In order to use the stiffness method for trusses, I need to extract certain elements from a large global stiffness matrix.
Say I have a 9 x 9 matrix K representing a three-member truss. This means that the first 3 rows and columns correspond to the first node, the second set of three rows and columns with the second node, and the third with the third node. In the code is a vector zDisp that corresponds to each node that has zero displacement. On paper, a zero displacement of a node means you would cross out the rows and columns corresponding to that displacement, leaving you with a smaller and easier to work with K matrix. So if the first and third nodes have zero displacement, you would be left with a 3 x 3 matrix corresponding to the intersection of the middle three rows and the middle three columns.
I thought I could accomplish this one node at a time with a function like so:
function [ B ] = deleteNode( B, node )
%deleteNode removes the corresponding rows and vectors to a node that has
% zero deflection from the global stiffness matrix
% --- Problem line - this gets the first location in the matrix corresponding to the node
start = 3*node- 2;
for i = 0 : 2
B(start+i,:) = [];
B(:,start+i) = [];
end
end
So my main project would go something like
% Zero displacement nodes
zDisp = [1;
3;
];
% --- Create 9 x 9 global matrix Kg ---
% Make a copy of the global matrix
S = Kg;
for(i = 1 : length(zDisp))
S = deleteNode(S, zDisp(i));
end
This does not work because once the loop executes for node 1 and removes the first 3 rows and columns, the problem line in the function no longer works to find the correct location in the smaller matrix to find the node.
So I think this step needs to be executed all at once. I am thinking I may need to instead input which nodes are NOT zero displacement, and create a submatrix based off of that. Any tips on this? Been thinking on it awhile. Thanks all.
In your example, you want to remove rows/columns 1, 2, 3, 7, 8, and 9, so if zDisp=[1;3],
remCols=bsxfun(#plus,1:3,3*(zDisp-1))
If I understand correctly, you should just be able to first remove the columns given by zDisp:
S(remCols(:),:)=[]
then remove the rows:
S(:,remCols(:))=[]
Related
I have a matrix where each element is a single unit of a 2d coordinate. As such, each element in any given row are paired, where elements in the first column are paired with those in the second, and elements in the third column paired with the fourth. All possible combinations of the 4 numbers are present in the matrix.
What I need to do is depup the matrix by removing rows where the first set of coordinates (e.g columns 1 and 2 in a row) are swapped with the second set of coordinates. For example if one row contains the value "3, 4, 2, 1" then I would need to remove "2, 1, 3, 4" from else where in the matrix.
An example of this could be seen below, where I would want to remove the last row, as it is the reverse of the first row;
3 3 1 1
1 2 2 3
3 4 1 2
4 4 3 1
4 1 4 4
1 1 3 3
I'm quite stumped as to how to do this, and my previous attempts have all failed. Whilst it may not be useful to the answer, I have included my code showing how I am constructing the initial matrix below;
%create list of all piece coordinates
p1_row_index=(1:n);
p1_column_index=(1:n);
p2_row_index=(1:n);
p2_column_index=(1:n);
% get all possible combinations of these variables
[p1_row_index,p1_column_index,p2_row_index,p2_column_index]=BalanceFactors(1,1,1:n,1:n,1:n,1:n);
pc_list(:,1)=p1_row_index; % piece1 coordiantes for rows
pc_list(:,2)=p1_column_index; % piece1 coordiantes for columns
pc_list(:,3)=p2_row_index; % piece2 coordiantes for rows
pc_list(:,4)=p2_column_index; % piece2 coordiantes for columns
Thank you for your time.
Many thanks,
Matt
Complex numbers come in handy for this:
[~, ind] = unique(sort(M(:,[1 3])+1j*M(:,[2 4]), 2), 'rows', 'stable');
result = M(ind, :);
The code works as follows:
M(:,[1 3])+1j*M(:,[2 4]) creates a complex matrix with half the columns, where each pair of coordinates of the original matrix becomes a complex number.
sort(..., 2) sorts each row. Rows that originally were shuffled versions of each other now become identical.
[~, ind] = unique(..., 'rows', 'stable') gives the index of the first occurrence of each unique (complex, sorted) row.
M(ind, :) selects the desired rows from M.
So I have written this:
HSRXdistpR = squeeze(comDatape_m1(2,7,1,:,isubj));
HSRXdistpL = squeeze(comDatape_m1(2,4,1,:,isubj));
TocomXdistp = squeeze(comDatape_m1(2,10,1,:,isubj));
for i = 1:2;
HSRXp = NaN(8,3*i);
HSRXp(:,i*3) = [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)];
end
In the first part I am just selecting data from a 5-D matrix, nothing special. All that's important here is that it creates an 8x2 matrix per line (isubj=2). Now I want to add the first column of each matrix into an 8x3 matrix, and then the second column of each matrix into the same matrix (creating an 8x6 matrix). Since the number of my subjects will vary, I want to do this in a for loop. This way, if the isubj increases to 3, it should go on to create an 8x9 matrix.
