New here, i cannot figure it out ( Index exceeds the number of array elements (2). )

i=imread('1_crack.png');
bw=rgb2gray(i);
[S T]=graythresh(bw);
bw=imbinarize(bw,S);
bwn1=bw_filter(bw,15);
f = ones(3,3) / 9;
smooth_0 = imfilter(bw,f);
for i = 1:5
    smooth_0 = imfilter(smooth_0,f);
end
[canny_smooth_0,thres_0] = edge(smooth_0,'canny');

[canny_smooth_0,thres_0] = edge(smooth_0,'canny',thres_0);
canny_smooth_0 = uint8(canny_smooth_0) * 255; %把遮罩轉換成影像
bwn1=canny_smooth_0;
bwn1 = imfill(bwn1, 'holes');

labeledImage = bwlabel(bwn1);
% % Measure the area
measurements_o = regionprops(labeledImage, 'Area');
[~, columns] = size(bw);
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
boundaries = bwboundaries(bw);  %跟蹤二值圖像中的區域邊界
numberOfBoundaries = size(boundaries, 1);
for blobIndex_o = 1 : numberOfBoundaries
    thisBoundary = boundaries{blobIndex_o};
    x_o = thisBoundary(:, 2); % x = columns.
    y_o = thisBoundary(:, 1); % y = rows.
    
    % Find which two bounary points are farthest from each other.
    maxDistance_o = 0;
    for k = 1 : length(x_o)
        distances = sqrt( (x_o(k) - x_o) .^ 2 + (y_o(k) - y_o) .^ 2 );
        [thisMaxDistance, indexOfMaxDistance] = max(distances);
        if thisMaxDistance > maxDistance_o
            maxDistance_o = thisMaxDistance;
            index1_o = k;
            index2_o = indexOfMaxDistance;
        end
    end
    
    
    % Find the midpoint of the line.
    xMidPoint_o = mean([x_o(index1_o), x_o(index2_o)]);
    yMidPoint_o = mean([y_o(index1_o), y_o(index2_o)]);
    longSlope = (y_o(index1_o) - y_o(index2_o)) / (x_o(index1_o) - x_o(index2_o));
    perpendicularSlope = -1/longSlope;
    % Use point slope formula (y-ym) = slope * (x - xm) to get points
    y1_o = perpendicularSlope * (1 - xMidPoint_o) + yMidPoint_o;
    y2_o = perpendicularSlope * (columns - xMidPoint_o) + yMidPoint_o;
    
    % Get the profile perpendicular to the midpoint so we can find out when if first enters and last leaves the object.
    [cx_o,cy_o,c] = improfile(bw,[1, columns], [y1_o, y2_o], 1000);
    % Get rid of NAN's that occur when the line's endpoints go above or below the image.
    c(isnan(c)) = 0;
    firstIndex_o = find(c, 1, 'first');
    lastIndex_o = find(c, 1, 'last');
    % Compute the distance of that perpendicular width.
    perpendicularWidth_o = sqrt( (cx_o(firstIndex_o) - cx_o(lastIndex_o)) .^ 2 + (cy_o(firstIndex_o) - cy_o(lastIndex_o)) .^ 2 );
    % Get the average perpendicular width.  This will approximately be the area divided by the longest length.
    averageWidth_o = measurements_o(blobIndex_o).Area / maxDistance_o;
end

error : Index exceeds the number of array elements (2).

Answer 1

[Simon Chan]

The labeledImage is based on 'bwn1' and I think you should also use 'bwn1' instead of 'bw' when finding their boundaries.

[~, columns] = size(bwn1);
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
boundaries = bwboundaries(bwn1);  %跟蹤二值圖像中的區域邊界

Answer 2

[yanqi liu]

clc; clear all; close all;
i=imread('image.png');
bw=rgb2gray(i);
[S T]=graythresh(bw);
bw=imbinarize(bw,S);
% bwn1=bw_filter(bw,15);
f = ones(3,3) / 9;
smooth_0 = imfilter(bw,f);
for i = 1:5
    smooth_0 = imfilter(smooth_0,f);
end
[canny_smooth_0,thres_0] = edge(smooth_0,'canny');

[canny_smooth_0,thres_0] = edge(smooth_0,'canny',thres_0);
canny_smooth_0 = uint8(canny_smooth_0) * 255; %把遮罩轉換成影像
bwn1=canny_smooth_0;
bwn1 = imfill(bwn1, 'holes');

labeledImage = bwlabel(bwn1);
% % Measure the area
measurements_o = regionprops(labeledImage, 'Area');
[~, columns] = size(bw);
% bwboundaries() returns a cell array, where each cell contains the row/column coordinates for an object in the image.
boundaries = bwboundaries(~bw);  %跟蹤二值圖像中的區域邊界
numberOfBoundaries = size(boundaries, 1);
for blobIndex_o = 1 : size(boundaries,1)
    thisBoundary = boundaries{blobIndex_o};
    x_o = thisBoundary(:, 2); % x = columns.
    y_o = thisBoundary(:, 1); % y = rows.
    
    % Find which two bounary points are farthest from each other.
    maxDistance_o = 0;
    for k = 1 : length(x_o)
        distances = sqrt( (x_o(k) - x_o) .^ 2 + (y_o(k) - y_o) .^ 2 );
        [thisMaxDistance, indexOfMaxDistance] = max(distances);
        if thisMaxDistance > maxDistance_o
            maxDistance_o = thisMaxDistance;
            index1_o = k;
            index2_o = indexOfMaxDistance;
        end
    end
    
    if index2_o > length(x_o)
        continue;
    end
    % Find the midpoint of the line.
    xMidPoint_o = mean([x_o(index1_o), x_o(index2_o)]);
    yMidPoint_o = mean([y_o(index1_o), y_o(index2_o)]);
    longSlope = (y_o(index1_o) - y_o(index2_o)) / (x_o(index1_o) - x_o(index2_o));
    perpendicularSlope = -1/longSlope;
    % Use point slope formula (y-ym) = slope * (x - xm) to get points
    y1_o = perpendicularSlope * (1 - xMidPoint_o) + yMidPoint_o;
    y2_o = perpendicularSlope * (columns - xMidPoint_o) + yMidPoint_o;
    
    % Get the profile perpendicular to the midpoint so we can find out when if first enters and last leaves the object.
    [cx_o,cy_o,c] = improfile(bw,[1, columns], [y1_o, y2_o], 1000);
    % Get rid of NAN's that occur when the line's endpoints go above or below the image.
    c(isnan(c)) = 0;
    firstIndex_o = find(c, 1, 'first');
    lastIndex_o = find(c, 1, 'last');
    % Compute the distance of that perpendicular width.
    perpendicularWidth_o = sqrt( (cx_o(firstIndex_o) - cx_o(lastIndex_o)) .^ 2 + (cy_o(firstIndex_o) - cy_o(lastIndex_o)) .^ 2 );
    % Get the average perpendicular width.  This will approximately be the area divided by the longest length.
    bwi = ~bw; 
    bwi = bwselect(bwi, round(thisBoundary(:, 2)), round(thisBoundary(:, 1)));
    measurements_oi = regionprops(bwi, 'Area');
    if isempty(measurements_oi)
        continue;
    end
    averageWidth_o = measurements_oi.Area / maxDistance_o;
end