In Scala, is it possible to get number of days between XMLGregorianCalendar? I cannot find any methods in this class that gets the range of two dates. If not how are you guys doing it?
Ah I guess not I just went with this instead and it works fine:
val range:Long = (toDate.toGregorianCalendar.getTimeInMillis - fromDate.toGregorianCalendar.getTimeInMillis) / (1000 * 60 * 60 * 24)
Related
Very strange issue I am noticing... This link says the numeric data type should be able to precisely handle 16383 digits after the decimal: https://www.postgresql.org/docs/10/datatype-numeric.html
So can someone plz explain to me why this function returns 9499.99999999999905:
(((60000::numeric / 50500 * 50500) - 50500) * (50500::numeric / 50500))::numeric
The correct answer is 9500.
When I use this function I get the right answer:
(((60000::numeric(20,11) / 50500 * 50500) - 50500) * (50500::numeric(20,11) / 50500))::numeric(20,11)
and this gives wrong answer:
(((60000::numeric(25,16) / 50500 * 50500) - 50500) * (50500::numeric(25,16) / 50500))::numeric(25,16) = 9499.9999999999990500
The odd thing is this same issue is happening on this website: https://web2.0calc.com/
if you paste the formula:
((60000.0000000 / 50500 * 50500) - 50500) * (50500.0000000 / 50500) = 9500.
But if I instead add an extra 0 to each of those:
((60000.00000000 / 50500 * 50500) - 50500) * (50500.00000000 / 50500) = 9499.99999999999999999999999999999999999999999999999999999999999905.
Even weirder, for both postgres and this website, if I break down the formula into two executions like this:
((60000.00000000 / 50500 * 50500) - 50500) = 9500
(50500.00000000 / 50500) = 1
9500 * 1 = 9500.
What the heck is going on here?
That you get the right answer with numeric(20,11) and the wrong one with numeric(25,16) is a coincidence: both will have rounding errors, because they calculate only to a limited precision, but in the first case the rounded result happens to be the correct one.
The same is the case for 60000.0000000 and 60000.00000000: they are interpreted as numeric values with different scale.
SELECT scale(60000.0000000), scale(60000.00000000);
scale | scale
-------+-------
7 | 8
(1 row
The only thing that is not obvious is why you get such a bad scale when you cast to numeric without any scale or precision.
The scale of 60000::numeric is 0, and 50500 is also converted to a numeric with scale 0. Now if you divide two numeric values, the resulting scale is calculated by the function select_div_scale, and a comment there clarifies the matter:
/*
* The result scale of a division isn't specified in any SQL standard. For
* PostgreSQL we select a result scale that will give at least
* NUMERIC_MIN_SIG_DIGITS significant digits, so that numeric gives a
* result no less accurate than float8; but use a scale not less than
* either input's display scale.
*/
NUMERIC_MIN_SIG_DIGITS has the value 16. The problem that this heuristic solves is that the scale of the division cannot be determined by looking at the scale of the arguments. So PostgreSQL chooses a value of 16, unless one of the arguments has a bigger scale. This avoids ending up with extremely large result values, unless someone explicitly asks for it.
In Pine Script, how do I find the price based on a certain number of days ago? I've tried something like this...
// Find the price 90 days ago
target = time - 90 * 60 * 60 * 24 * 1000
valuewhen(time < target, close, 1)
...however time < target never seems to return true – presumably because the current bar's time cannot also be in the past at the same time. Perhaps valuewhen() wasn't designed to be used with dynamic values that change on every bar?
Do I need to use a loop instead, and scan through every past bar until I find the date I'm looking for?
Perhaps there's a better way, but the workaround I'm using currently using is a function with a for loop, scanning backwards until the appropriate date is found. Here is my function:
priceXDaysAgo(numDays) =>
targetTimestamp = time - numDays*60*60*24*1000
// Declare a result variable with a "void" value
float result = if false
1
// We'll scan backwards through the preceding bars to find the first bar
// earlier than X days ago (it might be a little greater than X days if
// there was a break in trading: weekend, public holiday, etc.)
for i = 1 to 1000
if time[i] < targetTimestamp
result := close[i]
break
result
You can then call the function anywhere in your script:
priceXDaysAgo(90)
I've searched for this but i didn't find anything, i hope this is not a doubled question.
I'm doing a formula in TSQL like this:
#Temp = SQRT((((#Base1 - 1) * (#StDev1 * #StDev1))
+ ((#AvgBase - 1) * (#AvgStDev * #AvgStDev)))
* ((1 / #Base1) + (1 / #AvgBase))
/ (#Base1 + #AvgBase - 2))
But it always returns me a 0.
#Base1 and #AvgBase are int, the rest of parameters are float, but i've also tried with decimal(15,15).
