Nt = N0e-λt
N0 is the initial quantity
Nt is the quantity that still remains after a time t,
t1/2 is the half-life
τ is the mean lifetime
λ is the decay constant
I am pretty stuck on how to make this into a formula for objective-c and I require it.
double sourceStart = [textField.text doubleValue];
double sourceNow = 0;
double daysBetween = 8;
if (textField.text.length > 0) {
//Find how many half lives have been accumulated
double totalNumberOfHalfLives = daysBetween / sourceHalfLife;
//Find the factor
double reductionFactor = pow(0.5, totalNumberOfHalfLives);
//Find the source strength now
double sourceNow = sourceStart * reductionFactor;
}
I'm assuming I need something a long the lines of this? or completely wrong.
Then, I also need to be able to find how many days have passed between a certain period of days, for instance. Start Date = Apr 15th Now Date = Apr 25th, 10 days between.. How do I work that out in objective c? As well as my original question.
This will do it (in straight C, but it'll work as-is in Objective-C, and you can extract the logic easily enough):
#include <stdio.h>
#include <math.h>
int main(void) {
double start_quantity = 100;
double half_life = 8;
double days = 16;
double end_quantity = start_quantity * pow(0.5, days / half_life);
printf("After %.1f days with a half life of %.1f days, %.1f decays to %.1f.\
n",
days, half_life, start_quantity, end_quantity);
return 0;
}
and outputs:
paul#local:~/src$ ./halflife
After 16.0 days with a half life of 8.0 days, 100.0 decays to 25.0.
paul#local:~/src$
For the second part of your question, you can store dates in an NSDate, and use the timeIntervalSinceDate: method to get the time between them in seconds. Something like this:
#import <Foundation/Foundation.h>
int main(int argc, const char * argv[])
{
#autoreleasepool {
const int kSecsInADay = 86400;
NSDate * startDate = [NSDate dateWithTimeIntervalSinceNow:-16 * kSecsInADay];
NSDate * endDate = [NSDate date];
NSTimeInterval seconds_diff = [endDate timeIntervalSinceDate:startDate];
double days_diff = seconds_diff / kSecsInADay;
NSLog(#"There are %.1f days between %# and %#.", days_diff,
[startDate description], [endDate description]);
}
return 0;
}
which outputs:
There are 16.0 days between 2014-04-09 21:41:12 +0000 and 2014-04-25 21:41:12 +0000.
Notes:
[NSDate date] returns an NSDate object representing the current time.
[NSDate dateWithTimeIntervalSinceNow:seconds] returns an NSDate object that's seconds seconds away from the current time. In this case, I've created it exactly 16 days before the current date. Based on the comments, in your case you'll also want to create startDate with [NSDate date], and then store it somewhere so you can calculate the difference between it and the current time at some point in the future.
Related
I am trying to parse the pwdLastSet value from NSTask response when I do an ldapsearch. I've successfully extracted the value (129875475241190194) and I am trying to convert it to an NSDate Object.
Reference: http://www.chrisnowell.com/information_security_tools/date_converter/Windows_active_directory_date_converter.asp
I tried to extract the Javascript code from the page above and convert it but I am getting a different date.
int iYearsFrom1601to1970 = 1970 - 1601;
int iDaysFrom1601to1970 = iYearsFrom1601to1970 * 365;
iDaysFrom1601to1970 += (int)(iYearsFrom1601to1970 / 4); // leap years
iDaysFrom1601to1970 -= 3; // non-leap centuries (1700,1800,1900). 2000 is a leap century
float iSecondsFrom1601to1970 = iDaysFrom1601to1970 * 24 * 60 * 60;
int iTotalSecondsSince1601 = (int)(129875475241190194 / 10000000);
float iTotalSecondsSince1970 = iTotalSecondsSince1601 - iSecondsFrom1601to1970;
NSDate *date = [NSDate dateWithTimeIntervalSince1970:iTotalSecondsSince1970];
Any help would be appreciated.
Thanks!
