Say I have the following basic if-statement:
if (A ~= 0)
% do something like divide your favorite number by A
else
% do something like return NaN or infinity
end
The problem is that A is not a simple number but a vector. Matlab returns true if no element in A is 0. What I am looking for is a vectorized? way of perforimg the if-statement above for each element in A.
Actually, I simply want to do this as fast as possible.
Vectorized ifs don't exist, but there are some options. If you want to test for all or any elements true, use the all or any function.
Here's one example of conditionally modifying values of a matrix:
b = A ~= 0; % b is a boolean matrix pointing to nonzero indices
% (b could be derived from some other condition,
% like b = sin(A)>0
A(b) = f(A(b)) % do something with the indices that pass
A(~b) = g(A(~b)) % do something else with the indices that fail
B = zeros(size(A));
B(A~=0) = FAV./A(A~=0);
B(A==0) = NaN;
In general, to perform one operation on some elements of a matrix and another operation on the remaining elements, a one-line solution is:
Z = B .* X + ~B .* Y;
where B is a logical matrix. As an example,
Z = (A == 0) .* -1 + (A ~= 0) .* A;
copies A but assigns -1 everywhere that A is zero.
However, because the question deals with infinity or NaNs, it can be done even more succinctly:
Z = FAV ./ A; % produces inf where A == 0
Z = (A ~= 0) .* FAV ./ A; % produces NaN where A == 0
Are you looking for all non-zero elements? You can do this a couple of ways.
nonzero = find(A); % returns indicies to all non-zero elements of A
y = x./A(nonzero); % divides x by all non-zero elements of A
% y will be the same size as nonzero
Or for a one-liner, you can use a conditional in place of indicies
y = x./A(A~=0); % divides x by all non-zero elements of A
What you need to do is identify the elements you want to operate on. I would use FIND. I store the results in VI (Valid Indicies) and use that to populate the matrix.
clear
clc
den = [2 0 2; 0 2 0; -2 -2 -2]
num = ones(size(den));
frac = nan(size(den));
vi = (den ~=0)
frac(vi) = num(vi)./den(vi)
vi = (den == 0)
frac(vi) = nan %just for good measure...
Related
Consider a Matlab matrix B which lists all possible unordered pairs (without repetitions) from [1 2 ... n]. For example, if n=4,
B=[1 2;
1 3;
1 4;
2 3;
2 4;
3 4]
Note that B has size n(n-1)/2 x 2
I want to take a random draw of m rows from B and store them in a matrix C. Continuing the example above, I could do that as
m=2;
C=B(randi([1 size(B,1)],m,1),:);
However, in my actual case, n=371293. Hence, I cannot create B and, then, run the code above to obtain C. This is because storing B would require a huge amount of memory.
Could you advise on how I could proceed to create C, without having to first store B? Comments on a different question suggest to
Draw at random m integers between 1 and n(n-1)/2.
I=randi([1 n*(n-1)/2],m,1);
Use ind2sub to obtain C.
Here, I'm struggling to implement the second step.
Thanks to the comments below, I wrote this
n=4;
m=10;
coord=NaN(m,2);
R= randi([1 n^2],m,1);
for i=1:m
[cr, cc]=ind2sub([n,n],R(i));
if cr>cc
coord(i,1)=cc;
coord(i,2)=cr;
elseif cr<cc
coord(i,1)=cr;
coord(i,2)=cc;
end
end
coord(any(isnan(coord),2),:) = []; %delete NaN rows from coord
I guess there are more efficient ways to implement the same thing.
You can use the function named myind2ind in this post to take random rows of all possible unordered pairs without generating all of them.
