For a vector v (e.g. v=[1,2,3,4,5]), and two index vectors (e.g. a=[1,1,1,2,3] and b=[3,4,5,5,5], with all a(i)<b(i)), I would like to construct w=sum(v(a:b)), which gives the values
w = zeros(length(a),1);
for i = 1:length(a)
w(i)=sum(v(a(i):b(i)));
end
It is slow when length(a) is large. Can I compute w without the for loop?
Yes! The nth element of cumsum(v) is the sum of the first n elements in v, so just take that and subtract the sum of the elements that you don't want to include:
v=[1,2,3,4,5]
a=[1,1,1,2,3]
b=[3,4,5,5,5]
C=cumsum(v)
C(b)-C(a)+v(a)
%// or alternatively
C=cumsum([0 v])
C(b+1)-C(a)
The following code works, but it is of course much less readable:
% assume v is a column vector
units = 1:length(v); units = units'; %units is a column vector
units_matrix = repmat(units, [1 length(a)]);
a_matrix = repmat(a, [length(v) 1]); % assuming a is is a row vector
b_matrix = repmat(b, [length(v) 1]);
weights = (units_matrix>=a_matrix) & (units_matrix<=b_matrix);
v_matrix = repmat(v, [1 length(a)]);
w = sum(v_matrix.*weights);
Explanation:
v_matrix contains copies of v. The summation will be done along
the column, so we need to prepare the other needed information in
vectorized form. units_matrix contains the indexes in v along the
columns. The columns are identical. a_matrix and b_matrix, in
each of their column, contains the indexes that are relevant for each
partial summation. All rows are identical. weights is a logical
matrix where, for each column, the indexes contained in
units_matrix between the corresponding a and b are 1 (true),
and the rest is 0. The element-wise multiplication thus filters the
"right" values, and all the indexes outside the range (again, for
each different column) is multiplied by zero. w is then he result
of the sum function, i.e. a row vector (every column of the
"filtered" matrix is summed).
Related
I have two matrices A and B. A(:,1) corresponds to an x-coordinate, A(:,2) corresponds to a y-coordinate, and A(:,3) corresponds to a certain radius. All three values in a row describe the same circle. Now let's say...
A =
[1,4,3]
[8,8,7]
[3,6,3]
B =
[1,3,3]
[1, 92,3]
[4,57,8]
[5,62,1]
[3,4,6]
[9,8,7]
What I need is to be able to loop through matrix A and determine if there are any rows in matrix B that are "similar" as in the x value is within a range (-2,2) of the x value of A (Likewise with the y-coordinate and radius).If it satisfies all three of these conditions, it will be added to a new matrix with the values that were in A. So for example I would need the above data to return...
ans =
[1,4,3]
[8,8,7]
Please help and thank you in advance to anyone willing to take the time!
You can use ismembertol.
result = A(ismembertol(A,B,2,'ByRows',1,'DataScale',1),:)
Manual method
A = [1,4,3;
8,8,7;
3,6,3];
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]; % example matrices
t = 2; % desired threshold
m = any(all(abs(bsxfun(#minus, A, permute(B, [3 2 1])))<=t, 2), 3);
result = A(m,:);
The key is using permute to move the first dimension of B to the third dimension. Then bsxfun computes the element-wise differences for all pairs of rows in the original matrices. A row of A should be selected if all the absolute differences with respect to any column of B are less than the desired threshold t. The resulting variable m is a logical index which is used for selecting those rows.
Using pdist2 (Statistics and Machine Learning Toolbox)
m = any(pdist2(A, B, 'chebychev')<=t, 2);
result = A(m,:);
Ths pdist2 function with the chebychev option computes the maximum coordinate difference (Chebychev distance, or L∞ metric) between pairs of rows.
With for loop
It should work:
A = [1,4,3;
8,8,7;
3,6,3]
B = [1,3,3;
1,92,3;
4,57,8;
5,62,1;
3,4,6;
9,8,7]
index = 1;
for i = 1:size(A,1)
C = abs(B - A(i,:));
if any(max(C,[],2)<=2)
out(index,:) = A(i,:);
index = index + 1
end
end
For each row of A, computes the absolute difference between B and that row, then checks if there exists a row in which the maximum is less than 2.
