I have a data file matrix.txt, it has three columns. The first column stores the row index, the second column stores the column index, the third column stores the value. How do I read these into a matrix called mat. To be explicit, suppose our mat is a n*n square matrix, let n=2 for instance. In the text file, it has:
0 0 10
1 1 -10
The element in mat not specified is 0. Thus mat is supposed to be:
mat = 10 0
0 -10
How do I achieve this?
This should work for the generic 2-D case.
% Read in matrix specification
fID = fopen('matrix.txt');
tmp = fscanf(fID, '%u%u%f', [3 inf])';
fclose(fID);
% Use the maximum row and column subscripts to obtain the matrix size
tmp(:, 1:2) = tmp(:, 1:2) + 1; % MATLAB doesn't use 0-based indexing
matsize = [max(tmp(:,1)), max(tmp(:,2))];
% Convert subscripts to linear indices
lidx = sub2ind(matsize, tmp(:,1), tmp(:,2));
mat = zeros(matsize); % Initialize matrix
mat(lidx) = tmp(:,3); % Assign data
Using a sample matrix.txt:
0 0 10
1 1 -10
1 2 20
We receive:
>> mat
mat =
10 0 0
0 -10 20
Since in MATLAB, indices begin with 1 (not zero), we should add 1 to our indices in code.
r and c stand for row and column.
Alsom and n is for m by n zero matrix
A = importdata('matrix.txt');
r = A(:, 1)';
c = A(:, 2)';
m = max(r);
n = max(c);
B = zeros(m + 1, n + 1);
for k = 1:size(A,1);
B(r(k) + 1, c(k) + 1) = A(k, 3);
end
Result:
B =
10 0
0 -10
I see I am too slow, but I decided post my answer anyway...
I initialized matrix A as a vector, and used reshape:
%Load all file to matrix at once
%You may consider using fopen and fscanf, in case Matrix.txt is not ordered perfectly.
row_column_val = load('Matrix.txt', '-ascii');
R = row_column_val(:, 1) + 1; %Get vector of row indexes (add 1 - convert to Matalb indeces).
C = row_column_val(:, 2) + 1; %Get vector of column indexes (add 1 - convert to Matalb indeces).
V = row_column_val(:, 3); %Get vector of values.
nrows = max(R); %Number of rows in matrix.
ncols = max(C); %Number of columns in matrix.
A = zeros(nrows*ncols, 1); %Initialize A as a vector instead of a matrix (length of A is nrows*ncols).
%Put value v in place c*ncols + r for all elements of V, C and R.
%The formula is used for column major matrix (Matlab stored matrices in column major format).
A((C-1)*nrows + R) = V;
A = reshape(A, [nrows, ncols]);
Related
I want to generalise to any n the Matlab code below.
Let A be an n-dimensional array:
clear
rng default
n=4;
A=randn(n,n,n,n);
n=5;
A=randn(n,n,n,n,n);
Note that A is composed of n^(n-2) 2-dimensional matrices, each of size nxn.
For example, when n=4 these matrices are A(:,:,1,1),...,A(:,:,4,1),A(:,:,1,2),...,A(:,:,4,4).
Suppose I'm interested in a code which:
1) deletes the last column and row in each of the n^(n-2) 2-dimensional matrices
%when n=4
A(n,:,:,:)=[];
A(:,n,:,:)=[];
%when n=5
A(n,:,:,:,:)=[];
A(:,n,:,:,:)=[];
2) deletes the 2-dimensional matrices with the 3-th,4-th,5-th,n-th index equal to n.
%when n=4
A(:,:,n,:)=[];
A(:,:,:,n)=[];
%when n=5
A(:,:,n,:,:)=[];
A(:,:,:,n,:)=[];
A(:,:,:,:,n)=[];
Question: could you help me to generalise the code above to any n? I cannot see how to proceed.
You can index your matrix with a cell containing multiple elements. Each element will be interpreted as a new index (more information here):
%Example 1: A(:,:,1:3,1:3,1:3}
%elements per dimension
n = 4;
%number of dimension
d = 5;
%random matrix
repdim = repmat({n},d,1)
A = rand(repdim{:});
%We want A(:,:,1:3,1:3,1:3}, so we create c = {1:3,1:3,1:3}
c = repmat({1:n-1},d-2,1);
%Get the new matrix
A = A(:,:,c{:});
%Example 2: A(1:3,1:3,:,:,:}
%elements per dimension
n = 4;
%number of dimension
d = 5;
%random matrix
repdim = repmat({n},d,1)
A = rand(repdim{:});
%We want A(1:3,1:3,:,:,:}, so we create c1 = {1:3,1:3} and c2 = {':',':',':'}
c1 = repmat({1:n-1},2,1);
c2 = repmat({':'},d-2,1); %thanks to #LuisMendo for the suggestion.