function [R , C] = myind2ind(ii, N)
jj = N * (N - 1) / 2 + 1 - ii;
r = (1 + sqrt(8 * jj)) / 2;
R = N -floor(r);
idx_first = (floor(r + 1) .* floor(r)) / 2;
C = idx_first-jj + R + 1;
end
I=randi([1 n*(n-1)/2],m,1);
[C1 C2] = myind2ind (I, n);
If you look at the odds, for i=1:n-1, the number of combinations where the first value is equal to i is (n-i) and the total number of cominations is n*(n-1)/2. You can use this law to generate the first column of C. The values of the second column of C can then be generated randomly as integers uniformly distributed in the range [i+1, n]. Here is a code that performs the desired tasks:
clc; clear all; close all;
% Parameters
n = 371293; m = 10;
% Generation of C
R = rand(m,1);
C = zeros(m,2);
s = 0;
t = n*(n-1)/2;
for i=1:n-1
if (i<n-1)
ind_i = R>=s/t & R<(s+n-i)/t;
else % To avoid rounding errors for n>>1, we impose (s+n-i)=t at the last iteration (R<(s+n-i)/t=1 always true)
ind_i = R>=s/t;
end
C(ind_i,1) = i;
C(ind_i,2) = randi([i+1,n],sum(ind_i),1);
s = s+n-i;
end
% Display
C
Output:
C =
84333 266452
46609 223000
176395 328914
84865 94391
104444 227034
221905 302546
227497 335959
188486 344305
164789 266497
153603 354932
Good luck!
I am trying to implement a simplex algorithm following the rules I was given at my optimization course. The problem is
min c'*x s.t.
Ax = b
x >= 0
All vectors are assumes to be columns, ' denotes the transpose. The algorithm should also return the solution to dual LP. The rules to follow are:
Here, A_J denotes columns from A with indices in J and x_J, x_K denotes elements of vector x with indices in J or K respectively. Vector a_s is column s of matrix A.
Now I do not understand how this algorithm takes care of condition x >= 0, but I decided to give it a try and follow it step by step. I used Matlab for this and got the following code.
X = zeros(n, 1);
Y = zeros(m, 1);
% i. Choose starting basis J and K = {1,2,...,n} \ J
J = [4 5 6] % for our problem
K = setdiff(1:n, J)
% this while is for goto
while 1
% ii. Solve system A_J*\bar{x}_J = b.
xbar = A(:,J) \ b
% iii. Calculate value of criterion function with respect to current x_J.
fval = c(J)' * xbar
% iv. Calculate dual solution y from A_J^T*y = c_J.
y = A(:,J)' \ c(J)
% v. Calculate \bar{c}^T = c_K^T - u^T A_K. If \bar{c}^T >= 0, we have
% found the optimal solution. If not, select the smallest s \in K, such
% that c_s < 0. Variable x_s enters basis.
cbar = c(K)' - c(J)' * inv(A(:,J)) * A(:,K)
cbar = cbar'
tmp = findnegative(cbar)
if tmp == -1 % we have found the optimal solution since cbar >= 0
X(J) = xbar;
Y = y;
FVAL = fval;
return
end
s = findnegative(c, K) %x_s enters basis
% vi. Solve system A_J*\bar{a} = a_s. If \bar{a} <= 0, then the problem is
% unbounded.
abar = A(:,J) \ A(:,s)
if findpositive(abar) == -1 % we failed to find positive number
disp('The problem is unbounded.')
return;
end
% vii. Calculate v = \bar{x}_J / \bar{a} and find the smallest rho \in J,
% such that v_rho > 0. Variable x_rho exits basis.
v = xbar ./ abar
rho = J(findpositive(v))
% viii. Update J and K and goto ii.
J = setdiff(J, rho)
J = union(J, s)
K = setdiff(K, s)
K = union(K, rho)
end
Functions findpositive(x) and findnegative(x, S) return the first index of positive or negative value in x. S is the set of indices, over which we look at. If S is omitted, whole vector is checked. Semicolons are omitted for debugging purposes.
The problem I tested this code on is
c = [-3 -1 -3 zeros(1,3)];
A = [2 1 1; 1 2 3; 2 2 1];
A = [A eye(3)];
b = [2; 5; 6];
The reason for zeros(1,3) and eye(3) is that the problem is inequalities and we need slack variables. I have set starting basis to [4 5 6] because the notes say that starting basis should be set to slack variables.
Now, what happens during execution is that on first run of while, variable with index 1 enters basis (in Matlab, indices go from 1 on) and 4 exits it and that is reasonable. On the second run, 2 enters the basis (since it is the smallest index such that c(idx) < 0 and 1 leaves it. But now on the next iteration, 1 enters basis again and I understand why it enters, because it is the smallest index, such that c(idx) < 0. But here the looping starts. I assume that should not have happened, but following the rules I cannot see how to prevent this.
I guess that there has to be something wrong with my interpretation of the notes but I just cannot see where I am wrong. I also remember that when we solved LP on the paper, we were updating our subjective function on each go, since when a variable entered basis, we removed it from the subjective function and expressed that variable in subj. function with the expression from one of the equalities, but I assume that is different algorithm.
