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how can i change the value of red in a pixel in a photo?

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hi voici le code
clc; clear all; close all;
im=imread('photo.jpg');
b=1;
delta=20;
wi=0
l=impixel(im,1,5);
f=l(1)
m=floor(f/((2^b)*delta));
r=f-(2^b)*m*delta;
f1=(2^b)*m*delta+wi*delta+r/(2^b)
impixel(im,1,5)(1)=f1
le dernier ligne a généré cette erreur
??? Error: File: Untitledtest.m Line: 14 Column: 1
()-indexing must appear last in an index expression.
comment faire pour changer un seul valeur parmi le 3 valeurs dans un pixel choisi??
merci

Risposta accettata

Image Analyst
Image Analyst il 6 Giu 2015
You cannot do this:
impixel(im,1,5)(1)=f1
If you want to set row 1, column 5 of image im to a value of f1, you need to do this:
im(1, 5) = f1;
  7 Commenti
amira hl
amira hl il 6 Giu 2015
Modificato: Walter Roberson il 6 Giu 2015
clc; clear all; close all;
im=imread('photo.jpg');
l=impixel(im,1,5)
im(1,5,1)=10;
impixel(im,1,5)
executing =>
l =
232 239 255
ans =
232 239 255
nothing change
Image Analyst
Image Analyst il 6 Giu 2015
You need to make sure whether you're dealing with x,y or row, column. Looking at the help to make sure.....
% Read in RGB image.
im = imread('peppers.png');
% Define a row and column.
c = 1;
r = 5;
% Check value of image at row 5, column 1.
rgbValues1 = impixel(im, c, r)
im(r,c,1)=10;
rgbValues2 = impixel(im, c, r)
You get:
rgbValues1 =
63 31 62
rgbValues2 =
10 31 62
Note, impixel takes x,y (which is the same as col, row), whereas indexing an array is always row, col. So you need to make sure you're using the right form (order) in the right place.

Accedi per commentare.

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