根据第一列中的数字拆分矩阵 [英] Split matrix based on number in first column

问题描述

我有一个矩阵，其格式如下:

I have a matrix which has the following form:

M = 
[1 4 56 1;
 1 3 5  1;
 1 3 6  4;
 2 3 5  0;
 2 0 0  0;
 3 1 2  3;
 3 3 3  3]

我想根据第一列中给出的数字拆分此矩阵.所以我想将矩阵拆分成这样:

I want to split this matrix based on the number given in the first column. So I want to split the matrix into this:

A = 
[1 4 56 1;
 1 3 5  1;
 1 3 6  4]

B = 
[2 3 5  0;
 2 0 0  0]

C =
[3 1 2  3;
 3 3 3  3]

我通过以下循环进行了尝试，但这为我提供了所需的矩阵(带有零行):

I tried this by making the following loop, but this gave me the desired matrices with rows of zeros:

for i = 1:length(M)
    if (M(i,1) == 1)
        A(i,:) = M(i,:);
    elseif (M(i,1) == 2)
        B(i,:) = M(i,:);
    elseif (M(i,1) == 3)
        C(i,:) = M(i,:);
    end
end

则矩阵C的结果例如为:

The result for matrix C is then for example:

C = 
[0 0 0 0;
 0 0 0 0;
 0 0 0 0;
 2 3 5 0;
 2 0 0 0]

我应该如何解决这个问题?

How should I solve this issue?

其他信息:
实际数据的第一列中的日期为yyyymmdd格式.该数据集跨越数年，我想将此数据集分解为每年的矩阵，然后将其分解为每个月的矩阵.

Additional information:
The actual data has a date in the first column in the form yyyymmdd. The data set spans several years and I want to split this dataset in matrices for each year and after that for each month.

推荐答案

您可以使用 arrayfun 解决此任务:

You can use arrayfun to solve this task:

M = [
1 4 56 1;
 1 3 5  1;
 1 3 6  4;
 2 3 5  0;
 2 0 0  0;
 3 1 2  3;
 3 3 3  3]


A = arrayfun(@(x) M(M(:,1) == x, :), unique(M(:,1)), 'uniformoutput', false)

结果A是一个单元格数组，可以按以下方式访问其内容:

The result A is a cell array and its contents can be accessed as follows:

>> a{1}

ans =

     1     4    56     1
     1     3     5     1
     1     3     6     4

>> a{2}

ans =

     2     3     5     0
     2     0     0     0

>> a{3}

ans =

     3     1     2     3
     3     3     3     3

要在第一列中基于yyyymmdd格式拆分数据，可以使用以下命令:

To split the data based on an yyyymmdd format in the first column, you can use the following:

yearly = arrayfun(@(x) M(floor(M(:,1)/10000) == x, :), unique(floor(M(:,1)/10000)), 'uniformoutput', false)

monthly = arrayfun(@(x) M(floor(M(:,1)/100) == x, :), unique(floor(M(:,1)/100)), 'uniformoutput', false)

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