Adding BigDecimal hours to DateTime is wrong by 1 second - ruby-on-rails

This happens with the update to ActiveSupport 6
start_time = DateTime.now.beginning_of_day
start_time + BigDecimal(2).hours #=> Wed, 11 Sep 2019 01:59:59 +0000
Oddly enough this works fine with Time
start_time = Time.now.beginning_of_day
start_time + BigDecimal(2).hours #=> 2019-09-11 02:00:00 +0000
Can anybody explain why?

Ultimately, it boils down to floating point errors in some of the math that ActiveSupport does internally.
Notice that using Rational instead of BigDecimal works:
DateTime.now.beginning_of_day + Rational(2, 1).hours
# => Mon, 02 Dec 2019 02:00:00 -0800
Time.now.beginning_of_day + Rational(2, 1).hours
# => 2019-12-02 02:00:00 -0800
Here's the relevant code from Time/DateTime/ActiveSupport:
class DateTime
def since(seconds)
self + Rational(seconds, 86400)
end
def plus_with_duration(other) #:nodoc:
if ActiveSupport::Duration === other
other.since(self)
else
plus_without_duration(other)
end
end
end
class Time
def since(seconds)
self + seconds
rescue
to_datetime.since(seconds)
end
def plus_with_duration(other) #:nodoc:
if ActiveSupport::Duration === other
other.since(self)
else
plus_without_duration(other)
end
end
def advance(options)
unless options[:weeks].nil?
options[:weeks], partial_weeks = options[:weeks].divmod(1)
options[:days] = options.fetch(:days, 0) + 7 * partial_weeks
end
unless options[:days].nil?
options[:days], partial_days = options[:days].divmod(1)
options[:hours] = options.fetch(:hours, 0) + 24 * partial_days
end
d = to_date.gregorian.advance(options)
time_advanced_by_date = change(year: d.year, month: d.month, day: d.day)
seconds_to_advance = \
options.fetch(:seconds, 0) +
options.fetch(:minutes, 0) * 60 +
options.fetch(:hours, 0) * 3600
if seconds_to_advance.zero?
time_advanced_by_date
else
time_advanced_by_date.since(seconds_to_advance)
end
end
end
class ActiveSupport::Duration
def since(time = ::Time.current)
sum(1, time)
end
def sum(sign, time = ::Time.current)
parts.inject(time) do |t, (type, number)|
if t.acts_like?(:time) || t.acts_like?(:date)
if type == :seconds
t.since(sign * number)
elsif type == :minutes
t.since(sign * number * 60)
elsif type == :hours
t.since(sign * number * 3600)
else
t.advance(type => sign * number)
end
else
raise ::ArgumentError, "expected a time or date, got #{time.inspect}"
end
end
end
end
What's happening in your case is on the t.since(sign * number * 3600) line, number is BigDecimal(2), and DateTime.since does Rational(seconds, 86400). So the whole expression when using DateTime is Rational(1 * BigDecimal(2) * 3600, 86400).
Since a BigDecimal is used as an argument to Rational, the result isn't rational at all:
Rational(1 * BigDecimal(2) * 3600, 86400)
# => 0.83333333333333333e-1 # Since there's no obvious way to coerce a BigDecimal into a Rational, this returns a BigDecimal
Rational(1 * 2 * 3600, 86400)
# => (1/12) # A rational, as expected
This value makes it back to Time#advance. Here are the results of the calculations it makes:
options[:days], partial_days = options[:days].divmod(1)
# => [0.0, 0.83333333333333333e-1] # 0 days, 2 hours
options[:hours] = options.fetch(:hours, 0) + 24 * partial_days
# => 0.1999999999999999992e1 # juuuust under 2 hours.
And finally, 0.1999999999999999992e1 * 3600 = 7199.9999999999999712, which gets floored when it's finally converted back to a time/datetime.
This doesn't happen with Time, since Time doesn't ever need to pass the duration's value into a Rational.
I don't think this should be considered a bug, since if you're passing a BigDecimal then is how you should expect the code to treat your data: As a number with a decimal component, rather than as a ratio. That is, when you use BigDecimals you open yourself up to floating point errors.

