汇总矩阵列中数字的共同概率-MATLAB [英] Tallying co-incidences of numbers in columns of a matrix - MATLAB

问题描述

我有一个(A)矩阵(实际上更大):

I have a matrix (A) in the form of (much larger in reality):

205   204   201
202   208   202

我如何才能在列与列之间计算数字的重合度，然后将其输出到矩阵中?

How can I tally the co-incidence of numbers on a column-by-column basis and then output this to a matrix?

我希望最终矩阵从min(A):max(A)(或能够指定特定范围)遍历顶部和底部，并使其与数字中的共同点相符每列.使用上面的示例:

I'd want the final matrix to run from min(A):max(A) (or be able to specify a specific range) across the top and down the side and for it to tally co-incidences of numbers in each column. Using the above example:

    200 201 202 203 204 205 206 207 208
200  0   0   0   0   0   0   0   0   0
201  0   0   1   0   0   0   0   0   0
202  0   0   0   0   0   1   0   0   0 
203  0   0   0   0   0   0   0   0   0
204  0   0   0   0   0   0   0   0   1
205  0   0   0   0   0   0   0   0   0
206  0   0   0   0   0   0   0   0   0
207  0   0   0   0   0   0   0   0   0
208  0   0   0   0   0   0   0   0   0

(不需要矩阵标签)

两个要点:计数必须是不可重复的，并且应按数字顺序进行.例如，包含以下内容的列:

Two important points: The tallying needs to be non-duplicating and occur in numerical order. For example a column containing:

205
202

将其与202一起出现在205(如上面的矩阵所示)中，而不是205与202一起出现-互为倒数.在决定使用哪个数字作为参考时，它应该是最小的.

Will tally this as a 202 occurring with 205 (as shown in the above matrix) but NOT 205 with 202 - the duplicate reciprocal. When deciding what number to use as the reference, it should be the smallest.

推荐答案

进行救援！

让您的数据和所需范围定义为

Let your data and desired range be defined as

A = [ 205   204   201
      202   208   202 ]; %// data. Two-row matrix
limits = [200 208]; %// desired range. It needn't include all values of A

然后

lim1 = limits(1)-1;
s = limits(2)-lim1;
cols = all((A>=limits(1)) & (A<=limits(2)), 1);
B = sort(A(:,cols), 1, 'descend')-lim1;
R = full(sparse(B(2,:), B(1,:), 1, s, s));

给予

R =
     0     0     0     0     0     0     0     0     0
     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     1     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     1
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0

或者，您可以省去sort并使用矩阵加法后跟triu来获得相同的结果(可能更快):

Alternatively, you can dispense with sort and use matrix addition followed by triu to obtain the same result (possibly faster):

lim1 = limits(1)-1;
s = limits(2)-lim1;
cols = all( (A>=limits(1)) & (A<=limits(2)) , 1);
R = full(sparse(A(2,cols)-lim1, A(1,cols)-lim1, 1, s, s));
R = triu(R + R.');

这两种方法都可以处理重复的列(直至排序)，从而正确地增加其计数.例如，

Both approaches handle repeated columns (up to sorting), correctly increasing their tally. For example,

A = [205   204   201
     201   208   205]

给予

R =
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     2     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     1
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0

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