如何将 3 维每日数据转换为每月数据？

Question

我有十年（2001-2010）的 3 维数据矩阵。每个文件中的数据矩阵为 180 x 360 x 365/366（纬度 x 经度 x 日降雨量）。例如：2001：180x360x365、2002：180x360x365、2003：180x360x365、2004：180x360x366....................... 2010：180x360x365

现在我想将这个每日降雨量转换为每月降雨量（通过求和）并将所有年份合并到一个文件中。

所以我的最终输出将是 180x360x120（纬度 x 经度 x 十年间的月降雨量）。

Answer 1

这可能很耗时，但我想您可以使用某种形式的循环来每月迭代每年的数据，为每个月挑选出适当数量的数据点，然后将其添加到最终数据集。下面的（非常粗略的）代码可能会起作用：

years = ['2001','2002,'2003',...,'2010'];

months = ['Jan','Feb','Mar',...,'Dec'];

finalDataset=[];
for i=1:length(years)
    year = years(i);
    yearData=%% load in dataset for that year %%
    for j=1:length(months)
        month = months(j);
        switch month
             case {'Jan','Mar'}
                  days=30;
             case 'Feb'
                  days=28'
                  if(year=='2004' || year=='2008')
                     days=29;
                  end
              % then continue with cases to include each month
         end
         monthData=yearData(:,:,1:days) % extract the data for those months
         yearData(:,:,1:days)=[]; % delete data already extracted
         summedRain = % take mean of rainfall data
         monthSummed = % replace daily rainfall data with monthly rainfall, but keep latitude and longitude data
         finalDataset=[finalDataset; monthSummed];
      end
  end

抱歉，它非常简陋，我没有包含一些索引细节，但我希望这至少有助于说明一个想法？我也不完全确定 'if' 语句是否在 'switch' 语句中工作，但如果没有，可以在其他地方添加天数修正。

Answer 2

我相信您可以将其矢量化以更快地工作，但它应该可以完成这项工作。没有正确测试

% range of years
years = 2000:2016;
leap_years = [2000 2004 2008 2012 2016];

% Generating random data
nr_of_years = numel(years);
rainfall_data = cell(nr_of_years, 1);
for i=1:nr_of_years
    nr_of_days = 365;
    if ismember(years(i), leap_years);
        nr_of_days = 366;
    end
    rainfall_data{i} = rand(180, 360, nr_of_days);
end

您需要的实际代码如下

% fixed stuff
months = 12;
nr_of_days = [31 28 31 30 31 30 31 31 30 31 30 31];
nr_of_days_leap = [31 29 31 30 31 30 31 31 30 31 30 31];

% building vectors of month indices for days
month_indices = [];
month_indices_leap = [];
for i=1:months
    month_indices_temp = repmat(i, nr_of_days(i), 1);
    month_indices_leap_temp = repmat(i, nr_of_days_leap(i), 1);
    month_indices = [month_indices; month_indices_temp];
    month_indices_leap = [month_indices_leap; month_indices_leap_temp];
end

% the result will be stored here
result = zeros(size(rainfall_data{i}, 1), size(rainfall_data{i}, 2), months*nr_of_years);

for i=1:nr_of_years
    % determining which indices to use depending if it is a leap year
    month_indices_temp = month_indices;
    if size(rainfall_data{i}, 3)==366
        month_indices_temp = month_indices_leap;
    end

    % data for the current year
    current_data = rainfall_data{i};
    % this holds the data for current year
    monthy_sums = zeros(size(rainfall_data{i}, 1), size(rainfall_data{i}, 2), months);
    for j=1:months
        monthy_sums(:,:,j) = sum(current_data(:,:,j==month_indices_temp), 3);
    end
    % putting it into the combined matrix
    result(:,:,((i-1)*months+1):(i*months)) = monthy_sums;
end

您可能可以使用内置 datetime、datestr 和 datenum 实现更优雅的解决方案，但我不确定它们会更快或更短。

编辑：使用内置日期函数的替代方法

months = 12;
% where the result will be stored
result = zeros(size(rainfall_data{i}, 1), size(rainfall_data{i}, 2), months*nr_of_years);
for i=1:nr_of_years
    current_data = rainfall_data{i};
    % first day of the year
    year_start_timestamp = datenum(datetime(years(i), 1, 1));

    % holding current sums
    monthy_sums = zeros(size(current_data, 1), size(current_data, 2), months);

    % finding the month indices vector
    datetime_obj = datetime(datestr(year_start_timestamp:(year_start_timestamp+size(current_data, 3)-1)));
    month_indices = datetime_obj.Month;

    % summing
    for j=1:months
        monthy_sums(:,:,j) = sum(current_data(:,:,j==month_indices), 3);
    end 

    % result
    result(:,:,((i-1)*months+1):(i*months)) = monthy_sums;
end

第二个解决方案对我来说需要 1.45 秒，而第一个解决方案需要 1.2 秒。两种情况的结果相同。希望这会有所帮助。