How to split monthly value into days

Hello everyone

I have a little issue here. Solution must be easy, but i cant figure it out.

I have 12 values (each represent ideal rain in every month) IR = [0 0 0 70 83 100 110 100 70 0 0 0]'

Then i have long datetime (10 years) with daily time step.

Need to find out value for each day.

Example: IR for 27.Sept = IR(Sept)/number of days in Sept

Thanks for every help

Answer 1

[Andrei Bobrov]

Let dates - your long datetime (10 years) with daily time step.

out = IR(month(dates))./day(dates,"dayofmonth")

[Jan Koncel]

Thank you very much for your answer. It almost works, but i dont want to divide it by number of day in month. For example: Since April has 30 days and IR for April is 70, I want to have value 70/30 on every day in April. Now it gives me for 1.Apr 70/1, for 2.Apr 70/2...

[Jan Koncel]

I found way it works out for me, your advice was very useful, thanks

IR = [0 0 0 70 83 100 110 100 70 0 0 0]';
days_in_month = days(dateshift(Date,'end','month')-dateshift(Date,'start','month')+1);

IR_d = IR(month(Date))./days_in_month;