I tried also changing the self multiplication with the function POWER() but the only problem i can't solve is this part: (1 / #Base1) + (1 / #AvgBase), because #Base1 and #AvgBase are so big, and the result of the calc is 0,0001... and some more numbers. How can i force the engine to not round the result to 0? Thanks
EDIT: I solved it changing the #AvgBase and the #Base1 to float type. I guess that the result 1/#param, with #param -> int gives you the rounded result and when you go for casting it or whatever, you are working on a rounded result anyway.
have you tried to create a #INVBase1 = (1/#Base1) ? will this also be rounded to 0? what happens when you play around with the data format of this new variable?
alternatively have you tried
/ ((#Base1) + (#AvgBase))
instead of
* ((1 / #Base1) + (1 / #AvgBase))
I am using following function to convert timestamp in format (e.g.) 02:49:02.506 to milliseconds in perl.
sub to_millis {
my( $hours, $minutes, $seconds, $millis) = split /:/, $_[0];
$millis += 1000 * $seconds;
$millis += 1000 * 60 * $minutes;
$millis += 1000 * 60 * 60 * $hours;
return $millis;
}
I am then using the milliseconds generated from above routine to calculate the time difference between two timestamps in milliseconds. This works fine all day but gets messed up around midnight, when the timestamp changes to 00:00:00.000. So any logs generated for 1 hr (between 12am to 1am) gets me values in negative for the timestamp difference. Since my timestamp doesn't have a date in it, how do I fix this problem? I am trying to do this on a mobile device, which doesn't have many perl modules installed. So I don't have the liberty of using all the perl modules available.
If you know the ordering of your two timestamps, and if you know that they're no more than 24 hours apart, if you get a negative difference add 24 hours (86,400,000 milliseconds).
If you don't have that information, then you won't be able to distinguish between a 2-minute span and a span of 24 hours and 2 minutes.
I assume that your timestamps will never be more than 23 hours 59 minutes apart?
Let's take two time stamps A and B. I am assuming that A happens before B.
Normally, if A is less than B, I know I can get my time by subtracting A from B. However, in this case, A is bigger than B. I now have to assume that I've wrapped around midnight. What do I do?
I know that the difference between A and B is A going to midnight, PLUS B.
For example, A is 11:58:30 and B is 00:02:00
I know that A will be 90 seconds before midnight, and B will add another 120 seconds to that time. Thus, the total difference will be 90 + 120 = 210 seconds.
Using your routine:
my $midnight = to_millis( "23:59:00:000" ); # Need the time at midnight
my $a_time = to_millis( $a_timestamp );
my $b_time = to_millis( $b_timestamp );
my $time_diff;
if ( $a_time < $b_time ) { # Normal timestamp issue
$time_diff = $b_time - $a_time;
}
else { # We wrapped around midnight!
my $first_part = $midnight - $a_time; # Time from A to midnight
$time_diff = $first_part + $b_time # We add the time from midnite to B
}
You have two timestamps, A and B. If B is always conceptually "after" A but the interval from A to B could cross a date boundary, then do
if (B < A) B+=86400000
and then do the subtraction. Or equivalently
diff = B - A
if (diff < 0) diff+=86400000
If, however you are not guaranteed that B will always be "after" A, you have to decide what is the acceptable range of positive and negative values for the difference. If it's more than half a day you're out of luck, there's no way to solve the problem as you cannot tell if a negative interval represents a real negative interval or a positive one that happened to cross a day boundary.
To handle the wrap around at midnight:
$elapsed_ms = $t2_ms - $t1_ms;
if ($elapsed_ms < 0) $elapsed_ms += (24 * 60 * 60 * 1000);
Trying to think of a way to subtract 5 minutes from 2 hours.
It doesn't make sense to subtract 5 from 2, because we end up with -3 generic time units, which is useless. But if "hour" is a subtype of "minute", we could convert 2 hours to 120 minutes, and yield 115 minutes, or 1 hour and 55 minutes.
Similarly, if we want to add 5 apples to 5 oranges, we cannot evaluate this in terms of apples, but might expect to end up with 10 fruit.
It seems in the above examples, and generally when using a number as an adjective, the integers need to be parameterized by the type of object they describing. I think it would be very useful if instead of declaring
val hours = 2
val minutes = 5
you could do something like
val hours = 2[Hour]
val minutes = 5[Minute]
val result = hours - minutes
assert (result == 115[Minute])
Does anything like this exist, would it be useful, and is it something that could be implemented?
EDIT: to clarify, the time example above is just a random example I thought up. My question is more whether in general the idea of parameterized Numerics is a useful concept, just as you have parameterized Lists etc. (The answer might be "no", I don't know!)
You can accomplish this by having two classes for Hours and Minutes, along with an implicit conversion function from hours to minutes
trait TimeUnit
case class Hour(val num: Int) extends TimeUnit
case class Minute(val num: Int) extends TimeUnit {
def - (sub: Minute) = Minute(num - sub.num)
}
implicit def hour2Minute(hour: Hour) = Minute(hour.num * 60)
This allows you to do something like
val h = Hour(2) - Minute(30) //returns Minute(90)
You can find some examples for this in the lift framework (spec).
import net.liftweb.utils.TimeHelpers._
3.minutes == 6 * 30.seconds
(Note: it seems you need to have reasonable numbers for correct comparison. Eg. There may be no more than 60 seconds.)
You might try scala-time, which is a wrapper around Joda Time and makes it a bit more idiomatic for Scala, including some DSL to do time period computations, similar to what Brian Agnew suggested in his answer.
For instance,
2.hours + 45.minutes + 10.seconds
creates a Joda Period.
It seems to me a DSL would be of use here. So you could write
2.hours - 5.minutes
and the appropriate conversions would take place to convert 2 hours into a Hours object (value 2) etc.
Lots of resources exist describing Scala's DSL capabilities. e.g. see this from O'Reilly