I'm writing a piece of code for submission through an online grader, as showcased below. B is some given array filled any/all integers 1 through K, and I want to extract the corresponding logical indices of matrix X and perform some operations on those elements, to be put into a return array:
for i = 1:K
A = X(B == i, :);
returnArr(i, :) = sum(A) / length(A);
end
This did not pass the grader at all, and so I looked to change my approach, instead indexing array X indirectly via first using the "find" function, as below:
for i = 1:K
C = find(B == i);
returnArr(i,:) = sum(X(C,:)) / length(C);
end
To my surprise, this code passed the grader without any issues. I know there are a plethora of variations between graders, and one might handle a particular function differently than another, but from a MATLAB functionality/coding perspective, what am I missing in terms of discrepancies between the two approaches? Thanks!
I think the problem is that:
length(C) == sum(B == i)
while
length(A) == max([sum(B == i) , size(X , 2)])
In other words, to obtain the same result of the second example with the first one, you should modify it like this:
A = X(B == i , :);
returnArr(i, :) = sum(A) / size(A,1);
The function length returns the length of largest array dimension
Related
Below is a MATLAB code (recursion) which inputs a vector (l_1,l_2,...,l_r) of non negative integers and an integer m prints all sequences (m_1,m_2,...,m_r) satisfying:
0 <= m_i <= l_i for all 1 <= i <= r and m_1 + m_2 + ... + m_r = m
The r is captured in the function definition by calling the size of the (l_i) array below:
function arr=sumseq(m,lims)
arr=[];
r=size(lims,2);
if r==0 || m<0
arr=[];
elseif r==1 && lims(1)>=m
arr=[m]; %#ok<NBRAK>
else
for i=0:lims(1)
if(lims(1)<0)
arr=[];
else
v=sumseq(m-i,lims(2:end));
arr=[arr;[i*ones(size(v,1),1) v]];
end
end
end
end
Here what I have done is, stored a whole array of them and made it my output. Instead I want to only print them one by one and not store them in an array. This seems simple enough as there is not much choice in which line(s) I need to change (I believe it is the contents of the else block inside the for loop), but I get into a fix every time I try to achieve it.
(Also, MATLAB warned me that if I kept re-initializing the array with a larger array like in the statement:
arr=[arr;[i*ones(size(v,1),1) v]];
it reallocates a fresh array for all the contents of arr and spends a 'lot' of time doing so.)
In short: recursion or not, I want to save the trouble of storing it, and need an algorithm which is as efficient as or more efficient than what I have here.
find() function returns the indices where the elements are non-zero. I tried with different array sizes but both give error :
In an assignment A(I) = B, the number of elements in B and I must be the same.
I am confused because when the array size is same, still I am getting this error.
This is just to understand what went wrong:
LEt,
Example 1: Same array size
A = [20;21;3;45;5;19;1;8;2;1];
B = A;
for i =1:length(B)
pos(i) = find(A == B(i));
end
I should have got pos = [1,2,3,4,5,6,7,8,9,10]. But the loops exits after i = 7, giving `pos = [1,2,3,4,5,6]'
Example 2: Dissimilar array size
C = [20;1;10;3];
for i =1:length(C)
pos(i) = find(A == C(i));
end
Can somebody please explain what is wrong in my understanding and an illustration of how I can work with same and different array length of A and B? Thank you.
The problem is that find(A == 1) returns two indexes, both 7 and 10, and that can't be stored in pos(i), since pos(i) can only hold a single number.
Unfortunately, the generic error message happened to have the same name for the matrices as two of your matrices, which can be confusing before you'we seen it a few times.
I tried to check if a 3D array is not all zeros using the next code:
notAll_n0_GreaterThan_ni=1;
while notAll_n0_GreaterThan_ni
notAll_n0_GreaterThan_ni=0;
mask=(n0<ni);
numDimensions=ndims(mask);
for dim_ind=1:numDimensions
if any(mask,dim_ind)
notAll_n0_GreaterThan_ni=1;
break;
end
end
if notAll_n0_GreaterThan_ni
n0(mask)=n0(mask)+1;
end
end
It seems I have error in the code because at the end I get for example: n_0(11,3,69)=21 while ni(11,3,69)=21.1556.
