I'm optimizing my codes. Now I have an MxN matrix, and I want to generate a mean MxN matrix which is the mean of other rows
for example: if i have matrix A:
1 2 3
3 4 5
2 3 2
In the new matrix B, I want each one is the mean of other rows.
mean(row2,row3)
mean(row1,row3)
mean(row1,row2)
I've think of many ways, but can't avoid a loop
for row=1:3
temp = A;
temp(row,:) = [];
B(row,:) = mean(temp);
end
any thoughts?
Simple with bsxfun -
B = (bsxfun(#minus,sum(A,1),A))./(size(A,1)-1)
The trick is to subtract the current row from the sum all rows with bsxfun in a vectorized manner, thus giving us the sum of all rows except the current one. Finally, get the average values by dividing them by the number of rows minus 1.
Related
I want search indexes maximum value/s in each row. If row has more than one maximum, so I want to save both indexes.
For example:
X = [5 6 8
1 2 3
4 4 0];
And I need indexes
inds = [1 3
2 3
3 1
3 2];
I wanted to use function max but this function only saves one index.
You can use max to compute the max for each row and then compare the elements in each row to it's row-wise max using bsxfun and eq. Then you can find the row/column positions of these maxima. We use a transpose in there (.') to ensure that we get the ordering of the output that you expect.
[c,r] = find(bsxfun(#eq, d, max(d, [], 2)).')
output = [r,c];
Another way to do it, would be to use max and repmat.
First you find the maximum of each row using
rowMaximum=max(X,[],2);
Then you replicate the maximum so that it has the same dimension as your input and compare it to the input
logicalMaximum=repmat(rowMaximum,1,size(X,2))==X;
And the last thing you wanna do is converting this logical array into your desired indexes
[columns,rows]=find(logicalMaximum);
result=[rows,columns];
I'm trying to construct a matrix in Matlab where the sum over the rows is constant, but every combination is taken into account.
For example, take a NxM matrix where M is a fixed number and N will depend on K, the result to which all rows must sum.
For example, say K = 3 and M = 3, this will then give the matrix:
[1,1,1
2,1,0
2,0,1
1,2,0
1,0,2
0,2,1
0,1,2
3,0,0
0,3,0
0,0,3]
At the moment I do this by first creating the matrix of all possible combinations, without regard for the sum (for example this also contains [2,2,1] and [3,3,3]) and then throw away the element for which the sum is unequal to K
However this is very memory inefficient (especially for larger K and M), but I couldn't think of a nice way to construct this matrix without first constructing the total matrix.
Is this possible in a nice way? Or should I use a whole bunch of for-loops?
Here is a very simple version using dynamic programming. The basic idea of dynamic programming is to build up a data structure (here S) which holds the intermediate results for smaller instances of the same problem.
M=3;
K=3;
%S(k+1,m) will hold the intermediate result for k and m
S=cell(K+1,M);
%Initialisation, for M=1 there is only a trivial solution using one number.
S(:,1)=num2cell(0:K);
for iM=2:M
for temporary_k=0:K
for new_element=0:temporary_k
h=S{temporary_k-new_element+1,iM-1};
h(:,end+1)=new_element;
S{temporary_k+1,iM}=[S{temporary_k+1,iM};h];
end
end
end
final_result=S{K+1,M}
This may be more efficient than your original approach, although it still generates (and then discards) more rows than needed.
Let M denote the number of columns, and S the desired sum. The problem can be interpreted as partitioning an interval of length S into M subintervals with non-negative integer lengths.
The idea is to generate not the subinterval lengths, but the subinterval edges; and from those compute the subinterval lengths. This can be done in the following steps:
The subinterval edges are M-1 integer values (not necessarily different) between 0 and S. These can be generated as a Cartesian product using for example this answer.
Sort the interval edges, and remove duplicate sets of edges. This is why the algorithm is not totally efficient: it produces duplicates. But hopefully the number of discarded tentative solutions will be less than in your original approach, because this does take into account the fixed sum.
Compute subinterval lengths from their edges. Each length is the difference between two consecutive edges, including a fixed initial edge at 0 and a final edge at S.
Code:
%// Data
S = 3; %// desired sum
M = 3; %// number of pieces
%// Step 1 (adapted from linked answer):
combs = cell(1,M-1);
[combs{end:-1:1}] = ndgrid(0:S);
combs = cat(M+1, combs{:});
combs = reshape(combs,[],M-1);
%// Step 2
combs = unique(sort(combs,2), 'rows');
%// Step 3
combs = [zeros(size(combs,1),1) combs repmat(S, size(combs,1),1)]
result = diff(combs,[],2);
The result is sorted in lexicographical order. In your example,
result =
0 0 3
0 1 2
0 2 1
0 3 0
1 0 2
1 1 1
1 2 0
2 0 1
2 1 0
3 0 0
Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
In MATLAB have a large matrix with transition probabilities transition_probs, and an adjacency matrix adj_mat. I want to compute the cumulative sum of the transition matrix along the columns and then element wise multiply it against the adjacency matrix which acts as a mask in this way:
cumsumTransitionMat = cumsum(transition_probs,2) .* adj_mat;
I get a MEMORY error because with the cumsum all the entries of the matrix are then non-zero.
I would like to avoid this problem by only having the cumulative sum entries where there are non zero entries in the first place. How can this be done without the use of a for loop?
when CUMSUM is applied on rows, for each row it will go and fill with values starting with the first nonzero column it finds up until the last column, thats what it does by definition.
The worst case in terms of storage is when the sparse matrix contains values at the first column, the best case is when all nonzero values occur at the last column. Example:
% worst case
>> M = sparse([ones(5,1) zeros(5,4)]);
>> MM = cumsum(M,2); % completely dense matrix
>> nnz(MM)
ans =
25
% best case
>> MM = cumsum(fliplr(M),2);
If the resulting matrix does not fit in memory, I dont see what else you can do, except maybe use a for-loop over the rows, and process the matrix is smaller batches...
Note that you cannot apply the masking operation before computing the cumulative sum, since this will alter the results. So you cant say cumsum(transition_probs .* adj_mat, 2).
You can apply cumsum on the non-zero elements only. Here is some code:
A = sparse(round(rand(100,1))); %some sparse data
A_cum = A; %instantiate A_cum by copy A
idx_A = find(A); %find non-zeros
A_cum(idx_A) = cumsum(A(idx_A)); %cumsum on non-zeros elements only
You can check the output with
B = cumsum(A);
A_cum B
1 1
0 1
0 1
2 2
3 3
4 4
5 5
0 5
0 5
6 6
and isequal(A_cum(find(A_cum)), B(find(A_cum))) gives 1.
I have a 7x10 matrix, from this matrix I want to randomly select 4 rows without any repetition and this selection will include the index values of the selected rows. So, my question is: how to get a randomly selected matrix with no repetition of rows along with its index values of the original matrix from where it is selected?
Is this what you are after?
B = A(randperm(size(A,1),4),:)
Update: (thanks to federico)
idx = randperm(size(A,1),4);
B = A(idx,:)
Now idx will be a set of 4 integers between 1 and n, where n is the number of rows in A.
A(idx,:) gives you the elements in the rows represented by idx.