I have this method to calculate the standard deviation of an array of NSNumber integers, given a mean. The calculation uses NSDecimals to retain the highest resolution. This is currently demanding many cpu cycles, any help to speed it up while retaining the resolution required is appreciated! Thank you.
-(NSDecimal)standardDeviationOf:(NSMutableArray *)array withMean:(NSDecimal)mean {
if (![array count]) return CPTDecimalFromInt(0);
NSDecimal sumOfSquaredDifferences = CPTDecimalFromInt(0);
for (NSNumber *number in array) {
NSDecimal valueOfNumber = CPTDecimalFromInt([number intValue]);
NSDecimal difference = CPTDecimalSubtract(valueOfNumber, mean);
sumOfSquaredDifferences = CPTDecimalAdd(sumOfSquaredDifferences, CPTDecimalMultiply(difference, difference));
}
return CPTDecimalFromDouble(
sqrt(
CPTDecimalDoubleValue(sumOfSquaredDifferences) / [[NSNumber numberWithInt:[array count]] doubleValue]
)
);
}
An NSDecimal has 38 digits of precision, whereas double has roughly 16 digits of precision. But at the end of your loop, when you convert sumOfSquaredDifferences to double for the sqrt function, all the extra precision you had in the NSDecimal is "lost". You might as well perform the arithmetic of your inner loop using double, which should be much faster than NSDecimal:
double sumOfSquaredDifferences = 0;
double valueOfMean = [mean doubleValue];
for (NSNumber *number in array) {
double valueOfNumber = [number intValue];
double difference = valueOfNumber - valueOfMean;
sumOfSquaredDifferences += difference * difference;
}
return CPTDecimalFromDouble(sqrt(sumOfSquaredDifferences /
double([array count])));
Related
I want to round a double to one decimal place in Objective-C.
In Swift I can do it with an extension:
public extension Double {
/// Rounds the double to decimal places value
func rounded(toPlaces places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
However, apparently you cannot call extensions on primitives from Objective-C so I can't use the extension.
I would be happy to do the rounding either on the double directly or as a string, however, neither of the following is working:
double mydub = 122.12022222223322;
NSString *axtstr = [NSString stringWithFormat:#"%2f", mydub]; //gives 122.120222
double rounded = (round(mydub*10)) / 10.0; //gives 122.100000
How do I convert 122.12022222223322; into 122.1?
You need to put a decimal between the % and 2f
[NSString stringWithFormat:#"%.2f", mydub];
double mydouble = 122.12022222223322;
NSString *str = [NSString stringWithFormat:#"%.2f", mydouble];
// = #"122.12"
.. will not round mydouble. Instead it will only apply format to the output as string.
double d = 122.49062222223322;
NSString *dStr = [NSString stringWithFormat:#"%.f %.1f %.2f %.3f", d, d, d, d];
// = #"122 122.5 122.49 122.491"
As Objective-C shares the language rules from C you can round safely with
#include <math.h>
double rounded = round(mydouble);
// = 122.000000
of course you can shift comma with multiplication and dividing the power of ten you want.
double commashifted = round(mydouble*100.0)/100.0;
// = 122.120000;
If you are really into Objective-C Classes to do same in deluxe have a look into 'NSDecimal.h' in the Foundation Framework.
Last but not least you can do the same with C as you did with swift.
double roundbycomma(int commata, double zahl) {
double divisor = pow(10.0, commata);
return round(zahl * divisor) / divisor;
}
I have two NSInteger variables called "domande" and "corrette". I have to execute this operation with them: corrette*10/domande. I want the result to be a float variable, so I declared a "voto" variable as so: "float voto = corrette*10/domande;" . When I output the value of "voto" with NSLog I get an approximated value of the result followed by ".000000".
Here's the code:
NSInteger domande = [numDomande integerValue];
NSInteger corrette = [numRisposteCorrette integerValue];
float voto = corrette*10/domande;
NSLog(#"float value is: %f", voto);
When I assign to "domande" a value of 7, and to "corrette" a value of 4: voto=5.000000
Instead it should be voto=5.71...
How can I have the division return not an integer type converted to float, but directly a float type?
Simplest way is to do:
float voto = 10.0f * corrette / domande;
By making the first argument a float, you guarantee that the others will be promoted as well and that intermediate and final results will not suffer truncation.
