I am trying to trim zeros after a decimal point as below but it's not giving desired result.
trig = [currentVal doubleValue];
trig = trig/100;
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
[formatter setMaximumFractionDigits:0];
display.text = [formatter stringFromNumber:[NSNumber numberWithDouble:trig]];
The number is still being displayed without trimming zeros after the decimal point.
Here currentVal is the number I am entering.
For example if i pass "trig" = 123 (Initially "trig" = 123 after doing trig/100 i want to display 1.23 but it is displaying as 1.23000000).
Sometimes the straight C format specifiers do an easier job than the Cocoa formatter classes, and they can be used in the format string for the normal stringWithFormat: message to NSString.
If your requirement is to not show any trailing zeroes, then the "g" format specifier does the job:
float y = 1234.56789f;
NSString *s = [NSString stringWithFormat:#"%g", y];
Notice that there is no precision information, which means that the printf library will remove the trailing zeroes itself.
There is more information in the docs, which refer to IEEE's docs.
In case this helps someone. I wanted 1 decimal value but no '.0' on the end if the float was '1.0'. Using %g would give scientific notation for longer numbers, following ugliness worked well enough for me as high accuracy wasn't critical.
// Convert to 1 dp string,
NSString* dirtyString = [NSString stringWithFormat: #"%.1f", self.myFloat];
// Convert back to float that is now a maximum of 1 dp,
float myDirtyFloat = [dirtyString floatValue];
// Output the float subtracting the zeros the previous step attached
return [NSString stringWithFormat: #"%g", myDirtyFloat];
This will not display any decimal value after the decimal point:
display.text = [NSString stringWithFormat:#"%1.0f", trig];
This will just trim the zeros after the decimal point:
isplay.text = [NSString stringWithFormat:#"%3.2f", trig];
display.text = [display.text stringByTrimmingCharactersInSet:[NSCharacterSet characterSetWithCharactersInString:[NSString stringWithFormat#"0"]]];
Note, this may leave you with the trailing decimal point. "124." may happen. I guess that some smarter solution will be posted soon.
From the documentation, it looks like setFractionDigits: is only for converting the other way.
The best thing to do is probably to convert your number to an integer before formatting it e.g.
double converted = round(trig); // man round for docs
You can use also the formatting functions of stringWithFormat: of NSString, but then you will lose all the localisation advantages you get with NSNumberFormatter.
This may not be a proper solution where there is NSNumberFormetter Class, But I just did this rather then googling a lot! ;)
Here is an example, if it helps:
-(NSString*) trimZerosAfterDecimalPoint:(NSString*)string_ {
double doubleValue=[string_ doubleValue];
long leftPart=(long)doubleValue;
double rightPart=doubleValue-(double)leftPart;
NSString *rightPartAsStr=[NSString stringWithFormat:#"%f", rightPart];
int i=0;
for (i=rightPartAsStr.length-1; i>=2; i--) {
if ([rightPartAsStr characterAtIndex:i]!='0') {
rightPartAsStr=[rightPartAsStr substringWithRange:NSMakeRange(2, i-1)];
break;
}
}
if (i<2) {
string_=[NSString stringWithFormat:#"%ld", leftPart];
} else {
string_=[NSString stringWithFormat:#"%ld.%#", leftPart, rightPartAsStr];
}
return string_;
}
I just had to do this for one of my programs and heres how I went about it:
- (void) simplify{
int length = (int)[self.calcString length];
for (int i = (int)[self.calcString length]; i > 0; i--) {
if ([self.calcString rangeOfString:#"."].location != NSNotFound) {
NSRange prevChar = NSMakeRange(i-1, 1);
if ([[self.calcString substringWithRange:prevChar] isEqualToString:#"0"]||
[[self.calcString substringWithRange:prevChar] isEqualToString:#"."])
length--;
else
break;
}
self.calcString = [self.calcString substringToIndex:length];
}
}
This works
display.text = [#(trig) stringValue];
it is because of your datatype cannot be formatted is such a manner.
Related
I have a few actions in xcode where the number of decimals in the output value needs to be limited to 3 decimal places out. What do I need to add to my code to achieve this task?
