Finding the numerator of a float's fractional value, given a specific denominator

Question

Possible Duplicates:
How to convert floats to human-readable fractions?
Convert decimal to fraction in Objective-C?

I want to take a decimal, 5.50, which is in a variable, and divide only the fractional part by 0.0625, which is my accuracy point. This would give me 8, as in 8/16 or 1/2.

Then I would like to display that answer as 5 8/16 or 1/2 in a text field. I know some answers will return a decimal still and not a whole number when dividing by .0625, but I would round that answer, which would still give me to the nearest 16th. What would be the best way to do this? I would like to make a function so I can reuse it. Yes, I posted a similar question that was answered, but wasn't able to get it to work. I'm thinking there's a better and easier way to do this, so I've posted this. Thanks in advance.

No, I seen that too. This is much more simple, or it should be I would think. just need to cast everything after decimal to NSInteger, divide by .0625 and then just display everything together as a fraction. Right? — Jason, May 19 '11 at 16:59
With enough reputation users have the ability to edit questions and answers. You can see who and what edits have been made by looking at the bottom of the question - the second box that says 'edited x minutes ago' — Abizern, May 19 '11 at 17:50
@Jason: I edited your question to clarify it before making a decision as to whether it duplicated your earlier question. I tried to make the tile more specific, but I still believe that the answer to your earlier question also answers this one, so I voted to close as a duplicate. If you disagree, please edit your question and explain the specific reason why the other answer (which you accepted) is unsuitable in this case. — jscs, May 19 '11 at 18:03
Because it seems like too much code for what I'm doing. I want to find a more simple and easy to understand way to do this. — Jason, May 19 '11 at 18:48
The preferred way to handle that situation is to _not_ accept the answer if it doesn't actually help you, and edit the old question to get an answer that does. — jscs, May 19 '11 at 20:29

Anne · Accepted Answer · 2011-05-19T22:27:08.130

4

Function:

- (NSString *)theFunction:(float)input {

    NSArray * array = [NSarray initWithObjects:nil,nil@"1/8",nil,@"1/4",]

    int fractions = lroundf((input - (int)input)/((float)1/(float)16));
    if(fractions == 0 || fractions == 16) {
        return [NSString stringWithFormat:@"%d",lroundf(input)];
    } else {
        return [NSString stringWithFormat:@"%d %d/16",(int)input,fractions];
    }

}

Note:

The if statement converts 5 0/16 into 5 and 5 16/16 into 6.
If you prefer the 5 0/16 and 5 16/16 notation, replace the if statement by:

return  [NSString stringWithFormat:@"%d %d/16",(int)input,fractions];

EDIT: (by Jason)

//Just to make it a little sweeter!
- (NSString *)theFunction:(float)input {
    int fractions = lroundf((input - (int)input)/((float)1/(float)16));
    if(fractions == 0 || fractions == 16) {
        return  [NSString stringWithFormat:@"%d",lroundf(input)];
    } else if(fractions == 2) {
        return  [NSString stringWithFormat:@"%d 1/8",(int)input];
    } else if(fractions == 4) {
        return  [NSString stringWithFormat:@"%d 1/4",(int)input];
    } else if(fractions == 6) {
        return  [NSString stringWithFormat:@"%d 3/8",(int)input];
    } else if(fractions == 8) {
        return  [NSString stringWithFormat:@"%d 1/2",(int)input];
    } else if(fractions == 10) {
        return  [NSString stringWithFormat:@"%d 5/8",(int)input];
    } else if(fractions == 12) {
        return  [NSString stringWithFormat:@"%d 3/4",(int)input];
    } else if(fractions == 14) {
        return  [NSString stringWithFormat:@"%d 7/8",(int)input];
    } else {
        return  [NSString stringWithFormat:@"%d %d/16",(int)input,fractions];
    }
}

EDIT (Response to edit by Jason)

I optimized your code, this way it's much cleaner.
Also check the code below, I think it's more efficient to use an array.

