I'm looking to find the median of just a specific row of an m x n array. I couldn't find anything useful in the Matlab help section. E.g. if I had a small array
[1 2 4; 2 3 4; 6 2 8]
how would I find the median of row 2? Many thanks.
Search for median in google or matlab (doc median) and you will find the median function.
If you want to find the median of row 2, just use row,column indexing syntax, where : means all entries:
A = [1 2 4; 2 3 4; 6 2 8];
median(A(2,:))
You could try to use:
A = [1 2 4; 2 3 4; 6 2 8];
medianRows = median(A, 2); % to find every rows' median
medianRows(2)
Or:
medianRow2 = median(A(2, :)); % every column of 2nd row
Related
What I'm trying to do: given a 2D matrix, get the column indices of the elements in each row that satisfy some particular condition.
For example, say my matrix is
M = [16 2 3 13; 5 11 10 8; 9 7 6 12; 4 14 15 1]
and my condition is M>6. Then my desired output would be something like
Indices = {[1 4]'; [2 3 4]'; [1 2 4]'; [2 3]';}
After reading the answers to this similar question I came up with this partial solution using find and accumarray:
[ix, iy] = find(M>6);
Indices = accumarray(ix,iy,[],#(iy){iy});
This gives very nearly the results I want -- in fact, the indices are all right, but they're not ordered the way I expected. For example, Indices{2} = [2 4 3]' instead of [2 3 4]', and I can't understand why. There are 3 occurrences of 2 in ix, at indices 3, 6, and 9. The corresponding values of iy at those indices are 2, 3, and 4, in that order. What exactly is creating the observed order? Is it just arbitrary? Is there a way to force it to be what I want, other than sorting each element of Indices afterwards?
Here's one way to solve it with arrayfun -
idx = arrayfun(#(x) find(M(x,:)>6),1:size(M,1),'Uni',0)
Display output wtih celldisp(idx) -
idx{1} =
1 4
idx{2} =
2 3 4
idx{3} =
1 2 4
idx{4} =
2 3
To continue working with accumarray, you can wrap iy with sort to get your desired output which doesn't look too pretty maybe -
Indices = accumarray(ix,iy,[],#(iy){sort(iy)})
Output -
>> celldisp(Indices)
Indices{1} =
1
4
Indices{2} =
2
3
4
Indices{3} =
1
2
4
Indices{4} =
2
3
accumarray is not guaranteed to preserve order of each chunk of its second input (see here, and also here). However, it does seem to preserve it when the first input is already sorted:
[iy, ix] = find(M.'>6); %'// transpose and reverse outputs, to make ix sorted
Indices = accumarray(ix,iy,[],#(iy){iy}); %// this line is the same as yours
produces
Indices{1} =
1
4
Indices{2} =
2
3
4
Indices{3} =
1
2
4
Indices{4} =
2
3
I have a matrix that looks like this:
M = [2 3 4; 2 4 6; 1 5 3]
How do I find the minimum value in each row? For example, I would first want to find the minimum value of the first row? Thanks!
Use min
>> M = [2 3 4; 2 4 6; 1 5 3];
>> X=min(M')
X =
2 2 1 % row-wise
OR
>> [E,I] = min(M,[],2) ; % E- Elements, I- Index
You also can try
M = min(A(n,:)); %where n is the row you want to look at
I have a m-by-n matrix and I want to shift each row elements k no. of times (" one resultant matrix for each one shift so a total of k matrices corresponding to each row shifts ")(k can be different for different rows and 0<=k<=n) and want to index all the resultant matrices corresponding to each individual shift.
Eg: I have the matrix: [1 2 3 4; 5 6 7 8; 2 3 4 5]. Now, say, I want to shift row1 by 2 times (i.e. k=2 for row1) and row2 by 3times (i.e. k=3 for row2) and want to index all the shifted versions of matrices (It is similar to combinatorics of rows but with limited and diffeent no. of shifts to each row).