You could achieve a similar result by casting corrette to a float but I tend to prefer simplicity where possible.
Rather than converting integers to floats, you could just get floats in the first place:
CGFloat domandeFloat = [numDomande floatValue];
CGFloat corretteFloat = [numRisposteCorrette floatValue];
CGFloat voto = (corretteFloat / domandeFloat) * 10.0f
NSLog(#"float value is: %f", voto);
NSInteger does not have a method called floatValue. NSInteger is just an int. Instead, the solution would be:
CGFloat domandeFloat = [[NSNumber numberWithInt: numDomande] floatValue];
CGFloat domandeFloat = [[NSNumber numberWithInt: numRisposteCorrette] floatValue];
CGFloat voto = (corretteFloat / domandeFloat) * 10.0f;
Try to convert the NSIntegers to a float type first:
float voto = (float)corrette*10/(float)domande;
you can cast "10" from int to float by writing it as "10.0"
float voto = corrette*10.0/domande;
or
float voto = ((float)corrette*10) / (float)domande;
Operation "/" is returning type of it operands - 5/4 will return int result 1 because 5 and 4 are int, and 5.0/4.0 will return 1.25, because 5.0 and 4.0 are interpreted as float values. So you should manually cast type of input variables corrette and domande to float
I want to convert a decimal NSNumber to an int or other form which I can do math with. Here's the annotated code for my project:
NSNumber *Left = [left valueForOutputKey:#"Y"];
This line gets a Quartz Composer Outlet, usually with a value around 0.512.
Basically, I want to multiply this by 10, and then do some operations like greater than and less than to see which range it is in.
Since it looks like you're dealing with a fractional component, you want to convert it to a float or a double to perform your operations, depending on how big you expect that value to be. A float should be fine unless you're dealing with ridiculously large or precise numbers. Here's how it would look, for example:
float lValue = [[left valueForOutputKey:#"Y"] floatValue];
lValue *= 10;
if (lValue < 10) {
// do whatever
}
else if (lValue > 50) {
// do whatever
}
Then to store the value back in your outlet or whatever, you pack it back into a NSNumber:
NSNumber *newValue = [NSNumber numberWithFloat:lValue];
[left setValue:newValue forKey:#"Y"];
You may have to convert newValue into a string to display it in a control, just use [newValue stringValue] to do that.
Use one of the methods of NSNumber:
int leftInt = [Left intValue];
float leftFloat = [Left floatValue];
double leftDouble = [Left doubleValue];
Here is my situation. Its driving me nuts:
I have an NSMutableArray with a count value of 517. I have a double value that is my multiplier.
double multiplier = 0.1223;
double result = [myArray count] * multiplier; // 63 even (wrong!)
In fact it should be 63.2291. If I go:
double result = [myArray count] * 0.1223; // 63.2291 (right!)
or..
double result = 517 * multiplier; // 63.2291 (right!)
Does this make any sense to anyone?
Addendum:
here is my actual function:
- (double) getValueForPercentage:(double)percVal
{
int adjustedCount = [originalData count] - 1;
double final = percVal * (double)adjustedCount;
return final;
}
I never get any digits beyond the decimal point when I do this. It does however work if I get rid of the "-1", a-la:
- (double) getValueForPercentage:(double)percVal
{
int adjustedCount = [originalData count];
double final = percVal * (double)adjustedCount;
return final;
}
Of course, I need to have the -1.
Second addendum:
Another interesting thing I noted was, if I pass a hard-coded number to this function it works fine, but if I pass the double value that I need to use, it fails:
int pointCount = [srcData getDayCount];
for (int i = 0; i < pointCount; i++) {
double progress = (double)i/(double)(pointCount - 1);
double satv = [srcData getValueForPercentage:progress];
// satv is always a number without any digits beyond the decimal
}
Well, when I started to have these issues i looked around a bit and found no reason or explanation.
What I do now is make everything become an NSNumber and then call doubleValue on it. This should yield the results you're looking for:
NSNumber * pointCount = [NSNumber numberWithUnsignedInt: [srcData getDayCount]];
for (NSInteger i = 0; i < [pointCount intValue]; i++) {
NSNumber * count = [ NSNumber numberWithInt: i ];
double progress = [count doubleValue]/[pointCount doubleValue] - 1.0;
double satv = [srcData getValueForPercentage:progress];
// satv is always a number without any digits beyond the decimal
}
Hope it helps.
