The first 3 lines re-create the input matrix named m. See inline comments to understand the rest.
It produces a table T summarizing the results.
% recreate input data (m)
d = datetime(1883,9,1) : days(1) : datetime(1950,12,31);
[yr,mo,dy] = datevec(d);
m = [yr(:),mo(:),dy(:),rand(size(yr(:)))*50+40];
% Identify winter months
winterMonths = [11, 12, 1]; % any order but must be consecutive months
isWinter = ismember(m(:,2), winterMonths);
% group the consecutive winter months
L = bwlabel(isWinter); %requires image processing toolbox
% -------------------------------------------------------
% % If you don't have image processing tool box: %
% isWinterCS = cumsum(diff([0;isWinter]) == 1); %
% L = zeros(size(isWinter)); %
% L(isWinter) = isWinterCS(isWinter); %
% -------------------------------------------------------
% Compute max dischage per annual winter
maxDischarge = grpstats(m(:,4),L,'max'); %requires stats & ML toolbox
% Organize data by year (using the earliest year in each winter)
maxDischargeYear = grpstats(m(:,1),L,'min'); %requires stats & ML toolbox
% -------------------------------------------------------
% % If you don't have stats & ML tool box: %
% groupID = findgroups(L); %
% maxDischarge = splitapply(@max,m(:,4),groupID); %
% maxDischargeYear = splitapply(@min,m(:,1),groupID); %
% -------------------------------------------------------
maxDischargeYear(1) = []; % Remove group 0 (non-winter months)
maxDischarge(1) = []; % Remove group 0 (non-winter months)
T = table(maxDischargeYear(:),maxDischarge(:),'VariableNames',{'Year','MaxDischarge'});
% Show first few rows
head(T)
% Year MaxDischarge
% ____ ____________
%
% 1883 89.476
% 1884 89.657
% 1885 89.815
% 1886 89.866
% 1887 89.884
% 1888 88.552
% 1889 89.299
% 1890 89.571
