I have a sequence where values are generated at random times (real time stock market prices). I have a requirement to find the highest and lowest value of the sequence between a one minute period. I know you can use something like Buffer for this.
But the minute window should start with 00 seconds and finish at 59 seconds. e.g. the minute should start from 8:00:00 and finish at 8:00:59 the second minute should start from 8:01:00 to 8:01:59. Can we do this with Rx? Thanks. Vipter
I believe this would work:
var query =
source
.Publish(ss =>
ss
.GroupByUntil(
x => x.Timestamp.ToUnixTimeSeconds() / 60,
x => x.Value,
x => ss.Where(s => x.Key != s.Timestamp.ToUnixTimeSeconds() / 60))
.Select(gxs => gxs.ToArray().Select(xs => new
{
min = xs.Min(), max = xs.Max()
})));
Related
Is it possible to automatically check time then execute certain codes?
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer == "18:04:40" then
print("hello")
end
I am trying to print hello on "18:04:40" everyday (os.date's time) without setting up a timer (which counts how much time past since the program's initiation) as I can't run the program 24 hours non-stop...
Thanks for reading.
This may not be the best solution but, when using a library like love2d for example you could run something like this:
function love.update(dt)
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer >= value then
--stuff here
end
end
Or if you wanna make it so you have a whole number something like
tick = 0
function love.update(dt)
tick = tick + dt
if tick > 1 then
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer >= value then
--stuff here
end
end
end
Lua has to check the time in some way.
Without a loop that can be realized with debug.sethook().
Example with Lua 5.1 typed in an interactive Lua (lua -i)...
> print(_VERSION)
Lua 5.1
> debug.sethook() -- This clears a defined hook
> -- Next set up a hook function that fires on 'line' events
> debug.sethook(function() local hour, min, sec = 23, 59, 59 print(os.date('%H:%M:%S', os.time({year = 2021, month = 12, day = 11, hour = hour, min = min, sec = sec}))) end, 'l')
-- just hit return/enter or do other things
23:59:59
5.9 - The Debug Library
https://www.lua.org/manual/5.1/manual.html#5.9
Is there a way to delay or offset a scheduled command from the proposed frequency options?
e.g:
$schedule->command('GetX')->everyTenMinutes(); --> run at 9:10, 9:20, 9:30
$schedule->command('GetY')->everyTenMinutes(); --> run at 9:15, 9:25, 9:35
There are no delay function when scheduling tasks.
But when method can be used to schedule a task every 10 minutes, delay 5 minutes:
// this command is scheduled to run if minute is 05, 15, 25, 35, 45, 55
// the truth test is checked every minute
$schedule->command('foo:bar')->everyMinute()->when(function () {
return date('m') - 5 % 10 == 0;
});
Follow this rule, you can schedule a task every x minutes, delay y minutes
$schedule->command('foo:bar')->everyMinute()->when(function () {
return date('m') - y % x == 0;
});
If it gets difficult, the direct way you can simply write a custom Cron schedule. It is the easier way to understand without getting headache when you read the code later.
$schedule->command('foo:bar')->cron('05,15,25,35,45,55 * * * *');
There is the following task: I need to get minutes between one time and another one: for example, between "8:15" and "7:45". I have the following code:
(Time.parse("8:15") - Time.parse("7:45")).minute
But I get result as "108000.0 seconds".
How can I fix it?
The result you get back is a float of the number of seconds not a Time object. So to get the number of minutes and seconds between the two times:
require 'time'
t1 = Time.parse("8:15")
t2 = Time.parse("7:45")
total_seconds = (t1 - t2) # => 1800.0
minutes = (total_seconds / 60).floor # => 30
seconds = total_seconds.to_i % 60 # => 0
puts "difference is #{minutes} minute(s) and #{seconds} second(s)"
Using floor and modulus (%) allows you to split up the minutes and seconds so it's more human readable, rather than having '6.57 minutes'
You can avoid weird time parsing gotchas (Daylight Saving, running the code around midnight) by simply doing some math on the hours and minutes instead of parsing them into Time objects. Something along these lines (I'd verify the math with tests):
one = "8:15"
two = "7:45"
h1, m1 = one.split(":").map(&:to_i)
h2, m2 = two.split(":").map(&:to_i)
puts (h1 - h2) * 60 + m1 - m2
If you do want to take Daylight Saving into account (e.g. you sometimes want an extra hour added or subtracted depending on today's date) then you will need to involve Time, of course.
