[python] add st, nd, rd and th to day of the month

 

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.

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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"

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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

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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

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If you want it embedded in datetime

from datetime import date

today = date.today()
print(ordinal(today.day))

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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.

 

 

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