Without for loop
ind = any(max(abs(B - permute(A,[3 2 1])),[],2)<=2);
out = A(ind(:),:);
I have sum of 3 cell arrays
A=72x1
B=72x720
C=72x90
resultant=A+B+C
size of resultant=72x64800
now when I find the minimum value with row and column indices I can locate the row element easily but how can I locate the column element in variables?
for example
after dong calculations for A,B,C I added them all and got a resultant in from of <72x(720x90)> or can say a matrix of integers of size <72x64800> then I found the minimum value of resultant with row and column index using the code below.
[minimumValue,ind]=min(resultant(:));
[row,col]=find(result== minimumValue);
then row got 14 and column got 6840 value..
now I can trace row 14 of all A,B,C variables easily but how can I know that the resultant column 6480 belongs to which combination of A,B,C?
Instead of using find, use the ind output from the min function. This is the linear index for minimumValue. To do that you can use ind2sub:
[r,c] = ind2sub(size(resultant),ind);
It is not quite clear what do you mean by resultant = A+B+C since you clearly don't sum them if you get a bigger array (72x64800), on the other hand, this is not a simple concatenation ([A B C]) since this would result in a 72x811 array.
However, assuming this is a concatenation you can do the following:
% get the 2nd dimension size of all matrices:
cols = cellfun(#(x) size(x,2),{A,B,C})
% create a vector with reapiting matrices names for all their columns:
mats = repelem(['A' 'B' 'C'],cols);
% get the relevant matrix for the c column:
mats(c)
so mats(c) will be the matrix with the minimum value.
EDIT:
From your comment I understand that your code looks something like this:
% arbitrary data:
A = rand(72,1);
B = rand(72,720);
C = rand(72,90);
% initializing:
K = size(B,2);
N = size(C,2);
counter = 1;
resultant = zeros(72,K*N);
% summing:
for k = 1:K
for n = 1:N
resultant(:,counter) = A + B(:,k) + C(:,n);
counter = counter+1;
end
end
% finding the minimum value:
[minimumValue,ind] = min(resultant(:))
and from the start of the answer you know that you can do this:
[r,c] = ind2sub(size(resultant),ind)
to get the row and column of minimumValue in resultant. So, in the same way you can do:
[Ccol,Bcol] = ind2sub([N,K],c)
where Bcol and Ccol is the column in B and C, respectively, so that:
minimumValue == A(r) + B(r,Bcol) + C(r,Ccol)
To see how it's working imagine that the loop above fills a matrix M with the value of counter, and M has a size of N-by-K. Because we fill M with a linear index, it will be filled in a column-major way, so the row will correspond to the n iterator, and the column will correspond to the k iterator. Now c corresponds to the counter where we got the minimum value, and the row and column of counter in M tells us the columns in B and C, so we can use ind2sub again to get the subscripts of the position of counter. Off course, we don't really need to create M, because the values within it are just the linear indices themselves.
I have a quite big (107 x n) matrix X. Within these n columns, each three columns belong to each other. So, the first three columns of matrix X build a block, then columns 4,5,6 and so on.
Within each block, the first 100 row elements of the first column are important X(1:100,1:3:end). Whenever in this first column the number of zeros or NaNs is greater or equal 20, it should delete the whole block.
Is there a way to do this without a loop?
Thanks for any advice!
Assuming the number of columns of the input to be a multiple of 3, there could be two approaches here.
Approach #1
%// parameters
rl = 100; %// row limit
cl = 20; %// count limit
X1 = X(1:rl,1:3:end) %// Important elements from input
match_mat = isnan(X1) | X1==0 %// binary array of matches
match_blk_id = find(sum(match_mat)>=cl) %// blocks that satisfy requirements
match_colstart = (match_blk_id-1).*3+1 %// start column indices that satisfy
all_col_ind = bsxfun(#plus,match_colstart,[0:2]') %//'columns indices to be removed
X(:,all_col_ind)=[] %// final output after removing to be removed columns
Or if you prefer "compact" codes -
X1 = X(1:rl,1:3:end);
X(:,bsxfun(#plus,(find(sum(isnan(X1) | X1==0)>=cl)-1).*3+1,[0:2]'))=[];
Approach #2
X1 = X(1:rl,1:3:end)
match_mat = isnan(X1) | X1==0 %// binary array of matches
X(:,repmat(sum(match_mat)>=cl,[3 1]))=[] %// Find matching blocks, replicate to
%// next two columns and remove them from X
Note: If X is not a multiple of 3, use this before using the codes - X = [X zeros(size(X,1) ,3 - mod(size(X,2),3))].
