Suppose I have a 4 x n array:
A = [1 2 3 4; ...
2 4 8 9; ...
6 7 9 4; ...
1 8 3 4];
I want to filter the whole array based on the content of the first two columns.
For example, if I want to return array rows which contain a 2 in the first two columns, the answer I'm looking for isL
R = [1 2 3 4;...
2 4 8 9];
Or, if I want to return rows containing a 1 in the first two columns, the answer I'm looking for is...
A = [1 2 3 4;...
1 8 3 4];
I'm sure it's obvious but how can I do this in MATLAB? Filtering the whole array based on find or evaluation commands (e.g. A == 2) is totally fine. It's the filtering based on multiple columns in any order I can't figure out.
To check for a given number, just apply any along 2nd dimension restricted to the desired columns, and use that as a logical index to select the desired rows:
cols = [1 2]; %// columns to look at
val = 1; %// value to look for
R = A(any(A(:, cols)==val, 2), :);
If you want to look for several values, for example, select all rows that contain either 2 or 3 in columns 1 or 2: use ismember instead of ==:
cols = [1 2]; %// columns to look at
vals = [2 3]; %// values to look for
R = A(any(ismember(A(:, cols), vals), 2), :);
If you want to check if the numbers are within a range:
cols = [1 2]; %// columns to look at
v1 = 6; %// numbers should be greater or equal to this...
v2 = 8; %// ...and less than this
R = A(any(A(:, cols)>=v1, 2) & any(A(:, cols)<v2, 2), :);
Related
I have a 2D matrix where the № of columns is always a multiple of 3 (e.g. 250×27) - due to a repeating organisation of the results (A,B,C, A,B,C, A,B,C, and so forth). I wish to reshape this matrix to create a new matrix with 3 columns - each containing the aggregated data for each type (A,B,C) (e.g. 2250×3).
So in a matrix of 250×27, all the data in columns 1,4,7,10,13,16,19,22,25 would be merged to form the first column of the resulting reshaped matrix.
The second column in the resulting reshaped matrix would contain all the data from columns 2,5,8,11,14,17,20,23,26 - and so forth.
Is there a simple way to do this in MATLAB? I only know how to use reshape if the columns I wanted to merge were adjacent (1,2,3,4,5,6) rather than non-adjacent (1,4,7,10,13,16) etc.
Shameless steal from #Divakar:
B = reshape( permute( reshape(A,size(A,1),3,[]), [1,3,2]), [], 3 );
Let A be your matrix. You can save every third column in one matrix like:
(Note that you don't have to save them as matrices separately but it makes this example easier to read).
A = rand(27); %as test
B = A(:,1:3:end);
C = A(:,2:3:end);
D = A(:,3:3:end);
Then you use reshape:
B = reshape(B,[],1);
C = reshape(C,[],1);
D = reshape(D,[],1);
And finally put it all together:
A = [B C D];
You can just treat every set of columns as a single item and do three reshapes together. This should do the trick:
[save as "reshape3.m" file in your Matlab folder to call it as a function]
function out = reshape3(in)
[~,C]=size(in); % determine number of columns
if mod(C,3) ~=0
error('ERROR: Number of rows must be a multiple of 3')
end
R_out=numel(in)/3; % number of rows in output
% Reshape columns 1,4,7 together as new column 1, column 2,5,8 as new col 2 and so on
out=[reshape(in(:,1:3:end),R_out,1), ...
reshape(in(:,2:3:end),R_out,1), ...
reshape(in(:,3:3:end),R_out,1)];
end
Lets suppose you have a 3x6 matrix A
A = [1 2 3 4 5 6;6 5 4 3 2 1;2 3 4 5 6 7]
A =
1 2 3 4 5 6
6 5 4 3 2 1
2 3 4 5 6 7
you extract the size of the matrix
b =size(A)
and then extract each third column for a single row
c1 = A((1:b(1)),[1:3:b(2)])
c2 = A((1:b(1)),[2:3:b(2)])
c3 = A((1:b(1)),[3:3:b(2)])
and put them in one matrix
A_result = [c1(:) c2(:) c3(:)]
A_result =
1 2 3
6 5 4
2 3 4
4 5 6
3 2 1
5 6 7
My 2 cents:
nRows = size(matrix, 1);
nBlocks = size(matrix, 2) / 3;
matrix = reshape(matrix, [nRows 3 nBlocks]);
matrix = permute(matrix, [1 3 2]);
matrix = reshape(matrix, [nRows * nBlocks 1 3]);
matrix = reshape(matrix(:), [nRows * nBlocks 3]);
Here's my 2 minute take on it:
rv = #(x) x(:);
ind = 1:3:size(A,2);
B = [rv(A(:,ind)) rv(A(:,ind+1)) rv(A(:,ind+2))];
saves a few ugly reshapes, may be a bit slower though.