- (NSString *)theFunction:(float)input {

    NSArray * array = [[NSArray alloc] initWithObjects: @"",@"",@"1/8",@"",@"1/4",@"",@"3/8",@"",@"1/2",@"",@"5/8",@"",@"3/4",@"",@"3/4",@"",@"7/8",@"",nil];

    int fractions = lroundf((input - (int)input)/((float)1/(float)16));
    if(fractions == 0 || fractions == 16) {
        return [NSString stringWithFormat:@"%d",lroundf(input)];
    } else {
        if([[array objectAtIndex:fractions] isEqualToString:@""]) {
            return [NSString stringWithFormat:@"%d %d/16",(int)input,fractions];
        } else {
            return [NSString stringWithFormat:@"%d %@",(int)input,[array objectAtIndex:fractions]];
        }
    }

}

edited May 19 '11 at 22:27

answered May 19 '11 at 17:45

Anne

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is "input" where the variable would go? wouldn't you need to set the text of variable? – Jason May 19 '11 at 17:52
+1 for code. I don't have an Objective-C compiler in front of me, and would have been too lazy anyway. – gatkin May 19 '11 at 18:04
@Jason. You can simply paste the function inside your implementation file (the .m file) and then call it like `NSString * result = [self theFunction:5.03];`. – Anne May 19 '11 at 18:08
Added snippet how to run the code without function. – Anne May 19 '11 at 18:12
So if I have your function, would this be right calling it and displaying? `NSString * result = [self theFunction:myVariable]; [myTextField setText:[NSString stringWithFormat:@"%d %d/16", (result)]];` – Jason May 19 '11 at 18:43
Call the function like this: `NSString * result = [self theFunction:myVariable];`. Next update the text field like this: `[myTextField setText:result];`. You don't need to format the result because it's already formatted by the function. You can also merge both lines into `[myTextField setText:[self theFunction:myVariable]];`. Make sure myVariable is a float. Hopefully this helps! – Anne May 19 '11 at 19:04
OMG thank you!!!! Been trying to figure this out for the longest time, I'm still pretty new to programming, but I knew it had to be easier than the answer to my other post. You are the MAN!! (Figuratively speaking) :) – Jason May 19 '11 at 19:05
I'm happy it worked :) Don't hesitate asking when you're struggling! – Anne May 19 '11 at 19:10
Thank you so much!! It's discouraging when you ask and people and they don't really give you the whole answer, I learn by seeing the code first, then breaking down how it works. I understand people want you to try and figure out things for yourself, which is fine. But I knew there was shorter and easier to understand code to get the result I was looking for. I can't thank you enough!! I SEE how it works now!! Thank You!! – Jason May 19 '11 at 19:15
Optimized your code and added an example with array. – Anne May 19 '11 at 21:45
Awesome! Is one better on performance or memory than the other? – Jason May 20 '11 at 02:48
_Sorry for the late response_ - Generally it's frowned upon to use many `} else if() {` statements. For small denominators like 4 you may use multiple `} else if() {` statements. But for larger denominators I strongly recomment using one big array instead of many `} else if() {` statements. It makes your code much more efficient, maintainable and readable. There is absolutely no significant difference in terms performance or memory usage. – Anne May 20 '11 at 21:44
I have a question, if you want I can make a new thread. If I have a textfeld formatted as "00ft 00 1/16in" from 3 pickers, and I'm trying to grab the value and convert all to decimal before calculation, could you help me with that? – Jason Jun 09 '11 at 19:12

score 2 · Answer 2 · answered May 19 '11 at 17:03

2

I think your problem is much simpler than the one @Caleb linked to.

Divide by 0.0625 to get the number of sixteenths. Round the result to the nearest integer i. Use integer division i / 16 to get the number of whole units and use the modulo operator j = i % 16 to get the fractional units. To reduce the fractions, use j as an index into an array you create ahead of time as { "", "1/16", "1/8", "3/16", ... }.

You can also divide by 16 and get the modulo 16 using bitwise operators, but that will make the code harder to read for not much benefit.

answered May 19 '11 at 17:03

gatkin

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I think you understand what I'm trying to do best. It sounds like you hit the nail on the head...is there anyway you would be willing to write a function here as a template? – Jason May 19 '11 at 17:11
1

He's just given you the steps you need to take quite clearly - what are you having difficulty with? – Abizern May 19 '11 at 17:55

Finding the numerator of a float's fractional value, given a specific denominator

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