So I tried to create a matrix that will grow by 3 for each iteration of i, which selects the ith column of each of the 3 matrices and then puts them in there.
However I get the following error:
Subscripted assignment dimension mismatch.
Is it possible to let a matrix grow by more than one in a for loop? Or how should it be done otherwise?
Here is your problem:
HSRXp(:,i*3) = [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)];
You're trying to assign an n x 3 matrix (RHS) into an n x 1 vector (LHS). It would be easier to simply use horizontal concatenation:
HSRXp = [HSRXp, [HSRXdistpR(:,i) HSRXdistpL(:,i) TocomXdistp(:,i)]];
But that would mean reallocation at each step, which might slow your code down if the matrix becomes large.
For my experiment I have 20 categories which contain 9 pictures each. I want to show these pictures in a pseudo-random sequence where the only constraint to randomness is that one image may not be followed directly by one of the same category.
So I need something similar to
r = randi([1 20],1,180);
just with an added constraint of two numbers not directly following each other. E.g.
14 8 15 15 7 16 6 4 1 8 is not legitimate, whereas
14 8 15 7 15 16 6 4 1 8 would be.
An alternative way I was thinking of was naming the categories A,B,C,...T, have them repeat 9 times and then shuffle the bunch. But there you run into the same problem I think?
I am an absolute Matlab beginner, so any guidance will be welcome.
The following uses modulo operations to make sure each value is different from the previous one:
m = 20; %// number of categories
n = 180; %// desired number of samples
x = [randi(m)-1 randi(m-1, [1 n-1])];
x = mod(cumsum(x), m) + 1;
How the code works
In the third line, the first entry of x is a random value between 0 and m-1. Each subsequent entry represents the change that, modulo m, will give the next value (this is done in the fourth line).
The key is to choose that change between 1 and m-1 (not between 0 and m-1), to assure consecutive values will be different. In other words, given a value, there are m-1 (not m) choices for the next value.
After the modulo operation, 1 is added to to transform the range of resulting values from 0,...,m-1 to 1,...,m.
Test
Take all (n-1) pairs of consecutive entries in the generated x vector and count occurrences of all (m^2) possible combinations of values:
count = accumarray([x(1:end-1); x(2:end)].', 1, [m m]);
imagesc(count)
axis square
colorbar
The following image has been obtained for m=20; n=1e6;. It is seen that all combinations are (more or less) equally likely, except for pairs with repeated values, which never occur.
You could look for the repetitions in an iterative manner and put new set of integers from the same group [1 20] only into those places where repetitions have occurred. We continue to do so until there are no repetitions left -
interval = [1 20]; %// interval from where the random integers are to be chosen
r = randi(interval,1,180); %// create the first batch of numbers
idx = diff(r)==0; %// logical array, where 1s denote repetitions for first batch
while nnz(idx)~=0
idx = diff(r)==0; %// logical array, where 1s denote repetitions for
%// subsequent batches
rN = randi(interval,1,nnz(idx)); %// new set of random integers to be placed
%// at the positions where repetitions have occured
r(find(idx)+1) = rN; %// place ramdom integers at their respective positions
end
I am looking to solve a Traveling Salesman type problem using a matrix in order to find the minimum time between transitions. The matrix looks something like this:
A = [inf 4 3 5;
1 inf 3 5;
4 5 inf 3;
6 7 1 inf]
The y-axis represents the "from" node and the x-axis represents the "to" node. I am trying to find the optimal time from node 1 to node 4. I was told that there is a Matlab function called "TravellingSalesman". Is that true, and if not, how would I go about solving this matrix?
Thanks!
Here's an outline of the brute-force algorithm to solve TSP for paths from node 1 to node n:
C = inf
P = zeros(1,n-2)
for each permutation P of the nodes [2..n-1]
// paths always start from node 1 and end on node n
C = A(1,P(1)) + A(P(1),P(2)) + A(P(2),P(3)) + ... +
A(P(n-3),P(n-2)) + A(P(n-2),n)
if C < minCost
minCost = C
minPath = P
elseif C == minCost // you only need this part if you want
minPath = [minPath; P] // ALL paths with the shortest distance
end
end
Note that the first and last factors in the sum are different because you know beforehand what the first and last nodes are, so you don't have to include them in the permutations. So in the example given, with n=4, there are actually only 2!=2 possible paths.
The list of permutations can be precalculated using perms(2:n-1), but that might involve storing a large matrix (n! x n). Or you can calculate the cost as you generate each permutation. There are several files on the Mathworks file exchange with names like nextPerm that should work for you. Either way, as n grows you're going to be generating a very large number of permutations and your calculations will take a very long time.
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))].