Why it is showing error?

2 vues (au cours des 30 derniers jours)
Kartikey Rai
Kartikey Rai le 11 Oct 2019
Commenté : Walter Roberson le 10 Avr 2021
function letter = readLetter(snap)
load NewTemplates
snap = imresize(snap,[42 24]);
rec = [ ];
for n = 1:length(NewTemplates)
cor = corr2(NewTemplates{1,n},snap);
rec = [rec cor];
end
ind = find(rec == max(rec));
display(find(rec == max(rec)));
% Alphabets listings.
if ind == 1 || ind == 2
letter = 'A';
elseif ind == 3 || ind == 4
letter = 'B';
elseif ind == 5
letter = 'C';
elseif ind == 6 || ind == 7
letter = 'D';
elseif ind == 8
letter = 'E';
elseif ind == 9
letter = 'F';
elseif ind == 10
letter = 'G';
elseif ind == 11
letter = 'H';
elseif ind == 12
letter = 'I';
elseif ind == 13
letter = 'J';
elseif ind == 14
letter = 'K';
elseif ind == 15
letter = 'L';
elseif ind == 16
letter = 'M';
elseif ind == 17
letter = 'N';
elseif ind == 18 || ind == 19
letter = 'O';
elseif ind == 20 || ind == 21
letter = 'P';
elseif ind == 22 || ind == 23
letter = 'Q';
elseif ind == 24 || ind == 25
letter = 'R';
elseif ind == 26
letter = 'S';
elseif ind == 27
letter = 'T';
elseif ind == 28
letter = 'U';
elseif ind == 29
letter = 'V';
elseif ind == 30
letter = 'W';
elseif ind == 31
letter = 'X';
elseif ind == 32
letter = 'Y';
elseif ind == 33
letter = 'Z';
% Numerals listings.
elseif ind == 34
letter = '1';
elseif ind == 35
letter = '2';
elseif ind == 36
letter = '3';
elseif ind == 37 || ind == 38
letter = '4';
elseif ind == 39
letter = '5';
elseif ind == 40 || ind == 41 || ind == 42
letter = '6';
elseif ind == 43
letter = '7';
elseif ind == 44 || ind == 45
letter = '8';
elseif ind == 46 || ind == 47 || ind == 48
letter = '9';
else
letter = '0';
end
end
Screenshot (39).png

Réponses (1)

Geoff Hayes
Geoff Hayes le 11 Oct 2019
Kartikey - there appears to be a hidden character at your line 70 which is the "blank" line in
% Numerals listings.
elseif ind == 34
If, on Linux, I do a cat -t readLetter.m, I see a question mark at the blank line
elseif ind == 33
letter = 'Z';
?
% Numerals listings.
elseif ind == 34
Just delete the blank line and the line with the comment to ensure that this special character is removed.
  7 commentaires
JIAHONG LIM
JIAHONG LIM le 10 Avr 2021
Hi GeiffHayes and Image Analyst, may I know why some letters need only 1 ind to determine but some need two? Thank you very much
if ind==1 || ind==2
letter='A';
elseif ind==3 || ind==4
letter='B';
elseif ind==5
letter='C';
Walter Roberson
Walter Roberson le 10 Avr 2021
This is taken from https://circuitdigest.com/tutorial/vehicle-number-plate-detection-using-matlab-and-image-processing which has a mismatch between the template building code and the template data file.
The characters that have two entries are the ones that have a "hole" in them. In the first version of the template, the hole is empoty, and in the second version of the template, the hole is filled.

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