如何计算矩阵中 1 和 0 的个数? [英] How to count number of 1 and 0 in the matrix?

问题描述

我有一张图片，我只剪掉了一列.之后我把它变得合乎逻辑，所以这个专栏里只有 0 和 1.

I have an image of which I cut out only one column. After that I made it to be logical so there are be only 0 and 1 in this column.

假设我在此列中的值是

Suppose my values in this column are

1111000110000000000000011111111

我想计算每个 1 块或每个 0 块的长度.

I want to count the length of each block of ones or each block of zeros.

结果是

1 - 4 (first 1)
0 - 3 (first 0)
1 - 2 
and so on...

我只知道整个列的计数，但我不能对每个不同的块进行计数.请任何人帮助我.

I know only count for the entire column but I can't do it for each distinct block. Anyone please help me.

推荐答案

设 vec 为 row 向量 (1-by-n) 的零和一，那么您可以使用以下代码

Let vec be a row vector (1-by-n) of zeros and ones, then you can use the following code

rl = ( find( vec ~= [vec(2:end), vec(end)+1] ) );
data =  vec( rl );
rl(2:end) = rl(2:end) - rl(1:end-1);

rl 会给你连续的 0 和 1 的数量，而 data 会告诉你每个块是 0 还是 1.

rl will give you the number of consecutive zeros and ones, while data will tell you for each block if it is zero or one.

这个问题与运行长度编码密切相关.

演示:

vec = [1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1];
rl = ( find( vec ~= [vec(2:end), vec(end)+1] ) );
data =  vec( rl ),
rl(2:end) = rl(2:end) - rl(1:end-1),

data =
     1     0     1     0     1

rl =
     4     3     2    14     8

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