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I am new to Matlab and for the values of a, l and w i need to find all the values for l in the data set and the corresponding w values.
a=10;
l=(0:10)
w=(0:10)
for l,d
if a == l.*w
disp(l)
disp(w)
end
end
Not sure what you want to do, but I think your code could be put as follows:
a = 10;
l = 0:a; %// actually, it would suffice to check numbers up to floor(a/2)
ind = rem(a,l)==0; %// logical index that tells if remainder is zero or not
disp([l(ind); a./l(ind)])
Result:
1 2 5 10
10 5 2 1
You could do it more directly with Matlab's factor function:
f = factor(a);
disp([f; a./f])
Result:
2 5
5 2
Related
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I have several struct of different dimensions in Matlab. Let us suppose that they were 5, the first is 1x100, the second 1x250, the third 1x200, the fourth 1x100, and the fifth 1x150. I want to assemble a new structure containing only the last element of the each previous structures, that is, I want to get a struct of the 1x5 form. How can I do this?
Create 5 row vectors, then make a row vector from the last elements like this:
>> a=[1:100];
>> b=[1:250];
>> c=[1:200];
>> d=[1:100];
>> e=[1:150];
>> the_lasts = [a(end), b(end), c(end), d(end), e(end)]
sthe_last =
100 250 200 100 150
>>
That also can be generalized into a function, using a file named:
ends_of.m :
function lasts = ends_of (varargin)
% prepare a result row vector with nargin dimention
lasts = zeros(1,nargin);
for i = 1:nargin
element = varargin(i); % get the arg
last = element{}(end); % find the last element
lasts(i) = last; % save it as index i
endfor
endfunction
Then it can be called like this :
>> ends_of ([1,2,3],[4,5,6],[1:44], [3:33], [1,2,3,2,1])
ans =
3 6 44 33 1
>>
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Following advise from another post, I created a basic 3x3 averaging filter as follows:
blurP = zeros(512, 512);
for i = 1:510
for j = 1:510
sum = 0;
for k = i:i+2
for l = j:j+2
sum = sum+P(k,l);
end
end
blurP(i+1,j+1) = mean2(P(i:i+2,j:j+2));
end
end
imshow(P), figure, imshow(blurP, []);
I need to create a weighted filter with two options: One counts the center element (of the 3x3 grid) twice, thus giving me ten elements in total. The other has sixteen elements in total, with the center element counted four times and the adjacent ones counted twice—only the corner elements of the 3x3 grid are counted once.
The easiest option is to just use convolution (via the conv2 function). Then designing your kernels is a simple as writing out exactly what you described:
kernel1 = [1 1 1
1 2 1
1 1 1]/10;
blurP1 = conv2(P, kernel1, 'same');
and
kernel2 = [1 2 1
2 4 2
1 2 1]/16;
blurP2 = conv2(P, kernel2, 'same');
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I have a loop like this :
for i = 1:3
M = 1
for M = M:12
while (S(M) == i)
M = M+1
end
end
end
Now for the new incremented value of M in while loop the 'for' loop is not working for this new value.
Any solutions?
The code doesn't make any sense!
You should start to use different names for your parameters,
And note that:
for index = values
program statements
:
end
Avoid assigning a value to the index variable within the body of a loop. The for statement overrides any changes made to the index within the loop.
For the second loop for M = M : 12 is the same as for M = 1 : 12.
MATLAB takes the index values at their first definition, for example,
a = [1 2];
for i = a
disp(i);
a = [1 2 3];
end
You will see that i won't accept the value 3 because at the first use of for, i is set over [1 2].
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I have a vector [1 2 3 4 ... 100] and another vector storing the numbers I want to delete like [2 3 4]. After deleting, the numbers in original vector should be mapped to another order, for example, 1->1, 5->2, 6->3, etc. Is there any efficient way to do this? Thanks a lot!
I'd use setdiff:
% original vector
A = 1:100;
% elements to be removed
B = [2 3 4 18 21];
% new order (indices)
C = randperm(numel(A)-numel(B));
% Step 1) remove the elements
[D,I] = setdiff(A,B); % ordered
D = A(I); % restore original order
% Step 2) re-order the elements
D = D(C)
You can do:
original_vector = 1:100;
delete_vector = [2 3 4];
for ii = 1:length(delete_vector)
original_vector(original_vector==delete_vector(ii)) = [];
end
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I'm trying to understand why we're iterating through Mandelbrot points until |z| < 4.
why 4? is there somekind of a law? or is it based on statistical measurements?
thanks, igal
Consider the Mandelbrot set with along y=0, which would would be z(i) = z(i-1)^2 + c.
Consider when c = (x=-2, y=0)
z(0) = 0
z(1) = 0^2 + -2 = -2
z(2) = (-2)^2 + -2 = 4 - 2 = 2
z(3) = 2^2 + -2 = 4 - 2 = 2
z(...) = 2^2 + -2 = 4 - 2 = 2
This example (x=-2,y=0) is the point with the greatest magnitude that will never blow up. Thus when z^2 > 4, there is no point in further iteration since you already know it will blow up.
All other points where the magnitude of the point >= 2 will blow up.