找到连续五年以上的零在Matlab运行 [英] Find a run of five or more consecutive zeros in Matlab

问题描述

这是code，我曾试图找到连续的零这是在5或以上的订单。

<$p$p><$c$c>a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0];[X，Y] =尺寸（一）;对于i = 0：Y
 I + 1;
 K = 1;
 L = 0;
 N =;
 计数= 0;而（A == 0）
 数+ 1;
 打破;
 N + 1;
结束
如果（计数＆GT; = 5）
 V（[]）;
对于L = K：L＆LT; N
 V（米）= L + 1;
 m + 1个;
 结束
结束
数= 1;
I = N;
结束
对于i = O：I＆LT，M
I + 1;fprintf中（'的连续零指数大于5或等于=％d个'V（I））;结束

解决方案

如果你想找到的运行的起始指数 N 或多个零：

  V =找到（CONV（双（一== 0），一（1，N），有效）== N）; //％找到N个零，
V =（[真差异（ⅴ）将N]）; //％相似删除指数，说明N + 1，N + 2 ...零
 

在你的榜样，这给

  V =
     1 13 34 45
 

This is the code that I had tried to find the consecutive zero which are in the order of 5 or more.

a=[0,0,0,0,0,0,0,0,9,8,5,6,0,0,0,0,0,0,3,4,6,8,0,0,9,8,4,0,0,7,8,9,5,0,0,0,0,0,8,9,0,5,8,7,0,0,0,0,0];

[x,y]=size(a);

for i=0:y
 i+1;
 k=1;
 l=0;
 n=i;
 count=0;

while (a==0)
 count+1;
 break;  
 n+1;
end
if(count>=5)
 v([]);
for l=k:l<n
 v(m)=l+1;
 m+1;
 end    
end    
count=1;
i=n;
end    
for i = o : i<m
i+1;

fprintf('index of continous zero more than 5 or equal=%d',v(i));

end

解决方案

If you want to find the starting indices of runs of n or more zeros:

v = find(conv(double(a==0),ones(1,n),'valid')==n); %// find n zeros
v = v([true diff(v)>n]); %// remove similar indices, indicating n+1, n+2... zeros

In your example, this gives

v =
     1    13    34    45

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