I am taking masses (as g, mg, µg, ng & kg) and volumes (as ml, µl & l) as input to a chemistry app.
Currently I convert all masses to grams and volumes to litres, save in core data and perform any calculations as doubles.
Finally results are then converted back into a meaningful unit
ie 0.000034litres is more useful expressed as 34µl for my customers
What is best practice for working between different units?
There may be a few libraries out there but what you are doing is specific so I doubt there is a specific "best practice".
You may want to investigate the properties of NSDouble, CGFloat however since they might be more suited across devices and give you more options them the primitive double.
There are no unit data types or many built in native converter functions. A number is a number its up to the programmer to give that number meaning in the app's context.
I found a suitable engineering formatter at https://github.com/dhoerl/EngineeringNotationFormatter
In addition I created a simple version:
-(NSString *)engineeringFormat:(double)value digits:(int)digits {
//calculate exponent in step of 3
int perMill = trunc(log10(value)/3);
if (value<1) {
perMill -= 1;
}
//calculate mantissa format range of 1 to 1000
double corrected = value;
while (corrected<1) {
corrected = corrected*1000;
}
while (corrected>=1000) {
corrected=corrected/1000;
}
//format number of significant digits
NSNumberFormatter *numberFormatDigits = [[NSNumberFormatter alloc] init];
numberFormatDigits.usesSignificantDigits = YES;
numberFormatDigits.maximumSignificantDigits = digits;
NSString *mantissa = [numberFormatDigits stringFromNumber:[NSNumber numberWithDouble:corrected]];
//select engineering notation prefix
NSArray *engSuffix = #[#"T",#"G",#"M",#"k",#"",#"m",#"µ",#"n",#"p"];
int index = 4 - perMill;
NSString *result;
if ((index > engSuffix.count-1) || (index<0)) {
result = #"Out of range";
} else {
result = [NSString stringWithFormat:#"%# %#",mantissa,[engSuffix objectAtIndex:index]];
}
return result;
}
Related
I need to convert the results of calculations performed in a double, but I cannot use decimalNumberByMultiplyingBy or any other NSDecimalNumber function. I've tried to get an accurate result in the following ways:
double calc1 = 23.5 * 45.6 * 52.7; // <-- Correct answer is 56473.32
NSLog(#"calc1 = %.20f", calc1);
-> calc1 = 56473.32000000000698491931
NSDecimalNumber *calcDN = (NSDecimalNumber *)[NSDecimalNumber numberWithDouble:calc1];
NSLog(#"calcDN = %#", [calcDN stringValue]);
-> calcDN = 56473.32000000001024
NSDecimalNumber *testDN = [[[NSDecimalNumber decimalNumberWithString:#"23.5"] decimalNumberByMultiplyingBy:[NSDecimalNumber decimalNumberWithString:#"45.6"]] decimalNumberByMultiplyingBy:[NSDecimalNumber decimalNumberWithString:#"52.7"]];
NSLog(#"testDN = %#", [testDN stringValue]);
-> testDN = 56473.32
I understand that this difference is related to the respective accuracies.
But here's my question: How can I round this number in the most accurate way possible regardless of what the initial value of double may be? And if a more accurate method exists to do the initial calculation, what is that method?
Well, you can either use double to represent the numbers and embrace inaccuracies or use some different number representation, such as NSDecimalNumber. It all depends on what are the expected values and business requirements concerning accuracy.
If it is really crucial not to use arithmetic methods provided by NSDecimalNumber, than the rounding behaviour is best controlled using NSDecimalNumberHandler, which is a concrete implementation of NSDecimalNumberBehaviors protocol. The actual rounding is performed using decimalNumberByRoundingAccordingToBehavior: method.
Here comes the snippet - it's in Swift, but it should be readable:
let behavior = NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundPlain,
scale: 2,
raiseOnExactness: false,
raiseOnOverflow: false,
raiseOnUnderflow: false,
raiseOnDivideByZero: false)
let calcDN : NSDecimalNumber = NSDecimalNumber(double: calc1)
.decimalNumberByRoundingAccordingToBehavior(behavior)
calcDN.stringValue // "56473.32"
I do not know of any method of improving the accuracy of the actual computations when using double representation.
