How to split monthly value into days

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Jan Koncel
Jan Koncel 2022 年 3 月 27 日
コメント済み: Jan Koncel 2022 年 3 月 28 日
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

採用された回答

Andrei Bobrov
Andrei Bobrov 2022 年 3 月 27 日
Let dates - your long datetime (10 years) with daily time step.
out = IR(month(dates))./day(dates,"dayofmonth")
  2 件のコメント
Jan Koncel
Jan Koncel 2022 年 3 月 27 日
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
Jan Koncel 2022 年 3 月 28 日
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;

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