I have a start_at, a decimal quantity and an interval which is one of day | week | month | year.
start_at = Time.parse('2016-01-01 00:00:00 UTC') # leap year
quantity = BigDecimal.new('1.998') # changed from 2.998, should end on 2/29/16 sometime
interval = 'month' # could be any of day|week|month|year
With whole numbers, I've used duration i.e. 1.month, and I looked at Date#advance, though it only recognizes integer values.
It would seem simple but I cannot find anything in the standard libraries or in ActiveSupport.
References:
SO answer potentially used for input to Date#advance?
SO explanation of duration
Question
How can I establish the end_at date from a decimal?
Why? What purpose?
Proration to the second for a given amount and given interval.
Expectations
I'm looking for an end_at to the second as accurate as possible with respect to advancing the next interval(s) by the decimal quantity. Given interval = 'month', for the fractional part, when you pass the start of the month, means you are in that month and using it's duration. For example, January 2016 is 31 days, while February (leap) is 29 days (only in the leap year).
I'd say your best option is to use Ruby's date methods to advance time based on the whole number of the decimal, then calculate how many seconds your fraction is of your current interval.
So for 1.998 months, advance time 1 month. Find the current month you are in and get the .998 of the seconds in that month (i.e. for July, 31x24x60x60x.998) and then advance time that many seconds.
What does advancing time a fractional month mean?
Lets say we have the following date 2015-01-01 00:00:00 UTC. It is easy to advance exactly 1 whole month, we simply increment the number that represents months: 2015-02-01 00:00:00 UTC. Alternatively, we could view this as adding 31 days, which we know is the number of days in January.
But what if we want to advance 0.5 months from 2015-01-01 00:00:00 UTC?
We can't just increment like we did when advancing a whole month. Since we know January has 31 days, perhaps we could just advance 15.5 days: 2015-01-16 12:00:00 UTC. That sort of works.
How about 1.5 months from 2015-01-01 00:00:00 UTC? If we combine our previous approaches, we'd first increment, getting us to 0.5 left to advance and 2015-02-01 00:00:00 UTC. Then we'd take half of 28 and get to 2015-02-15 00:00:00 UTC.
But wait, what if instead we took the total number of days between the two months and then took 3/4 of that? Like 2(month) * (3/4), which would simplify to (3(month)) / 2, or 1.5(month). Lets try it.
(28 days + 31 days) * 0.75 = 44.25 days
Now adding that to 2015-01-01 00:00:00 UTC we get 2015-02-14 06:00:00 UTC. That's three-quarters of a day off from our other answer.
The problem here is that the length of a month varies. So fractional months are not consistently definable.
Imagine you have two oranges. One contains a little bit more juice than the other (perhaps 31ml and 29ml of juice). Your recipe calls for the juice of 1.5 oranges. Depending on which one you decide to cut in half, you could have either 44.5 ml or 45.5 ml. But if your recipe calls for 40 ml of orange juice, you can pretty consistently measure that. Much like you can consistently (kind of) increment a date by 40 days.
Time is really tricky. We have leap seconds, leap years, inconsistent units (months), timezones, daylight saving time, etc... to take into account. Depending on your use case, you could attempt to approximate fractional months, but I'd highly recommend trying to avoid the need for dealing with fractional months.
Related
I want to count say how many Mondays we have from 2022-02-01 - 2022-03-01. I found smth like this:
=SUMPRODUCT(WEEKDAY(B4:C4)=2) - B4 and C4 are the dates
But it returns 0. I assume it only checks if specific date is the specific day. Any ideas how I can do this but for a date range? So how count how many Mondays there are in February
I also found this
=NETWORKDAYS.INTL(B4;C4;"1000000")
but this returns 25
You can take advantage of the NETWORKDAYS.INTL function by using the string method to make all the days as weekend except for Monday.
The String method states:
weekends can be specified using seven 0’s and 1’s, where the first number in the set represents Monday and the last number is for Sunday. A zero means that the day is a work day, a 1 means that the day is a weekend. For example, “0000011” would mean Saturday and Sunday are weekends.
In this case since you only want to know the Mondays, the string would be "0111111" and the function would look like:
=NETWORKDAYS.INTL(StartDate,EndDate,"0111111")
I think this is right. It's counting inclusively so you would get five Mondays starting on Monday 7th Feb 2022 and ended on Monday 7th March 2022 for example.
=floor((B2-(A2+7-weekday(A2,12)))/7)+1
where A2 and B2 contain the start date end end date.
Obvs nul points for me again but for the record this could be generalised if you put the day number in C2 (e.g. 1 if you want to find Sundays, 2 for Mondays):
=floor((B2-(A2+7-weekday(A2,10+C2)))/7)+1
I want to take a time range and convert it to minutes within a table of 30 minute segments. For example, Bob works from 3:35 AM to 5:00 AM. So within the table from 3:30 AM - 4:00 AM should represent 25 minutes, because Bob was working a total of 25 minutes during that time slot. Then the time slots for 4:00 - 4:30 AM and 4:30 - 5:00 AM would both have 30 in their corresponding cells.
