Split a target date interval into seasons and find the percentile of days for each season

I have the following seasons:

• Spring: {01-March , 31-May}
• Summer: {01-June, 31-August}
• Autumn: {01-Sep, 30-Nov}
• Winter: {01-Dec, 28-Feb}

and I also have as a data a reference period, e.g. 01-Jan until 15-Sep (254 days).

What I want is to compute what is the percentile of the days of reference period that belong to each season.

For example, in the present case, we have:

• Winter: 01-Jan until 28-Feb --> 59 days / 254 days = 23.5%
• Spring: 01-Mar until 31-May --> 90 days / 254 days = 35%
• Summer: 01-Jun until 31-Aug --> 90 days / 254 days = 35%
• Autumn: 01-Sep until 15-Sep --> 15 days /254 days = 6.5%

So the requested values would be {23.5, 35, 35, 6.5}.

Answer 1

[Scott MacKenzie]

There might be a way to shorten this, but I think it achieves what you are after. You didn't mention the year, so I set this up as a variable (change, as necessary). This script accommodates leap years. BTW, the number of reference days below is 259. Not sure why this differs from your count of 254.

yr = 2020; % change, as necessary

% reference days of interest 
r1= datetime(yr,1,1):datetime(yr,2,eomday(yr,2));
r2= datetime(yr,3,1):datetime(yr,5,31);  
r3= datetime(yr,6,1):datetime(yr,8,31);
r4= datetime(yr,9,1):datetime(yr,9,15);

n = length([r1 r2 r3 r4]) % number of reference days

n = 259

percentDays = [length(r1), length(r2), length(r3), length(r4)] /n * 100

percentDays = 1×4

23.1660 35.5212 35.5212 5.7915

[DIMITRIS GEORGIADIS]

Thank you very much Scott for your answer. It was simpler than I expected. Actually, the year is irrelevant for me. Is there any way to get rid of it? I want to check onle the relevant days and monts.

[DIMITRIS GEORGIADIS]

Can we generalize the algorithm for a random reference period?

[Scott MacKenzie]

@DIMITRIS GEORGIADIS You're welcome. Glad to help. Concerning your first question, you can't really get rid of the yr variable because of leap years. If don't care about Feb 29, then sure. Just substitute any year (e.g., 1) for yr in the script.

In terms of generalizing this, sure that can be done. You could write a function that receives as input the month and day of the begining of each reference period. You'd also need the month and day of end of the last reference period. The function would return the percentages, as in my example script. To shorten the code and allow for any number of reference periods (in case you somtimes need >4), you could use structures or arrays to hold the month+day inputs and the percentages output.

[DIMITRIS GEORGIADIS]

Thank you again @Scott MacKenzie. I did generalize it.

Answer 2

[Seth Furman]

To add to Scott's answer, this kind of grouped calculation on timestamped data lends itself well to timetable and groupsummary.

https://www.mathworks.com/help/matlab/timetables.html

https://www.mathworks.com/help/matlab/ref/double.groupsummary.html

isWinter = @(dt) dt.Month == 12 | dt.Month <= 2;
isSpring = @(dt) 3 <= dt.Month & dt.Month <= 5;
isSummer = @(dt) 6 <= dt.Month & dt.Month <= 8;
isAutumn = @(dt) 9 <= dt.Month & dt.Month <= 11;

referencePeriod = timetable('RowTimes',datetime(2020,1,1):caldays(1):datetime(2020,9,15));
referencePeriod.Season(isWinter(referencePeriod.Time)) = categorical("Winter");
referencePeriod.Season(isSpring(referencePeriod.Time)) = categorical("Spring");
referencePeriod.Season(isSummer(referencePeriod.Time)) = categorical("Summer");
referencePeriod.Season(isAutumn(referencePeriod.Time)) = categorical("Autumn");

counts = groupsummary(referencePeriod,"Season")

counts = 4×2 table
    Season    GroupCount
    ______    __________

    Winter        60    
    Spring        92    
    Summer        92    
    Autumn        15    

counts.Percentages = counts.GroupCount ./ sum(counts.GroupCount)

counts = 4×3 table
    Season    GroupCount    Percentages
    ______    __________    ___________

    Winter        60          0.23166  
    Spring        92          0.35521  
    Summer        92          0.35521  
    Autumn        15         0.057915  

[DIMITRIS GEORGIADIS]

Dear @Seth Furman thank you very much for your very interesting and elegant answer!

[Peter Perkins]

The general idea in Seth's suggestion is a good one, and his solution works for multiple years of data. If the data are always within one year, discretize would make this even simpler. Obviously winter crosses a year boundary, but it's not hard to use discretize in a way that handles that:

dt = datetime(2022,1,1):datetime(2022,9,15);
edges = datetime(2022,[1,3,6,9,12,13],[1,1,1,1,1,1]); % winter at both ends
season = discretize(dt,edges,"categorical",["Winter" "Spring" "Summer" "Autumn" "Winter"])

season = 1×258 categorical array
     Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Winter      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Spring      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Summer      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn      Autumn 

summary(season)

     Winter      Spring      Summer      Autumn 
     59          92          92          15     

Using groupsummary gives you a nice output table; creating season would work as a lead-up to that too.

Answer 3

[Peter Perkins]

Another possibility:

>> yr = 2020;
>> edges = datetime(yr,[1 3 6 9 12,12],[1 1 1 1 1 32])
edges = 
  1×6 datetime array
   01-Jan-2020   01-Mar-2020   01-Jun-2020   01-Sep-2020   01-Dec-2020   01-Jan-2021
>> t = datetime(yr,1,1:259)';
>> tf = isbetween(t,edges(1:end-1),edges(2:end),'openright'); % uses implicit expansion
>> tf(:,1) = tf(:,1) + tf(:,end);
>> tf(:,end) = [];
>> 100*sum(tf,1)/length(t)
ans =
       23.166       35.521       35.521       5.7915