I have a 2D matrix, say M=zeros(10,10);
I have another column matrix, V=[1;2;3;4;5;6;5;4;3;2];
I would like to be able to set M(i,j) = 1 for all j >= V(i)
I know I can do this in a loop
for i=1:10
M(i,V(i):10) = 1;
end
but it seems that it would be possible to use some form of Matlab indexing to avoid using a loop. For example something like :
M(:,V:10)=1;
or
M(:,V(:):10)=1;
but neither of these produces the expected result.
Is there some syntactic sugar I can use to achieve this or should I revert to looping?
Since you're looking for syntactic sugar, here's a sort of esoteric way of doing it. :)
Assuming the length of V is the size of both dimensions in the desired matrix M, first create an identity matrix of the same size, then index appropriately and take cumsum:
V = [1;2;3;4;5;6;5;4;3;2]; #% 10x1 vector
E = eye(length(V), length(V)); #%10x10 identity matrix
M = cumsum(E(V,:),2)
M =
1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1
0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 1 1 1 1 1
0 0 0 0 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
Ok, now: less fun, but (on my machine) faster than any of the other options that have been tested so far:
n=10000;
V = randi(n-1, 1, n); #% as in #KevinRatelle's answer (but not transposed)
tic;
Vlinear = reshape(V + (0:n-1)*n, 1, []); #% find linear indices of first "ones"
M = zeros(n);
M(Vlinear)=1;
M=cumsum(M);
toc
Its hardly subtle and not really better than a loop I don't think but:
[J,I] = meshgrid(1:10,1:10);
V = [1;2;3;4;5;6;5;4;3;2];
M = J>V(I);
Enjoy.
I tried the loop method, and the 'meshgrid' method. I was wondering about the time to compute for large matrices (since the issue with loops in matlab is usually the time).
First, I optimised the code to look like this :
V = V*ones(1,n);
N = ones(1,n)'*(1:n);
M = N>=V;
Actually, N is a meshgrid, but it seems way faster to do it this way...
I tried this :
n = 10000;
V = randi(n-1,1,n)';
tic;
M = zeros(n);
for i=1:n
M(i,V(i):n) = 1;
end
toc
tic;
[J,I] = meshgrid(1:n,1:n);
M = J>=V(I);
toc
tic;
V = V*ones(1,n);
N = ones(1,n)'*(1:n);
M = N>=V;
toc
And results were :
Elapsed time is 1.726872 seconds.
Elapsed time is 5.206657 seconds.
Elapsed time is 1.548600 seconds.
But methods using matrices instead of loops are memory consuming for large n. I would personnaly stick to the loop.
try this:
v = [1;2;3;4;5;6;5;4;3;2];
n = 10;
M = repmat((1:n)', 1, numel(v)) > repmat(v', n, 1);
Related
I have a 2D matrix composed of ones and zeros.
mat = [0 0 0 0 1 1 1 0 0
1 1 1 1 1 0 0 1 0
0 0 1 0 1 1 0 0 1];
I need to find all consecutive repetitions of ones in each row and replace all ones with zeros only when the sequence size is smaller than 5 (5 consecutive ones):
mat = [0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0];
Any suggestion on how to approach this problem would be very welcome.
You can use diff to find the start and end points of the runs of 1, and some logic based on that to zero out the runs which are too short. Please see the below code with associated comments
% Input matrix of 0s and 1s
mat = [0 0 0 0 1 1 1 0 0
1 1 1 1 1 0 0 1 0
0 0 1 0 1 1 0 0 1];
% Minimum run length of 1s to keep
N = 5;
% Get the start and end points of the runs of 1. Add in values from the
% original matrix to ensure that start and end points are always paired
d = [mat(:,1),diff(mat,1,2),-mat(:,end)];
% Find those start and end points. Use the transpose during the find to
% flip rows/cols and search row-wise relative to input matrix.
[cs,r] = find(d.'>0.5); % Start points
[ce,~] = find(d.'<-0.5); % End points
c = [cs, ce]; % Column number array for start/end
idx = diff(c,1,2) < N; % From column number, check run length vs N
% Loop over the runs which didn't satisfy the threshold and zero them
for ii = find(idx.')
mat(r(ii),c(ii,1):c(ii,2)-1) = 0;
end
If you want to throw legibility out of the window, this can be condensed for a slightly faster and denser version, based on the exact same logic:
[c,r] = find([mat(:,1),diff(mat,1,2),-mat(:,end)].'); % find run start/end points
for ii = 1:2:numel(c) % Loop over runs
if c(ii+1)-c(ii) < N % Check if run exceeds threshold length
mat(r(ii),c(ii):c(ii+1)-1) = 0; % Zero the run if not
end
end
The vectorized solution by #Wolfie is nice and concise, but a bit hard to understand and far from the wording of the problem. Here is a direct translation of the problem using loops. It has the advantage of being easier to understand and is slightly faster with less memory allocations, which means it will work for huge inputs.
