从30年日常数据的巨大矩阵中选择最大值

Question

假设我在矩阵中有30年的每日数据.为简单起见,假设它只有1列,10957行表示30年的天数.这一年从2010年开始.我想找到每年的最大值,以便输出为1列和30行.有没有自动化的方法在Matlab中编程？目前我正在手动做我做的是:

%for the first year
  max(RAINFALL(1:365);
.
.
%for the 30th of year
  max(RAINFALL(10593:10957);

手动完成并且我有相当少的相同数据集是很累人的.我用下面的代码计算了30年的平均值和标准差.我尝试修改代码以适用于上面的任务但我无法成功.希望任何人都可以修改代码或向我提出新的建议.

data = rand(32872,100); % replace with your data matrix

[nDays,nData] = size(data);

% let MATLAB construct the vector of dates and worry about things like leap
% year.
dayFirst = datenum(2010,1,1);

dayStamp = dayFirst:(dayFirst + nDays - 1);
dayVec = datevec(dayStamp);

year = dayVec(:,1);

uniqueYear = unique(year);

K = length(uniqueYear);

a = nan(1,K);
b = nan(1,K);

for k = 1:K
   % use logical indexing to pick out the year
   currentYear = year == uniqueYear(k);
   a(k) = mean2(data(currentYear,:));
   b(k) = std2(data(currentYear,:));
end 

Answer 1

一种可能的方法:

1. 创建一个包含每个数据值年份的列,使用datenum和datevec处理闰年.

2. 找到每年的最大值accumarray.

码:

%// Example data: 
RAINFALL = rand(10957,1); %// one column
start_year = 2010;        %// data starts on January 1st of this year

%// Computations:
[year, ~] = datevec(datenum(start_year,1,1) + (0:size(RAINFALL,1)-1)); %// step 1
result = accumarray(year.'-start_year+1, RAINFALL.', [], @max);        %// step 2

作为奖励:如果您@max在步骤2中更改任何一个@mean或者@std,请猜测您得到的...比您的代码简单得多.