It's off by one second, not milisecond. Why not use 2.hours instead of BigDecimal(2).hours?

Related

Need to convert to integer and then sum of the Plucked data ruby 2.6

I need SUM of the plucked timings.
I did
#dailystatus_infos.task_times.pluck(:total_min)
I got following
["00:00:00", "1:52:00", "00:00:00", "00:02:28", "1:54:00"]
output. [Hour:Minute:Second] format
Now I need to convert those Minutes to integer and sum of it.
I need SUM of the plucked timings
I'd start by writing a helper method to convert the hh:mm:ss string to seconds. A regular expression would work:
def to_seconds(string)
string.match(/(?<hours>\d+):(?<minutes>\d+):(?<seconds>\d+)/) do |m|
m[:hours].to_i * 3600 + m[:minutes].to_i * 60 + m[:seconds].to_i
end
end
to_seconds('00:00:12') #=> 12
to_seconds('00:01:00') #=> 60
to_seconds('00:01:12') #=> 72
Now you can sum the seconds via:
total_mins = ["00:00:00", "1:52:00", "00:00:00", "00:02:28", "1:54:00"]
total_mins.sum { |str| to_seconds(str) }
#=> 13708
And, if necessary, convert that back to h:mm:ss via divmod:
seconds = 13708
hours, seconds = seconds.divmod(3600)
minutes, seconds = seconds.divmod(60)
format('%d:%02d:%02d', hours, minutes, seconds)
#=> "3:48:28"
We first determine the total seconds:
arr = ["00:00:00", "1:52:00", "00:00:00", "00:02:28", "1:54:00"]
s = arr.sum do |str|
str.split(':').reduce(0) { |t, s| t * 60 + s.to_i }
end
#=> 13708
and then manipulate s as desired. The number of minutes, for example, equals
s.fdiv(60)
#=> 228.46666666666667
which might be rounded or truncated.
Something like this
total_minutes = #dailystatus_infos.task_times.pluck(:total_min)
total_minutes.map { |t| (::Time.parse(t).seconds_since_midnight / 60).to_i }.sum
arr = ["00:00:00", "1:52:00", "00:00:00", "00:02:28", "1:54:00"]
arr.map { |t| ::Time.parse(t).seconds_since_midnight / 60 }.sum.to_i
=> 228
You get the minute by splitting on :. Then map and sum.
ary = ["00:00:00", "1:52:00", "00:00:00", "00:02:28", "1:54:00"]
ary.map{ |s| s.split(':')[1].to_i }.sum #=> 108 (0 + 52 + 0 + 2 + 54)

Any option to post-process returned value in long conditional, other than setting variables for each statement?

def some_method(x)
if x == 1
date = Date.today
elsif x == 5
date = Date.today + 2
else
date = Date.today - 2
end
date + 20
end
For visual clarity, is it possible somehow to omit date = for each statement and catch whatever the returned value is from the conditional and add 20 to it?
(The code is for example purpose, my own code has 10 if-statements.)
def some_method(x)
date = if x == 1
Date.today
elsif x == 5
Date.today + 2
else
Date.today - 2
end
date + 20
end
If you have 10 if statements it is probably better to refactor code using case-when like this:
def some_method(x)
date = case x
when 1; Date.today
when 5; Date.today + 2
else; Date.today - 2
end
date + 20
end

How to determine the nth weekday for a given month?

