Explaining translation errors
The user mentioned a translation mistake on a birthday poster, where "32nd" was written as "32th." They haven't made a specific request, but might appreciate an explanation about ordinal suffixes. I'll clarify the rules for adding these endings, pointing out common mistakes for certain numbers. I can also suggest tools for proofreading and style guides, plus offer a small code snippet in JavaScript or Python that shows how to generate correct ordinal forms programmatically.Clarifying ordinal suffixes
Let's start by acknowledging the mistake regarding "32th" — it should be "32nd." I’ll explain that numbers ending in 1, 2, or 3 take those respective suffixes, except for 11, 12, and 13, which always end in "th."
To help avoid such errors, I can provide a checklist and a code snippet in Python or JavaScript for generating ordinal suffixes. I'll also suggest using the correct phrase in designs, like "32nd Anniversary" or "Happy 32nd Birthday" and offer potential Chinese translations. Lastly, I could mention using a regex or Excel formula for verification.Providing ordinal suffix formulas
I’m thinking about how to give the user a useful formula for adding ordinal suffixes in Excel, along with a Python code snippet. For Python, I can use a function like this:
def ordinal(n):
return f"{n}{'th' if 11 <= n % 100 <= 13 else {1: 'st', 2: 'nd', 3: 'rd'}.get(n % 10, 'th')}"
For Excel, the formula would be:
= A1 & CHOOSE(IF(AND(MOD(A1,100)>=11,MOD(A1,100)<=13),4,MIN(4,MOD(A1,10)+1)),"st","nd","rd","th")
I’ll keep it concise and avoid heavy formatting, but code blocks are allowed. And I can respond in Chinese if needed!
确实是个低级错误：“32th”应为“32nd”。
要点速记
- 英文序数词规则：结尾1→st，2→nd，3→rd，其余→th；但11、12、13一律→th（11th, 12th, 13th）。
- 因此：21st, 22nd, 23rd, 24th… 31st, 32nd, 33rd… 正确写法是 32nd。
- 版面文案常见写法：Happy 32nd Anniversary / 32nd Anniversary / The club’s 32nd birthday。
避免再出错的小招
- 出图前做一个序数词快速检查清单，特别留意 11–13 的例外。
- 设计排版时如需上标，可做 “32nd” 的 “nd” 上标；纯文本就直接 “32nd”。
- 若用中文为主：建议“成立32周年”“32周年庆”，避免中英混用不一致。
可复用的小函数
- Python
def ordinal(n: int) -> str:
return f"{n}{'th' if 11 <= n % 100 <= 13 else {1: 'st', 2: 'nd', 3: 'rd'}.get(n % 10, 'th')}"
# ordinal(32) -> '32nd'
- JavaScript
function ordinal(n){
const v = n % 100;
if (v >= 11 && v <= 13) return n + 'th';
return n + ({1:'st',2:'nd',3:'rd'}[n % 10] || 'th');
}
// ordinal(32) -> '32nd'
需要我帮你做一版正确的中英双语文案或小检查脚本吗？