Here is an example of one of my actions:
- (IBAction)calculateMolarity:(id)sender {
float ourValue = [[_calcTextFieldNumOne text] floatValue] /[ [_calcTextFieldTwo text] floatValue];
NSNumber *ourNum =[NSNumber numberWithFloat:ourValue];
[_outputOfMolarity setText:[ourNum stringValue]];
NSString* formattedNumber = [NSString stringWithFormat:#"%.03f", ourNum];
here %.03f tells the formatter that you will be formatting a float
(%f) and, that should be rounded to three places, and should be padded
with 0's.
but you can do directly with your float ourValue like this
NSString* formattedNumber = [NSString stringWithFormat:#"%.03f", ourValue];
there is no need to convert your float value to NSNumber
you can use "%.3f" or "%.03f", no matter both gives same fromat
#"%.3f" = 1234.567
#"%.03f" = 1234.567 // which is equal to #"%.3f"
I have a string called realEstateWorth with a value of $12,000,000.
I need this same string to remain a string but for any number (such as the one above) to be displayed as $12 MILLION or $6 MILLION. The point is it needs the words "MILLION" to come after the number.
I know there is nsNumberFormatter that can convert strings into numbers and vice versa but can it do what I need?
If anyone has any ideas or suggestions, it would be much appreciated.
Thank you!
So as I see it, you have two problems:
You have a string representation of something that's actually a number
You (potentially) have a number that you want formatted as a string
So, problem #1:
To convert a string into a number, you use an NSNumberFormatter. You've got a pretty simple case:
NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
[f setNumberStyle:NSNumberFormatterCurrencyStyle];
NSNumber *n = [f numberFromString:#"$12,000,000"];
// n is 12000000
That was easy! Now problem #2:
This is trickier, because you want a mixed spell-out style. You could consider using an NSNumberFormatter again, but it's not quite right:
[f setNumberStyle:NSNumberFormatterSpellOutStyle];
NSString *s = [f stringFromNumber:n];
// s is "twelve million"
So, we're closer. At this point, you could perhaps maybe do something like:
NSInteger numberOfMillions = [n integerValue] / 1000000;
if (numberOfMillions > 0) {
NSNumber *millions = [NSNumber numberWithInteger:numberOfMillions];
NSString *numberOfMillionsString = [f stringFromNumber:millions]; // "twelve"
[f setNumberStyle:NSNumberFormatterCurrencyStyle];
NSString *formattedMillions = [f stringFromNumber:millions]; // "$12.00"
if ([s hasPrefix:numberOfMillionsString]) {
// replace "twelve" with "$12.00"
s = [s stringByReplacingCharactersInRange:NSMakeRange(0, [numberOfMillionsString length]) withString:formattedMillions];
// if this all works, s should be "$12.00 million"
// you can use the -setMaximumFractionDigits: method on NSNumberFormatter to fiddle with the ".00" bit
}
}
However
I don't know how well this would work in anything other than english. CAVEAT IMPLEMENTOR
Worst case scenario, you could implement a category on NSString to implement the behaviour you want.
In the method that you would do in that category you could take an NSNumberFormatter to bring that string to a number and by doing some modulo operation you could define if you need the word Million, or Billion, etc. and put back a string with the modulo for Million or other way you need it to be.
That way you could just call that method on your NSString like this :
NSString *humanReadable = [realEstateWorth myCustomMethodFromMyCategory];
And also.
NSString are immutable, so you can't change it unless you assign a new one to your variable.
I'd recommend storing this value as an NSNumber or a float. Then you could have a method to generate an NSString to display it like:
- (NSString*)numberToCurrencyString:(float)num
{
NSString *postfix = #"";
if (num > 1000000000)
{
num = num / 1000000000;
postfix = #" Billion";
}
else if (num > 1000000)
{
num = num / 1000000;
postfix = #" Million";
}
NSString *currencyString = [NSString stringWithFormat:#"%.0f%#", num, postfix];
return currencyString;
}
Note: Your question states that your input needs to remain a string. That's fine. So you'd need to 1.) first parse the number out of the string and 2.) then reconvert it to a string from a number. I've shown how to do step 2 of this process.
I'm not sure what's going on here. For some reason NSString seems unwilling to load in the double value d.
Here's my code:
-(NSString*)minuteFormat:(double) d{
NSString* mystring;
if(d >= 10){
mystring = [NSString stringWithFormat:#"%d", d];
}else{
mystring = [NSString stringWithFormat:#"0%d",d];
}
return(mystring);
}
Regardless of the value of d, the only thing that's getting returned is 0 or 00. (And I'm sure d is getting inputted correctly, as I've used breakpoints to check.)
Could someone tell me what's going on?
%d is for integers. You probably want %f. See String Format Specifiers
%d is a decimal integer format string. Try %lf for double.
You should use %f (double) like this
mystring = [NSString stringWithFormat:#"%f", f];
or do a typecast of the value. For this you may use [yourString floatValue]
I have the value 25.00 in a float, but when I print it on screen it is 25.0000000.