This snippet of code determines the date for the nth weekday of a given month.
Example: for the 2nd Tuesday of December 2013:
>> nth_weekday(2013,11,2,2)
=> Tue Nov 12 00:00:00 UTC 2013
Last Sunday of December 2013:
>> nth_weekday(2013,12,'last',0)
=> Sun Dec 29 00:00:00 UTC 2013
I was not able to find working code for this question so I'm sharing my own.
If you are using Rails, you can do this.
def nth_weekday(year, month, n, wday)
first_day = DateTime.new(year, month, 1)
arr = (first_day..(first_day.end_of_month)).to_a.select {|d| d.wday == wday }
n == 'last' ? arr.last : arr[n - 1]
end
> n = nth_weekday(2013,11,2,2)
# => Tue, 12 Nov 2013 00:00:00 +0000
You can use:
require 'date'
class Date
#~ class DateError < ArgumentError; end
#Get the third monday in march 2008: new_by_mday( 2008, 3, 1, 3)
#
#Based on http://forum.ruby-portal.de/viewtopic.php?f=1&t=6157
def self.new_by_mday(year, month, weekday, nr)
raise( ArgumentError, "No number for weekday/nr") unless weekday.respond_to?(:between?) and nr.respond_to?(:between?)
raise( ArgumentError, "Number not in Range 1..5: #{nr}") unless nr.between?(1,5)
raise( ArgumentError, "Weekday not between 0 (Sunday)and 6 (Saturday): #{nr}") unless weekday.between?(0,6)
day = (weekday-Date.new(year, month, 1).wday)%7 + (nr-1)*7 + 1
if nr == 5
lastday = (Date.new(year, (month)%12+1, 1)-1).day # each december has the same no. of days
raise "There are not 5 weekdays with number #{weekday} in month #{month}" if day > lastday
end
Date.new(year, month, day)
end
end
p Date.new_by_mday(2013,11,2,2)
This is also available in a gem:
gem "date_tools", "~> 0.1.0"
require 'date_tools/date_creator'
p Date.new_by_mday(2013,11,2,2)
# nth can be 1..4 or 'last'
def nth_weekday(year,month,nth,week_day)
first_date = Time.gm(year,month,1)
last_date = month < 12 ? Time.gm(year,month+1)-1.day : Time.gm(year+1,1)-1.day
date = nil
if nth.class == Fixnum and nth > 0 and nth < 5
date = first_date
nth_counter = 0
while date <= last_date
nth_counter += 1 if date.wday == week_day
nth_counter == nth ? break : date += 1.day
end
elsif nth == 'last'
date = last_date
while date >= first_date
date.wday == week_day ? break : date -= 1.day
end
else
raise 'Error: nth_weekday called with out of range parameters'
end
return date
end

How do I convert a string into a Time object?

I searched for my problem and got a lot of solutions, but unfortunately none satisfy my need.
My problem is, I have two or more strings, and I want to convert those strings into times, and add them:
time1 = "10min 43s"
time2 = "32min 30s"
The output will be: 43min 13s
My attempted solution is:
time1 = "10min 43s"
d1=DateTime.strptime(time1, '%M ')
# Sat, 02 Nov 2013 00:10:00 +0000
time2 = "32min 30s"
d2=DateTime.strptime(time2, '%M ')
# Sat, 02 Nov 2013 00:32:00 +0000
Then I can't progress.
There are many ways to do this. Here's another:
time1 = "10min 43s"
time2 = "32min 30s"
def get_mins_and_secs(time_str)
time_str.scan(/\d+/).map(&:to_i)
#=> [10, 43] for time_str = time1, [32, 30] for time_str = time2
end
min, sec = get_mins_and_secs(time1)
min2, sec2 = get_mins_and_secs(time2)
min += min2
sec += sec2
if sec > 59
min += 1
sec -= 60
end
puts "#{min}min #{sec}sec"
Let's consider what's happening here. Firstly, you need to extract the minutes and seconds from the time strings. I made a method to do that:
def get_mins_and_secs(time_str)
time_str.scan(/\d+/).map(&:to_i)
#=> [10, 43] for time_str = time1, [32, 30] for time_str = time2
end
For time_str = "10min 43s", we apply the String#scan method to extract the two numbers as strings:
"10min 43s".scan(/\d+/) # => ["10", "43"]
Array#map is then used to convert these two strings to integers
["10", "43"].map {|e| e.to_i} # => [10, 43]
This can be written more succinctly as
["10", "43"].map(&:to_i} # => [10, 43]
By chaining map to to scan we obtain
"10min 43s".scan(/\d+/).map(&:to_i} # => [10, 43]
The array [10, 43] is returned and received (deconstructed) by the variables min and sec:
min, sec = get_mins_and_secs(time_str)
The rest is straightforward.
Here's a simple solution assuming that the format stays the same:
time1 = "10min 43s"
time2 = "32min 30s"
strings = [time1, time2]
total_time = strings.inject(0) do |sum, entry|
minutes, seconds = entry.split(' ')
minutes = minutes.gsub("min", "").to_i.send(:minutes)
seconds = seconds.gsub("s", "").to_i.send(:seconds)
sum + minutes + seconds
end
puts "#{total_time/60}min #{total_time%60}s"
Something like the following should do the trick:
# split the string on all the integers in the string
def to_seconds(time_string)
min, sec = time_string.gsub(/\d+/).map(&:to_i)
min.minutes + sec.seconds
end
# Divide the seconds with 60 to get minutes and format the output.
def to_time_str(seconds)
minutes = seconds / 60
seconds = seconds % 60
format("%02dmin %02dsec", minutes, seconds)
end
time_in_seconds1 = to_seconds("10min 43s")
time_in_seconds2 = to_seconds("32min 30s")
to_time_str(time_in_seconds1 + time_in_seconds2)
My solution that takes any number of time strings and return the sum in the same format:
def add_times(*times)
digits = /\d+/
total_time = times.inject(0){|sum, entry|
m, s = entry.scan(digits).map(&:to_i)
sum + m*60 + s
}.divmod(60)
times.first.gsub(digits){total_time.shift}
end
p add_times("10min 43s", "32min 55s", "1min 2s") #=> "44min, 40s"
p add_times("10:43", "32:55") #=> "38:43"

