I'm using SAS 9.3
I need to create a way to sum up by week total, and I have no idea how to do it. So basically I have a year list of dates (left column below) with a total from that date (the right column). Our week goes from Friday to the previous Thursday (e.g. Thursday Oct 17 through Friday the Oct 25th).
An issue I also have is as you see the dates on the left are not completely daily and don't always have a Thursday date before the last Friday date. Would any know a way to add these weeks up - Week 1, Week 2, etc etc ...?
Thanks for any help that can be provided
2013-01-01 3
2013-01-02 8
2013-01-03 8
2013-01-04 10
2013-01-06 1
2013-01-07 10
2013-01-08 14
2013-01-09 12
2013-01-10 8
2013-01-11 9
2013-01-12 1
2013-01-14 12
2013-01-15 8
2013-01-16 5
2013-01-17 15
2013-01-18 7
2013-01-20 1
Trivial way:
data want;
set have;
weekno = ceil((date-'03JAN2013'd)/7);
run;
IE, subtract the first thursday and divide by 7, (so 1/1-1/3 is weekno=0).
INTCK function is also adept at calculating this. The basic structure is
weekno=intck('WEEK.5','04JAN2013'd,date); *the second argument is when you want your first week to start;
WEEK means calculate weeks, # on left side of decimal is multiple week groups (2 week periods is WEEK2.), on right side is shift index (from the default sunday-saturday week).
You could also create a format that contained your weeks, and use that.
Related
I need help on calculating my start date for my report date parameters.
The end date will always be the last Sunday, here: =DateAdd("d", 1 - WeekDay(Today(), 1), Today())
What I need help with is how to write a formula to go back 6 months from today and pick the 1st Saturday in that range..
Thanks in advance.
Assuming your start day is a Sunday then you can use this...
=DATEADD(
DateInterval.Day,
7 - WEEKEDAY( DATEADD(DateInterval.Month,-6,Today()), FirstDayOfWeek.Sunday),
DATEADD(DateInterval.Month,-6,Today())
)
This works as follows
WEEKEDAY( DATEADD(DateInterval.Month,-6,Today()), FirstDayOfWeek.Sunday)
Takes today's date, subtracts 6 months and then finds out what daynumber that is. Running that today (2018-11-08) gives use (2018-05-10) which is a Thursday, this is day number 5
Saturdays are day number 7 (if your first day of week is a Sunday). As there can be no higher number than 7 we can do a simple subtraction of 7 minus the day number we landed on (from above) which gives us a required adjustment of 2 days.
Finally the outer DATEADD function simply says add our calculated 2 days to the date 6 months ago.
Hope that makes sense!?
If the first day of the week is not a Sunday for you then you may have to do some Mod% calc on the second argument to calculate the correct number of days to adjust by.
Would it be possible to get same day of week last year using Excel? please below example:
Input: Monday 9 Nov 2015 | Output: 10 Nov 2014
Thanks
Simply subtract 52 full weeks with 7 days = 364 days. So if the date is in A1, the formula =A1-364 will get the date exactly 52 weeks before, which is the same day of week in the year before.
To show that it works even for leap years, try the following:
You see the formula date - 364 (=A2-364, =A3-364, ...) always gets the same day of week a year before. That is because it gets the day minus 52 full weeks (52 * 7 days) before. In leap years it gets a different day but the same day of week.
Try this:
=DATE(YEAR(A1)-1,MONTH(A1),DAY(A1))+WEEKDAY(A1)-WEEKDAY(DATE(YEAR(A1)-1,MONTH(A1),DAY(A1)))
It returns the closest date within a week. A1 is the cell with this year's date.
I need to convert the week field into Date(mm/dd/yyyy).
Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8 Week 9 Week 10 Week 11 Week 12 Week 13 Week 14 Week 15 Week 16 Week 17 Week 18 Week 19 Week 20 Week 21 Week 22
Actually I have dragged the date field into Columns and I selected the week.
I need to convert this into Date (mm/dd/yyyy). How to achieve this?
Directly in your sheet, if you right click on your date field, you'll get a menu with the following options: "Filter", "Show quick filter", etc. You'll also see an option which says "Day.... May 8, 2011". That gives you a full date.
But if you want that date to have the "mm/dd/yyyy" format, you'll need to change it from the left pane. Right clicking on your date dimension provides the following option: "Default Properties/Date Format". You have a few default formats in there, including a custom formatting where you could make it mm/dd/yyyy.
Olivier
I think you want this formula: DATETRUNC('week', [day], 'Monday'), assuming your week starts on Monday. You may also need to change the format to m/d/y in the left-hand pane, per Jim Dehner: https://community.tableau.com/thread/234590
For dates 1/2 - 1/9/2017 this will return (Monday) 1/2/2017. For 1/9 - 1/16, it will return 1/9/2017, etc.