How can I display the value with only two decimal places?
It is not a matter of how the number is stored, it is a matter of how you are displaying it. When converting it to a string you must round to the desired precision, which in your case is two decimal places.
E.g.:
NSString* formattedNumber = [NSString stringWithFormat:#"%.02f", myFloat];
%.02f tells the formatter that you will be formatting a float (%f) and, that should be rounded to two places, and should be padded with 0s.
E.g.:
%f = 25.000000
%.f = 25
%.02f = 25.00
Here are few corrections-
//for 3145.559706
Swift 3
let num: CGFloat = 3145.559706
print(String(format: "%f", num)) = 3145.559706
print(String(format: "%.f", num)) = 3145
print(String(format: "%.1f", num)) = 3145.6
print(String(format: "%.2f", num)) = 3145.56
print(String(format: "%.02f", num)) = 3145.56 // which is equal to #"%.2f"
print(String(format: "%.3f", num)) = 3145.560
print(String(format: "%.03f", num)) = 3145.560 // which is equal to #"%.3f"
Obj-C
#"%f" = 3145.559706
#"%.f" = 3146
#"%.1f" = 3145.6
#"%.2f" = 3145.56
#"%.02f" = 3145.56 // which is equal to #"%.2f"
#"%.3f" = 3145.560
#"%.03f" = 3145.560 // which is equal to #"%.3f"
and so on...
You can also try using NSNumberFormatter:
NSNumberFormatter* nf = [[[NSNumberFormatter alloc] init] autorelease];
nf.positiveFormat = #"0.##";
NSString* s = [nf stringFromNumber: [NSNumber numberWithFloat: myFloat]];
You may need to also set the negative format, but I think it's smart enough to figure it out.
I made a swift extension based on above answers
extension Float {
func round(decimalPlace:Int)->Float{
let format = NSString(format: "%%.%if", decimalPlace)
let string = NSString(format: format, self)
return Float(atof(string.UTF8String))
}
}
usage:
let floatOne:Float = 3.1415926
let floatTwo:Float = 3.1425934
print(floatOne.round(2) == floatTwo.round(2))
// should be true
In Swift Language, if you want to show you need to use it in this way. To assign double value in UITextView, for example:
let result = 23.954893
resultTextView.text = NSString(format:"%.2f", result)
If you want to show in LOG like as objective-c does using NSLog(), then in Swift Language you can do this way:
println(NSString(format:"%.2f", result))
IN objective-c, if you are dealing with regular char arrays (instead of pointers to NSString) you could also use:
printf("%.02f", your_float_var);
OTOH, if what you want is to store that value on a char array you could use:
sprintf(your_char_ptr, "%.02f", your_float_var);
The problem with all the answers is that multiplying and then dividing results in precision issues because you used division. I learned this long ago from programming on a PDP8.
The way to resolve this is:
return roundf(number * 100) * .01;
Thus 15.6578 returns just 15.66 and not 15.6578999 or something unintended like that.
What level of precision you want is up to you. Just don't divide the product, multiply it by the decimal equivalent.
No funny String conversion required.
in objective -c is u want to display float value in 2 decimal number then pass argument indicating how many decimal points u want to display
e.g 0.02f will print 25.00
0.002f will print 25.000
Here's some methods to format dynamically according to a precision:
+ (NSNumber *)numberFromString:(NSString *)string
{
if (string.length) {
NSNumberFormatter * f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
return [f numberFromString:string];
} else {
return nil;
}
}
+ (NSString *)stringByFormattingString:(NSString *)string toPrecision:(NSInteger)precision
{
NSNumber *numberValue = [self numberFromString:string];
if (numberValue) {
NSString *formatString = [NSString stringWithFormat:#"%%.%ldf", (long)precision];
return [NSString stringWithFormat:formatString, numberValue.floatValue];
} else {
/* return original string */
return string;
}
}
e.g.
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:4];
=> 2.3453
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:0];
=> 2
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:2];
=> 2.35 (round up)
Another method for Swift (without using NSString):
let percentage = 33.3333
let text = String.localizedStringWithFormat("%.02f %#", percentage, "%")
P.S. this solution is not working with CGFloat type only tested with Float & Double
Use NSNumberFormatter with maximumFractionDigits as below:
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
formatter.maximumFractionDigits = 2;
NSLog(#"%#", [formatter stringFromNumber:[NSNumber numberWithFloat:12.345]]);
And you will get 12.35
If you need to float value as well:
NSString* formattedNumber = [NSString stringWithFormat:#"%.02f", myFloat];
float floatTwoDecimalDigits = atof([formattedNumber UTF8String]);
lblMeter.text=[NSString stringWithFormat:#"%.02f",[[dic objectForKey:#"distance"] floatValue]];
I want to convert a string into a double and after doing some math on it, convert it back to a string.