I'd recommend rounding the number based on the number of digits in your double so that the NSDecimalNumber is truncated to only show the appropriate number of digits, thus eliminating the digits formed by potential error, ex:
// Get the number of decimal digits in the double
int digits = [self countDigits:calc1];
// Round based on the number of decimal digits in the double
NSDecimalNumberHandler *behavior = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundDown scale:digits raiseOnExactness:NO raiseOnOverflow:NO raiseOnUnderflow:NO raiseOnDivideByZero:NO];
NSDecimalNumber *calcDN = (NSDecimalNumber *)[NSDecimalNumber numberWithDouble:calc1];
calcDN = [calcDN decimalNumberByRoundingAccordingToBehavior:behavior];
I've adapted the countDigits: method from this answer:
- (int)countDigits:(double)num {
int rv = 0;
const double insignificantDigit = 18; // <-- since you want 18 significant digits
double intpart, fracpart;
fracpart = modf(num, &intpart); // <-- Breaks num into an integral and a fractional part.
// While the fractional part is greater than 0.0000001f,
// multiply it by 10 and count each iteration
while ((fabs(fracpart) > 0.0000001f) && (rv < insignificantDigit)) {
num *= 10;
fracpart = modf(num, &intpart);
rv++;
}
return rv;
}
I have IDs in JSON file and some of them are really big but they fit inside bounds of unsigned long long int.
"id":9223372036854775807,
How to get this large number from JSON using objectForKey:idKey of NSDictionary?
Can I use NSDecimalNumber? Some of this IDs fit into regular integer.
Tricky. Apple's JSON code converts integers above 10^18 to NSDecimalNumber, and smaller integers to plain NSNumber containing a 64 bit integer value. Now you might have hoped that unsignedLongLongValue would give you a 64 bit value, but it doesn't for NSDecimalNumber: The NSDecimalNumber first gets converted to double, and the result to unsigned long long, so you lose precision.
Here's something that you can add as an extension to NSNumber. It's a bit tricky, because if you get a value very close to 2^64, converting it to double might get rounded to 2^64, which cannot be converted to 64 bit. So we need to divide by 10 first to make sure the result isn't too big.
- (uint64_t)unsigned64bitValue
{
if ([self isKindOfClass:[NSDecimalNumber class]])
{
NSDecimalNumber* asDecimal = (NSDecimalNumber *) self;
uint64_t tmp = (uint64_t) (asDecimal.doubleValue / 10.0);
NSDecimalNumber* tmp1 = [[NSDecimalNumber alloc] initWithUnsignedLongLong:tmp];
NSDecimalNumber* tmp2 = [tmp1 decimalNumberByMultiplyingByPowerOf10: 1];
NSDecimalNumber* remainder = [asDecimal decimalNumberBySubtracting:tmp2];
return (tmp * 10) + remainder.unsignedLongLongValue;
}
else
{
return self.unsignedLongLongValue;
}
}
Or process the raw JSON string, look for '"id" = number; '. With often included white space, you can find the number, then over write it with the number quoted. You can put the data into a mutable data object and get a char pointer to it, to overwrite.
[entered using iPhone so a bit terse]
I am looking to write my own power function to work with NSDecimalNumbers and exponents that are not whole numbers. I first tried to use a combination of newtons method and the built in integer power method, but due to newtons method i am getting overflow errors when I have exponents with more than 2 decimals. So I thought maybe the float value pow function might serve as a good model for my own function. So I was wondering if anyone knows where I can fond some sort of documentation on the inner workings of the pow function?
Edit:
#wombat57, those links look like they could be what I am looking for however I have no idea to read them. The algorithm you suggest is in fact what I am using. the overflow comes from newtons method due to very large exponents. Because I am getting exponents in decimal form I have to convert it to a fraction first. the only way of ding this in code, as far as I know, multiplying the decimal by ten until you have a whole number, and using that as the numerator. Doing this you get exponents of 100+ for numbers with 3 or more decimals. this causes an overflow error.
EDIT 1: Here are links to the actual source
http://opensource.apple.com/source/Libm/Libm-2026/Source/Intel/expf_logf_powf.c
http://opensource.apple.com/source/Libm/Libm-315/Source/ARM/powf.c
I got the links from this question, which has a bunch of relevant discussion
self made pow() c++
This page describes an algorithm: Link.
x^(1/n) = the nth root of x, and x^mn = (x^m)^n. Thus, x^(m/n) = (the nth root of x)^m. Arbitrary roots can be calculated with Newton's method. Integer powers can be calculated with Exponentiation by squaring. For irrational exponents, you can use increasingly accurate rational approximations until you get the desired number of significant digits.
