I'm working on MATLAB. I have the following matrices
A = [
1 2 3 4
5 6 7 8
1 5 2 3
6 7 8 9
1 3 6 2
6 3 1 6
9 7 4 7
];
B = [
1 5 2 3
6 7 8 9
];
I want to find A-B
so that the answer should be like,
ans = [
1 2 3 4
5 6 7 8
1 3 6 2
6 3 1 6
9 7 4 7
];
Use setdiff with the 'rows' and 'stable' options:
>> C = setdiff(A,B,'rows','stable')
C =
1 2 3 4
5 6 7 8
1 3 6 2
6 3 1 6
9 7 4 7
Use ismember to find the common rows and neglect those in the final output.
Code
out = A(~ismember(A,B,'rows'),:)
Output
out =
1 2 3 4
5 6 7 8
1 3 6 2
6 3 1 6
9 7 4 7
clear;
s=0;
A = [
1 2 3 4
5 6 7 8
1 5 2 3
6 7 8 9
1 3 6 2
6 3 1 6
9 7 4 7
];
B = [
1 5 2 3
6 7 8 9
];
for i=1:size(B)
s=s+(ismember(A, B(i,:), 'rows'))
end
A_B = A(s==0,:)
#Divakar or #chappjc's answers are the way to go.
But I can't help inviting bsxfun to the party:
C = A(~any(squeeze(all(bsxfun(#eq, A.', permute(B, [2 3 1])))).'),:);
And its friend pdist2 is coming too:
C = A(all(pdist2(A, B, 'hamming').'),:);
Related
I have a matrix A ,and vector x as following (left side)
where S0, H0,...is row number of each block. I want to exchange these blocks such that S0 and S1; H0 and H1 are near together as right side. This is my code
S0=3;
H0=2;
N0=2;
S1=4;
H1=5;
N1=4;
Cols=5;
Rows=S0+H0+N0+S1+H1+N1;
A=randi(10,[ Rows Cols]);
x=randi(10,[Rows 1]);
%% Exchange two block
temp=A(S0+H0+1:S0+H0+N0,1:end);
A(S0+H0+1:S0+H0+H1,1:end)=A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end);
A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end)=temp;
%% How exchange x
The above code is not work. How can I fixed it in MATLAB? Thank in advance.
One approach with mat2cell and cell2mat -
grps = [S0,H0,N0,S1,H1,N1]
new_pattern = [1 4 2 5 3 6]
celldata_roworder = mat2cell((1:size(A,1))',grps); %//'
newx = cell2mat(celldata_roworder(new_pattern)).'; %//'
newA = A(newx,:)
Sample run -
Input :
A =
6 8 9 8 7
4 8 8 3 4
3 8 2 1 10
5 2 6 8 3
5 7 4 7 7
4 5 6 8 7
6 3 4 7 4
8 1 5 5 2
5 9 2 4 1
5 2 3 9 5
2 2 1 4 2
1 7 10 9 8
3 9 7 8 4
4 6 10 9 9
7 8 2 6 8
10 2 10 7 6
10 10 8 10 2
5 6 6 5 10
3 7 5 1 3
8 1 3 9 10
grps =
3 2 2 4 5 4
new_pattern =
1 4 2 5 3 6
Output:
newx =
1 2 3 8 9 10 11 4 5 12 ...
13 14 15 16 6 7 17 18 19 20
newA =
3 3 2 5 8
4 3 3 7 7
1 5 2 8 1
4 6 4 1 4
7 1 5 8 8
4 9 10 10 8
7 10 10 4 3
7 3 1 6 9
2 9 2 6 10
1 1 7 10 3
10 10 10 4 7
9 1 8 9 5
8 7 4 5 7
9 8 7 5 3
1 10 7 6 8
8 1 10 6 1
4 6 3 3 2
7 9 3 2 9
6 9 7 4 8
6 7 6 8 10
I assume you are using a 2-dimensional matrix with Row rows and Cols columns.
You can use the colon : as a second index to address a full row, e.g. for the third row:
A(3, :)
(equal to A(3, 1:end) but little bit clearer).
So you could split your matrix into lines and re-arrange them like this (putting back together the lines to a two-dimensional matrix):
A = [ A(3:4, :); A(1:2, :); A(5:end, :) ]
This moves rows 3 and 4 at the beginning, then old lines 1 and 2 and then all the rest. Does this help you?
Hint: you can use eye for experimenting.
for eg if i have a matrix
4 5 9 8 3 8
3 2 4 10 1 3
1 9 9 6 7 7
2 1 7 4 6 7
2 6 3 5 4 2
7 2 2 9 3 4
How do I calculate the sum of the diagonal of the element 10 if I have its row and column indices?
So the output should be 9 + 10 + 7 + 7.