If you have the Image Processing Toolbox, im2col is a very handy solution:
out = im2col(A,[1 4], 'distinct').'
Try Matlab function mat2cell, I think this form is allowed.
X is the "start matrix"
C = mat2cell(X, [n], [3, 3, 3]); %n is the number of rows, repeat "3" as many times as you nedd
%extract every matrix
C1 = C{1,1}; %first group of 3 columns
C2 = C{1,2}; %second group of 3 columns
%repeat for all your groups
%join the matrix with vertcat
Cnew = vertcat(C1,C2,C3); %join as many matrix n-by-3 as you have
I've always been told that almost all for loops can be omitted in MATLAB and that they in general slow down the process. So is there a way to do so here?:
I have a cell-array (tsCell). tsCell stores time-arrays with varying length. I want to find an intersecting time-array for all time-arrays (InterSection):
InterSection = tsCell{1}.time
for i = 2:length{tsCell};
InterSection = intersect(InterSection,tsCell{i}.time);
end
Here's another way. This also assumes there are no duplicates within each original vector.
tsCell_time = {[1 6 4 5] [4 7 1] [1 4 3] [4 3 1 7]}; %// example data (from Divakar)
t = [tsCell_time{:}]; %// concat into a single vector
u = unique(t); %// get unique elements
ind = sum(bsxfun(#eq, t(:), u), 1)==numel(tsCell_time); %// indices of unique elements
%// that appear maximum number of times
result = u(ind); %// output those elements
Here's a vectorized approach using unique and accumarray, assuming there are no duplicates within each cell of the input cell array -
[~,~,idx] = unique([tsCell_time{:}],'stable')
out = tsCell_time{1}(accumarray(idx,1) == length(tsCell_time))
Sample run -
>> tsCell_time = {[1 6 4 5],[4 7 1],[1 4 3],[4 3 1 7]};
>> InterSection = tsCell_time{1};
for i = 2:length(tsCell_time)
InterSection = intersect(InterSection,tsCell_time{i});
end
>> InterSection
InterSection =
1 4
>> [~,~,idx] = unique([tsCell_time{:}],'stable');
out = tsCell_time{1}(accumarray(idx,1) == length(tsCell_time));
>> out
out =
1 4
I have a matrix where each row is a combination of two numbers, like A = [1 2; 2 5; 3 4; 4 6; 5 6]
A is built so that, for each row, the first elements is always smaller than the second one.
I need to return, from A, the lists of chained elements (in the case above, the lists of chained elements are 1 2 5 6 and 3 4 6). These lists are essentially built by considering a row, and checking is the last number is the first number of another row. Do you have any suggestion on how to do this?
If I got the question correctly, assuming A as the input array, you can use bsxfun -
mask = bsxfun(#eq,A(:,1),A(:,2).');
out = unique(A(any(mask,1).' | any(mask,2),:))
Sample run -
>> A
A =
1 2
3 4
2 5
5 6
>> mask = bsxfun(#eq,A(:,1),A(:,2).');
>> unique(A(any(mask,1).' | any(mask,2),:))
ans =
1
2
5
6
You can also use ismember, like so -
out = unique(A(ismember(A(:,1),A(:,2)) | ismember(A(:,2),A(:,1)),:))
Third option would be to use intersect to solve it, like so -
[~,idx1,idx2] = intersect(A(:,1),A(:,2));
out = unique(A([idx1,idx2],:))
The following seems to work. It builds a matrix (B) that tells which elements are connected (by 1 step). It then extends that matrix ( C) to include 0-step, 1-step, ..., (n-1)-step connections, where n is the number of nodes.
