Why it is showing error?

Question

Kartikey Rai 2019-10-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/484827-why-it-is-showing-error

评论： Walter Roberson 2021-4-10

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

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Answer 1

Geoff Hayes 2019-10-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/484827-why-it-is-showing-error#answer_395932

在 MATLAB Online 中打开

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 个评论
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Terry Lim 2020-5-26

在 MATLAB Online 中打开

Hi, Geoff Hayes

I am facing a problem which almost similar to his/her code. Which my error is:

Not enough input arguments.
Error in Letter_detection (line 4)
snap=imresize(snap,[42 24]);

below is the code for detect letter:

function letter=Letter_detection(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(ind);
% 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

For the template creation. The code is shown as below:

%CREATE TEMPLATES 
%Alphabets
A=imread('alpha/A.bmp');B=imread('alpha/B.bmp');C=imread('alpha/C.bmp');
D=imread('alpha/D.bmp');E=imread('alpha/E.bmp');F=imread('alpha/F.bmp');
G=imread('alpha/G.bmp');H=imread('alpha/H.bmp');I=imread('alpha/I.bmp');
J=imread('alpha/J.bmp');K=imread('alpha/K.bmp');L=imread('alpha/L.bmp');
M=imread('alpha/M.bmp');N=imread('alpha/N.bmp');O=imread('alpha/O.bmp');
P=imread('alpha/P.bmp');Q=imread('alpha/Q.bmp');R=imread('alpha/R.bmp');
S=imread('alpha/S.bmp');T=imread('alpha/T.bmp');U=imread('alpha/U.bmp');
V=imread('alpha/V.bmp');W=imread('alpha/W.bmp');X=imread('alpha/X.bmp');
Y=imread('alpha/Y.bmp');Z=imread('alpha/Z.bmp');
%Natural Numbers
one=imread('alpha/1.bmp');two=imread('alpha/2.bmp');
three=imread('alpha/3.bmp');four=imread('alpha/4.bmp');
five=imread('alpha/5.bmp'); six=imread('alpha/6.bmp');
seven=imread('alpha/7.bmp');eight=imread('alpha/8.bmp');
nine=imread('alpha/9.bmp'); zero=imread('alpha/0.bmp');
%Creating Array for Alphabets
letter=[A B C D E F G H I J K L M N O P Q R S T U V W X Y Z];
%Creating Array for Numbers
number=[one two three four five six seven eight nine zero];
NewTemplates=[letter number];
save ('NewTemplates','NewTemplates')
clear all

I had tried my best to solve this problem maybe there is some other things or method i havent tried. by grabbing this oppotunity, I hope you could help me to solve this problem.

JIAHONG LIM 2021-4-10

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 2021-4-10

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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