Any remarks or help will be highly appreciated.
The problem has been solved. Turned out that the point 7 in the notes was wrong. Instead, point 7 should be
Here is a question about whether we can use vectorization type of operation in matlab to avoid writing for loop.
I have a vector
Q = [0.1,0.3,0.6,1.0]
I generate a uniformly distributed random vector over [0,1)
X = [0.11,0.72,0.32,0.94]
I want to know whether each entry of X is between [0,0.1) or [0.1,0.3) or [0.3,0.6), or [0.6,1.0) and I want to return a vector which contains the index of the maximum element in Q that each entry of X is less than.
I could write a for loop
Y = zeros(length(X),1)
for i = 1:1:length(X)
Y(i) = find(X(i)<Q, 1);
end
Expected result for this example:
Y = [2,4,3,4]
But I wonder if there is a way to avoid writing for loop? (I see many very good answers to my question. Thank you so much! Now if we go one step further, what if my Q is a matrix, such that I want check whether )
Y = zeros(length(X),1)
for i = 1:1:length(X)
Y(i) = find(X(i)<Q(i), 1);
end
Use the second output of max, which acts as a sort of "vectorized find":
[~, Y] = max(bsxfun(#lt, X(:).', Q(:)), [], 1);
How this works:
For each element of X, test if it is less than each element of Q. This is done with bsxfun(#lt, X(:).', Q(:)). Note each column in the result corresponds to an element of X, and each row to an element of Q.
Then, for each element of X, get the index of the first element of Q for which that comparison is true. This is done with [~, Y] = max(..., [], 1). Note that the second output of max returns the index of the first maximizer (along the specified dimension), so in this case it gives the index of the first true in each column.
For your example values,
Q = [0.1, 0.3, 0.6, 1.0];
X = [0.11, 0.72, 0.32, 0.94];
[~, Y] = max(bsxfun(#lt, X(:).', Q(:)), [], 1);
gives
Y =
2 4 3 4
Using bsxfun will help accomplish this. You'll need to read about it. I also added a Q = 0 at the beginning to handle the small X case
X = [0.11,0.72,0.32,0.94 0.01];
Q = [0.1,0.3,0.6,1.0];
Q_extra = [0 Q];
Diff = bsxfun(#minus,X(:)',Q_extra (:)); %vectorized subtraction
logical_matrix = diff(Diff < 0); %find the transition from neg to positive
[X_categories,~] = find(logical_matrix == true); % get indices
% output is 2 4 3 4 1
EDIT: How long does each method take?
I got curious about the difference between each solution:
Test Code Below:
Q = [0,0.1,0.3,0.6,1.0];
X = rand(1,1e3);
tic
Y = zeros(length(X),1);
for i = 1:1:length(X)
Y(i) = find(X(i)<Q, 1);
end
toc
tic
result = arrayfun(#(x)find(x < Q, 1), X);
toc
tic
Q = [0 Q];
Diff = bsxfun(#minus,X(:)',Q(:)); %vectorized subtraction
logical_matrix = diff(Diff < 0); %find the transition from neg to positive
[X_categories,~] = find(logical_matrix == true); % get indices
toc
Run it for yourself, I found that when the size of X was 1e6, bsxfun was much faster, while for smaller arrays the differences were varying and negligible.
SAMPLE: when size X was 1e3
Elapsed time is 0.001582 seconds. % for loop
Elapsed time is 0.007324 seconds. % anonymous function
Elapsed time is 0.000785 seconds. % bsxfun
Octave has a function lookup to do exactly that. It takes a lookup table of sorted values and an array, and returns an array with indices for values in the lookup table.
octave> Q = [0.1 0.3 0.6 1.0];
octave> x = [0.11 0.72 0.32 0.94];
octave> lookup (Q, X)
ans =
1 3 2 3
The only issue is that your lookup table has an implicit zero which be fixed easily with:
octave> lookup ([0 Q], X) # alternatively, just add 1 at the results
ans =
2 4 3 4
You can create an anonymous function to perform the comparison, then apply it to each member of X using arrayfun:
compareFunc = #(x)find(x < Q, 1);
result = arrayfun(compareFunc, X, 'UniformOutput', 1);
The Q array will be stored in the anonymous function ( compareFunc ) when the anonymous function is created.
