Why it is showing error?

1 visualizzazione (ultimi 30 giorni)
Kartikey Rai
Kartikey Rai il 11 Ott 2019
Commentato: Walter Roberson il 10 Apr 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

Risposte (1)

Geoff Hayes
Geoff Hayes il 11 Ott 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 Commenti
JIAHONG LIM
JIAHONG LIM il 10 Apr 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 il 10 Apr 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.

Accedi per commentare.

Categorie

Scopri di più su Denoising and Compression in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by