Ruby: given a date find the next 2nd or 4th Tuesday

I can't seem to find an elegant way to do this...
Given a date how can I find the next Tuesday that is either the 2nd or the 4th Tuesday of the calendar month?
For example:
Given 2012-10-19 then return 2012-10-23
or
Given 2012-10-31 then return 2012-11-13
October November
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 1 2 3
7 8 9 10 11 12 13 4 5 6 7 8 9 10
14 15 16 17 18 19 20 11 12 13 14 15 16 17
21 22 23 24 25 26 27 18 19 20 21 22 23 24
28 29 30 31 25 26 27 28 29 30
Scroll to the bottom if you just want to see what the end result can look like..
Using code snippets from some date processing work I've done recently in ruby 1.9.3.
Some upgrades to DateTime:
require 'date'
class DateTime
ALL_DAYS = [ 'sunday', 'monday', 'tuesday',
'wednesday', 'thursday', 'friday', 'saturday' ]
def next_week
self + (7 - self.wday)
end
def next_wday (n)
n > self.wday ? self + (n - self.wday) : self.next_week.next_day(n)
end
def nth_wday (n, i)
current = self.next_wday(n)
while (i > 0)
current = current.next_wday(n)
i = i - 1
end
current
end
def first_of_month
self - self.mday + 1
end
def last_of_month
self.first_of_month.next_month - 1
end
end
method_missing Tricks:
I have also supplemented the class with some method missing tricks to map calls from next_tuesday to next_wday(2) andnth_tuesday(2)tonth_wday(2, 2)`, which makes the next snippet easier on the eyes.
class DateTime
# ...
def method_missing (sym, *args, &block)
day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
dindex = ALL_DAYS.include?(day) ? ALL_DAYS.index(day.downcase) : nil
if (sym =~ /^next_[a-zA-Z]+$/i) && dindex
self.send(:next_wday, dindex)
elsif (sym =~ /^nth_[a-zA-Z]+$/i) && dindex
self.send(:nth_wday, dindex, args[0])
else
super(sym, *args, &block)
end
end
def respond_to? (sym)
day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
(((sym =~ /^next_[a-zA-Z]+$/i) || (sym =~ /^nth_[a-zA-Z]+$/i)) && ALL_DAYS.include?(day)) || super(sym)
end
end
Example:
Given a date:
today = DateTime.now
second_tuesday = (today.first_of_month - 1).nth_tuesday(2)
fourth_tuesday = (today.first_of_month - 1).nth_tuesday(4)
if today == second_tuesday
puts "Today is the second tuesday of this month!"
elsif today == fourth_tuesday
puts "Today is the fourth tuesday of this month!"
else
puts "Today is not interesting."
end
You could also edit method_missing to handle calls such as :second_tuesday_of_this_month, :fourth_tuesday_of_this_month, etc. I'll post the code here if I decide to write it at a later date.
Take a look at Chronic or Tickle, both are gems for parsing complex times and dates. Tickle in particular will parse recurring times (I think it uses Chronic as well).
Check out this gem, you might be able to figure out your answer
https://github.com/mojombo/chronic/
Since you already use Rails, you don't need the includes, but this works in pure Ruby as well for reference.