%Get the new matrix
A = A(c1{:},c2{:});
Suppose I have a matrix A of dimension Nx(N-1) in MATLAB, e.g.
N=5;
A=[1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16;
17 18 19 20 ];
I want to transform A into an NxN matrix B, just by adding a zero diagonal, i.e.,
B=[ 0 1 2 3 4;
5 0 6 7 8;
9 10 0 11 12;
13 14 15 0 16;
17 18 19 20 0];
This code does what I want:
B_temp = zeros(N,N);
B_temp(1,:) = [0 A(1,:)];
B_temp(N,:) = [A(N,:) 0];
for j=2:N-1
B_temp(j,:)= [A(j,1:j-1) 0 A(j,j:end)];
end
B = B_temp;
Could you suggest an efficient way to vectorise it?
You can do this with upper and lower triangular parts of the matrix (triu and tril).
Then it's a 1 line solution:
B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
Edit: benchmark
This is a comparison of the loop method, the 2 methods in Sardar's answer, and my method above.
Benchmark code, using timeit for timing and directly lifting code from question and answers:
function benchie()
N = 1e4; A = rand(N,N-1); % Initialise large matrix
% Set up anonymous functions for input to timeit
s1 = #() sardar1(A,N); s2 = #() sardar2(A,N);
w = #() wolfie(A,N); u = #() user3285148(A,N);
% timings
timeit(s1), timeit(s2), timeit(w), timeit(u)
end
function sardar1(A, N) % using eye as an indexing matrix
B=double(~eye(N)); B(find(B))=A.'; B=B.';
end
function sardar2(A,N) % similar to sardar1, but avoiding slow operations
B=1-eye(N); B(logical(B))=A.'; B=B.';
end
function wolfie(A,N) % using triangular parts of the matrix
B = [tril(A,-1) zeros(N, 1)] + [zeros(N,1) triu(A)];
end
function user3285148(A, N) % original looping method
B = zeros(N,N); B(1,:) = [0 A(1,:)]; B(N,:) = [A(N,:) 0];
for j=2:N-1; B(j,:)= [A(j,1:j-1) 0 A(j,j:end)]; end
end
Results:
Sardar method 1: 2.83 secs
Sardar method 2: 1.82 secs
My method: 1.45 secs
Looping method: 3.80 secs (!)
Conclusions:
Your desire to vectorise this was well founded, looping is way slower than other methods.
Avoiding data conversions and find for large matrices is important, saving ~35% processing time between Sardar's methods.
By avoiding indexing all together you can save a further 20% processing time.
Generate a matrix with zeros at diagonal and ones at non-diagonal indices. Replace the non-diagonal elements with the transpose of A (since MATLAB is column major). Transpose again to get the correct order.
B = double(~eye(N)); %Converting to double since we want to replace with double entries
B(find(B)) = A.'; %Replacing the entries
B = B.'; %Transposing again to get the matrix in the correct order
Edit:
As suggested by Wolfie for the same algorithm, you can get rid of conversion to double and the use of find with:
B = 1-eye(N);
B(logical(B)) = A.';
B = B.';
If you want to insert any vector on a diagonal of a matrix, one can use plain indexing. The following snippet gives you the indices of the desired diagonal, given the size of the square matrix n (matrix is n by n), and the number of the diagonal k, where k=0 corresponds to the main diagonal, positive numbers of k to upper diagonals and negative numbers of k to lower diagonals. ixd finally gives you the 2D indices.
function [idx] = diagidx(n,k)
% n size of square matrix
% k number of diagonal
if k==0 % identity
idx = [(1:n).' (1:n).']; % [row col]
elseif k>0 % Upper diagonal
idx = [(1:n-k).' (1+k:n).'];
elseif k<0 % lower diagonal
idx = [(1+abs(k):n).' (1:n-abs(k)).'];
end
end
Usage:
n=10;
k=3;
A = rand(n);
idx = diagidx(n,k);
A(idx) = 1:(n-k);
I have 2 matrices A and B.
I find the max values in the columns of A, and keep their indices in I. So far so good.
Now, I need to choose those arrays of B with the same index as stored in I. I don't know how to do this.
See below:
A = [1,2,3; 0,8,9]
B = [0,1,2; 4,2,3]
[~,I] = max(A)
h = B(I)
I need to get these values of B:
h = [0 2 3]
But the code results in a different one. How can I fix it?
