I have start dates with times and end dates with times tracking when work was entered into a status and when it was completed. I'd like to know how many working hours were spent doing the task.
For example:
Start Date
End Date
Working Hours (Time Spent between 9 AM - 5 PM)
8/1/2022 10:03 AM
8/5/2022 4:43 PM
?
8/8/2022 9:03 AM
8/10/2022 3:34 PM
?
8/10/2022 11:13 AM
8/15/2022 4:57 PM
?
I'd like to know how many hours were spent working between 9 AM and 5 PM, excluding weekends and public holidays.
How can I do this calculation in Google Sheets?
Try:
=(NETWORKDAYS(A2,B2)-1)*("5:00 PM"-"9:00 AM") +IF(NETWORKDAYS(B2,B2),MEDIAN(MOD(B2,1),"5:00 PM","9:00 AM"),"5:00 PM") -MEDIAN(NETWORKDAYS(A2,A2)*MOD(A2,1),"5:00 PM","9:00 AM")
Drag down to column.
Result:
Using NETWORKDAYS() function to only count the working days excluding weekends, then multiply it by the number of hours between 9AM and 5PM.
This will return a number which is in a form of the total number of days. You then format the cell to Duration.
To format the cell:
Highlight the cells -> Then on the menu click Format -> Number -> Duration
Final Result:
References:
NETWORKDAYS()
Get Work hours between dates
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 am new to google sheets, but I have this very simple and basic tracking system for my work schedule. I wish to make a formula so it automatically finds the number of hours on nights and weekends column, after a specific time.
So ex. night hours between 9 pm-11 pm Monday - Friday and weekend hours between 2 pm-8 pm on Saturday and Sunday.
Check if you get the expected hours (sorry for the layout a bit different!)
https://docs.google.com/spreadsheets/d/1YuY_8XgMUsPtBPJlBgxDFgz8JKB3lCmn2e6vVEutvBY/edit?usp=sharing
test if we =WEEKDAY(A4,2)>5
day on monday-friday =if(D4,, max(0,min(21/24,C4)-B4))
night on monday-friday =if(D4, ,max(0,C4-max(21/24,B4)))
If I have week 7 in 2017 what week date is the Monday in that week in Google Sheets?
=DATE(B9,1,1)-WEEKDAY(DATE(B9,1,1),3)+7*(WEEKDAY(DATE(B9,1,1),3)>3)+7*(A9-1)
is the least complicated formula I know which works for week numbers in Sweden (i.e. Monday first day of week, ISO rules for what is week 1).
Short answer (A1==Week, B1==Year):
=DATE(B1;1;1)+((A1-1)*7)-WEEKDAY(DATE(B1;1;1);3)
Long answer:
DATE(<year>;1;1) // days since 1970 until the frist day of the year
plus
((<week number>-1)*7) // how many days into the year is this week
minus
WEEKDAY(DATE(<year>;1;1);3) // how many extra days from previous year in first week
PS:
This assumes monday as the first day of week you have to change the arguments for WEEKDAY to change it to sunday
Because of this definition (https://en.wikipedia.org/wiki/Week) the 4th of January must be used instead the 1st. The 4th of January is the first day which is always in the week 1.
=DATE(B1;1;4)+((A1-1)*7)-WEEKDAY(DATE(B1;1;4);3)
If you are using ISO weeks, the accepted answer doesn't account for weeks overlapping on 2 technical years like 2020-w53, which is from 28 Dec 2020 until 3 Jan 2021.
Therefore I'm using this formula instead:
=DATE(K2,1,1)-WEEKDAY(DATE(K2,1,1),2)+7*(WEEKDAY(DATE(K2,1,1),2)>3)+7*(L2-1) +1
Where K is the Year, and L is the Week number (split in 2 columns from yyyy-ww)
to have it in an arrayformula:
=ArrayFormula(if(K2:K="",, DATE(K2:K,1,1)-WEEKDAY(DATE(K2:K,1,1),2)+7*(WEEKDAY(DATE(K2:K,1,1),2)>3)+7*(L2:L-1) +1 ))
You can use =ArrayFormula(if(E2:E="",,split(E2:E,"-"))) to split yyyy-ww in two columns.
NOTE: This formula would return the Monday (Which is the first day of the week in international standard, ISO)
Worked this up for 2023. It will work through end of 2024 too .. that said the AND logic is flawed .. feel free to suggest something to make this better
=IFS(
AND(ISOWEEKNUM(A8)=52,YEAR(A8)<>YEAR(A7)),
DATE(YEAR(A8-1),1,1)-WEEKDAY(DATE(YEAR(A8-1),1,1),3)+7*(WEEKDAY(DATE(YEAR(A8-1),1,1),3)>3)+7*(ISOWEEKNUM(A8)-1),
DATE(YEAR(A8),1,1)-WEEKDAY(DATE(YEAR(A8),1,1),3)+7*(WEEKDAY(DATE(YEAR(A8),1,1),3)>3)+7*(ISOWEEKNUM(A8)-1)
)
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.
Looking for a formula (for google spreadsheets) that shows either the days past in the current month if we're not past the last day of the current month, or the total days in that month if that month is behind us.
So, if today's March 25th, the formula would output 25 ... if today's April 1st, though, the formula would output 31.
Based on your sample, this should work (assuming the date is in A1):
=if(day(A1)=1,A1-date(year(A1),month(A1)-1,1),A1-date(year(A1),month(A1),1)+1)