Apologies to leave this so broad but I honestly have no idea where to start and this forum has been immensely helpful.
https://docs.google.com/spreadsheets/d/1YpHU-UHlqXL6c8I27zSDZaRu72ViUw5W6RPru-HE3Iw/edit#gid=0
Any help is appreciated.
For each 30-minute interval, you have to check whether these two conditions are met:
The interval start time (3:30) falls between the employee start and end time (3:35 and 5:00).
The interval end time (4:00) falls between the employee start and end time.
If any of these conditions are met, the working time for that interval won't be 0. The working time will be the difference between the minimum of both end times (4:00 and 5:00) and the maximum of both start times (3:30 and 3:35).
Translated to sheets functions, you could do something like this:
=IF(MIN($D2,O$1)-MAX($C2,N$1)>0,TIMEVALUE(MIN($D2,O$1)-MAX($C2,N$1))*24*60,0)
Or, alternatively, this:
=IF(OR(AND($C2<N$1,N$1<$D2),AND($C2<O$1,O$1<$D2)),TIMEVALUE(MIN($D2,O$1)-MAX($C2,N$1))*24*60,0)
I have a Date variable with a person's date birthday. I would like to know how many days remains before this person next birthday. It should be calculated from today date to current year birthday date.
How can this be done with Swift? Also it will be great to consider February 29 in leap years.
To the guys who tried to close this: This is about birthday which has totally different rules from days.
Birthdays are complicated. Your birthday was the date of the moment when you were born, in the timezone where you were born. Considering that Samoa = UTC+14 and Baker Island = UTC-12, it is possible that people born at the same moment in time have birthdays that are two days apart.
So to store somebody's birthday, not the moment of birth, you either store year/month/day, or if you want to store it as a point in time, you store the beginning of that day in UTC, with the understanding that this is to specify a day, and must not be converted to local time.
Now when does your birthday repeat? If the person is born on D/M/Y and D/M is not February 29th, then the next birthday is either D/M/current year or D/M/next year. It is D/M/current year if that date is in the future, otherwise D/M/next year.
If the person is born on February 29th, then you have to determine when officially the next birthday is if the year is not a leap year - this will be February 28th or March 1st, depending on which rules apply.
We also need to clarify what "number of days" means. If my birthday is on April 1st, and now it is March 31st, one second to midnight, my birthday will be one second from now. However, I will assume that the result is supposed to be "one day from now".
So here is the algorithm:
Step 1: Find day/month/year when the person was born.
Step 2: Determine the current time, and convert it to local day/month/year. Determine the current time only once to avoid problems if this calculation is done nanoseconds before midnight.
Step 3: Determine the year when the birthday repeats: If day/month of birthday is greater than current day/month, then the year when the birthday repeats is the current year, otherwise the next year. This is also correct if the birthday was on Feb. 29th.
Step 4: Determine the day/month when the birthday repeats: This is the same as the day/month of the birthday, unless the birthday was on Feb. 29th and the year when the birthday repeats is not a leap year. In that case, the birthday repeats on Feb 28th or March 1st, depending on which rules you decide to apply.
Step 5: Convert the current day/month/year + 12 hours to UTC. Convert the date when the birthday repeats + 12 hours to UTC. Calculate the difference in seconds (which the OS should do for you). Divide by 86,400, then round to the nearest integer. The "+12 hours" and "round to nearest integer" make sure that you have no problems with daylight savings time, leap seconds etc.
Writing this code in Swift or any other language should be no problem.
It depends on what you are looking to use the days value for but here is a small function that will return a Double describing the amount of days until a given Date. Martin R gave a really good answer here and my answer is mainly based on theirs with a little bit of documentation added.
/// This function takes a `Date` parameter and returns an `Int` describing how many days away the `Date` is into the future (accounting for leap years by making 2/29 dates land on 3/1 in non-leap years).
/// - Parameter date: The `Date` object to be evaluated.
func daysUntil(birthday: Date) -> Int {
let cal = Calendar.current
let today = cal.startOfDay(for: Date())
let date = cal.startOfDay(for: birthday)
let components = cal.dateComponents([.day, .month], from: date)
let nextDate = cal.nextDate(after: today, matching: components, matchingPolicy: .nextTimePreservingSmallerComponents)
return cal.dateComponents([.day], from: today, to: nextDate ?? today).day ?? 0
}
There's the subtract() method but the documentation says it's not aware of daylight savings which makes it pretty much useless in this case, or in any other case except where the programmer doesn't know how many milliseconds there are in 24 hours.
I'm thinking of two ways:
get the day of the month, and then subtract N from it and if it's less than 1 then subtract the month and the year if appropriate and set the day for the last day of whichever month it turns out to be
OR
subtract N days from the noon of the current day and then get start of the day for the resulting day
Is there some easier/better way to do this?
You should probably try to convert both DateTimes to UTC (standardize), then call difference(). That converts it to a nice, easy Duration, which you can convert as necessary to hours, days, months, or whatever else.
