如何计算重复元素的最大数量？

Question

Grid_outage(:,1) = 1;
Grid_outage(:,2) = 1;
Grid_outage(:,3) = 1;
Grid_outage(:,4) = 0;
Grid_outage(:,5) = 0;
Grid_outage(:,6) = 0;
Grid_outage(:,7) = 0;
Grid_outage(:,8) = 0;
Grid_outage(:,9) = 0;
Grid_outage(:,10) = 0;
Grid_outage(:,11) = 0;
Grid_outage(:,12) = 1;
Grid_outage(:,13) = 0;
Grid_outage(:,14) = 1;
Grid_outage(:,15) = 0;
Grid_outage(:,16) = 0;
Grid_outage(:,17) = 1;
Grid_outage(:,18) = 0;
Grid_outage(:,19) = 1;
Grid_outage(:,20) = 0;
Grid_outage(:,21) = 0;
Grid_outage(:,22) = 0;
Grid_outage(:,23) = 1;
Grid_outage(:,24) = 0;

我想计算序列中出现的最大零数，例如我希望上面的结果是8，它显示了一起出现的最大零数，而不是16的零的总数。 我如何在matlab中编写代码

Answer 1

max(diff(find(diff(Grid_outage))))

1. 使用diff
查找连续数字序列的变化位置
2. 使用find
获取发生这种情况的实际索引号
3. 再次使用diff来计算每个＆＃34;转换＆＃34;
之间的元素数量
4. 最后拨打max以获取最大的连续数字序列。

请注意，如果最大的序列出现在边缘，则可能会遇到问题，在这种情况下，我建议您先添加并向您的矩阵附加一个倒置位，如下所示：[1-Grid_outage(1), Grid_outage, 1-Grid_outage(end)];

Answer 2

为了透明的目的，这是一个非常简单的算法。

假设您可以将序列放在向量X：

中

% Create X containing some zeros.
X = round(rand(30,1));

% Use a counter to count the number of sequential zeros.
count = 0;
% Use a variable to keep the maximum.
max_count = 0;

% Loop over every element
for ii=1:length(X);
    % If a zero is encountered increase the counter
    if(X(ii)==0)
        count=count+1;
    % If no zero is encountered check if the number of zeros in the last sequence was largest.
    elseif count>max_count
        max_count=count;
        count=0;
    % Else just reset the counter
    else
        count=0;
    end
end
% Check if the last number of the vector exceeded the largest sequence.
if(count>max_count)
   max_count=count;
end

编辑：Dan的解决方案从大约200个元素开始提高效率。

Answer 3

为了查找参数x在序列中一起出现的次数，您可以使用：

if a(:) == x
    result = length(a); % Whole vector has the x parameter
else
    result = max(find(a~=x,1,'first') - 1, length(a) - find(a~=x,1,'last')); % Maximum difference in the edge   
    if ~isempty(max(diff(find(a~=x))) - 1)
        if isempty(result)
            result = max(diff(find(a~=x))) - 1; % Maximum difference in the body
        elseif result < max(diff(find(a~=x))) - 1
            result = max(diff(find(a~=x))) - 1; % Maximum difference in the body
        end;
    end;
end;