I am pretty new to Matlab, now i want to use the matlab to do some clustering job.
if I have 3 columns values
id1 id2 distvalue1
id1 id3 distvalue2
....
id2 id4 distvalue i
.....
5000 ids in total, but some ids pairs are missing the distance value
in python I can make loops to import these distance value into a matrix form. How I can do it in matlab?
and also let the matlab knows id1,...idx are identifies and the third column is the value
Thanks!
Based on the comments, you know how to get the data into the form of an N x 3 matrix, called X, where X(:,1) is the first index, X(:,2) is the second index, and X(:,3) is the corresponding distance.
Let's assume that the indices (id1... idx) are arbitrary numeric labels.
So then we can do the following:
% First, build a list of all the unique indices
indx = unique([X(:,1); X(:,2)]);
Nindx = length(indx);
% Second, initialize an empty connection matrix, C
C = zeros(Nindx, Nindx); %or you could use NaN(Nindx, Nindx)
% Third, loop over the rows of X, and map them to points in the matrix C
for n = 1:size(X,1)
row = find(X(n,1) == indx);
col = find(X(n,2) == indx);
C(row,col) = X(n,3);
end
This is not the most efficient method (that would be to remap the indices of X to the range [1... Nindx] in a vectorized manner), but it should be fine for 5000 ids.
If you end up dealing with very large numbers of unique indices, for which only very few of the index-pairs have assigned distance values, then you may want to look at using sparse matrices -- try help sparse -- instead of pre-allocating a large zero matrix.
I have a matrix and 2 parameters. The first parameter n is the number of elements to be selected. The second is the window size m.
I want to select n number of elements from every m-by-m window from the matrix. As a result we will have a p-by-q cell array where p is matrix_height/m and q is matrix_width/m.
Each element of the cell array contains the n greatest numbers from the corresponding window. It does not necessarily need to be a cell array, it can be anything that will store the necessary data.
One way to do this is by first finding all the unique m-by-m submatrices of your matrix using the function IM2COL, then sorting each column in descending order using the function SORT, and finally extracting the top n rows. If your initial matrix is A and your output matrix is B, this is what it would look like:
B = sort(im2col(A,[m m],'distinct'),1,'descend');
B = B(1:n,:); %# Get the top n values
Note that B will be an n-by-m^2 matrix. If you want to turn this into a p-by-q cell array, you can do this using the functions NUM2CELL and RESHAPE:
nBlocks = ceil(size(A)./m); %# The number of blocks in each dimension
B = reshape(num2cell(B,1),nBlocks(1),nBlocks(2));
EDIT:
If you also want to get the indices of each value with respect to the input matrix A, that is a bit more complicated. You can do it by getting the second output from SORT, which in this case will be the linear indices of the values within each m-by-m submatrix. You can convert these to subscripts using the function IND2SUB, then shift the row and column indices to account for the position of each m-by-m block:
[B,index] = sort(im2col(A,[m m],'distinct'),1,'descend');
B = B(1:n,:); %# Get the top n values
index = index(1:n,:); %# Get the top n values
[r,c] = ind2sub([m m],index); %# Convert linear indices to subscripts
nBlocks = size(A)./m; %# The number of blocks in each dimension
r = r+repmat(0:m:(nBlocks(1)-1)*m,n,nBlocks(2)); %# Shift the row indices
c = c+kron(0:m:(nBlocks(2)-1)*m,ones(n,nBlocks(1))); %# Shift the column indices
And now you can collect the row indices, column indices, and values together into a cell array using the functions MAT2CELL and RESHAPE:
B = mat2cell([r(:) c(:) B(:)],n.*ones(1,size(B,2)));
B = reshape(B,nBlocks(1),nBlocks(2));
Alternatively, you can create a structure array instead of a cell array using the functions NUM2CELL, STRUCT, and RESHAPE:
B = struct('rowIndices',num2cell(r,1),...
'colIndices',num2cell(c,1),...
'values',num2cell(B,1));
B = reshape(B,nBlocks(1),nBlocks(2));
NOTE:
The function IM2COL will pad partial blocks with zeroes in the event that a dimension of A is not an even multiple of m. If any of this zero-padding appears in the top n values for a block, one or both of the corresponding row and column indices will be out of range (i.e. have a value larger than the size of that dimension for the matrix A). Thus by checking that the row and column indices are in range you can make sure you are not including any of the zero-padding in your subsequent analysis.