Or, as one line (Uniform Output is the default behavior of arrayfun):
result = arrayfun(#(x)find(x < Q, 1), X);
Octave does a neat auto-vectorization trick for you if the vectors you have are along different dimensions. If you make Q a column vector, you can do this:
X = [0.11, 0.72, 0.32, 0.94];
Q = [0.1; 0.3; 0.6; 1.0; 2.0; 3.0];
X <= Q
The result is a 6x4 matrix indicating which elements of Q each element of X is less than. I made Q a different length than X just to illustrate this:
0 0 0 0
1 0 0 0
1 0 1 0
1 1 1 1
1 1 1 1
1 1 1 1
Going back to the original example you have, you can do
length(Q) - sum(X <= Q) + 1
to get
2 4 3 4
Notice that I have semicolons instead of commas in the definition of Q. If you want to make it a column vector after defining it, do something like this instead:
length(Q) - sum(X <= Q') + 1
The reason that this works is that Octave implicitly applies bsxfun to an operation on a row and column vector. MATLAB will not do this until R2016b according to #excaza's comment, so in MATLAB you can do this:
length(Q) - sum(bsxfun(#le, X, Q)) + 1
You can play around with this example in IDEOne here.
Inspired by the solution posted by #Mad Physicist, here is my solution.
Q = [0.1,0.3,0.6,1.0]
X = [0.11,0.72,0.32,0.94]
Temp = repmat(X',1,4)<repmat(Q,4,1)
[~, ind]= max( Temp~=0, [], 2 );
The idea is that make the X and Q into the "same shape", then use element wise comparison, then we obtain a logical matrix whose row tells whether a given element in X is less than each of the element in Q, then return the first non-zero index of each row of this logical matrix. I haven't tested how fast this method is comparing to other methods
I have a function replace_me which is defined like as: function w = replace_me(v,a,b,c). The first input argument v is a vector, while a, b, and c are all scalars. The function replaces every element of v that is equal to a with b and c. For example, the command
x = replace_me([1 2 3],2,4,5); returns x as [1 4 5 3].
The code that I have created is
function w = replace_me(v,a,b,c)
[row,column]=size(v);
new_col=column+1;
w=(row:new_col);
for n=(1:column)
if a==v(n)
v(n)=b;
o=n;
d=n-1;
u=n+1;
for z=1:d
w(z)=v(z);
end
for z=u:column
w(z+1)=v(z);
end
w(o)=b;
w(o+1)=c;
end
end
end
It works perfectly fine for x = replace_me([1 2 3],2,4,5); I get required output but when I try x = replace_me([1 2 3], 4, 4, 5) my function fails.
To resolve this problem I want to use an if else statements having conditions that if a is equal to any element of vector v we would follow the above equation else it returns back the vector.
I tried to use this as if condition but it didn't worked
if v(1:column)==a
Any ideas
I'm not entirely sure if I understand what you are trying to achieve, but form what I understand you're looking for something like this:
function [v] = replace_me(v,a,b,c)
v = reshape(v,numel(v),1); % Ensure that v is always a column vector
tol = 0.001;
aPos = find( abs(v-a) < tol ); % Used tol to avoid numerical issues as mentioned by excaza
for i=numel(aPos):-1:1 % Loop backwards since the indices change when inserting elements
index = aPos(i);
v = [v(1:index-1); b; c; v(index+1:end)];
end
end
function w = move_me(v,a)
if nargin == 2
w=v(v~=a);
w(end+1:end+(length(v)-length(w)))=a;
elseif isscalar(v)
w=v;
else
w=v(v~=0);
w(end+1)=0;
end
end
I want to vectorize the following MATLAB code. I think it must be simple but I'm finding it confusing nevertheless.
r = some constant less than m or n
[m,n] = size(C);
S = zeros(m-r,n-r);
for i=1:m-r+1
for j=1:n-r+1
S(i,j) = sum(diag(C(i:i+r-1,j:j+r-1)));
end
end
The code calculates a table of scores, S, for a dynamic programming algorithm, from another score table, C.
The diagonal summing is to generate scores for individual pieces of the data used to generate C, for all possible pieces (of size r).
Thanks in advance for any answers! Sorry if this one should be obvious...