EDIT 2:
Newton's method involves raising your current guess to the power of the root that you're trying to find. If that power is large, and the guess is even a little too high, this can result in overflow. One solution here is to identify this case. If overflow ever occurs, this means that the guess was too high. You can solve the problem by (whenever a guess results in overflow), setting the current guess to a value between the last guess that did not overflow and the current guess (you may have to do this several times). That is, whenever Newton's method overflows, do a binary search down toward the last guess that did not overflow. Here's some python that implements all of this:
def nroot(n, b, sig_figs = 10):
g1 = 1.0
g2 = 1.0
while True:
done = False
while not done:
try:
g3 = g2 - ((g2**b) - n) / (b * (g2**(b-1)))
done = True
except OverflowError:
g2 = (g1 + g2) / 2.0
if abs(g2 - g3) < 1.0 / (10**sig_figs):
return g3
g1 = g2
g2 = g3
def npowbysqr(n, p):
if p == 0:
return 1.0
if p % 2 == 0:
v = npowbysqr(n, p/2)
return v*v
else:
return n*npowbysqr(n, p-1)
def npow(n, p):
return npowbysqr(nroot(n, 1000000), int(p*1000000))
print npow(5, 4.3467)
print 5**4.3467
I should add that there are probably much better solutions. This does seem to work, however
I happened to need something like this a while ago. Thankfully, Dave DeLong had been tinkering with this in his DDMathParser, so I built off of that. He yanked his implementation from his code in this commit, but I took that and modified it. This is my version of his NSDecimal power function:
extern NSDecimal DDDecimalPower(NSDecimal d, NSDecimal power) {
NSDecimal r = DDDecimalOne();
NSDecimal zero = DDDecimalZero();
NSComparisonResult compareToZero = NSDecimalCompare(&zero, &power);
if (compareToZero == NSOrderedSame) {
return r;
}
if (DDDecimalIsInteger(power))
{
if (compareToZero == NSOrderedAscending)
{
// we can only use the NSDecimal function for positive integers
NSUInteger p = DDUIntegerFromDecimal(power);
NSDecimalPower(&r, &d, p, NSRoundBankers);
}
else
{
// For negative integers, we can take the inverse of the positive root
NSUInteger p = DDUIntegerFromDecimal(power);
p = -p;
NSDecimalPower(&r, &d, p, NSRoundBankers);
r = DDDecimalInverse(r);
}
} else {
// Check whether this is the inverse of an integer
NSDecimal inversePower = DDDecimalInverse(power);
NSDecimalRound(&inversePower, &inversePower, 34, NSRoundBankers); // Round to 34 digits to deal with cases like 1/3
if (DDDecimalIsInteger(inversePower))
{
r = DDDecimalNthRoot(d, inversePower);
}
else
{
double base = DDDoubleFromDecimal(d);
double p = DDDoubleFromDecimal(power);
double result = pow(base, p);
r = DDDecimalFromDouble(result);
}
}
return r;
}
It tries to identify common cases and use more precise calculations for those. It does fall back on pow() for things that don't fit in these cases, though.
The rest of the NSDecimal functions I use can be found here and here.
I have come up with a function that suits my needs and will hopefully suit the needs of many others. the following method is fully annotated and works for any power function that has a real value. This method also only uses NSDecimalNumbers meaning you will not loose any precision due to float rounding error. This method takes two arguments one for the base and one for the power, and both are NSDecimalNumbers. So here it is:
//these are constants that will be used
NSDecimalNumber *ten = [NSDecimalNumber decimalNumberWithString:#"10"];
NSDecimalNumber *one = NSDecimalNumber.one;
//these will together hold the power in fractional form
NSDecimalNumber *numerator = power, *denominator = one;
//this will hold the final answer and all previous guesses the first guess is set to be the base
NSDecimalNumber *powAns = base;
//this will hold the change in your guess, also serves as an idea of how large the error is
NSDecimalNumber *error = one;
//part1 holds f(x) and part2 holds f'(x)
NSDecimalNumber *part1, *part2;
//if the base is < 0 and the power is not whole, answer is not real
if ([base doubleValue] < 0 && [[power stringValue] rangeOfString:#"."].location != NSNotFound)
return NSDecimalNumber.notANumber;
//converts power to a fractional value
while ([[numerator stringValue] rangeOfString:#"."].location != NSNotFound) {
numerator = [numerator decimalNumberByMultiplyingBy:ten];
denominator = [denominator decimalNumberByMultiplyingBy:ten];
}
//conditions here are the precision you wish to get
while ([error compare:[NSDecimalNumber decimalNumberWithString:#"1e-20"]] == NSOrderedDescending ||
[error compare:[NSDecimalNumber decimalNumberWithString:#"-1e-20"]] == NSOrderedAscending) {
//if this catches an overflow error it is set to be a very large number otherwise the value cannot be a number, however no other error should be returned.