[m,n] = size(mat);
for i = 1:m
j = 1;
while j <= n
seqSum = 1;
if mat(i,j) == 1
for k = j+1:n
if mat(i,k) == 1
seqSum = seqSum + 1;
else
break
end
end
if seqSum < 5
mat(i,j:j+seqSum-1) = 0;
end
end
j = j + seqSum;
end
end
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
I have a loop that iterates over a matrix and sets all rows and columns with only one non-zero element to all zeroes.
so for example, it will transform this matrix:
A = [ 1 0 1 1
0 0 1 0
1 1 1 1
1 0 1 1 ]
to the matrix:
A' = [ 1 0 1 1
0 0 0 0
1 0 1 1
1 0 1 1 ]
row/column 2 of A only has 1 non zero element in it, so every element in row/column 2 is set to 0 in A'
(it is assumed that the matrices will always be diagonally symmetrical)
here is my non-vectorised code:
for ii = 1:length(A)
if nnz(A(ii,:)) == 1
A(ii,:) = 0;
A(:,ii) = 0;
end
end
Is there a more efficient way of writing this code in MATLAB?
EDIT:
I have been asked in the comments for some clarification, so I will oblige.
The purpose of this code is to remove edges from a graph that lead to a vertex of degree 1.
if A is the adjacency matrix representing a undirected graph G, then a row or column of that matrix which only has one non-zero element indicates that row/column represents a vertex of degree one, as it only has one edge incident to it.
My objective is to remove such edges from the graph, as these vertices will never be visited in a solution to the problem I am trying to solve, and reducing the graph will also reduce the size of the input to my search algorithm.
#TimeString, i understand that in the example you gave, recursively applying the algorithm to your matrix will result in a zero matrix, however the matrices that I am applying it to represent large, connected graphs, so there will never be a case like that. In response to your question as to why I only check for how many elements in a row, but the clear both columns and rows; this is because the matrix is always diagonally symmetrical, so i know that if something is true for a row, so it will be for the corresponding column..
so, just to clarify using another example:
I want to turn this graph G:
represented by matrix:
A = [ 0 1 1 0
1 0 1 0
1 1 0 1
0 0 1 0 ]
to this graph G':
represented by this matrix:
A' = [ 0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 0 ]
(i realise that this matrix should actually be a 3x3 matrix because point D has been removed, but i already know how to shrink the matrix in this instance, my question is about efficiently setting columns/rows with only 1 non-zero element all to 0)
i hope that is a good enough clarification..
Not sure if it's really faster (depends on Matlab's JIT) but you can try the following:
To find out which columns (equivalently, rows, since the matrix is symmetric) have more than one non zero element use:
sum(A ~= 0) > 1
The ~= 0 is probably not needed in your case since the matrix consists of 1/0 elements only (graph edges if I understand correctly).
Transform the above into a diagonal matrix in order to eliminate unwanted columns:
D = diag(sum(A~=0) > 1)
And multiply with A from left to zero rows and from right to zero columns:
res = D * A * D
Thanks to nimrodm's suggestion of using sum(A ~= 0) instead of nnz, i managed to find a better solution than my original one
to clear the rows with one element i use:
A(sum(A ~= 0) == 1,:) = 0;
and then to clear columns with one element:
A(:,sum(A ~= 0) == 1) = 0;
for those of you who are interested, i did a 'tic-toc' comparison on a 1000 x 1000 matrix:
% establish matrix
A = magic(1000);
rem_rows = [200,555,950];
A(rem_rows,:) = 0;
A(:,rem_rows) = 0;
% insert single element into empty rows/columns
A(rem_rows,500) = 5;
A(500,rem_rows) = 5;
% testing original version
A_temp = A;
for test = 1
tic
for ii = 1:length(A_temp)
if nnz(A_temp(ii,:)) == 1
A_temp(ii,:) = 0;
A_temp(:,ii) = 0;
end
end
toc
end
Elapsed time is 0.041104 seconds.
% testing new version
A_temp = A;
for test = 1
tic
A_temp(sum(A_temp ~= 0) == 1,:) = 0;
A_temp(:,sum(A_temp ~= 0) == 1) = 0;
toc
end
Elapsed time is 0.010378 seconds
% testing matrix operations based solution suggested by nimrodm
A_temp = A;
for test = 1
tic
B = diag(sum(A_temp ~= 0) > 1);
res = B * A_temp * B;
toc
end
Elapsed time is 0.258799 seconds
so it appears that the single line version that I came up with, inspired by nimrodm's suggestion, is the fastest
thanks for all your help!