Thanks!
column = 4;
row = 2;
output = sum(diag(A, column - row));
Here you go:
>> x = [4,5,9,8,3 ,8
3,2,4,10,1, 3
1,9,9,6,7 ,7
2,1,7,4,6 ,7
2,6,3,5,4 ,2
7,2,2,9,3 ,4]
x =
4 5 9 8 3 8
3 2 4 10 1 3
1 9 9 6 7 7
2 1 7 4 6 7
2 6 3 5 4 2
7 2 2 9 3 4
>> xsum = sum(diag(x,4-2));
>> xsum
xsum = 33
parameterize the indices in case you need to use it more than once.
I have a matrix for which I extract each column and do repmat function for each of them to build another matrix. Since i I have to do this for a large number of vectors(each column of my first matrix) it takes so long(relative to which I expect). If I do this for the whole matrix and then do something to build them, does it takes less time?
Consider this as an example:
A=[1 4 7;2 5 8;3 6 9]
I want to produce these
A1=[1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3]
A2=[4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6]
A3=[7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9]
As an alternative to #thewaywewalk's answer and using kron and repmat:
clear
A=[1 4 7;2 5 8;3 6 9];
B = repmat(kron(A',ones(3,1)),1,3);
A1 = B(1:3,:)
A2 = B(4:6,:)
A3 = B(7:end,:)
Which results in the following:
A1 =
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
A2 =
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
A3 =
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
Or as #Divakar pointed out, it would be advisable to create a single 3D array and store all your data in it (general solution):
n = 3; %// # of times you want to repeat the arrays.
A=[1 4 7;2 5 8;3 6 9];
B = repmat(kron(A',ones(n,1)),1,n);
C = zeros(n,n*size(A,2),3);
C(:,:,1) = B(1:n,:);
C(:,:,2) = B(n+1:2*n,:);
C(:,:,3) = B(2*n+1:end,:);
Try if this fits your needs:
A = [1 4 7;2 5 8;3 6 9];
n = 3; %// size(A,1)
cellArrayOutput = arrayfun(#(x) repmat( A(:,x).',n,n ), 1:size(A,2), 'uni',0)
instead of different variable names, everything is stored in a cell array.
if you insist on different names, I'd recommend to use structs:
A = [1 4 7;2 5 8;3 6 9];
n = 3;
structOutput = struct;
for ii = 1:size(A,2)
structOutput.(['A' num2str(ii)]) = repmat( A(:,ii).', n, n );
end
which gives you:
>> structOutput.A1
ans =
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
and so on.
I don't expect to much performance plus, you should share your full code for further help.
I have random matrix(A) and I find a result I'd like to use later for my code
A=randint(5,7,[1,9])
ans A =
8 1 2 2 6 7 7
9 3 9 4 1 7 1
2 5 9 9 8 4 3
9 9 5 8 9 6 1
6 9 8 9 7 2 1
How can I now get:
A = [8,1,2,2,6,7,7;9,3,9...7,2,1];
without having to type it myself.
MATLAB has a function for that: MAT2STR
>> A = randi([1,9],[5,7]);
>> mat2str(A)
ans =
[5 5 7 5 3 2 5;5 6 5 3 8 4 1;9 8 8 1 7 9 6;1 5 5 1 8 6 3;3 4 5 8 9 9 5]
This is suitable for use with EVAL
Just thought of another way. Your goal is to have A in your script - right?
You can just paste it as follows:
A = [
8 1 2 2 6 7 7
9 3 9 4 1 7 1
2 5 9 9 8 4 3
9 9 5 8 9 6 1
6 9 8 9 7 2 1
]
(note the square brackets)
It will evaluate to your original matrix.
Make the string yourself:
Str = ['[' sprintf('%i',A(1)) sprintf(',%i',A(2:end)) ']']
Note this string does not contain any ; as in your example. so when you evaluate it you will get a 1x35 vector (instead of the original 5x7matrix)
So easiest way to fix this would be to add after you evaluate the string.
A = reshape(A,5,7)
It will look like
B = [....
B = reshape(B,5,7)
I am trying to create a matrix that is 3 x n, with each of the columns being the same. What's the easiest way of achieving it? Concatenation?
After
n=7
x=[1;2;3]
it's either
repmat(x,[1 n])
or
x(:,ones(1,n))
(Octave can be considered as an open source/free version of MATLAB)
octave-3.0.3:2> rowvec = [1:10]
rowvec =
1 2 3 4 5 6 7 8 9 10
octave-3.0.3:3> [rowvec; rowvec; rowvec]
ans =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Use repmat if the number of rows is large.
octave-3.0.3:7> repmat(rowvec, 10, 1)
ans =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Use multiplication with a 1 x 3 matrix of ones
eg, x * [1 1 1]
Edit:
In Octave:
octave-3.0.3.exe:1> x = [1;2;3;4]
x =
1
2
3
4
octave-3.0.3.exe:5> x * [1 1 1]
ans =
1 1 1
2 2 2
3 3 3
4 4 4