#try {
part1 = [powAns decimalNumberByRaisingToPower:[denominator intValue]];
}
#catch (NSException *exception) {
if ([exception.name isEqual: NSDecimalNumberOverflowException])
part1 = [NSDecimalNumber decimalNumberWithString:#"10e127"];
else
return NSDecimalNumber.notANumber;
}
part1 = [part1 decimalNumberBySubtracting:base];
//if this catches an overflow error it is set to be a very large number otherwise the value cannot be a number, however no other error should be returned.
#try {
part2 = [powAns decimalNumberByRaisingToPower:[denominator intValue]-1];
part2 = [part2 decimalNumberByMultiplyingBy:denominator];
}
#catch (NSException *exception) {
if ([exception.name isEqual: NSDecimalNumberOverflowException])
part2 = [NSDecimalNumber decimalNumberWithString:#"10e127"];
else
return NSDecimalNumber.notANumber;
}
//error is the change in the estimated value or y - f(x)/f'(x)
error = [part1 decimalNumberByDividingBy:part2];
powAns = [powAns decimalNumberBySubtracting: error];
}
//if the numerator value is negative it must be made positive and the answer is then inverted
if ([numerator intValue] < 0) {
powAns = [powAns decimalNumberByRaisingToPower:abs([numerator intValue])];
powAns = [one decimalNumberByDividingBy:powAns];
}
else
powAns = [powAns decimalNumberByRaisingToPower:[numerator intValue]];
return powAns;
If anyone has any questions about my code I am happy to answer them.
I'm writing a program to compute the conversion of number system. My goal is to let the program compute very large number at a short time (<0.1s) and I made a NSString category to let me compute very large number (+ - * / %), a function to convert any base number to decimal, I think it's fast enough.
The problem is here: convert decimal to different bases. It is never fast enough, my algorithm even take me >10s to convert from decimal to binary or memory error.
- (NSString *)decimalToBase:(int)baseOfNewNumber
{
NSString *quotient, *remainder, *result;
result = #"";
quotient = number;
if (number != 0 && baseOfNewNumber != 10) {
for (int count = 0; [quotient compare:[NSString stringWithFormat:#"%i", baseOfNewNumber] options:NSNumericSearch] == (NSOrderedDescending|NSOrderedSame); count++) {
remainder = [quotient stringByModuloByString:[NSString stringWithFormat:#"%i", baseOfNewNumber]];
quotient = [quotient stringByDividedByString:[NSString stringWithFormat:#"%i", baseOfNewNumber]];
result = [remainder stringByAppendingString:result];
}
result = [[quotient stringByModuloByString:[NSString stringWithFormat:#"%i", baseOfNewNumber]] stringByAppendingString:result];
return result;
} else {
return number;
}
}
Lastly, this website converting the very very large number very fast, I saw its source code, it's likely simply using the usual method like me!
http://www.convertworld.com/en/numerals/
Why does the %g format for strings only handle six numbers in a float and after that it turns into scientific notation? Is there any other way of displaying a float with something similar to the %g format but allows more than six numbers?
EDIT: I have figured out %g with precision i.e turning %g into %.Xg where x is the specified number of significant digits. But it doesnt help me in this situation:
-(IBAction)numberPressed:(id)sender {
if (decimalChecker == 1) {
currentDecimal = currentDecimal*10+ (float)[sender tag];
decimaledNumberString = [[NSString alloc] initWithFormat:#"%.17g.%.17g", currentNumber, currentDecimal];
calculatorScreen.text = decimaledNumberString;
currentDecimaledNumber = [decimaledNumberString floatValue];
NSLog(#"regular");
} else {
currentNumber = currentNumber*10+ (float)[sender tag];
calculatorScreen.text = [[NSString alloc] initWithFormat:#"%.17g", currentNumber];
NSLog(#"regular");
}
}
If I press "5" eight times instead of 55555555, I get 55551782 or something similar. How can I fix it to where I get the desired eight fives instead of the crazy number?
Insert a period and a numeral to specify the maximum number of significant digits you would like displayed, such as %.17g for 17 significant digits. As you discovered, the default is six.
According to http://developer.apple.com/library/ios/#documentation/cocoa/Conceptual/Strings/Articles/FormatStrings.html#//apple_ref/doc/uid/20000943, iOS string formatting uses the same placeholders as C's printf(), which specifies g/G as representing FP values with exponential notation for very large/small values while f only uses non-exponential representation.
http://en.wikipedia.org/wiki/Printf_format_string#Format_placeholders