Time subtraction returns the value in seconds. So divide by 60 to get the answer in minutes:
=> (Time.parse("8:15") - Time.parse("7:45")) / 60
#> 30.0
I want to trigger a notification for all my users at a specific time in their time zone. I want to compute the delay the server should wait before firing the notification. I can compute the time at the users Time Zone using Time.now.in_time_zone(person.time_zone)
I can strip out the hours, minutes and seconds from that time and find out the seconds remaining to the specific time. However, I was wondering if there's a more elegant method where I could set 9:00 AM on today and tomorrow in a timezone and compare it with Time.now.in_time_zone(person.time_zone) and just find out the number of seconds using arithmetic operations in the ruby Time Class.
Or in short my question is: (was: before the downvote!)
How do I compute the number of seconds to the next 9:00 AM in New York?
What about this
next9am = Time.now.in_time_zone(person.time_zone).seconds_until_end_of_day + 3600 * 9
next9am -= 24 * 60 * 60 if Time.now.in_time_zone(person.time_zone).hour < 9
NOTIFICATION_HOUR = 9
local_time = Time.now.in_time_zone(person.time_zone)
desired_time = local_time.hour >= NOTIFICATION_HOUR ? local_time + 1.day : local_time
desired_time = Time.new(desired_time.year, desired_time.month, desired_time.day, NOTIFICATION_HOUR, 0, 0, desired_time.utc_offset)
return desired_time - local_time
How can I determine the number of days between two Time instances in Ruby?
> earlyTime = Time.at(123)
> laterTime = Time.now
> time_difference = laterTime - earlyTime
I'd like to determine the number of days in time_difference (I'm not worried about fractions of days. Rounding up or down is fine).
Difference of two times is in seconds. Divide it by number of seconds in 24 hours.
(t1 - t2).to_i / (24 * 60 * 60)
require 'date'
days_between = (Date.parse(laterTime.to_s) - Date.parse(earlyTime.to_s)).round
Edit ...or more simply...
require 'date'
(laterTime.to_date - earlyTime.to_date).round
earlyTime = Time.at(123)
laterTime = Time.now
time_difference = laterTime - earlyTime
time_difference_in_days = time_difference / 1.day # just divide by 1.day
[1] pry(main)> earlyTime = Time.at(123)
=> 1970-01-01 01:02:03 +0100
[2] pry(main)> laterTime = Time.now
=> 2014-04-15 11:13:40 +0200
[3] pry(main)> (laterTime.to_date - earlyTime.to_date).to_i
=> 16175
To account for DST (Daylight Saving Time), you'd have to count it by the days. Note that this assumes less than a day is counted as 1 (rounded up):
num = 0
cur = start_time
while cur < end_time
num += 1
cur = cur.advance(:days => 1)
end
return num
Here is a simple answer that works across DST:
numDays = ((laterTime - earlyTime)/(24.0*60*60)).round
60*60 is the number of seconds in an hour
24.0 is the number of hours in a day. It's a float because some days are a little more than 24 hours, some are less. So when we divide by the number of seconds in a day we still have a float, and round will round to the closest integer.
So if we go across DST, either way, we'll still round to the closest day. Even if you're in some weird timezone that changes more than an hour for DST.
in_days (Rails 6.1+)
Rails 6.1 introduces new ActiveSupport::Duration conversion methods like in_seconds, in_minutes, in_hours, in_days, in_weeks, in_months, and in_years.
As a result, now, your problem can be solved as:
date_1 = Time.parse('2020-10-18 00:00:00 UTC')
date_2 = Time.parse('2020-08-13 03:35:38 UTC')
(date_2 - date_1).seconds.in_days.to_i.abs
# => 65
Here is a link to the corresponding PR.
None of these answers will actually work if you don't want to estimate and you want to take into account daylight savings time.
For instance 10 AM on Wednesday before the fall change of clocks and 10 AM the Wednesday afterwards, the time between them would be 1 week and 1 hour. During the spring it would be 1 week minus 1 hour.
In order to get the accurate time you can use the following code
def self.days_between_two_dates later_time, early_time
days_between = (later_time.to_date-early_time.to_date).to_f
later_time_time_of_day_in_seconds = later_time.hour*3600+later_time.min*60+later_time.sec
earlier_time_time_of_day_in_seconds = early_time.hour*3600+early_time.min*60+early_time.sec
days_between + (later_time_time_of_day_in_seconds - early_time_time_of_day_in_seconds)/1.0.day
end