How can I in a smart way get the first business day of a month after the weekend?
To get the first business day of the month (given dateInput), we can do:
firstBusdayMonth = fbusdate(year(dateInput),month(dateInput));
As an example, for November, using the above function will return Thursday November 1 as the first business day. However, the first business day of the month after the first weekend is Monday November 5. How can I get this latter date?
Notes:
The weekend does not have to be in the same month.
If the Monday is not a working day, then I would like it to return the next business day
This function will do the trick. Here is the logic:
Create a datetime array for all days in the given month.
Get the day numbers.
Create a logical array, true from the first Monday onwards (so after the first weekend, accounting for the last day of the previous month being a Sunday).
Create another logical array using isbusday to exclude Mondays which aren't working days.
Finding the first day number where these two logical arrays are true, therefore the first business day after the weekend.
Code:
function d = fbusdateAferWE( y, m )
% Inputs: y = year, m = month
% Outputs: day of the month, first business day after weekend
% Create array of days for the given month
dates = datetime( y, m, 1 ):days(1):datetime( y, m, eomday( y, m ) );
% Get the weekday numbers to find first Monday, 1 = Sunday
dayNum = weekday( dates );
% Create the logical array to determine days from first Monday
afterFirstWeekend = ( cumsum(dayNum==2) > 0 ).';
% Get first day which is afterFirstWeekend and a business day.
d = find( afterFirstWeekend & isbusday( dates ), 1 );
end
You could probably speed this up (although it will be pretty rapid already) by not looking at the whole month, but say just 2 weeks. I used eomday to get the last day of the month, which means I don't have to make assumptions about a low number of holiday days in the first week or anything.
Edit: Working with datenum speeds it up by half (C/O JohnAndrews):
function d = fbusdateAferWE( y, m )
% Inputs: y = year, m = month
% Outputs: day of the month, first business day after weekend
% Create array of days for (first 2 weeks of) the given month
dates = datenum(datetime(y,m,1)):datenum(datetime(y,m,eomday(y,m)))-14;
% Get the weekday numbers to find first Monday, 1 = Sunday
dayNum = weekday( dates );
% Create the logical array to determine days from first Monday
afterFirstWeekend = ( cumsum(dayNum==2) > 0 ).';
% Get first day which is afterFirstWeekend and a business day.
d = find( afterFirstWeekend & isbusday( dates ), 1 );
end
I would add something like this after your statement:
[DayNumber, DayName] = weekday(firstBusdayMonth);
if DayNumber > 2
day = 10 - DayNumber;
else
This works because 'weekday' will return a number between 1 (Sunday) and 7 (Saturday).
The fbusdate() function will never return a 1 or 7, so we can ignore those cases.
If weekday(fbusdate()) == 2, the first is on a Monday and the firstBusdayMonth variable doesn't need to be changed.
If weekday(firstBusdayMonth) returns between 2 and 6, we need to skip to the next week, so we subtract the weekday value from 10 to find the next Monday.
It might not be the most elegant solution, but should work.
Related
I use Jiffy for calculate the difference in between 2 date in month.
But I don't the good result.
For end = 2/4/2023 and start = 1/4/2022 I have 12 months.
For end = 1/4/2023 and start = 1/4/2022 I have 11 months (error: expected 12).
Thanks,
num month = Jiffy(end).diff(Jiffy(start), Units.MONTH);
Right now it is checking 12 at midnight to 12 at midnight of the end date which is 11 months 30 days. Now if the end date is one second greater than the correct date it should work. So a hackey solution is to add one day duration to the end date and check
end.add(duration(day:1))
What is the best way to calculate the numbered day of the week over an 8 week period?
For example, my app tracks a users progress over an 8 week course.
I'm stuck on how to determine the current numbered day of the week according to the weeks progress, for example, Week 6 - Day 5
I have the users course start date as DateTime from Firebase, and can obviously get DateTime.now() to calculate the difference between course start date and now. How do I go from this so that I can display for example:
Week 2 - Day 3 or Week 6 - Day 4
Let's say the course is started 15 days ago, you can get what you want by something like this:
DateTime now = DateTime.now();
DateTime start = now.subtract(Duration(days:15));
int allDays = now.difference(start).inDays;
int w = allDays ~/ 7 + 1;
int d = allDays % 7;
print('Week $w - day $d');
~/ is the Truncating Division Operator which gives you the Integer result of a division
We can use fbusdate to get the first business day of a month:
Date = fbusdate(Year, Month);
However, how do we get the first business day of a week?
As an example, during the week that I'm posting this, Monday 09/07/2017 was a holiday in the US:
isbusday(736942) % = 0
How do I determine that the first business day for this week would be the next day 736943?
