I have a time series data in a Year, Month and day format. Now I want to convert it to a decimal year format. How to do it in octave.
Any clue
Thy this
DateString = {'09/16/2007';'05/14/1996';'11/29/2010'};
formatIn = 'mm/dd/yyyy';
datenum(DateString,formatIn)
For more info : https://www.mathworks.com/help/matlab/ref/datenum.html
Related
Is there an inbuilt function in SAS that gives the text day of the week? Such as Monday, Tuesday etc from a date variable?
So far, I have just found the weekday function, that just gives the date as a number from 1-7.
If you want to get a text day of the week from a date, you can use DOWNAME. format.
data _null_;
result = put(today(), dowName.);
put result=;
run;
If you want to get a weekday name from a weekday number, I do not know specific function, which does it, but you can use the fact that 1-7 are also dates and 0 is Friday, 1st January 1960 and add 2 to your number:
data _null_;
do day = 1 to 7;
weekDay = put(day + 2, dowName.);
put weekDay=;
end;
run;
Which will give you:
weekDay=Monday
weekDay=Tuesday
weekDay=Wednesday
weekDay=Thursday
weekDay=Friday
weekDay=Saturday
weekDay=Sunday
How to find how many week in particular month and year using mysql, I have only year and month no date value.
Example find out number of week year = 2017 and month = August
I search many times but all examples are given with date.
I want to find weeks from a specific date
If date =26-07-2017
I want week numbers corresponds to the date
Start date = date and end date should be week date
If i understand your question, you want number of the week in year.
to do that you need to do something like this:
int weekNumber =
actualWeek.get(IsoFields.WEEK_OF_WEEK_BASED_YEAR);
with that number you can get dates limits of that week:
LocalDate week = actualWeek.with(ChronoField.ALIGNED_WEEK_OF_YEAR, weekNumber);
this.startDate = week.with(DayOfWeek.MONDAY);
this.endDate = startDate.plusDays(6);
where startDate and endDate is a predefined kind of LocalDate variables
I have date in the following format stored in Mailed_Date column
Mon, 09/20/10 01:04 PM
I have used a serde(csv-serde-1.1.2-0.11.0-all.jar) to get this date from a csv file and stored it as a String.
How can i get the time of day, day of week, Month from this date in Hive.
I tried using
select HOUR(Mailed_Date) from final3 limit 5;
but got a NULL as answer.
I think i found the answer to my query:-
select day(from_unixtime(unix_timestamp(substr(Mailed_Date, 6, 14), 'MM/dd/yy hh:mm'), 'yyyy-MM-dd hh:mm')) from final3 limit5
Here first I extract the date and then convert it to the required format which is acceptable by Hive to get the day, month or time from.
Got a hint from the below question:-
Convert mm/dd/yyyy to yyyy-mm-dd in Hive
I am relatively new to Access - in a Query how do I calculate the number of years between a historic date and today- date is in the format 00/00/0000
You are looking for DateDiff("yyyy", Now, TheDate) this will return the year number between TheDate and the current date.
You can use d or ww (instead of 'yyyy') if you want to learn the amount of day or week between the dates.
For detailed information you can look into this link:
https://support.office.com/en-us/article/DateDiff-Function-e6dd7ee6-3d01-4531-905c-e24fc238f85f