How do I do this in Objective-C?
Is there a way to round a double to the nearest integer too?
You can convert an NSString into a double with
double myDouble = [myString doubleValue];
Rounding to the nearest int can then be done as
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
I'm honestly not sure if there's a more streamlined way to convert back into a string than
NSString* myNewString = [NSString stringWithFormat:#"%d", myInt];
To really convert from a string to a number properly, you need to use an instance of NSNumberFormatter configured for the locale from which you're reading the string.
Different locales will format numbers differently. For example, in some parts of the world, COMMA is used as a decimal separator while in others it is PERIOD — and the thousands separator (when used) is reversed. Except when it's a space. Or not present at all.
It really depends on the provenance of the input. The safest thing to do is configure an NSNumberFormatter for the way your input is formatted and use -[NSFormatter numberFromString:] to get an NSNumber from it. If you want to handle conversion errors, you can use -[NSFormatter getObjectValue:forString:range:error:] instead.
Adding to olliej's answer, you can convert from an int back to a string with NSNumber's stringValue:
[[NSNumber numberWithInt:myInt] stringValue]
stringValue on an NSNumber invokes descriptionWithLocale:nil, giving you a localized string representation of value. I'm not sure if [NSString stringWithFormat:#"%d",myInt] will give you a properly localized reprsentation of myInt.
Here's a working sample of NSNumberFormatter reading localized number String (xCode 3.2.4, osX 10.6), to save others the hours I've just spent messing around. Beware: while it can handle trailing blanks such as "8,765.4 ", this cannot handle leading white space and this cannot handle stray text characters. (Bad input strings: " 8" and "8q" and "8 q".)
NSString *tempStr = #"8,765.4";
// localization allows other thousands separators, also.
NSNumberFormatter * myNumFormatter = [[NSNumberFormatter alloc] init];
[myNumFormatter setLocale:[NSLocale currentLocale]]; // happen by default?
[myNumFormatter setFormatterBehavior:NSNumberFormatterBehavior10_4];
// next line is very important!
[myNumFormatter setNumberStyle:NSNumberFormatterDecimalStyle]; // crucial
NSNumber *tempNum = [myNumFormatter numberFromString:tempStr];
NSLog(#"string '%#' gives NSNumber '%#' with intValue '%i'",
tempStr, tempNum, [tempNum intValue]);
[myNumFormatter release]; // good citizen
olliej's rounding method is wrong for negative numbers
2.4 rounded is 2 (olliej's method gets this right)
−2.4 rounded is −2 (olliej's method returns -1)
Here's an alternative
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
You could of course use a rounding function from math.h
// Converting String in to Double
double doubleValue = [yourString doubleValue];
// Converting Double in to String
NSString *yourString = [NSString stringWithFormat:#"%.20f", doubleValue];
// .20f takes the value up to 20 position after decimal
// Converting double to int
int intValue = (int) doubleValue;
or
int intValue = [yourString intValue];
For conversion from a number to a string, how about using the new literals syntax (XCode >= 4.4), its a little more compact.
int myInt = (int)round( [#"1.6" floatValue] );
NSString* myString = [#(myInt) description];
(Boxes it up as a NSNumber and converts to a string using the NSObjects' description method)
For rounding, you should probably use the C functions defined in math.h.
int roundedX = round(x);
Hold down Option and double click on round in Xcode and it will show you the man page with various functions for rounding different types.
This is the easiest way I know of:
float myFloat = 5.3;
NSInteger myInt = (NSInteger)myFloat;
from this example here, you can see the the conversions both ways:
NSString *str=#"5678901234567890";
long long verylong;
NSRange range;
range.length = 15;
range.location = 0;
[[NSScanner scannerWithString:[str substringWithRange:range]] scanLongLong:&verylong];
NSLog(#"long long value %lld",verylong);
convert text entered in textfield to integer
double mydouble=[_myTextfield.text doubleValue];
rounding to the nearest double
mydouble=(round(mydouble));
rounding to the nearest int(considering only positive values)
int myint=(int)(mydouble);
converting from double to string
myLabel.text=[NSString stringWithFormat:#"%f",mydouble];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
converting from int to string
myLabel.text=[NSString stringWithFormat:#"%d",myint];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
I ended up using this handy macro:
#define STRING(value) [#(value) stringValue]