I am confused by NSDecimalNumber and its "behaviors". I have an NSDecimalNumber that represents a dollar value, say $37.50. I'd like to find out how many times say 5.0 goes into that number and then know what's left over. I can get the straight division and get 7.50 but I want 7 mod 2.50. I could convert to an integer but need to save the "cents" so wondering if there's some tricks in the framework?
Using Peter Hoseys example, but with iOS code:
NSDecimalNumber *dividend = [NSDecimalNumber decimalNumberWithDecimal:[[NSNumber numberWithDouble:37.5] decimalValue]];
NSDecimalNumber *divisor = [NSDecimalNumber decimalNumberWithDecimal:[[NSNumber numberWithDouble:5.0] decimalValue]];
NSDecimalNumber *quotient = [dividend decimalNumberByDividingBy:divisor withBehavior:[NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundDown scale:0 raiseOnExactness:NO raiseOnOverflow:NO raiseOnUnderflow:NO raiseOnDivideByZero:NO]];
NSDecimalNumber *subtractAmount = [quotient decimalNumberByMultiplyingBy:divisor];
NSDecimalNumber *remainder = [dividend decimalNumberBySubtracting:subtractAmount];
Divide the dividend by the divisor, rounding down. This gets you the quotient.
Multiply the divisor by the quotient.
Subtract that from the dividend. This gets you the remainder.
(37.50 // 5) == 7; 7 * 5 == 35; 37.50 - 35 = 2.50.
(Note: // is Python's operator for integral division. I've borrowed it here. Obviously, you should not actually attempt to use // for division in Objective-C code.)
Here's a NSDecimalNumber category which also works with negative numbers and negative divisors:
- (BOOL)isNegative
{
return (NSOrderedDescending == [[NSDecimalNumber zero] compare:self]);
}
- (NSDecimalNumber *)invertedNumber
{
NSDecimalNumber *negOne = [NSDecimalNumber decimalNumberWithMantissa:1 exponent:0 isNegative:YES];
return [self decimalNumberByMultiplyingBy:negOne];
}
- (NSDecimalNumber *)moduloFor:(NSDecimalNumber *)divisor
{
NSRoundingMode roundingMode = ([self isNegative] ^ [divisor isNegative]) ? NSRoundUp : NSRoundDown;
NSDecimalNumberHandler *rounding = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:roundingMode
scale:0
raiseOnExactness:NO
raiseOnOverflow:NO
raiseOnUnderflow:NO
raiseOnDivideByZero:NO];
// divide and get the remainder
NSDecimalNumber *quotient = [self decimalNumberByDividingBy:divisor withBehavior:rounding];
NSDecimalNumber *subtract = [quotient decimalNumberByMultiplyingBy:divisor];
NSDecimalNumber *modulo = [self decimalNumberBySubtracting:subtract];
if ([divisor isNegative]) {
return [modulo invertedNumber];
}
return modulo;
}
Same solution in Swift 4:
let dividend = NSDecimalNumber(string: Price)
let divisor = NSDecimalNumber(decimal: 0.05)
let quotient = dividend.dividing(by: divisor, withBehavior: NSDecimalNumberHandler(roundingMode: .down, scale: 0, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: false))
let subtractAmount = quotient?.multiplying(by: divisor)
let remainder = dividend.subtracting(anAmount)
if remainder != 0
{ ...
You could theoretically do some tricks with the NSDecimalNumber methods decimalNumberByDividingBy:, decimalNumberByMultiplyingBy:, and decimalValue. Divide the numbers, grab the decimal number structure, figure out the remainder from the structure, then create a new decimal number with that remainder and multiply that remainder by the original number.
The easier way to do it, however, would probably be to figure out the maximum precision you want (call it N places after the decimal point), multiply the number by 10eN, then just grab the integer value and do your division and mod on that. You run the risk of losing data, especially with very large numbers, however, so check the biggest number you want and figure out what data type - if any - will work for you. NSNumber does support unsignedLongLongValue.