如何快速获取datenum日期范围内的唯一季度列表？

Hey there! I see you're looking to speed up generating a unique list of quarters between two datenum values, and the bottleneck is those slow string-to-double conversions. Let's ditch the inefficient string operations entirely and use a pure numeric approach—this will be lightning fast, even for large date ranges.

The Problem with Your Current Approach #

Your existing code likely loops through every date in the range, converts it to a string (with datestr), formats the quarter, then converts back to a number. This is slow because string operations are computationally expensive, especially when dealing with thousands of dates.

The Efficient Numeric Solution #

Instead of iterating over every date, we can calculate the start and end quarters directly, then generate the full sequence of quarters in between. Here's how:

1. Extract the year and month from the start/end datenum values using datevec (which returns numeric values, no strings involved).
2. Calculate the quarter number for each date (1-4, based on the month).
3. Generate the full sequence of quarters from the start quarter to the end quarter, handling cross-year ranges seamlessly.

Code Implementation #

function quarters = getUniqueQuarters(dateStart, dateEnd)
    % Extract year and month from start date
    [y1, m1] = datevec(dateStart);
    q1 = ceil(m1 / 3);  % Calculate quarter (1=Jan-Mar, 2=Apr-Jun, etc.)
    start_q = y1 * 100 + q1;
    
    % Extract year and month from end date
    [y2, m2] = datevec(dateEnd);
    q2 = ceil(m2 / 3);
    end_q = y2 * 100 + q2;
    
    if y1 == y2
        % Same year: just generate quarters from q1 to q2
        quarters = y1 * 100 + (q1:q2);
    else
        % Quarters remaining in the start year
        first_part = y1 * 100 + (q1:4);
        
        % All quarters for middle years (if any)
        middle_years = (y1 + 1):(y2 - 1);
        middle_part = reshape(middle_years' * 100 + repmat(1:4, length(middle_years), 1), 1, []);
        
        % Quarters up to q2 in the end year
        last_part = y2 * 100 + (1:q2);
        
        % Combine all parts
        quarters = [first_part, middle_part, last_part];
    end
end

Example Usage #

For your sample range: dateStart = datenum(2014, 1, 1) and dateEnd = datenum(2015, 2, 2):

quarters = getUniqueQuarters(datenum(2014,1,1), datenum(2015,2,2))

This returns:

ans =
       201401       201402       201403       201404       201501

(Note: Your sample output included 201502, which doesn't align with standard quarter definitions (Feb is Q1). If you use a custom quarter grouping, just adjust the ceil(m/3) calculation to match your needs—e.g., ceil(m/2) for 2-month "quarters".)

Why This Is Faster #

This approach avoids all string conversions and date-by-date iteration. It computes the quarter sequence in constant time, regardless of how large your date range is. For your test case (dateStart=735173, dateEnd=735538), this function will run in a fraction of a millisecond—way faster than your current 0.013 seconds.

内容的提问来源于stack exchange，提问作者WJA