Here's how I would do it:
NSDateComponents *base = [[NSDateComponents alloc] init];
[base setDay:1];
[base setMonth:1];
[base setYear:1601];
[base setEra:1]; // AD
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDate *baseDate = [gregorian dateFromComponents:base];
[base release];
[gregorian release];
NSTimeInterval timestamp = 129875475241190194.0 / 10000000.0;
NSDate *finalDate = [baseDate dateByAddingTimeInterval:timestamp];
This gives me a finalDate of 2012-07-24 03:58:22 +0000.
Since the timestamp is a time interval since Jan 1, 1601 at 00:00 UTC, you can use the -dateByAddingTimeInterval: method on NSDate to add the timestamp to the base date to get the final NSDate.
Once you've done that, you can run it through an NSDateFormatter to format it for display.
Assuming the, well, daring conversion between the basetimes is correct: actually looking at the warnings, instead of casting them away, might actually help:
int main(void)
{
int iTotalSecondsSince1601 = (129875475241190194 / 10000000);
return 0;
}
stieber#gatekeeper:~$ clang++ Test.cpp
Test.cpp:4:8: warning: implicit conversion from 'long' to 'int' changes value from 12987547524 to 102645636
....
That should account for a good deal of the difference...
Try this
NSTimeInterval value = 129875475241190194;
// instead of trying to compute seconds between 1601 and 1970
const NSTimeInterval EPOCH = 11644473600;
const NSTimeInterval NANO = 10000000;
NSTimeInterval seconds = value / NANO - EPOCH;
NSDate *answer = [NSDate dateWithTimeIntervalSince1970:seconds];
Also this is reason you don't want to calculate seconds since 1601: ...in the last millennium, 1600 and 2000 were leap years, but 1700, 1800 and 1900 were not. Excerpt from Wikipedia on Gregorian calendar.
The value for EPOCH is explained on Convert Active Directory "LastLogon:" time to (UNIX) readable time
.
Note: The information about accountExpires which starts from 12-31-1601 (11644473600). The values lastLogon and lastLogonTimeStamp however use 01-01-1601 as the date to calculate this value (11676009600).
From the iOS docs:
seconds
The number of seconds from the current date and time for the new date. Use a negative value to specify a date before the current date.
However this code gives dates in 2148:
#import <Foundation/Foundation.h>
#import <stdlib.h>
int main(int argc, const char * argv[])
{
#autoreleasepool {
for(int i = 0; i < 30; i++) {
NSDate* date = [NSDate dateWithTimeIntervalSinceNow:(NSTimeInterval)-((u_int32_t)i * 12 * 60 * 60 + arc4random_uniform(8 * 60 * 60))];
NSLog(#"%#", date);
}
}
return 0;
}
It should generates semi-random date and times from the near past.
As Hot Licks said, it was due to casting to NSTimeInterval after making the number negative instead of first negating the number and afterwards casting it.
The fixed version of the line will look like this:
NSDate* date = [NSDate dateWithTimeIntervalSinceNow:-(NSTimeInterval)((u_int32_t)i * 12 * 60 * 60 + arc4random_uniform(8 * 60 * 60))];
Once a IBAction is pressed, I want a counter which counts up.
But in this form
00:00:00
Hours:minutes:seconds
This is the code so far:
-(void)countUp {
mainInt += 1;
seconds.text = [NSString stringWithFormat:#"%02d", mainInt];
}
This only counts up in 00 format
Thanks for the help!
Just do the appropriate math to break the count up into its constituent parts:
NSString *timeString = [NSString stringWithFormat:#"%02d:%02d:%02d",
totalSeconds/3600, // hours
(totalSeconds/60)%60, // minutes
totalSeconds%3600] // seconds
For the sake of readability, it'd be nice to replace that inline math with macros or functions, for example:
#define secondsPerMinute 60
#define minutesPerHour 60
int hours(int secs) {
return secs/(minutesPerHour * secondsPerMinute);
}
int minutes(int secs) {
return (secs/secondsPerMinute) % minutesPerHour;
}
int seconds(int secs) {
return secs % (minutesPerHour * secondsPerMinute);
}
// ...
