如何计算重复出现的元素的最大数量？

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;

我想计算一个序列中出现的最大零点数，例如，上面的结果是8，它显示了一起发生的最大零数，而不是总零数为16。

Answer 1

为了提高透明度，一个非常简单的algoritm。

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，效率更高。

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Answer 2

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

1. 使用diff查找连续数字序列更改的位置
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“转换”之间的元素数
4. 最后，调用max以获得最大的连续数序列。

请注意，如果最大的序列发生在边缘，那么您可能会遇到麻烦，在本例中，我建议您首先在矩阵中添加一个倒置位，如：[1-Grid_outage(1), Grid_outage, 1-Grid_outage(end)];

Answer 3

为了找出参数x在一个序列中一起发生的次数，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;