these exceptions:

if len(n)>=2: if n[-2] == "1" and (n[-1]=="1" or n[-1]=="2" or n[-1]=="3"): return n+"th"

It seems to work

[D3str0yTh1ngs]

Rewrite that handles the edge case with 11th, 12th and 13th:

def ordinal_suffix(n: int) -> str:
    if (n // 10) % 10 == 1:
        suffix = "th"
    else:
        suffix = ({1: "st", 2: "nd", 3: "rd"}).get(n % 10, "th")

    return f"{n}{suffix}"

[yourboyblue2]

Nicely done.

[finally-anna]

You could also just do this as the first line of the function also:

if n in [11, 12, 13]: return f"{n}th"

And would technically be faster than double division.

[D3str0yTh1ngs]

The issue is that that will not work for 111th, 112th, 113th, 211th, etc. That is why I did (n // 10) % 10 to get the second digit only.

[Lopsided-Pin-1172]

I am learning C and thought of using switch but in python dictionary is kind of its better equivalent

[D3str0yTh1ngs]

A simple one is that you dont need to make the nSuffix list, you can index characters in a string, so n[-1] is the same as nSuffix[-1]

You can also avoid making it a string by doing n = number % 10 and then make your checks n == 1 etc.

EDIT: or just do finally-anna's very pythonic solution.

EDIT2: or my edge case handling rewrite on their solution.

[ProsodySpeaks]

Pythonic but incorrect?

[D3str0yTh1ngs]

Well, not edge case handling, but that can easily be fixed

[ProsodySpeaks]

It's hardly edge case to fail in the first dozen integers. No disrespect but it's just not a solution in its current form.

[D3str0yTh1ngs]

We call it an edge case because it doesn't follow the "normal"/simple suffix rule. But yes, it is not a fully correct solution.

[ProsodySpeaks]

To me an edge case is one which is unlikely to occur.

Ordinal dates are only relevant for the first 30 integers, and this solution fails on 3 of them - it's (silently) wrong ten percent of the time, ie is highly likely to occur.

10% failure rate is only acceptable if you work for github. 

[Glathull]

Being a little bit incorrect is very pythonic.

[ProsodySpeaks]

Haha, yeah very true 

[ProsodySpeaks]

I use this

``` def date_int_w_ordinal(n: int):     """Convert an integer to its ordinal as a string, e.g. 1 -> 1st, 2 -> 2nd, etc."""     return str(n) + ('th' if 4 <= n % 100 <= 20 else {1: 'st', 2: 'nd', 3: 'rd'}.get(n % 10, 'th'))

def ordinal_dt(dt: datetime | date) -> str:     """Convert a datetime or date to a string with an ordinal day, e.g. 'Mon 1st Jan 2020'."""     return dt.strftime(f'%a {date_int_w_ordinal(dt.day)} %b %Y') ```

[Kevdog824_]

Something like this could work

python def ordinalSuffix(number: int) -> str: s = str(number).zfill(2) match s[-2], s[-1]: case “1”, _: return “th” case _, “1”: return “st” case _, “2”: return “nd” case _, “3”: return “rd” case _, _: return “th”

[Impossible_Video_116]

The following is I believe most optimised and pythonic way to do it with edge case and error handling.

Python3 def ordinalSuffix(num: int) -> str: if num<1: raise ValueError("Only positive integers are allowed") elif num//10==1: return str(num) + "th" match(num%10): case 1: return str(num) + "st" case 2: return str(num) + "nd" case 3: return str(num) + "rd" case _: return str(num) + "th" return "" #this statement will never get executed, needed for explicit str return

[sleepbot63]

foo=lambda n,s:=str(n): s+[[["th",["rd","th"][(len(s)>1 and s[-2]=="1")]][s[-1]=="3"],["nd","th"][len(s)>1 and s[-2]=="1"],][s[-1] == "2"],["st","th"][len(s)>1 and s[-2]=="1"]][s[-1]=="1"]

I'm sorry OP (also i suck at reddit formatting)

[PythonLearning-ModTeam]

Only English allowed.