Note
The built-in conv2 turned out to be faster than convnfft, because my eye(r) is quite small ( 5 <= r <= 20 ). convnfft.m states that r should be > 20 for any benefit to manifest.
If I understand correctly, you're trying to calculate the diagonal sum of every subarray of C, where you have removed the last row and column of C (if you should not remove the row/col, you need to loop to m-r+1, and you need to pass the entire array C to the function in my solution below).
You can do this operation via a convolution, like so:
S = conv2(C(1:end-1,1:end-1),eye(r),'valid');
If C and r are large, you may want to have a look at CONVNFFT from the Matlab File Exchange to speed up calculations.
Based on the idea of JS, and as Jonas pointed out in the comments, this can be done in two lines using IM2COL with some array manipulation:
B = im2col(C, [r r], 'sliding');
S = reshape( sum(B(1:r+1:end,:)), size(C)-r+1 );
Basically B contains the elements of all sliding blocks of size r-by-r over the matrix C. Then we take the elements on the diagonal of each of these blocks B(1:r+1:end,:), compute their sum, and reshape the result to the expected size.
Comparing this to the convolution-based solution by Jonas, this does not perform any matrix multiplication, only indexing...
I would think you might need to rearrange C into a 3D matrix before summing it along one of the dimensions. I'll post with an answer shortly.
EDIT
I didn't manage to find a way to vectorise it cleanly, but I did find the function accumarray, which might be of some help. I'll look at it in more detail when I am home.
EDIT#2
Found a simpler solution by using linear indexing, but this could be memory-intensive.
At C(1,1), the indexes we want to sum are 1+[0, m+1, 2*m+2, 3*m+3, 4*m+4, ... ], or (0:r-1)+(0:m:(r-1)*m)
sum_ind = (0:r-1)+(0:m:(r-1)*m);
create S_offset, an (m-r) by (n-r) by r matrix, such that S_offset(:,:,1) = 0, S_offset(:,:,2) = m+1, S_offset(:,:,3) = 2*m+2, and so on.
S_offset = permute(repmat( sum_ind, [m-r, 1, n-r] ), [1, 3, 2]);
create S_base, a matrix of base array addresses from which the offset will be calculated.
S_base = reshape(1:m*n,[m n]);
S_base = repmat(S_base(1:m-r,1:n-r), [1, 1, r]);
Finally, use S_base+S_offset to address the values of C.
S = sum(C(S_base+S_offset), 3);
You can, of course, use bsxfun and other methods to make it more efficient; here I chose to lay it out for clarity. I have yet to benchmark this to see how it compares with the double-loop method though; I need to head home for dinner first!
Is this what you're looking for? This function adds the diagonals and puts them into a vector similar to how the function 'sum' adds up all of the columns in a matrix and puts them into a vector.
function [diagSum] = diagSumCalc(squareMatrix, LLUR0_ULLR1)
%
% Input: squareMatrix: A square matrix.
% LLUR0_ULLR1: LowerLeft to UpperRight addition = 0
% UpperLeft to LowerRight addition = 1
%
% Output: diagSum: A vector of the sum of the diagnols of the matrix.
%
% Example:
%
% >> squareMatrix = [1 2 3;
% 4 5 6;
% 7 8 9];
%
% >> diagSum = diagSumCalc(squareMatrix, 0);
%
% diagSum =
%
% 1 6 15 14 9
%
% >> diagSum = diagSumCalc(squareMatrix, 1);
%
% diagSum =
%
% 7 12 15 8 3
%
% Written by M. Phillips
% Oct. 16th, 2013
% MIT Open Source Copywrite
% Contact mphillips#hmc.edu fmi.
%
if (nargin < 2)
disp('Error on input. Needs two inputs.');
return;
end
if (LLUR0_ULLR1 ~= 0 && LLUR0_ULLR1~= 1)
disp('Error on input. Only accepts 0 or 1 as input for second condition.');
return;
end
[M, N] = size(squareMatrix);
if (M ~= N)
disp('Error on input. Only accepts a square matrix as input.');
return;
end
diagSum = zeros(1, M+N-1);
if LLUR0_ULLR1 == 1
squareMatrix = rot90(squareMatrix, -1);
end
for i = 1:length(diagSum)
if i <= M
countUp = 1;
countDown = i;
while countDown ~= 0
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
if i > M
countUp = i-M+1;
countDown = M;
while countUp ~= M+1
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
end
Cheers