NSString *timeString = [NSString stringWithFormat:#"%02d:%02d:%02d",
hours(totalSeconds),
minutes(totalSeconds),
seconds(totalSeconds)];
Often when implementing this sort of display, you don't want the colons to jump around as the elapsed time changes. Many fonts have fixed-width numerals, so it's not always a problem, but you might want to use three separate labels for the hours, minutes, and seconds, with unchanging labels in between for the colons.
Another approach to the math above is to store the seconds, minutes, and hours in three variables instead of one, and just be careful to increment minutes and reset seconds when seconds hit 60, and so on. To make this easier to use, encapsulate it in a class, like:
#interface Time : NSObject {
int seconds;
int minutes;
int hours;
}
- (void)countUp;
- (NSString*)timeString;
#end;
I'm creating a countdown timer and I need to printout the time left (hour:minute:seconds) until a specific date. I've found how to get the time interval between Now and the target date but I don't know how to format the time interval as a string. Does NSDateFormater work on NSTimeInterval?
NSTimeInterval is in seconds, use divide and remainder to break it up and format (code untested):
NSString *timeIntervalToString(NSTimeInterval interval)
{
long work = (long)interval; // convert to long, NSTimeInterval is *some* numeric type
long seconds = work % 60; // remainder is seconds
work /= 60; // total number of mins
long minutes = work % 60; // remainder is minutes
long hours = work / 60 // number of hours
// now format and return - %ld is long decimal, %02ld is zero-padded two digit long decimal
return [NSString stringWithFormat:#"%ld:%02ld:%02ld", hours, minutes, seconds];
}
You would first compare two NSDate objects to retrieve the difference in seconds between the two, the NSDate method you should use is
- (NSTimeInterval)timeIntervalSinceDate:(NSDate *)anotherDate
Then you could simply write a function to parse the seconds into hours/minutes/seconds, for example you could use this (untested):
-(NSDictionary*)createTimemapForSeconds:(int)seconds{
int hours = floor(seconds / (60 * 60) );
float minute_divisor = seconds % (60 * 60);
int minutes = floor(minute_divisor / 60);
float seconds_divisor = seconds % 60;
seconds = ceil(seconds_divisor);
NSDictionary * timeMap = [NSDictionary dictionaryWithObjects:[NSArray arrayWithObjects:[NSNumber numberWithInt:hours], [NSNumber numberWithInt:minutes], [NSNumber numberWithInt:seconds], nil] forKeys:[NSArray arrayWithObjects:#"h", #"m", #"s", nil]];
return timeMap;
}
This is code from my project:
-(NSString*)timeLeftString
{
long seconds = [self msLeft]/1000;
if( seconds == 0 )
return #"";
if( seconds < 60 )
return [NSString stringWithFormat:
pluralString(seconds,
NSLocalizedString(#"en|%ld second left|%ld seconds left", #"")), seconds];
long minutes = seconds / 60;
seconds -= minutes*60;
if( minutes < 60 )
return [NSString stringWithFormat:
NSLocalizedString(#"%ld:%02ld left",#""),
minutes, seconds];
long hours = minutes/60;
minutes -= hours*60;
return [NSString stringWithFormat:
NSLocalizedString(#"%ld:%02ld:%02ld left",#""),
hours, minutes, seconds];
}
msLeft --- my function that returns time in milliseconds
pluralString --- my function that provides different parts of format string depending on the value (http://translate.sourceforge.net/wiki/l10n/pluralforms)
Function returns different format for different timer values (1 second left, 5 seconds left, 2:34 left, 1:15:14 left).
In any case, progress bad should be visible during long operation
One more thought: In case that time left is "small" (less then a minute?), probably time left should not be shown --- just progress bar left to reduce interface "visual noise".
I've got an amount of seconds that passed from a certain event. It's stored in a NSTimeInterval data type.
I want to convert it into minutes and seconds.
