Add St, Nd, Rd and Th (Ordinal) Suffix to a Number

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:

NSNumberFormatter and 'th' 'st' 'nd' 'rd' (ordinal) number endings

Since the question asked for a number formatter, here's a rough one I made.

//
//  OrdinalNumberFormatter.h
//

#import <Foundation/Foundation.h>

@interface OrdinalNumberFormatter : NSNumberFormatter {

}

@end

and the implementation:

//
//  OrdinalNumberFormatter.m
//

#import "OrdinalNumberFormatter.h"

@implementation OrdinalNumberFormatter

- (BOOL)getObjectValue:(id *)anObject forString:(NSString *)string errorDescription:(NSString **)error {
    NSInteger integerNumber;
    NSScanner *scanner;
    BOOL isSuccessful = NO;
    NSCharacterSet *letters = [NSCharacterSet letterCharacterSet];

    scanner = [NSScanner scannerWithString:string];
    [scanner setCaseSensitive:NO];
    [scanner setCharactersToBeSkipped:letters];

    if ([scanner scanInteger:&integerNumber]){
        isSuccessful = YES;
        if (anObject) {
            *anObject = [NSNumber numberWithInteger:integerNumber];
        }
    } else {
        if (error) {
            *error = [NSString stringWithFormat:@"Unable to create number from %@", string];
        }
    }

    return isSuccessful;
}

- (NSString *)stringForObjectValue:(id)anObject {
    if (![anObject isKindOfClass:[NSNumber class]]) {
        return nil;
    }

    NSString *strRep = [anObject stringValue];
    NSString *lastDigit = [strRep substringFromIndex:([strRep length]-1)];

    NSString *ordinal;

    if ([strRep isEqualToString:@"11"] || [strRep isEqualToString:@"12"] || [strRep isEqualToString:@"13"]) {
        ordinal = @"th";
    } else if ([lastDigit isEqualToString:@"1"]) {
        ordinal = @"st";
    } else if ([lastDigit isEqualToString:@"2"]) {
        ordinal = @"nd";
    } else if ([lastDigit isEqualToString:@"3"]) {
        ordinal = @"rd";
    } else {
        ordinal = @"th";
    }

    return [NSString stringWithFormat:@"%@%@", strRep, ordinal];
}

@end

Instantiate this as an Interface Builder object and attach the Text Field's formatter outlet to it. For finer control (such as setting maximum and minimum values, you should create an instance of the formatter, set the properties as you wish and attach it to text field using it's setFormatter: method.

You can download the class from GitHub (including an example project)

Append ordinal suffixes ('-st', '-nd', '-rd', '-th') based on the item index

You may put necessary suffixes into array and pick the one that corresponds to your index:

const arr = [...'abcdefghijklmnopqrstuvwxyz'];const suffixes = ['th', 'st', 'nd', 'rd'];
const result = arr.map((item, i) =>   (idx = ~~(i % 10) + 1 > 3 || ~~(i / 10) == 1 ? 0 : ~~(i % 10) + 1,   `${i+1}-${suffixes[idx]} options is '${item}'`));
console.log(result);

.as-console-wrapper {min-width: 100%}

Is there an easy way in .NET to get st, nd, rd and th endings for numbers?

No, there is no inbuilt capability in the .NET Base Class Library.

How do I retrieve an integer's ordinal suffix in Perl (like st, nd, rd, th)

Try this:

my $ordinal;
if ($foo =~ /(?<!1)1$/) {
    $ordinal = 'st';
} elsif ($foo =~ /(?<!1)2$/) {
    $ordinal = 'nd';
} elsif ($foo =~ /(?<!1)3$/) {
    $ordinal = 'rd';
} else {
    $ordinal = 'th';
}

Format for ordinal dates (day of month with suffixes -st, -nd, -rd, -th)

Enjoy the power of lubridate:

library(lubridate)    
mdy(ord_dates)

[1] "2016-09-01" "2016-09-02" "2016-09-03" "2016-09-04"

Internally, lubridate doesn't have any special conversion specifications which enable this. Rather, lubridate first uses (by smart guessing) the format "%B %dst, %Y". This gets the first element of ord_dates.

It then checks for NAs and repeats its smart guessing on the remaining elements, settling on "%B %dnd, %Y" to get the second element. It continues in this way until there are no NAs left (which happens in this case after 4 iterations), or until its smart guessing fails to turn up a likely format candidate.

You can imagine this makes lubridate slower, and it does -- about half the speed of just using the smart regex suggested by @alistaire above:

set.seed(109123)
ord_dates <- sample(
  c("September 1st, 2016", "September 2nd, 2016",
    "September 3rd, 2016", "September 4th, 2016"),
  1e6, TRUE
  )

library(microbenchmark)

microbenchmark(times = 10L,
               lubridate = mdy(ord_dates),
               base = as.Date(sub("\\D+,", "", ord_dates),
                              format = "%B %e %Y"))
# Unit: seconds
#       expr      min       lq     mean   median       uq      max neval cld
#  lubridate 2.167957 2.219463 2.290950 2.252565 2.301725 2.587724    10   b
#       base 1.183970 1.224824 1.218642 1.227034 1.228324 1.229095    10  a

The obvious advantage in lubridate's favor being its conciseness and flexibility.

How do I print ordinal indicators in a C program? Can't print numbers with 'st', 'nd', 'rd'. (Beginner)

You can mod by 10 to get the last digit. Then based on that you can use "st", "nd", "rd", or "th". You'll also need special cases for 11, 12, and 13.

    if ((i % 10 == 1) && (i % 100 != 11))
        printf("%dst value : ", i);
    else if ((i % 10 == 2) && (i % 100 != 12))
        printf("%dnd value : ", i);
    else if ((i % 10 == 3) && (i % 100 != 13))
        printf("%drd value : ", i);
    else
        printf("%dth value : ", i);

How to append th, st, nd and rd for dates of the day in rdlc report?

Using the suggestion in Is there an easy way to create ordinals in C#? you can create a Custom Code for your rdlc report and use it with the expression.

In the rdlc properties add the following function in Code tab

Public Shared Function AddOrdinal(num As Integer) As String
    If num <= 0 Then
        Return num.ToString()
    End If
    Select Case num Mod 100
        Case 11, 12, 13
            Return num & "th"
    End Select
    Select Case num Mod 10
        Case 1
            Return num & "st"
        Case 2
            Return num & "nd"
        Case 3
            Return num & "rd"
        Case Else
            Return num & "th"
    End Select
End Function

Then modify your Expression as follows

=Code.AddOrdinal(Day(Fields!Date.value))

Add St, Nd, Rd and Th (Ordinal) Suffix to a Number