I'm not aware of a builtin function that returns the first working day of a week, but you can obtain it by requesting the next working day after Sunday:
busdate(736941); % 736941 = Sunday 09/03/2017
Your desired fbusdateweek function can be done in one line using just the function weekday to get the first Sunday of the week then busdate to get the next business day after that:
dn = 736942; % Date number for any day in a week
Date = busdate(dn-weekday(dn)+1);
Note: busdate uses the function holidays by default to get all holidays and special nontrading days for the New York Stock Exchange. If necessary, you can define an alternate set of holidays for busdate to use as follows:
holidayArray = ...; % Some set of date numbers, vectors, or datetimes
Date = busdate(dn-weekday(dn)+1, 1, holidayArray);
This way you can define a set of localized holidays.
Solved it. Here is a function that is based on the answer of #m7913d:
function Busday = fbusdateweek(date)
% Return the first business day after Sunday
% 'date' is a datenum input
dperiod = date-6:date;
sundays = weekday(dperiod)==1;
sunday = find(sundays==1,1,'first');
datesunday = dperiod(sunday);
% -->
Busday = busdate(datesunday);
end
I want to write a matlab code to get which day is the first day of year and also get n and display which is the nth day of that year.
And also i don't know why it can not compare a(for example sat) and w(i)
(W=[sat,sun,....,fri]
Please help me I really can't make it work!
This is what I have done so far:
First=input('sat,sun,...,fri');
day=('a number between 1and 365');
day=mod(day,7);
w=[sat,sun,....,fri];
for i=1:7
if first==w(i)
disp(mod(i+day,7))
end
end
Note the syntax in the code below. I noticed you had quite a lot of errors in your syntax. Also, I recommend you use datenumand datestr as in the code below. Run help datenum and help datestr to get more information about the functions.
% User selects a year as a double
year = input('Select a year: ');
% the first day of that year as a value
date = datenum([num2str(year),'-01-01']);
% Get the name of the first day and diplay it
first = datestr(date,'dddd');
disp(['The first day of ', num2str(year), ' was a ', first])
% get nth day from user
day = input('Choose a number between 1 and 365: ');
% Add this value to the value of 1st jan on the selected year
newDate = date + day-1;
% Turn this date into a string and display it
nth = datestr(newDate,'dddd-dd-mmmm');
disp(['Day ', num2str(day),' of ', num2str(year), ' was ', nth])
If I've got a time_t value from gettimeofday() or compatible in a Unix environment (e.g., Linux, BSD), is there a compact algorithm available that would be able to tell me the corresponding week number within the month?
Ideally the return value would work in similar to the way %W behaves in strftime() , except giving the week within the month rather than the week within the year.
I think Java has a W formatting token that does something more or less like what I'm asking.
[Everything below written after answers were posted by David Nehme, Branan, and Sparr.]
I realized that to return this result in a similar way to %W, we want to count the number of Mondays that have occurred in the month so far. If that number is zero, then 0 should be returned.
Thanks to David Nehme and Branan in particular for their solutions which started things on the right track. The bit of code returning [using Branan's variable names] ((ts->mday - 1) / 7) tells the number of complete weeks that have occurred before the current day.
However, if we're counting the number of Mondays that have occurred so far, then we want to count the number of integral weeks, including today, then consider if the fractional week left over also contains any Mondays.
To figure out whether the fractional week left after taking out the whole weeks contains a Monday, we need to consider ts->mday % 7 and compare it to the day of the week, ts->wday. This is easy to see if you write out the combinations, but if we insure the day is not Sunday (wday > 0), then anytime ts->wday <= (ts->mday % 7) we need to increment the count of Mondays by 1. This comes from considering the number of days since the start of the month, and whether, based on the current day of the week within the the first fractional week, the fractional week contains a Monday.
So I would rewrite Branan's return statement as follows:
return (ts->tm_mday / 7) + ((ts->tm_wday > 0) && (ts->tm_wday <= (ts->tm_mday % 7)));
If you define the first week to be days 1-7 of the month, the second week days 8-14, ... then the following code will work.
int week_of_month( const time_t *my_time)
{
struct tm *timeinfo;
timeinfo =localtime(my_time);
return 1 + (timeinfo->tm_mday-1) / 7;
}
Assuming your first week is week 1:
int getWeekOfMonth()
{
time_t my_time;
struct tm *ts;
my_time = time(NULL);
ts = localtime(&my_time);
return ((ts->tm_mday -1) / 7) + 1;
}
For 0-index, drop the +1 in the return statement.
Consider this pseudo-code, since I am writing it in mostly C syntax but pretending I can borrow functionality from other languages (string->int assignment, string->time conversion). Adapt or expand for your language of choice.
int week_num_in_month(time_t timestamp) {
int first_weekday_of_month, day_of_month;
day_of_month = strftime(timestamp,"%d");
first_weekday_of_month = strftime(timefstr(strftime(timestamp,"%d/%m/01")),"%w");
return (day_of_month + first_weekday_of_month - 1 ) / 7 + 1;
}
Obviously I am assuming that you want to handle weeks of the month the way the standard time functions handle weeks of the year, as opposed to just days 1-7, 8-13, etc.