Write a function to show a square “random” image. The function will take in “at least” one input argument which is the size of the image in pixels. As a default, the image will be shown in black-white. If a second input argument is provided, it will
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Caitlin Schmidt
el 21 de Jul. de 2019
Editada: John D'Errico
el 21 de Jul. de 2019
This is what I have so far:
function week7hw1(pixels, varargin)
%This function will receive the size of the pixels as an input
%if a second input is entered, the image will be the colormap
n=nargin;
mat=randi(pixels);
if n==0
image(mat)
elseif n==1;
colormap(varargin)
image(mat)
end
end
I'm not sure what I am doing wrong for this function.
0 comentarios
Respuesta aceptada
John D'Errico
el 21 de Jul. de 2019
Editada: John D'Errico
el 21 de Jul. de 2019
Your problem is that you misunderstand nargin.
By way of example, I wrote this very simple function.
function week7hw1(pixels, varargin)
nargin
numel(varargin)
So all it does is write out the number of elements in varargin, as well as write out the value of nargin. Now I'll try it out.
week7hw1(1,2)
ans =
2
ans =
1
Hmm. there were two inputs to the function.
Nargin was 2. But numel(varargin) was 1, as expected.
So, what does nargin do? It is NOT the number of elements in varargin. nargin is the TOTAL number of input arguments to the function. Similarly, if I provide no second argument, then what happens?
week7hw1(1)
ans =
1
ans =
0
Again, I provided no second argument. nargin was 1, since there was ONE argument. varargin was an empty array this time.
So, why does your code fail? You test to see if nargin is ZERO or ONE. But nargin will always be at least 1, because you always are providing the first argument. So you want to distinguish if there are 1 or 2 arguments provided. The necessary fix to your code is very simple.
2 comentarios
John D'Errico
el 21 de Jul. de 2019
Editada: John D'Errico
el 21 de Jul. de 2019
That part is easy. varargin is a CELL array. This happens because your additional arguments may be all sorts of things. So it must contain anything. But colormap cannot take a cell array as input. Instead, you need to extract that element from the cell array.
colormap(varargin{1})
You index a cell array using {} curly braces, in order to extract an element of that array. So the line as I wrote it extracts the first element of varargin, then passes that into colormap.
Más respuestas (1)
KALYAN ACHARJYA
el 21 de Jul. de 2019
Editada: KALYAN ACHARJYA
el 21 de Jul. de 2019
function image1=rand_image()
pixel_num=input('Enter the value ');
image1=logical(randi(2,[pixel_num pixel_num])-1); %@Image Analyst Answer
imshow(image1);
end
Command Window:
>> y=rand_image();
Enter the value 100
Figure Window:
Second Part is not complete-
If a second input argument is provided, it will
3 comentarios
KALYAN ACHARJYA
el 21 de Jul. de 2019
Editada: KALYAN ACHARJYA
el 21 de Jul. de 2019
function image1=rand_image(pixel_num)
image1=logical(randi(2,[pixel_num pixel_num])-1); %@Image Analyst Answer
imshow(image1);
end
Now pass the pixel value from here
y=rand_image(100)
Ver también
Categorías
Más información sobre Colormaps en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!