require 'rubygems'
require 'active_support/core_ext'
d = DateTime.parse('2012-10-19')
result = nil
valid_weeks = [d.beginning_of_month.cweek + 1, d.beginning_of_month.cweek + 3]
if valid_weeks.include?(d.next_week(:tuesday).cweek)
result = d.next_week(:tuesday)
else
result = d.next_week.next_week(:tuesday)
end
puts result
I think you should probably use a library if you're needing to branch out into more interesting logic, but if what you've described is all you need, the code below should help
SECONDS_PER_DAY = 60 * 60 * 24
def find_tuesday_datenight(now)
tuesdays = [*-31..62].map { |i| now + (SECONDS_PER_DAY * i) }
.select { |d| d.tuesday? }
.group_by { |d| d.month }
[tuesdays[now.month][1], tuesdays[now.month][-2], tuesdays[(now.month + 1) % 12][1]]
.find {|d| d.yday > now.yday }
end
Loop through the last month and next month, grab the tuesdays, group by month, take the 2nd and the 2nd last tuesday of the current month (If you actually do want the 4th tuesday, just change -2 to 3) and the 2nd tuesday of the next month and then choose the first one after the provided date.
Here's some tests, 4 tuesdays in month, 5 tuesdays in month, random, and your examples:
[[2013, 5, 1], [2013, 12, 1], [2012, 10, 1], [2012, 10, 19], [2012, 10, 31]].each do |t|
puts "#{t} => #{find_tuesday_datenight(Time.new *t)}"
end
which produces
[2013, 5, 1] => 2013-05-14 00:00:00 +0800
[2013, 12, 1] => 2013-12-10 00:00:00 +0800
[2012, 10, 1] => 2012-10-09 00:00:00 +0800
[2012, 10, 19] => 2012-10-23 00:00:00 +0800
[2012, 10, 31] => 2012-11-13 00:00:00 +0800
I'm sure it could be simplified, and I'd love to hear some suggestions :) (way too late &tired to even bother figuring out what the actual range should be for valid dates i.e. smaller than -31..62)
so here is the code that will resolve a weekday for a given week in a month (what you asked for with little sugar). You should not have problems if you are running inside rails framework. Otherwise make sure you have active_support gem installed. Method name is stupid so feel free to change it :)
usage: get_next_day_of_week(some_date, :friday, 1)
require 'rubygems'
require 'active_support/core_ext'
def get_next_day_of_week(date, day_name, count)
next_date = date + (-date.days_to_week_start(day_name.to_sym) % 7)
while (next_date.mday / 7) != count - 1 do
next_date = next_date + 7
end
next_date
end
I use the following to calculate Microsoft's patch Tuesday date. It was adapted from some C# code.
require 'date'
#find nth iteration of given day (day specified in 'weekday' variable)
findnthday = 2
#Ruby wday number (days are numbered 0-7 beginning with Sunday)
weekday = 2
today = Time.now
todayM = today.month
todayY = today.year
StrtMonth = DateTime.new(todayY,todayM ,1)
while StrtMonth.wday != weekday do
StrtMonth = StrtMonth + 1;
end
PatchTuesday = StrtMonth + (7 * (findnthday - 1))

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