matlab车牌投影分割法,基于垂直投影的车牌字符分割matlab程序-程序员宅基地

技术标签: matlab车牌投影分割法  

function [seg] = character_segmentation(bw, DIGIT_WIDTH, MIN_AREA);

% character_segmentation: Returns the digit segments in the supplied binary image.

% The function uses the "segment" function, keeping only the seven

% segments in the result with largest area, and in case less than seven

% segments were found, it attempts to recall the function, making the

% separation between the already found segments clearer (by cleaning the

% bits which are there.

seg = segment(bw, DIGIT_WIDTH, MIN_AREA);

[x y] = size(seg);

% If we got less than 7 digits, we try to make the sepration between them

% clearer by cleaning the bits between them, and we call the "segment"

% function again:

if x < 7

for i = 1 : x

bw(:,seg(i,2))=0;

end;

seg = segment(bw, DIGIT_WIDTH, MIN_AREA);

end;

% Keeping in the results the seven segments with the largest area:

area = [];

for i = 1 : x

pic = bw(:, seg(i,1) : seg(i,2), :);

area(i) = bwarea(pic);   %bwarea函数计算对象面积

end;

area1 = sort(area);

seg = seg';

for j = 1:(length(area1)-7)

i = find(area == area1(j));

len = length(area);

if i == 1

area = [area(2:len)];

seg = [seg(:,2:len)];

elseif i == len

area = [area(1:i-1)];

seg = [seg(:,1:i-1)];

else

area = [area(1:i-1) area(i+1:len)];

seg = [seg(:,1:i-1) seg(:,i+1:len)];

end;

end;

seg = seg';

return;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [segmentation] = segment(im, digit_width, min_area);

% segment: Segment the pictures in digit images according to the variable

% "digit_width" and returns a matrix containing the two bounds of the each

% digit segment. The function keeps in the result only segment whose

% "rectangular" areas is more than "min_area".

segmentation = [];

% Summing the colums of the pic:

t = sum(im);

% Getting the segments in the pic:

seg = clean(find_valleys(t, 2, 1, digit_width), 3);

% Keeping in the result only the segments whose rectangular areas is more than min_area:

j = 1;

for i = 1 : (length(seg) - 1)

band_width = seg(i+1) - seg(i);

maxi = max(t(1, seg(i):seg(i+1)));

if(maxi * band_width > min_area)

segmentation(j, 1) = seg(i);

segmentation(j, 2) = seg(i+1);

j = j + 1;

end;

end;

return;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [s] = find_valleys(t, val, offset, digit_width);

% find_valleys: Uses the method named peak-to-valleyin order to segment the

% pictures in digit images getting the two bounds of the each digit segment

% according to the statistical parameter digit_width = 18.

% The function is recursive; it uses the vector of the sums of the columns

% in the LP binary image supplied in the parameter "t".

% The function passes over the graph corresponding to this vector from left

% to right, bottom-up, incrementing at each recursive step the height that

% is examined on the graph (val). It checks the bandwidth of the first part

% of the signal: if it is greater than DIGIT_WIDTH, the function is

% recursively called after incrementing the height which is examined on

% the graph, (val). Otherwise, if the bandwidth is good, the two bounds of

% the signal with this bandwidth are taken as a digit segment, and the

% function is recursively called for the part of the image which is at

% the right side of the digit segment just found. This is done until the

% whole width of the picture has been passed over.

% Determining the points which are inferior to the examined hieght:

s = find(t < val);

% If no more than one point is found, incrementing val and recursively calling the function again.

if(length(s) < 2)

s = find_valleys(t, val + 1, offset, digit_width);

return;

end;

% If no point is found terminating:

if length(s) == 0

return;

end;

% Arranging the boundaries, so that if we have a big value at the beginning

% or the end of the picture the algorithm still works: in this case, the

% algorithm includes also those points.

if((t(1,1) >= val) & s(1) ~= 1)  %&  ? &&

s = [1 s];

end;

if((t(1, length(t)) >= val) & s(length(s)) ~= length(t))

s = [s length(t)];

end;

% Updating the real coordinates according to offset:

s = add(s, offset - 1);

% Cleaning points which are very close each other keeping only one of them.

s = clean(s, 3);

% While there is a bad segment in "s", (starting from the left side):

while bad_segm(s, digit_width) == 1

for i = 1: (length(s) - 1)

if (s(i + 1) - s(i)) > digit_width

% The subvector which does not correspond to a valid digit

% segemnt:

sub_vec = t(1, s(i) - offset + 1 : s(i+1) - offset + 1);

% Recursively, separating this bad segment in two or more valid

% digit segments:

s = [s(1 : i) find_valleys(sub_vec, val + 1, s(i), digit_width) s(i+1 : length(s))];

end;

end;

end;

return;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [bool] = bad_segm(s, digit_width);

% bad_segm: Returns true (1) iff there is a bad digit segment in s, namely,

% two points that ar distant one from the other by more than "digit_width".

if length(s) == 0

bool = 0;

return;

end;

tmp = s(1);

bool = 0;

for i = 2 : length(s)

if(s(i) - tmp) > digit_width

bool = 1;

return;

end;

tmp = s(i);

end;

return;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [t] = clean(s, val);

% clean: Cleans form the vector s all the poins which are distant the one

% from the other by less than "val" keeping only one of them.

t = [];

len = length(s);

i = 2;

j = 1;

while i <= len

while(s(i) - s(i-1) <= val)

i = i + 1;

if(i > len)

return;

end;

end;

if j == 1 | (s(i-1) - t(j-1)) > val   %|| ?  |

t(j) = s(i-1);

j = j + 1;

end;

t(j) = s(i);

j = j + 1;

i = i + 1;

end;

return;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function [t] = add(s, val); % add: Adds "val" to each one of the entries in the vector s and returns the new vector. len = length(s); t = []; for i = 1:len     t(i) = s(i) + val; end; return;

版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
本文链接:https://blog.csdn.net/weixin_39725594/article/details/115891958

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