I want to perform few calculations on float values, the issue is it returns NaN, here is what i am trying
float totalRatingCount = users.count;
float score = 0.0;
for (NSManagedObject *managedObj in users) {
User_agent_rating *user = (User_agent_rating*)managedObj;
score = score + [user.status.score floatValue];
}
//float scoreInFloat = (float)score/(float)totalRatingCount;
float scoreInFloat = score/totalRatingCount;
scoreInFloat returns NaN in almost all the cases, I am looking for a basic division of two values, what could be the issue here?
I refactor a little to use the wonderful operator on key value coding:
#include <math.h>
float scoreInFloat = [[users valueForKeyPath:#"#avg.status.score"] floatValue];
if (isnan(scoreInFloat)) {
// handle nan here
}
else {
}
I'm writing without checking so there could be few errors.
KVC has various operator such as: #sum,#count etc.
Your operation is correct, but this is more syntetic and more readable. Check more here.
Division by zero (some other cases as well) produces a NaN result (not a number).
To check for NaN you compare the same value with itself and will return false for NaN.
if(scoreInFloat != scoreInFloat) isNaN = YES
Related
I'm trying to output the value of a latitude and longitude, which I have queried from an sql table. My issue is that when the value is 0.000000 it outputs something along the lines of 1.04480899798659e-276.
I have checked the value of the output above these two line of code:
latitudeOutput.text = [[NSNumber numberWithDouble:[theGPS latitude]] stringValue];
longitudeOutput.text = [[NSNumber numberWithDouble:[theGPS longitude]] stringValue];
and it should definitely be outputting 0.000000 . Does anyone know why this is happening, and anyway I can fix it?
----Added Section----
Why does this if else statement carry straight on through to the else section, despite the latitude and longitude value being equal to 0.000000?
if(latitude == #"0.000000" && longitude == #"0.000000"){
latitudeOutput.textColor = [UIColor grayColor];
longitudeOutput.textColor = [UIColor grayColor];
latitudeOutput.text = #"Latitude";
longitudeOutput.text = #"Longitude";
}
else {
latitudeOutput.text = latitude;
longitudeOutput.text = longitude;
}
Memory addresses have format of 0xdeadbeef.
Your value is a double written in a "variation" of scientfic notation. It is something extremely close to 0.0 (wolfram alpha visualization)
This happens because of how computers store floating point values. You can read more here.
To fix your issue, I'd suggest formating the output to include only as much digits after decimal point as you need.
NSString* formattedNumber = [NSString stringWithFormat:#"%.02f", [theGPS latitude]];
This will give you 2 places after decimal point.
I'm trying to find a super fast way of getting the sign of each value in a vector. I was hoping to find a function in the accelerate framework to do this, but couldn't find one. Here's what it would do:
float *inputVector = .... // some audio vector
int length = ...// length of input vector.
float *outputVector = ....// result
for( int i = 0; i<length; i++ )
{
if( inputVector[i] >= 0 ) outputVector[i] = 1;
else outputVector[i] = -1;
}
Ok, I think I've found a way...
vvcopysignf() "Copies an array, setting the sign of each value based on a second array."
So, one method would be to make an array of 1s, then use this function to change the sign of the 1s based on an input array.
float *ones = ... // a vector filled with 1's
float *input = .... // an input vector
float *output = ... // an output vector
int bufferSize = ... // size of the vectors;
vvcopysignf(output, ones, input, &bufferSize);
//output now is an array of -1s and 1s based the sign of the input.
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;
}
How would I get the nearest float in my array to a float of my choice? Here is my array:
[1.20, 1.50, 1.75, 1.95, 2.10]
For example, if my float was 1.60, I would like to produce the float 1.50.
Any ideas? Thanks in advance!
You can do it by sorting the array and finding the nearest one.
For this you can use sortDescriptors and then your algorithm will go.
Even you can loop through, by assuming first as the required value and store the minimum absolute (abs()) difference, if next difference is lesser than hold that value.
The working sample, however you need to handle other conditions like two similar values or your value is just between two value like 2 lies between 1 and 3.
NSArray *array = #[#1.20, #1.50, #1.75, #1.95, #2.10];
double my = 1.7;
NSNumber *nearest = array[0];
double diff = fabs(my - [array[0] doubleValue]);
for (NSNumber *num in array) {
double d = [num doubleValue];
if (diff > fabs(my - d) ) {
nearest = num;
diff = my - d;
}
}
NSLog(#"%#", nearest);
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.