A =
1 2 3
0 8 9
B =
0 1 2
4 2 3
I =
1 2 2
h =
0 4 4
Thanks in advance
The max function how you used it works like
If A is a matrix, then max(A) is a row vector containing the maximum value of each column.
so M = max(A) is equivalent to M = max(A,[],1). But rather use the third input if you're not sure.
If you use max to find the maxima in the columns of the matrix, it returns the row indices. The column indices are for your case simply 1:size(A,2) = [1 2 3].
Now you need to convert your row and column indices to linear indices with sub2ind:
%// data
A = [1,2,3; 0,8,9]
B = [0,1,2; 4,2,3]
%// find maxima of each column in A
[~, I] = max( A, [], 1 ) %// returns row indices
%// get linear indices for both, row indices and column indices
I = sub2ind( size(A), I, 1:size(A,2) )
%// index B
h = B(I)
returns:
h =
0 2 3
Is there an easy way of making a 'random' sparse matrix with a specific number of nonzero entries?
Here is my attempt:
r = randperm(n,m) % n = size of matrix, m = number of nonzeros in each column
H = sparse(r, r,1,n,n);
But the matrix H doesn't have exactly m nonzeros in each column. For example if I use this to make a 100 x 100 matrix with 10 nonzeros in each column only 10 columns have exactly 10 1's in them.
I'm sure there's an easy way to do this but I can't see it.
This will generate a 100-by-100 matrix with exactly ten 1s per column:
n = 100;
m = 10;
nonzerosPerColumn = repmat(m, 1, n);
%%// Build vector of linear indices to nonzero entries
pos = cell2mat(arrayfun(#(i)randperm(n,nonzerosPerColumn(i))+(i-1)*n,1:n,'uni',0));
%%// Generate the matrix
M = reshape(sparse(pos,1,1,n*n,1),n,n);
Here's a vectorized approach:
r = 100; %// number of rows
c = 100; %// number of columns
m = 10; %// number of ones that there should be in each column
H = sparse([], [], [], r, c, c*m); %// preallocate and initiallize to zero
[~, ind] = sort(rand(r,c)); %// randomly generate...
ind = ind(1:m,:); %// ... m row indices per column
H(bsxfun(#plus, ind, (0:c-1)*r)) = 1; %// fill in ones, using linear indexing
I have a vector y of length n. y(i) is an integer in 1..m. Is there a simpler way to convert y into an n x m logical matrix yy, where yy(i, j) = 1 if y(i) = j, but 0 otherwise? Here's how I've been doing it:
% If m is known (m = 3 here), you could write it out all at once
yy = [y == 1; y== 2; y == 3];
yy = reshape(yy, n, 3);
or
% if m is not known ahead of time
yy = [ y == 1 ];
for i = 2:m;
yy = [ yy; y == i ];
end
yy = reshape(yy, n, m);
You can use bsxfun for this
yy = bsxfun(#eq,y(:),[1,2,3])
y is transformed (if necessary) to a column-vector, while the other vector is a row vector. bsxfun implicitly expands the m-by-1 and 1-by-n arrays so that the result becomes m-by-n.
If n*m is sufficiently large (and m is, by itself, sufficiently large), it is a good idea to create yy as a sparse matrix. Your y vector is really a special type of sparse matrix format, but we can translate it into the built-in sparse matrix format by doing the following.
yy = sparse(1:length(y), y, 1);
This will keep your storage to O(n). It is not going to be doing you a lot of favors if you are using yy for a lot of indexing. If that is the case you are better off using your original sparse structure (i.e., y).
A slight modification to your method:
% A n-dimensional vector y, with values in some range 1..m
m = 4;
n = 7;
y = randi([1 m], n, 1);
% Preallocating a n by m matrix of zeros
nXm = zeros(n, m);
% In each pass of this loop a single column of nXm is updated, where
% for each column index j in nXm, if y(i) = j then nXm(i,j) = 1
for j = 1:m;
nXm(:,j) = (y == j);
end
From Machine Learning on Coursera:
yy = eye(m)(y, :)
This requires that the list be a range 1:m (as OP stated). For an irregular list, like [2 3 5], do this
yy = eye(m)(:, [2 3 5])(y, :)
Note: not tested on MATLAB.
In octave you can write:
yy = y' == (1:m); % or y == (1:m)' for transposed
[1 2 1 3 2] == [1 2 3]' % = [1 0 1 0 0; 0 1 0 0 1; 0 0 0 1 0]