Num of max consective occurance of 'T'

Question

Sarvesh 2024-6-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2129231-num-of-max-consective-occurance-of-t

评论： Sarvesh 2024-6-24，23:37

Consective Occ of T.mat

Hi,

I have a set of cyclone data that I am trying to analyse. in the sample timetable, each system is described by unique id, for each id i want to count the MAX CONSECTIIVE occuance of 'T' in the 'thrsh' column. I also want to know the closest location of the next or previous 'T' after of before the max conseitive occuance for each system. the final table should be [id, maxOcc, nextLoc]. Some selected ids are shown below:

id = 1990S001, maxOcc = 2, nextT = 0 (no occurance of next or previous T after or before F)

id = 1990S005, maxOcc = 2, nextT = 0

id = 1990S014, maxOcc = 5, nextT = -8 (8th occurance of T before F)

id = 1990S019, maxOcc = 3, nextT = -2 (2nd occurance of T before F)

High appreciate your help.

Thanks

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Ganesh 2024-6-17

Hi @Sarvesh,

To clarify, 1990S005 has only 2 entries but you have mentioned that it has a maxOcc of 5. Does this mean that the consecutive Ts should be considered in the system 1990S006 too?

Sarvesh 2024-6-17

编辑：Sarvesh 2024-6-17

Hi @Ganesh,

apologies for that, I have corrected the numbers in the question. there was an error from my end. *005 should indeed have 2 maxOcc only and every unique ID should be considered as a unique system and not a combination.

thanks heaps

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Answer 1

Ganesh 2024-6-17

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2129231-num-of-max-consective-occurance-of-t#answer_1472751

编辑：Ganesh 2024-6-17

在 MATLAB Online 中打开

Consective Occ of T.mat

Hi @Sarvesh,

You can use the following code:

load('Consective Occ of T.mat')
[uniqueIDs, maxConsecutiveOnes, closestOneIndices] = findMaxConsecutiveOnesAndClosest(barra1.id,cell2mat(barra1.thrsh));
table(uniqueIDs,maxConsecutiveOnes,closestOneIndices)
ans = 49x3 table
    uniqueIDs    maxConsecutiveOnes    closestOneIndices
    _________    __________________    _________________

    1990S001             2                    Inf       
    1990S002             0                    Inf       
    1990S003             0                    Inf       
    1990S004             7                    Inf       
    1990S005             2                    Inf       
    1990S006             3                      3       
    1990S007             2                    Inf       
    1990S008             0                    Inf       
    1990S009             0                    Inf       
    1990S010             0                    Inf       
    1990S011             0                    Inf       
    1990S012             0                    Inf       
    1990S013             5                    Inf       
    1990S014             5                     -8       
    1990S015             0                    Inf       
    1990S016             0                    Inf       
function [uniqueIDs,maxConsecutiveOnes,closestOneIndices] = findMaxConsecutiveOnesAndClosest(ids, boolChars)
    
    uniqueIDs = unique(ids); 
    
    maxConsecutiveOnes = zeros(length(uniqueIDs),1); 
    closestOneIndices = zeros(length(uniqueIDs),1);
    
    for i = 1:length(uniqueIDs)
        id = uniqueIDs(i);
        boolValues = boolChars(ids == id) == 'T'; % Filters out only values for each ID
       
        [maxConsecutive, endIndex] = maxConsecutiveOnesHelper(boolValues);
        maxConsecutiveOnes(i) = maxConsecutive;
        
        if endIndex == 0 %If there are no consecutive 1s, skip
            closestOneIndex = inf;
        else
            closestOneIndex = findClosestOne(boolValues, endIndex, maxConsecutive);
        end
        closestOneIndices(i) = closestOneIndex;
            
    end
end
function [maxConsec, endIndex] = maxConsecutiveOnesHelper(boolValues)
    maxConsec = 0;
    currentConsec = 0; 
    endIndex = 0;
    
    % Manually checking the longetst consecutive 1s
    % Can be optimized
    for i = 1:length(boolValues)
        if boolValues(i) == true
            currentConsec = currentConsec + 1;
            if currentConsec > maxConsec
                maxConsec = currentConsec;
                endIndex = i;
            end
        else
            currentConsec = 0;
        end
    end
end
function closestIndex = findClosestOne(boolValues, maxEndIndex, maxConsecutive)
    if maxConsecutive == length(boolValues)
        closestIndex = inf; % I am using inf to denote no close element is present
        return;
    end
    
    startIndex = maxEndIndex - maxConsecutive + 1;
    
    % Find the closest distance before the consecutive 1s
    closestBefore = find(boolValues(1:startIndex-1), 1, 'last');
    if ~isempty(closestBefore)
        closestIndexBefore = closestBefore-length(boolValues(1:startIndex-1))-1;
    else
        closestIndexBefore = inf;
    end
    % Find the closest distance after the consective 1s
    closestAfter = find(boolValues(maxEndIndex+1:end), 1, 'first');
    if ~isempty(closestAfter)
        closestIndexAfter = closestAfter;
    else
        closestIndexAfter = inf;
    end
    
    % Find the minimum
    closestIndex = min(closestIndexBefore,closestIndexAfter);
end