For example I have: "326.4" seconds and I want to convert it into the following string:
"5:26".
What is the best way to achieve this goal?
Thanks.
Brief Description
The answer from Brian Ramsay is more convenient if you only want to convert to minutes.
If you want Cocoa API do it for you and convert your NSTimeInterval not only to minutes but also to days, months, week, etc,... I think this is a more generic approach
Use NSCalendar method:
(NSDateComponents *)components:(NSUInteger)unitFlags fromDate:(NSDate *)startingDate toDate:(NSDate *)resultDate options:(NSUInteger)opts
"Returns, as an NSDateComponents object using specified components, the difference between two supplied dates". From the API documentation.
Create 2 NSDate whose difference is the NSTimeInterval you want to convert. (If your NSTimeInterval comes from comparing 2 NSDate you don't need to do this step, and you don't even need the NSTimeInterval).
Get your quotes from NSDateComponents
Sample Code
// The time interval
NSTimeInterval theTimeInterval = 326.4;
// Get the system calendar
NSCalendar *sysCalendar = [NSCalendar currentCalendar];
// Create the NSDates
NSDate *date1 = [[NSDate alloc] init];
NSDate *date2 = [[NSDate alloc] initWithTimeInterval:theTimeInterval sinceDate:date1];
// Get conversion to months, days, hours, minutes
unsigned int unitFlags = NSHourCalendarUnit | NSMinuteCalendarUnit | NSDayCalendarUnit | NSMonthCalendarUnit;
NSDateComponents *conversionInfo = [sysCalendar components:unitFlags fromDate:date1 toDate:date2 options:0];
NSLog(#"Conversion: %dmin %dhours %ddays %dmoths",[conversionInfo minute], [conversionInfo hour], [conversionInfo day], [conversionInfo month]);
[date1 release];
[date2 release];
Known issues
Too much for just a conversion, you are right, but that's how the API works.
My suggestion: if you get used to manage your time data using NSDate and NSCalendar, the API will do the hard work for you.
pseudo-code:
minutes = floor(326.4/60)
seconds = round(326.4 - minutes * 60)
All of these look more complicated than they need to be! Here is a short and sweet way to convert a time interval into hours, minutes and seconds:
NSTimeInterval timeInterval = 326.4;
long seconds = lroundf(timeInterval); // Since modulo operator (%) below needs int or long
int hour = seconds / 3600;
int mins = (seconds % 3600) / 60;
int secs = seconds % 60;
Note when you put a float into an int, you get floor() automatically, but you can add it to the first two if if makes you feel better :-)
Forgive me for being a Stack virgin... I'm not sure how to reply to Brian Ramsay's answer...
Using round will not work for second values between 59.5 and 59.99999. The second value will be 60 during this period. Use trunc instead...
double progress;
int minutes = floor(progress/60);
int seconds = trunc(progress - minutes * 60);
If you're targeting at or above iOS 8 or OS X 10.10, this just got a lot easier. The new NSDateComponentsFormatter class allows you to convert a given NSTimeInterval from its value in seconds to a localized string to show the user. For example:
Objective-C
NSTimeInterval interval = 326.4;
NSDateComponentsFormatter *componentFormatter = [[NSDateComponentsFormatter alloc] init];
componentFormatter.unitsStyle = NSDateComponentsFormatterUnitsStylePositional;
componentFormatter.zeroFormattingBehavior = NSDateComponentsFormatterZeroFormattingBehaviorDropAll;
NSString *formattedString = [componentFormatter stringFromTimeInterval:interval];
NSLog(#"%#",formattedString); // 5:26
Swift
let interval = 326.4
let componentFormatter = NSDateComponentsFormatter()
componentFormatter.unitsStyle = .Positional
componentFormatter.zeroFormattingBehavior = .DropAll
if let formattedString = componentFormatter.stringFromTimeInterval(interval) {
print(formattedString) // 5:26
}
NSDateCompnentsFormatter also allows for this output to be in longer forms. More info can be found in NSHipster's NSFormatter article. And depending on what classes you're already working with (if not NSTimeInterval), it may be more convenient to pass the formatter an instance of NSDateComponents, or two NSDate objects, which can be done as well via the following methods.
