You are asking for ordinal day suffixes: 1st, 2nd, 3rd, 4th, … including the exception for 11, 12, 13.
This is a classic edge-case problem and should be implemented deterministically, not with ad-hoc conditionals.
Correct rule (non-negotiable)
-
If the day is 11, 12, or 13 →
"th" -
Otherwise:
-
Ends in 1 →
"st" -
Ends in 2 →
"nd" -
Ends in 3 →
"rd" -
Everything else →
"th"
-
Canonical Python implementation
def ordinal(day: int) -> str:if 11 <= day % 100 <= 13:suffix = "th"else:suffix = {1: "st", 2: "nd", 3: "rd"}.get(day % 10, "th")return f"{day}{suffix}"
Examples
for d in range(1, 32):print(ordinal(d))
Output (excerpt):
1st
2nd
3rd
4th
...
10th
11th
12th
13th
14th
...
21st
22nd
23rd
24th
...
31st
Why this is the correct approach
-
Uses mod 100 to correctly catch
11–13 -
Uses mod 10 only when safe
-
Avoids fragile chains of
if/elif -
Easily testable
-
Works for any positive integer, not just calendar days
If you want it embedded in datetime
from datetime import datetoday = date.today()
print(ordinal(today.day))
Blunt advice
If you ever see code that checks if day == 1, elif day == 2, etc., it is wrong, unscalable, and unacceptable in professional Python.
If you want variants (locale-aware, zero-padded, or formatted dates like March 21st, 2025), say so.
from ChatGPT