MATLAB，在更改每个步骤的矩阵中跟踪边界的最佳方法是什么？ [英] MATLAB, what is the best way to trace a boarder in a matrix that is changing each step?

问题描述

我想在这样的矩阵中计算每次迭代的平均斜率或梯度。

I want to calculate the average slope or gradient at each iteration in such a matrix.

a=[ 10 9 8 7 6 5 5;
9  9 8 7 8 5 5; 
8  8 7 7 5 5 5; 
7  7 7 6 5 5 5;
6  6 6.6 5 5 5 5;
6  6 6.1 5 5 5 5;
6.3  5 5 5 5 5 5]

我想要找到斜坡或每个步骤中a（1,1）位置之间的梯度和每个值为5的每个点之间的梯度。每次迭代5的位置发生变化，其他值也发生变化。

Where I am wanting to find the slope or gradient between the a(1,1) position during each step and at each point that boarders a value of 5. Each iteration the position of the 5's changes and so do the other values.

这样做之后我将平均斜率。我还没有遇到这样的问题，我找不到简化的Matlab命令。

After doing so I will then average the slope. I haven't encountered a problem like this yet and I could not find a Matlab command to simplify.

推荐答案

首先你必须找到海岸元素是哪个。从你的定义来看，如果一个元素是一个海岸元素（如右边）和一个5.如果海平面是5，并且是最低可能的值，即没有元素超出海平面，那么你必须首先找到所有的土地要素为，

First you must find out which the coast elements are. From your definition, an element is a coast element if it border (from the right) with a 5. If the sea level is 5, and is the lowest possible value i.e. no element goes beyond sea level, then you must first find all the land elements as,

land=a>5;

返回，

ans =

     1     1     1     1     1     0     0
     1     1     1     1     1     0     0
     1     1     1     1     0     0     0
     1     1     1     1     0     0     0
     1     1     1     0     0     0     0
     1     1     1     0     0     0     0
     1     0     0     0     0     0     0

现在，海岸元素是1，然后是0.取陆地矩阵的列差，

Now, the coast elements are 1s that are followed by a 0. Take the column difference of the land matrix,

coastTmp=diff(land,1,2);

返回，

ans =

     0     0     0     0    -1     0
     0     0     0     0    -1     0
     0     0     0    -1     0     0
     0     0     0    -1     0     0
     0     0    -1     0     0     0
     0     0    -1     0     0     0
    -1     0     0     0     0     0

并找到-1s，

coast=find(coastTmp==-1);

这是，

coast =

     7
    19
    20
    24
    25
    29
    30

从这里开始很简单。渐变是a（1,1）与所有海岸元素的差异，即

From here it is easy. The gradient is the difference of a(1,1) with all the coast elements, i.e.

slope=a(coast)-a(1,1);  % negative slope here

给予，

slope =

  -3.700000000000000
  -3.400000000000000
  -3.900000000000000
  -3.000000000000000
  -4.000000000000000
  -4.000000000000000
  -2.000000000000000

当然平均值是，

mean(slope);

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