Objective-C
NSString *formattedString = [componentFormatter stringFromDate:<#(NSDate *)#> toDate:<#(NSDate *)#>];
NSString *formattedString = [componentFormatter stringFromDateComponents:<#(NSDateComponents *)#>];
Swift
if let formattedString = componentFormatter.stringFromDate(<#T##startDate: NSDate##NSDate#>, toDate: <#T##NSDate#>) {
// ...
}
if let formattedString = componentFormatter.stringFromDateComponents(<#T##components: NSDateComponents##NSDateComponents#>) {
// ...
}
Brian Ramsay’s code, de-pseudofied:
- (NSString*)formattedStringForDuration:(NSTimeInterval)duration
{
NSInteger minutes = floor(duration/60);
NSInteger seconds = round(duration - minutes * 60);
return [NSString stringWithFormat:#"%d:%02d", minutes, seconds];
}
Here's a Swift version:
func durationsBySecond(seconds s: Int) -> (days:Int,hours:Int,minutes:Int,seconds:Int) {
return (s / (24 * 3600),(s % (24 * 3600)) / 3600, s % 3600 / 60, s % 60)
}
Can be used like this:
let (d,h,m,s) = durationsBySecond(seconds: duration)
println("time left: \(d) days \(h) hours \(m) minutes \(s) seconds")
NSDate *timeLater = [NSDate dateWithTimeIntervalSinceNow:60*90];
NSTimeInterval duration = [timeLater timeIntervalSinceNow];
NSInteger hours = floor(duration/(60*60));
NSInteger minutes = floor((duration/60) - hours * 60);
NSInteger seconds = floor(duration - (minutes * 60) - (hours * 60 * 60));
NSLog(#"timeLater: %#", [dateFormatter stringFromDate:timeLater]);
NSLog(#"time left: %d hours %d minutes %d seconds", hours,minutes,seconds);
Outputs:
timeLater: 22:27
timeLeft: 1 hours 29 minutes 59 seconds
Since it's essentially a double...
Divide by 60.0 and extract the integral part and the fractional part.
The integral part will be the whole number of minutes.
Multiply the fractional part by 60.0 again.
The result will be the remaining seconds.
Remember that the original question is about a string output, not pseudo-code or individual string components.
I want to convert it into the following string: "5:26"
Many answers are missing the internationalization issues, and most doing the math computations by hand. All just so 20th century...
Do not do the Math yourself (Swift 4)
let timeInterval: TimeInterval = 326.4
let dateComponentsFormatter = DateComponentsFormatter()
dateComponentsFormatter.unitsStyle = .positional
if let formatted = dateComponentsFormatter.string(from: timeInterval) {
print(formatted)
}
5:26
Leverage on libraries
If you really want individual components, and pleasantly readable code, check out SwiftDate:
import SwiftDate
...
if let minutes = Int(timeInterval).seconds.in(.minute) {
print("\(minutes)")
}
5
Credits to #mickmaccallum and #polarwar for adequate usage of DateComponentsFormatter
How I did this in Swift (including the string formatting to show it as "01:23"):
let totalSeconds: Double = someTimeInterval
let minutes = Int(floor(totalSeconds / 60))
let seconds = Int(round(totalSeconds % 60))
let timeString = String(format: "%02d:%02d", minutes, seconds)
NSLog(timeString)
Swift 2 version
extension NSTimeInterval {
func toMM_SS() -> String {
let interval = self
let componentFormatter = NSDateComponentsFormatter()
componentFormatter.unitsStyle = .Positional
componentFormatter.zeroFormattingBehavior = .Pad
componentFormatter.allowedUnits = [.Minute, .Second]
return componentFormatter.stringFromTimeInterval(interval) ?? ""
}
}
let duration = 326.4.toMM_SS()
print(duration) //"5:26"