在matlab中找到两个特定时间之间的范围

问题描述

我有一个年度数据集:

Jday = datenum('2009-01-01 00:00','yyyy-mm-dd HH:MM'):1/24:...
    datenum('2009-12-31 23:00','yyyy-mm-dd HH:MM');
DateV = datevec(Jday);
dat = 1+(10-1).*rand(length(Jday),10);
noise = 10*sin(2*pi*Jday/32)+20;
for i = 1:size(dat,2);
    dat2(:,i) = dat(:,i)+noise';
end

表示通过水柱记录的温度测量值.我想计算每天06:00至18:00之间的温度范围，以便最终得到每个深度的365个值，因此最终矩阵为365x10.

which represents temperature measurements recorded through a water column. I would like to calculate the temperature range between 06:00 and 18:00 for each day so that I end up with 365 values for each depth so a final matrix of 365x10.

我可以通过以下方式指定各天:

I can specify the individual days by:

[~,~,b] = unique(DateV(:,2:3),'rows');

但是我不知道如何只考虑上午06:00和下午18:00之间记录的值.谁能提供一些有关最佳方法的信息?

But I cant work out how to only consider values recorded between 06:00 am and 18:00 pm. Could anyone provide some information about the best possible way of doing this?

修正:答案稍长

Jday = datenum('2009-01-01 00:00','yyyy-mm-dd HH:MM'):1/24:...
    datenum('2009-12-31 23:00','yyyy-mm-dd HH:MM');
DateV = datevec(Jday);
dat = 1+(10-1).*rand(length(Jday),10);
noise = 10*sin(2*pi*Jday/32)+20;
for i = 1:size(dat,2);
    dat2(:,i) = dat(:,i)+noise';
end

for i = 1:size(dat2,2);
    data = [DateV(:,4),dat2(:,i)]; % obtain all hours from DateV array

    dd = find(data(:,1) == 6); % find 06:00
    dd(:,2) = find(data(:,1) == 18); % find 18:00

    for ii = 1:size(dd,1);
        result(ii,i) = range(data(dd(ii,1):dd(ii,2),2)); % calculate range
    end
end

推荐答案

如果 range 带有矩阵，它在每一列上独立运行.我们可以使用它来准备一个矩阵，其中每列包含06:00到18:00时间间隔的值，并应用range.

1. 棘手的部分是处理365x24 x 10矩阵以使列正确:

1. The tricky part is manipulating your 365x24-by-10 matrix to get the columns right:

A = permute(reshape(dat2', [10, 24, 365]), [2, 1, 3]);

这应生成3-D 24x10x365矩阵，可以从中提取第7至18行.

This should result in 3-D 24x10x365 matrix, from which rows 7 to 18 can be extracted.

下一步是应用range并将结果转换回二维365x10矩阵:

The next step is to apply range and transform the result back into a 2-D 365x10 matrix:

result = permute(range(A(7:18, :, :)), [3, 2, 1]);

因此，完整的解决方案是:

Therefore, the full solution is:

A = permute(reshape(dat2', [10, 24, 365]), [2, 1, 3]);
result = permute(range(A(7:18, :, :)), [3, 2, 1]);

编辑:
为了使解决方案更加通用，可以使用DateV获取所需行的索引:

EDIT:
To make the solution even more generic, you can use DateV to obtain the indices of the desired rows:

idx = DateV(1:24, 4) >= 6 & DateV(1:24, 4) < 18;

，然后像这样在解决方案中使用它:

and then use it in the solution like so:

A = permute(reshape(dat2', [10, 24, 365]), [2, 1, 3]);
result = permute(range(A(idx, :, :)), [3, 2, 1]);

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