计数由零分隔的点数

问题描述

我有一个这样的向量：

  A = [1 2 1 1 1 4 5 0 0 1 2 0 2 3 2 2 2 0 0 0 0 33] 
  

我想计算非零的GROUPS

4）：）

您能帮我吗？

感谢

解决方案

要计算您的组，使用是：

  count = sum （[A 0] == 0）== 1）
  

这假设 A 是一个行向量，如在您的示例中。这不需要零，全零，空向量，以及我尝试过的其他几个测试用例。

要获取你的值组，你可以使用一个变体：

  a0 =（A〜= 0）; 
 d = diff（a0）; 
 start = find（[a0（1）d] == 1）％每个组的起始索引
 len = find（[d -a0（end）] ==  -  1）-start + 1 ％长度，每组中的索引数量
  

在您的情况下， $ c> len 与

  finish = find（[d -a0（end）] == -1）％每个组的最后索引
  

长度 开始， len 和完成与 count 的值相同，所以你可以使用这个，如果你需要做分解。然后，您可以使用开始和 len （或完成）将组存储在单元格数组或结构或某些其他不规则数组中。例如：

  count = length（start）; 
 B = cell（count，1）; 
 for i = 1：count 
 B {i} = A（start（i）：finish（i））; 
 end 
  

I have a vector like this:

A = [1 2 1 1 1 4 5 0 0 1 2 0 2 3 2 2 2 0 0 0 0 33]

I would like to count how many GROUPS of non zero elements it contains and save them.

so I want to isolate:

and then count the groups (they should be 4) :)

Can you help me please?

Thanks

解决方案

To count your groups, a fast vectorized method using logical indexing is:

count = sum(diff([A 0]==0)==1)

This assumes that A is a row vector as in your example. This works with no zeros, all zeros, the empty vector, and several other test cases I tried.

To obtain your groups of values themselves, you can use a variation to my answer to a similar question:

a0 = (A~=0);
d = diff(a0);
start = find([a0(1) d]==1)           % Start index of each group
len = find([d -a0(end)]==-1)-start+1 % Length, number of indexes in each group

In your case it might make sense to replace len with

finish = find([d -a0(end)]==-1) % Last index of each group

The length of start, len, and finish should be the same as the value of count so you could just use this if you need to do the breaking up. You can then use start and len (or finish) to store your groups in a cell array or struct or some other ragged array. For example:

count = length(start);
B = cell(count,1);
for i = 1:count
    B{i} = A(start(i):finish(i));
end

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