Ordinal Numbers Replacement

Ordinal numbers replacement

The package number-parser can parse ordinal words ("first", "second", etc) to integers.

from number_parser import parse_ordinal
n = parse_ordinal("first")

To convert an integer to "1st", "2nd", etc, you can use the following (taken from Gareth on codegolf):

ordinal = lambda n: "%d%s" % (n,"tsnrhtdd"[(n//10%10!=1)*(n%10<4)*n%10::4])

This works on any number:

print([ordinal(n) for n in range(1,32)])

['1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th',
 '11th', '12th', '13th', '14th', '15th', '16th', '17th', '18th', '19th',
 '20th', '21st', '22nd', '23rd', '24th', '25th', '26th', '27th', '28th',
 '29th', '30th', '31st']

Find ordinal numbers with loop dynamically: find th - st - nd - rd

So, the problem is that 111 gets displayed as 111st instead of 111th.

You have 11 as ex1, I assume short for "exception 1", but your condition:

if number == ex1:

Clearly doesn't match 111.

Instead you could do:

if number % 100 == ex1:

Which will be true for 11, 111, 211 etc.

On a side note:

elif number % 10 == 1 or not ex1:

Clearly isn't what you intended. This is interpreted as:

elif (number % 10 == 1) or (not ex1):

not ex1 does not depend on number and will always evaluate the same way (False). But since you're already checking ex1 separately, it would be redundant to do it correctly here.

If you wanted to correct that, so that you don't need to check ex1 twice, you'd do this:

if number % 10 == 1 and number % 100 != 11:

I think in this case using != is clearer than not and I don't think there is any benefit from assigning a variable to 11.

Add st, nd, rd and th (ordinal) suffix to a number

The rules are as follows:

st is used with numbers ending in 1 (e.g. 1st, pronounced first)

nd is used with numbers ending in 2 (e.g. 92nd, pronounced ninety-second)

rd is used with numbers ending in 3 (e.g. 33rd, pronounced thirty-third)

As an exception to the above rules, all the "teen" numbers ending with 11, 12 or 13 use -th (e.g. 11th, pronounced eleventh, 112th,
pronounced one hundred [and] twelfth)

th is used for all other numbers (e.g. 9th, pronounced ninth).

The following JavaScript code (rewritten in Jun '14) accomplishes this:

function ordinal_suffix_of(i) {
    var j = i % 10,
        k = i % 100;
    if (j == 1 && k != 11) {
        return i + "st";
    }
    if (j == 2 && k != 12) {
        return i + "nd";
    }
    if (j == 3 && k != 13) {
        return i + "rd";
    }
    return i + "th";
}

Sample output for numbers between 0-115:

Is there a python function to convert ordinal number to word

You can extract your number from your string (i.e. extract 42 from 42nd) and then use num2words with ordinal=True:

import re
from num2words import num2words

ordinalAsNumber = "42nd"
number = re.search('\d+', ordinalAsNumber)

ordinalAsString = num2words(number[0], ordinal=True)

print( ordinalAsString ) # forty-second

You can extract all ordinals from a string like this:

text = "Text with several ordinals such as 42nd, 31st, and 5th as well as plain numbers such as 1, 2, 3."
numbers = re.findall('(\d+)[st|nd|rd|th]', text)

for n in numbers:
     ordinalAsString = num2words(n, ordinal=True)
     print( ordinalAsString )

Output:

forty-second
thirty-first
fifth

You can do replacements with re.sub():

text = "Text with several ordinals such as 42nd, 31st, and 5th as well as plain numbers such as 1, 2, 3."
numbers = re.findall('(\d+)[st|nd|rd|th]', text)

newText = text
for n in numbers:
     ordinalAsString = num2words(n, ordinal=True)
     newText=re.sub(r"\d+[st|nd|rd|th]", ordinalAsString, text, 1)

print ( text )
print( newText )

Output:

# Text with several ordinals such as 42nd, 31st, and 5th as well as plain numbers such as 1, 2, 3.
# Text with several ordinals such as forty-secondd, thirty-firstt, and fifthh as well as plain numbers such as 1, 2, 3.

Ordinal number to words in Python

Remove the ordinal endings from the strings:

import re

re.findall('\d+', stringValue)

and then use

num2words(foundInteger, ordinal=True)

How to get ordinal Numbers in flutter

Try with this

void main() {
   for(int i =1; i<=100;i++){
     print("$i${ordinal(i)}");
   }
}
String ordinal(int number) {
    if(!(number >= 1 && number <= 100)) {//here you change the range
      throw Exception('Invalid number');
    }

    if(number >= 11 && number <= 13) {
      return 'th';
    }

    switch(number % 10) {
      case 1: return 'st';
      case 2: return 'nd';
      case 3: return 'rd';
      default: return 'th';
    }
}

output:
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
..................

How to add ordinal numbers to the start of each line in text file

Something like this...

def ordinal_string(i):
   if i >= 10 and i <= 20:
      suffix = 'th'
   else:
     il = i % 10
     if il == 1:
        suffix = 'st'
     elif il == 2:
        suffix = 'nd'
     elif il == 3:
        suffix = 'rd'
     else:
        suffix = 'th'
   return str(i) + suffix + '. '

then:

for i, line in enumerate(sorted_modified_lines): print(ordinal_string(i+1) + line[1].strip()+"\n")

Ordinal Numbers Replacement