Can someone help to write up the code? (please help to write the general code but not for the example I mentioned here)
I found the following question useful to some extent, but it won't solve my problem as my problem looks like a special case of this problem:
Matlab: How to get all the possible different matrices by shifting it's rows (Update: each row has a different step)
See if this works for you -
%// Input m-by-n matrix
A = rand(2,5) %// Edit this to your data
%// Initialize shifts, k for each row. The number of elements would be m.
sr = [2 3]; %// Edit this to your data
[m,n] = size(A); %// Get size
%// Get all the shits in one go
sr_ind = arrayfun(#(x) 0:x,sr,'un',0); %//'
shifts = allcomb(sr_ind{:},'matlab')'; %//'
for k1 = 1:size(shifts,2)
%// Get shift to be used for each row for each iteration
shift1 = shifts(:,k1);
%// Get circularly shifted column indices
t2 = mod(bsxfun(#minus,1:n,shift1),n);
t2(t2==0) = n;
%// Get the linear indices and use them to index into input to get the output
ind = bsxfun(#plus,[1:m]',(t2-1)*m); %//'
all_matrices = A(ind) %// outputs
end
Please note that this code uses MATLAB file-exchange code allcomb.
If your problem in reality is not more complex than what you showed us, it can be done by a double loop. However, i don't like my solution, because you would need another nested loop for each row you want to shift. Also it generates all shift-combinations from your given k-numbers, so it has alot of overhead. But this can be a start:
% input
m = [1 2 3 4; 5 6 7 8; 2 3 4 5];
shift_times = {0:2, 0:3}; % 2 times for row 1, 3 times for row 2
% desird results
desired_matrices{1} = [4 1 2 3; 5 6 7 8; 2 3 4 5];
desired_matrices{2} = [3 4 1 2; 5 6 7 8; 2 3 4 5];
desired_matrices{3} = [1 2 3 4; 8 5 6 7; 2 3 4 5];
desired_matrices{4} = [4 1 2 3; 8 5 6 7; 2 3 4 5];
desired_matrices{5} = [3 4 1 2; 8 5 6 7; 2 3 4 5];
% info needed:
[rows, cols] = size(m);
count = 0;
% make all shift combinations
for shift1 = shift_times{1}
% shift row 1
m_shifted = m;
idx_shifted = [circshift([1:cols]',shift1)]';
m_shifted(1, :) = m_shifted(1, idx_shifted);
for shift2 = shift_times{2}
% shift row 2
idx_shifted = [circshift([1:cols]',shift2)]';
m_shifted(2, :) = m_shifted(r_s, idx_shifted);
% store them
store{shift1+1, shift2+1} = m_shifted;
end
end
% store{i+1, j+1} stores row 1 shifted by i and row 2 shifted by j
% example
all(all(store{2,1} == desired_matrices{1})) % row1: 1, row2: 0
all(all(store{2,2} == desired_matrices{4})) % row1: 1, row2: 1
all(all(store{3,2} == desired_matrices{5})) % row1: 2, row2: 1
suppose I have values need_find = [1 3 4] and a matrix A in the size of 3xK. I want to find the values of need_find on its corresponding row of A. How can I apply vectorized solution in matlab instead of iterate over each row?
For detailed example as expected;
A = [1 3 4; 1 3 5; 3 4 5];
my_method_do_what_I_want(A,need_find);
The method returns
ans = [1;2;2]
% so I find the index of each element of need_find at corresponding row at A
Long story short :seach 1 at row 1, search 3 at row2, search 4 at row3
Here's one way:
A = [1 3 4; 1 3 5; 3 4 5];
need_find = [1 3 4]
[~,idx] = find(bsxfun(#eq,A,need_find(:)))
which returns
idx =
1
2
2
This simple one-liner won't work if you have repeated values in the rows of A or if there are no matches at all, but I can only go by your example...
Suppose
A = [1 2 3 5 7 9
6 5 0 3 2 3]
I want to randomize the position of the matrix column position; to give B like so:
B = [3 9 1 7 2 5
0 3 6 2 5 3]
How can I do this Matlab?
Try
B = A(randperm(length(A)))
Consult the documentation for explanations.
Now that code is formatted properly, it's clear the OP wants column preservation, although randperm is still the easiest option as given by HPM's answer.
idx = randperm(size(A,2)); % Create list of integers 1:n, in random order,
% where n = num of columns
B = A(:, idx); % Shuffles columns about, on all rows, to indixes in idx
You'll want to use the randperm function to generate a random permutation of column indices (from 1 to the number of columns in A), which you can then index A with:
B = A(:, randperm(size(A, 2)));