I have a list of cost figures with start dates and end dates which I need to split between months, I have searched for the solution to this problem but cannot seem to find one that will work with partial months i.e.( startdate:01/01/2015 enddate: 15/04/2015 cost:10000) which would leave figures like Jan:2857, Feb:2857, Mar:2857, Apr:1429.
I have been trying to modify this example: http://www.excel-university.com/excel-formula-to-allocate-an-amount-into-monthly-columns/ but having no luck getting the partial months working.
Any suggestions or help would be most welcome. Thanks in Advance
if you calculate it on daily basis, would it be ok? the result would be:
01.01.2015 01.02.2015 01.03.2015 15.04.2015
2.857,14 2.857,14 2.857,14 1.428,57
your daily amount is:
=10.000/(DAYS360(startdate;enddate;TRUE)+1)
(be carefull of true and false argument)
under the dates or instead of 2.857,14 etc. insert the formula:
=IF(DAY("your date")>1;DAY("your date");30) * daily amount
This formula assumes that you want to have 30 days in each month:
=IF(DAY(01.01.2015)>1;DAY(01.01.2015);30)
result = 30
=IF(DAY(15.04.2015)>1;DAY(15.04.2015);30)
result = 15
so if months begins with a date different from the 1st it will give you the number of days.
if you want to match months with your startdate and enddate (if i understood your comment correctly), you could do:
=IF(OR(
AND(MONTH(startdate)=MONTH(your date);YEAR(startdate)=YEAR(your date));
AND(MONTH(enddate)=MONTH(your date);YEAR(enddate)=YEAR(your date))
);"match";"no match")
by this you make sure that month and year correspond.
If you want to get the number of days in a month automatically, you could use:
=DAY(DATE(YEAR("your date");MONTH("your date")+1;1)-1)
but this does not assume anymore 30 days, you can change it with if statement
I hope this helps,
Best - AB
Related
Testing one of my scripts it appears that Numpy busday_count includes Labor Day (September 2nd). Is there a way to disable the holidays?
print (np.busday_count('2019-08-31', '2019-09-02'))
Result = 0, but 9/2/2019 is clearly Labor Day and a Monday.
print (np.busday_count('2019-08-31', '2019-09-03'))
Result = 1, but it should be 2. As far as I know the formula excludes the first date but includes the second date. I also tried:
print (np.busday_count('2019-08-31', '2019-09-02', holidays=[]))
Result still = 0 despite attempt to clear any holidays. What am I missing?
As so often, the answer lies with the documentation and Python's love for half-open intervals:
Counts the number of valid days between begindates and enddates, not including the day of enddates.
I'm having trouble writing code in VBA that would allow me to input any given date then have an output of the current work week. I need a restriction of if the date is Sunday through Tuesday, it will keep that current work week and year but if the date is Wednesday through Saturday, then the next work week and year will show. For example, I'm looking to input (5/28/19) and have an output of 201922 or an input of (5/29/19) with an output of 201923 even though its technically the same work week.
Before getting too in depth, I do have a working function that provides the year and work week, but I'm trying to adapt the function or add a separate function that will change in to the next work week according to the given date.
I'm new to VBA but have tried to do a little research over the last few days. I was thinking that I could somehow have one input of the date then have two outputs where one would be the year and work week then the other would be the number associated with that date (1 for Sunday, 2 for Monday, and so on). I tried to create an if then statement that says if the number associated with that date is 1, 2, or 3 then the workweek would stay the same. If it was any other number then 1 would be added to that work week so it would move to the next one. I'm having trouble with trying to make two outputs and have them connected, if that makes any sense.
This is the code that I tried to create, but continuously failed at making. The function that gives the correct work week (without the adaptation of the work week based on the date weekday) is WWV1
Function WWV2(WeekdayName As Integer)
Dim WWV1 As Integer
If WeekdayName(Date) = 1 Or 2 Or 3 Then WWV1 = WWV1
Else: WWV1 = WWV1 + 1
End Function
This provides the cell with #NUM! when I use the function in that cell, which I assume is because I need to somehow connect the two functions.
how about this:
=YEAR(A1) & TEXT(WEEKNUM(A1,13),"00")
WeekNum returns the weeknumber with 13 saying it starts on Wednesday:
VBA
wkcd = Year(Range("A1")) & Format(Application.WeekNum(Range("A1")),"00")
I have this table in excel:
Date value
1/2/1970 100.00
1/5/1970 99.99
1/6/1970 100.37
1/7/1970 100.74
1/8/1970 101.26
1/9/1970 100.74
1/12/1970 100.79
1/13/1970 101.27
1/14/1970 101.95
1/15/1970 101.97
1/16/1970 101.76
1/19/1970 102.21
1/20/1970 102.70
1/21/1970 102.00
1/22/1970 101.46
1/23/1970 101.49
1/26/1970 100.97
1/27/1970 101.45
1/28/1970 101.70
1/29/1970 102.08
1/30/1970 102.19
2/2/1970 102.02
2/3/1970 101.85
These are values that I have daily, and I need to construct a sheet that takes a monthly index of the daily values, example below:
date index
1/31/1970 some_index
2/28/1970 some_index
3/31/1970 some_index
4/30/1970 some_index
I could only get this far when it came to getting the index of 30 days:
=AVERAGE(INDEX(B:B,1+30*(ROW()-ROW($C$1))):INDEX(B:B,30*(ROW()-ROW($C$1)+1)))
I'm just not sure how to structure this in the most efficient, yet correct way possible. Not all months are the same amount of days, so I was hoping to check to get all the next n rows where the date starts with a "1" for example, sometimes certain days are also missing. I can't think of a catch all approach.
With 1/31/1970 in C1 try this,
=averageifs(daily!b:b, daily!a:a, "<="&c1, daily!a:a, ">="&eomonth(c1, -1)+1)
A PivotTable might be more convenient:
Below excel formula is working fine but in some cases its not give me proper value .
Input:
19:20:42
24:58:36
26:11:18
After using this formula:
=IF(TIMEVALUE(K7)>TIMEVALUE("09:00:00"),TRUE,FALSE)
I got the below output:
FALSE
TRUE
TRUE
What I Observe if the time value is > or = 24:00:00 it will not give me the proper answer.
How do I fix this?
As an alternative to Captain's excellent answer, you could also use:
=IF(K7>(9/24),TRUE,FALSE)
DateTime values are internally stored as a number of days since 1/1/1900,
therefore 1 = 1 day = 24 hours. So 9/24 = 0.375 = 9 hours :-)
You can easily verify this by clearing the format of your DateTime cells.
Edit: note that such Boolean formula can be expressed in a shorter way without losing legibility:
=K7>(9/24)
When you go over 24 hours, Excel counts it as the next day... and then the TIMEVALUE is the time the next day (i.e. 00:58:36 and 02:11:18 in your examples) and can, therefore, be before 0900.
You could do DATEVALUE(K7)+TIMEVALUE(K7) to ensure that you count the day part too...
I have two columns with the format [$-409]d-mmm-yyyy;#, I want to calculate the difference of date of these two columns. Please help.
Thanks
The =datedif() formula will do that for you, as per http://www.cpearson.com/excel/datedif.aspx:
=datedif(FirstDate, LaterDate, Interval)
Where FirstDate is the earlier date, LaterDate is the later date, and interval is the return type. I usually use 'm' for months, since this can be divided by 12 to get decimal years. Lots of other options, though.