找出由其他指标控制的参数的平均值

问题描述

在这个问题上我需要一些帮助

I need some help in this problem

我在MATLAB中有这个矩阵:

I have this matrix in MATLAB:

A = [ 25    1.2    1
      28    1.2    2
      17    2.6    1
      18    2.6    2
      23    1.2    1
      29    1.2    2
      19    15     1
      22    15     2
      24    2.6    1
      26    2.6    2];  

第一列是一些温度测量值

1st column is some measured values for temperature

第二列是代表颜色(1.2:红色，....等)的索引代码

2nd column is an index code representing the color (1.2:red,.....etc)

第三列是采样时间.仅在1到2小时内

3rd column is the hour of taking the sample. Only at hours from 1 to 2

我希望矩阵由第二列控制，如下所示:

I want the matrix to be controlled by 2nd column as follows:

如果为1.2，程序将在第1小时找到所有温度的平均值

if it is 1.2, the program will find the average of all temperatures at hour 1 that

对应于1.2

所以，这里(25 + 23)/2 = 24

So, here ( 25 + 23 )/2 = 24

并找到第2小时所有温度的平均值，并与之对应

and also finds the average of all temperatures at hour 2 and that corresponds

至1.2，(28 + 29)/2 = 28.5

to 1.2, ( 28 + 29 ) /2 = 28.5

和该平均值:

                     [24 
                      28.5]

将替换第1和2小时的所有温度值

will replace all temperature values at hours 1 and 2

对应于1.2.

然后，它对索引2.6和15执行相同的操作

Then, it does the same thing for indices 2.6 and 15

因此，所需的输出将是:

So, the desired output will be:

B = [  24
       28.5

       15.5
       22

       24
       28.5

       19
       22

       15.5
       22]

我的问题是使用循环.我一次只能运行一个索引.

My problem is in using the loop. I could do it for only one index at one run.

例如，

T=[];
   index=1.2;

   for i=1:length(A)
       if A(i,2)==index
        T=[T A(i,1)];
    else
        T=[T 0];
       end
   end

所以，T是提取的T，它对应于1.2，其他条目为零

So, T is the extracted T that corresponds to 1.2 and other entries are zeros

然后，我写了很长的代码来找到平均值，最后我找到了矩阵

Then, I wrote long code to find the average and at the end I could find the matrix

仅与索引1.2相对应:

that corresponds to ONLY the index 1.2 :

B =   [24
       28.5

       0
       0

       24
       28.5

       0
       0

       0
       0]

但这仅用于一个索引，并且为其他索引分配零.我可以所有人做到这一点

But this is only for one index and it assigns zeros for the other indices. I can do this for all

在单独的运行中显示索引，然后添加B，但这距离我的真实情况要花费很长的时间

indices in separate runs and then add the B's but this will take very long time since my real

矩阵乘8是8760.

我敢肯定，这样做的方法更短.

I am sure that there is a shorter way to do that.

谢谢

致谢

推荐答案

尝试一下:

B = zeros(size(A, 1), 1);
C = unique(A(:, 2))';
T = [1 2];

for c = C,
  for t = T,
      I1 = find((A(:, 2) == c) & (A(:, 3) == t));
      B(I1) = mean(A(I1, 1));
  end
end

修改我认为您对c = 2.6和t = 1的预期答案是错误的...不应该是(17 + 24)/2 = 20.5吗?

EditI think your expected answer is wrong for c = 2.6 and t = 1... Shouldn't it be (17 + 24)/2 = 20.5?

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