From that matrix, groups of connected elements are obtained (R). Finally, only "maximal" groups are kept (that is, those, not contained in other groups).
A = [1 2; 3 4; 2 5; 4 6; 5 6]; %// data
n = max(A(:));
B = full(sparse(A(:,1), A(:,2), 1, n, n )); %// matrix of 1-step connections
C = eye(n) | B; %// initiallize with 0-step and 1-step connections
for k = 1:n-1
C = C | C*B; %// add k-step connections, up to k=n-1
end
[ii, jj] = find(C);
R = accumarray(ii, jj, [], #(x) {sort(x).'}); %'// all groups (maximal or not)
[xx, yy] = ndgrid(1:n);
C = cellfun(#(x,y) all(ismember(x, y)), R(xx), R(yy) ); %// group included in another?
result = R(all(~C | eye(n), 2)); %// keep only groups that are not included in others
This gives
>> result{:}
ans =
1 2 5 6
ans =
3 4 6
I've two matrix a and b and I'd like to combine the rows in a way that in the first row I got no duplicate value and in the second value, columns in a & b which have the same row value get added together in new matrix. i.e.
a =
1 2 3
8 2 5
b =
1 2 5 7
2 4 6 1
Desired outputc =
1 2 3 5 7
10 6 5 6 1
Any help is welcomed,please.
For two-row matrices
You want to add second-row values corresponding to the same first-row value. This is a typical use of unique and accumarray:
[ii, ~, kk] = unique([a(1,:) b(1,:)]);
result = [ ii; accumarray(kk(:), [a(2,:) b(2,:)].').'];
General case
If you need to accumulate columns with an arbitrary number of columns (based on the first-row value), you can use sparse as follows:
[ii, ~, kk] = unique([a(1,:) b(1,:)]);
r = repmat((1:size(a,1)-1).', 1, numel(kk));
c = repmat(kk.', size(a,1)-1, 1);
result = [ii; full(sparse(r,c,[a(2:end,:) b(2:end,:)]))];
Given a vector
X = [1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3]
I would like to generate a vector such
Y = [1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5]
So far what I have got is
idx = find(diff(X))
Y = [1:idx(1) 1:idx(2)-idx(1) 1:length(X)-idx(2)]
But I was wondering if there is a more elegant(robust) solution?
One approach with diff, find & cumsum for a generic case -
%// Initialize array of 1s with the same size as input array and an
%// intention of using cumsum on it after placing "appropriate" values
%// at "strategic" places for getting the final output.
out = ones(size(X))
%// Find starting indices of each "group", except the first group, and
%// by group here we mean run of identical numbers.
idx = find(diff(X))+1
%// Place differentiated and subtracted values of indices at starting locations
out(idx) = 1-diff([1 idx])
%// Perform cumulative summation for the final output
Y = cumsum(out)
Sample run -
X =
1 1 1 1 2 2 3 3 3 3 3 4 4 5
Y =
1 2 3 4 1 2 1 2 3 4 5 1 2 1
Just for fun, but customary bsxfun based alternative solution -
%// Logical mask with each column of ones for presence of each group elements
mask = bsxfun(#eq,X(:),unique(X(:).')) %//'
%// Cumulative summation along columns and use masked values for final output
vals = cumsum(mask,1)
Y = vals(mask)
Here's another approach:
Y = sum(triu(bsxfun(#eq, X, X.')), 1);
This works as follows:
Compare each element with all others (bsxfun(...)).
Keep only comparisons with current or previous elements (triu(...)).
Count, for each element, how many comparisons are true (sum(..., 1)); that is, how many elements, up to and including the current one, are equal to the current one.
Another method is using the function unique
like this:
[unqX ind Xout] = unique(X)
Y = [ind(1):ind(2) 1:ind(3)-ind(2) 1:length(X)-ind(3)]
Whether this is more elegant is up to you.
A more robust method will be:
[unqX ind Xout] = unique(X)
for ii = 1:length(unqX)-1
Y(ind(ii):ind(ii+1)-1) = 1:(ind(ii+1)-ind(ii));
end