I believe this post could help you out: https://community.tableau.com/s/question/0D54T00000C6Y9TSAV/how-to-convert-week-number-into-a-date-format
Basically you can use this formula:
DATETRUNC('week',DATE(DATEADD('week', WEEK(TODAY())-1, DATE("01/01/"+STR(YEAR(TODAY()))))))
Postgresql 8.4.
I'm new to this concept so if people could teach me I'd appreciate it.
For Obamacare, anyone that works 30 hours per week or more must be offered the same healthcare as is offered to any other worker. We can't afford that so we have to limit work hours for temp and part-timers. This is affecting the whole country.
I need to calculate the hours worked (doesn't matter if overtime,
regular time, double time, etc) between two dates, say Jan 1, 2014,
and Nov 1, 2014 (Saturday) for each custom week (which beings on Sunday), not the week as defined by Postgresql (which begins on Monday).
Each of my custom work weeks begins on Sunday and ends on Saturday.
I don't know if I have to include weeks where
they did not work at all in the average, but let's assume I do. Zero hours that week would draw down the average.
Table name is 'employeetime', date field is 'employeetime.stopdate', hours worked per day is in the field 'employeetime.hours', employeeid field is 'employeetime.empid'.
I'd prefer to do this in one query per employee and I will execute the query once per employee as I loop through employees. If not I'm open to suggestions. But I'd like to understand the SQL presented in the answer.
Currently EXTRACT(week from '2014-01-01') calculates the start of the week as a Monday, so that doesn't work for me. Link here.
How would I do that without doing, say a separate query for each week, per person? We have 200 people to process.
Thank you.
I have set up a table to match your format:
select * from employeetime order by date;
id date hours
1 2014-11-06 10
1 2014-11-07 3
1 2014-11-08 5
1 2014-11-09 3
1 2014-11-10 5
You can get the week starting on Sunday by shifting. Note, here the 9th is a Sunday, so that is where we want the boundary.
select *, extract(week from date + '1 day'::interval) as week
from employeetime
order by week;
id date hours week
1 2014-11-07 3 45
1 2014-11-06 10 45
1 2014-11-08 5 45
1 2014-11-09 3 46
1 2014-11-10 5 46
And now the week shifts on Sunday rather than Monday. From here, the query to get hours by week/employee would be simple:
select id, sum(hours) as hours, extract(week from date + '1 day'::interval) as week
from employeetime
group by id, week
order by id, week;
id hours week
1 18 45
1 8 46
I'm trying to separate weeks from timestamp per quarter so it should be between 1-13 week per quarter so I used function week() but it takes between 1-52 week as whole year so I made it to be divided by function of quarter like below
select Week (EVENTTIMESTAMP) / QUARTER (EVENTTIMESTAMP) from KAP
The thing here that results aren't accurate; for example it shows:
time stamp 2014-07-06 12:13:03.018
week number 9
which isn't correct because July is first month in Q3 and it's in the 6 days so it should be 1 week from Q3 not 9.
Any suggestion where it go wrong?
You want something like WEEK modulo 13 to get week number within a quarter. You will have to tinker with 'modulo 13 yields 0..12' by adding or subtracting one at appropriate points.
Some minimal Google searching using 'ibm db2 sql modulo' yields DB2 MOD function:
The MOD function divides the first argument by the second argument and returns the remainder.
Hence MOD(WEEK(...), 13), except you probably need MOD(WEEK(...)-1, 13) + 1, as intimated already.
You may need to watch for what the WEEK() function does at year ends:
The WEEK function returns an integer in the range of 1 to 54 that represents the week of the year. The week starts with Sunday, and January 1 is always in the first week.
I'm curious about how they can come up with week 54. I suppose it requires 1st January to be a Saturday (so 2nd January is the start of week 2) of a leap year, as in 2000 and 2028. Note that week 53 and (occasionally) week 54 will show up as weeks 1 and 2 of Q5 unless you do something. Also, Saturday 2000-03-25 would be the end of Q1 and Sunday 2000-03-26 would be the start of Q2 under the regime imposed by the WEEK() function and a simple MOD(WEEK(...), 13) calculation. You're likely to have to tune this to meet your real requirements.
There's also the WEEK_ISO() function:
The WEEK_ISO function returns an integer in the range of 1 to 53 that represents the week of the year. The week starts with Monday and includes seven days. Week 1 is the first week of the year that contains a Thursday, which is equivalent to the first week that contains January 4.
Note that under the ISO scheme, the 3rd of January can be in week 52 or 53 of the previous year, and the 29th of December can be in week 1 of the next year. Curiously, there doesn't seem to be a YEAR_ISO() function to resolve such ambiguities.
In a data warehouse, the proper solution to this is to create a time dimension that contains static mappings for days/weeks/months/quarters/years. This provides the ability to define these based on your business' fiscal calendar (if it is not following on the calendar year).
See: http://www.kimballgroup.com/1997/07/10/its-time-for-time/ for more information.