DateTime one = somedatetime.toUtc();
DateTime two = someotherdatetime.toUtc();
Duration diff = one.difference(two);
//Then just convert...
return diff.inDays;
Trying to parse and XLSX file using roo gem in a ruby script.
In excel dates are stored as floats or integers in the format DDDDD.ttttt, counting from 1900-01-00 (00 no 01). So in order to convert a date such as 40396 - you would take 1900-01-00 + 40396 and you should get 2010-10-15, but I'm getting 2010-08-08.
I'm using active_support/time to do calculation like so:
Time.new("1900-01-01") + 40396.days
Am I doing my calculation wrong or is there a bug in active support?
I'm running ruby 1.9.3-mri on Windows 7 + latest active_support gem (3.2.1)
EDIT
I was looking at the older file in Excel with the wrong data - my script / console were pulling the right data - hence my confusion - I was doing everything right, except for using the right file!!!! Damn the all-nighters!
Thanks to everyone replying, I will keep the question here in case somebody needs info on how to convert dates from excel using ruby.
Also for anyone else running into this - spreadsheet gem DOES NOT support reading XLSX files at this point (v 0.7.1) properly - so I'm using roo for reading, and axlsx for writing.
You have an off-by-one error in your day numbering - due to a bug in Lotus 1-2-3 that Excel and other spreadsheet programs have carefully maintained compatibility with for 30+ years.
Originally, day 1 was intended to be January 1, 1900 (which would, as you stated, make day 0 equal to December 31, 1899). But Lotus incorrectly considered 1900 to be a leap year, so if you use the Lotus numbers for the present and count backwards, correctly making 1900 a common year, the day numbers for everything before March 1st, 1900, are one too high. Day 1 becomes December 31st, 1899, and day 0 shifts back to the 30th. So the epoch for date arithmetic in Lotus-based spreadsheets is really Saturday, December 30th, 1899. (Modern Excel and some other spreadsheets extend the Lotus bug-compatibility far enough to show February 1900 actually having a 29th day, so they will label day 0 "December 31st" while agreeing that it was a Saturday! But other Lotus-based spreadsheets don't do that, and Ruby certainly doesn't either.)
Even allowing for this error, however, your stated example is incorrect: Lotus day number 40,396 is August 6th, 2010, not October 15th. I have confirmed this correspondence in Excel, LibreOffice, and Google sheets, all of which agree. You must have crossed examples somewhere.
Here's one way to do the conversion:
Time.utc(1899,12,30) + 40396.days #=> 2010-08-06 00:00:00 UTC
Alternatively, you could take advantage of another known correspondence. Time zero for Ruby (and POSIX systems in general) is the moment January 1, 1970, at midnight GMT. January 1, 1970 is Lotus day 25,569. As long as you remember to do your calculations in UTC, you can also do this:
Time.at( (40396 - 25569).days ).utc # => 2010-08-06 00:00:00 UTC
In either case, you probably want to declare a symbolic constant for the epoch date (either the Time object representing 1899-12-30 or the POSIX "day 0" value 25,569).
You can replace those calls to .days with multiplication by 86400 (seconds per day) if you don't need active_support/core_ext/integer/time for anything else, and don't want to load it just for this.
"Excel stores dates and times as a number representing the number of days since 1900-Jan-0, plus a fractional portion of a 24 hour day: ddddd.tttttt . This is called a serial date, or serial date-time." (http://www.cpearson.com/excel/datetime.htm)
If your column contains a date time, rather then just a date, the following code is useful:
dt = DateTime.new(1899, 12, 30) + excel_value.to_f
Also keep in mind that there are 2 modes of dates in an excel worksheet, 1900 based and 1904 based, which typically is enabled by default for spreadsheets created on the mac. If you consistently find your dates off by 4 years, you should use a different base date:
dt = DateTime.new(1904, 1, 1) + excel_value.to_f
You can enable/disable 1904 date mode for any spreadsheet, but the dates will then appear off by 4 years in the spreadsheet if you change the setting after adding data. In general you should always use 1900 date mode since most excel users in the wild are windows based.
Note: A gotcha with this method is that rounding might occur +/- 1 second. For me the dates I import are "close enough" but just something to keep in mind. A better solution might use rounding on fractional seconds to solve this issue.
You're doing your calculation wrong. How do you arrive at the expected result of 2010-10-15?
In Excel, 40396 is 2010-08-06 (not using the 1904 calendar, of course). To demonstrate that, type 40396 into an Excel cell and set the format to yyyy-mm-dd.
Alternatively:
40396 / 365.2422 = 110.6 (years -- 1900 + 110 = 2010)
0.6 * 12 = 7.2 (months -- January = 1; 1 + 7 = 8; 8 = August)
0.2 * 30 = 6 (days)
Excel's calendar incorrectly includes 1900-02-29; that accounts for one day's difference between your 2010-08-08 result; I'm not sure about the reason for the second day of difference.