I can't find the error. I'll appreciate if someone shows me where I'm wrong and also if there is a simpler way to check existence of nonzero elements in a 3D array.
Let x denote an n-dimensional array. To check if it contains at least one non-zero element, just use
any(x(:))
For example:
>> x = zeros(2,3,4);
>> any(x(:))
ans =
0
>> x(1,2,2) = 5;
>> any(x(:))
ans =
1
Other, more exotic possibilities include:
sum(abs(x(:)))>0
and
nnz(x)>0
This is what you looking for
B = any(your_Array_name_here(:) ==0); no need for loops
the (:) turns the elements of your_Array into a single column vector, so you can use this type of statement on an array of any size
I 've tested this and it works
A = rand(3,7,5) * 5;
B = any(A(:) ==0);
Is there a way to pass an argument multiple times to different arrays?
What I want to do is:
r = '1:10:end'; % This doesn't work for me
plot(x1(r), y1(r));
plot(x2(r), y2(r));
...
and pass r to different arrays (with different lengths) in many plot functions. I tried with [r] but no success.
As I understand it, you want to plot every 10th element of possibly different sized arrays. There are a few ways you could do this. One way would be to write a short function to filter your arrays for you, for instance:
plot_10 = #(x,y) plot(x(1:10:end),y(1:10:end));
plot_10(x1,y1);
plot_10(x2,y2);
...
EDIT: Just an additional thought. If you wanted to enable the extended functionality of plot (e.g. passing line/colour arguments, etc). You could do something like this:
plot_10 = #(x,y,varargin) plot(x(1:10:end),y(1:10:end),varargin{:});
plot_10(x1,t1,'k+');
To use the "end" operator, it needs to be inside an array access call;
n = 10;
r = 1 : 1 : n;
r(1:end) % is legal
r(1:floor(end/2)) % is legal
So you could do something like this:
s = rand(1,2*n);
s(r)
% to compare...
s( 1:1:n )
I want to store data coming from for-loops in an array. How can I do that?
sample output:
for x=1:100
for y=1:100
Diff(x,y) = B(x,y)-C(x,y);
if (Diff(x,y) ~= 0)
% I want to store these values of coordinates in array
% and find x-max,x-min,y-max,y-min
fprintf('(%d,%d)\n',x,y);
end
end
end
Can anybody please tell me how can i do that. Thanks
Marry
So you want lists of the x and y (or row and column) coordinates at which B and C are different. I assume B and C are matrices. First, you should vectorize your code to get rid of the loops, and second, use the find() function:
Diff = B - C; % vectorized, loops over indices automatically
[list_x, list_y] = find(Diff~=0);
% finds the row and column indices at which Diff~=0 is true
Or, even shorter,
[list_x, list_y] = find(B~=C);
Remember that the first index in matlab is the row of the matrix, and the second index is the column; if you tried to visualize your matrices B or C or Diff by using imagesc, say, what you're calling the X coordinate would actually be displayed in the vertical direction, and what you're calling the Y coordinate would be displayed in the horizontal direction. To be a little more clear, you could say instead
[list_rows, list_cols] = find(B~=C);
To then find the maximum and minimum, use
maxrow = max(list_rows);
minrow = min(list_rows);
and likewise for list_cols.
If B(x,y) and C(x,y) are functions that accept matrix input, then instead of the double-for loop you can do
[x,y] = meshgrid(1:100);
Diff = B(x,y)-C(x,y);
mins = min(Diff);
maxs = max(Diff);
min_x = mins(1); min_y = mins(2);
max_x = maxs(1); max_y = maxs(2);
If B and C are just matrices holding data, then you can do
Diff = B-C;
But really, I need more detail before I can answer this completely.
So: are B and C functions, matrices? You want to find min_x, max_x, but in the example you give that's just 1 and 100, respectively, so...what do you mean?