Bsxfuning it -
A(bsxfun(#or,(sum(A~=0,2)==1),(sum(A~=0,1)==1))) = 0
Sample run -
>> A
A =
1 0 1 1
0 0 1 0
1 1 1 1
1 0 1 1
>> A(bsxfun(#or,(sum(A~=0,2)==1),(sum(A~=0,1)==1))) = 0
A =
1 0 1 1
0 0 0 0
1 0 1 1
1 0 1 1
I'd like to create a 4x12 matrix which is very similar to a upper triangle matrix, it looks like this:
1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 1 1 1
So my question is. What is the most efficient way to create it? no loops, no cellfun. Thanks.
One vectorized approach -
nrows = 4;
ncols = 12;
row_idx = repmat(1:nrows,ncols/nrows,1)
out = bsxfun(#le,[1:nrows]',row_idx(:).')
The Matlab R2015a and later approach using the newly introduced repelem:
n = 4;
m = 3;
out = repelem(triu(ones(n)),1,m);
out =
1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 1 1 1
It even seems faster than the bsxfun approach, though I can't believe this ;)
Benchmark
Unfortunately I couldn't consider andrew's solution as it is not complete and I didn't got it totally.
function [t] = bench()
n = 4;
m = 12;
t = zeros(3,15);
for ii = 1:15
fcns = {
#() thewaywewalk(ii*n,ii*m);
#() Divakar(ii*n,ii*m);
#() LuisMendo(ii*n,ii*m);
};
% timeit
for jj = 1:100;
t(:,ii) = t(:,ii) + cellfun(#timeit, fcns);
end
end
plot(1:15,t(1,:)); hold on;
plot(1:15,t(2,:)); hold on;
plot(1:15,t(3,:)); hold on;
xlabel('Matrix size: n = x*4, m = x*12')
ylabel('timing')
legend({'thewaywewalk','Divakar','Luis Mendo'},'location','northwest')
end
function Z = thewaywewalk(n,m)
Z = repelem(triu(ones(n)),1,m/n);
end
function Z = Divakar(n,m)
row_idx = repmat(1:n,m/n,1);
Z = bsxfun(#le,[1:n]',row_idx(:).');
end
function Z = LuisMendo(n,m)
Z = reshape(repmat(permute(triu(ones(n,n)), [1 3 2]), [1 m/n 1]), [n m]);
end
First bottomline - small matrices:
The new repelem does a very good job, but also the reshape(repmat(permute... does not disappoint. The bsxfun approach stays a little behind for some medium-sized matrices, before it becomes the leader for large matrices:
Second bottomline - big matrices:
As predicted by Divakar, bsxfun is the fastest for larger matrices, actually as expected as bsxfun is always the fastest! Interesting that the other two align perfectly, on could guess they almost work the same internally.
Create an upper triangular matrix of ones, permute second and third dimensions, repeat along second dimension, and reshape into desired shape:
m = 4;
n = 12;
result = reshape(repmat(permute(triu(ones(m,m)), [1 3 2]), [1 n/m 1]), [m n]);
depending on your matlab version
m = 4;
n = 12;
dec2bin(bitshift(num,-1*[0:n/m:n-1])) %this prints out a string
these should be logical arrays (i don't have either of these so I cant test it)
decimalToBinaryVector(bitshift(num,-1*[0:n/m:n-1]))
de2bi(bitshift(num,-1*[0:n/m:n-1]))
Say I have a vector containing only logical values, such as
V = [1 0 1 0 1 1 1 1 0 0]
I would like to write a function in MATLAB which returns a 'streak' vector S for V, where S(i) represents the number of consecutive 1s in V up to but not including V(i). For the example above, the streak vector would be
S = [0 1 0 1 0 1 2 3 4 0]
Given that I have to do this for a very large matrix, I would very much appreciate any solution that is vectorized / efficient.
This should do the trick:
S = zeros(size(V));
for i=2:length(V)
if(V(i-1)==1)
S(i) = 1 + S(i-1);
end
end
The complexity is only O(n), which I guess should be good enough.
For your sample input:
V = [1 0 1 0 1 1 1 1 0 0];
S = zeros(size(V));
for i=2:length(V)
if(V(i-1)==1)
S(i) = 1 + S(i-1);
end
end
display(V);
display(S);
The result would be:
V =
1 0 1 0 1 1 1 1 0 0
S =
0 1 0 1 0 1 2 3 4 0
You could also do it completely vectorized with a couple intermediate steps:
V = [1 0 1 0 1 1 1 1 0 0];
Sall = cumsum(V);
stopidx = find(diff(V)==-1)+1;
V2=V;
V2(stopidx) = -Sall(stopidx)+[0 Sall(stopidx(1:end-1))];
S2 = cumsum(V2);
S = [0 S2(1:end-1)];
Afaik the only thing that can take a while is the find call; you can't use logical indexing everywhere and bypass the find call, because you need the absolute indices.
It's outside the box - but have you considered using text functions? Since strings are just vectors for Matlab it should be easy to use them.
Regexp contains some nice functions for finding repeated values.