I have a matrix
A= [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16; 17 18 19 20]
I want to do some calculation on this matrix. But actually I do not need all the rows. So I have to discard some of the rows from the above matrix before doing a calculation. After discarding 3 rows, we will have a new matrix.
B= [1 2 3 4; 9 10 11 12; 17 18 19 20];
Now I have to use B to make some other calculations. So how can I discard some of the unwanted rows from a matrix in matlab? Any suggestion will be helpful. Thanks.
Try this: (Use when no. of rows to keep is lesser)
%// Input A
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16; 17 18 19 20];
%// Rows (1-3,5) you wanted to keep
B = A([1:3, 5],:)
Output:
B =
1 2 3 4
5 6 7 8
9 10 11 12
17 18 19 20
Alternative: (Use when no. of rows to discard is lesser)
%// rows 2 and 3 discarded
A([2,3],:) = [];
Output:
>> A
A =
1 2 3 4
13 14 15 16
17 18 19 20
Note: Here (in the alternate method), the output replaces the original A. So you need to back up A if you need it afterwards. You could do this before discarding operation to backup Input matrix
%// Input A is backed up in B
B = A;
You can select the indices of the rows you want to keep:
A([1,3,5],:)
ans =
1 2 3 4
9 10 11 12
17 18 19 20
Related
Consider a row vector A and row vector B. For example:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
A has previously been checked to be a subset of B. By subset, I specifically mean that all elements in A can be found in B. I know that elements in A will not ever repeat. However, the elements in B are free to repeat as many or as few times as they like. I checked this condition using:
is_subset = all(ismember(A,B));
With all that out of the way, I need to know the indices of the elements of A within B including the times when these elements repeat within B. For the example A and B above, the output would be:
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
Use ismember to find the relevant logical indices. Then convert them to linear indices using find.
C = find(ismember(B,A));
You can find the difference of each element of A with B, and get the indices you want. Something like below:
A = [1 2 3 7 8 10 12];
B = [1 1 2 2 2 3 5 6 6 7 7 7 8 8 10 10 10 11 12 12 12 13 15 16 18 19];
C = [1 2 3 4 5 6 10 11 12 13 14 15 16 17 19 20 21];
tol = 10^-3 ;
N = length(A) ;
iwant = cell(N,1) ;
for i = 1:N
idx = abs(A(i)-B)<=tol ;
iwant{i} = find(idx) ;
end
iwant = [iwant{:}] ;
I have two matrices in Matlab A and B respectively of dimension MxN and GxN.
M can be greater or smaller than G.
A and B do not contain identical rows.
I want to construct a matrix C of dimension Hx(N+2) in the following way
C=[];
for i=1:size(A,1)
%if A(i,1:end-1) coincides with a row in B(:,1:end-1) (it can coincide with only one row at most)
%then C=[C;A(i,1:end) B(j,end)]; %where j is the index of the identical row in B
%otherwise impose C=[C;A(i,1:end) 0]
end
for i=1:size(B,1)
%if B(i,1:end-1) does not coincide with any row in A(:,1:end-1)
%then impose C=[C; B(i,1:end-1) 0 B(i,end)];
end
For example:
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104]
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104]
As M and G can be very high, I am looking for the fastest way to perform this.
You can use ismember + indexing to do your task:
[idx1,idx2] = ismember(A(:,1:end-1), B(:,1:end-1), 'rows');
idx3 = ~ismember(B(:,1:end-1), A(:,1:end-1), 'rows');
C(idx1,:) = [A(idx1,:) B(idx2(idx1),end)];
C(~idx1,:) = [A(~idx1,:) zeros(sum(~idx1),1)];
C=[C;B(idx3,1:end-1) zeros(sum(idx3),1) B(idx3,end)];
You could also use intersect with a bit of preallocation to speed up the assignment (if M or G gets really large).
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104];
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104];
[M,N] = size(A);
G = size(B,1);
[tmp, idxA, idxB] = intersect(A(:,1:end-1),B(:,1:end-1),'rows')
idxBnotA = setdiff([1:G],idxB);
H = M + G - length(idxA);
C1 = zeros(H,N+1);
C1(1:M,1:N) = A;
C1(idxA,end) = B(idxB,end);
C1(M+1:end,1:end-2) = B(idxBnotA,1:end-1);
C1(M+1:end,end) = B(idxBnotA,end)
The input is an N-by-1 matrix. I need to reshape it to L-by-M matrix. The following is an example.
Input:
b =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Set length = 18, Output:
X =
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
10 11 12
11 12 13
12 13 14
13 14 15
14 15 16
15 16 17
16 17 18
17 18 19
18 19 20
Because I have a very big matrix, using a loop to reshape is very inefficient. How can I improve the reshape speed?
Your example output matrix X is the perfect matrix to index a vector of length N to get what you want. It's also very easy to create using bsxfun:
N = 20;
b = rand(N,1);
M = 3; %// number of columns
L = N-M; %// Note that N-M is an upper limit for L!
idx = bsxfun(#plus, (0:L)', 1:M)
X = b(idx)
That's exactly what im2col (from the Image Processing Toolbox) does:
b = (1:20).'; %'// example data
L = 18; % // desired length of sliding blocks
x = im2col(b, [L 1]); % // result
I'd use horzcat. For example:
function X = reshaper(b,len)
diff = length(b) - len + 1;
X = b(1:len);
for i=2:diff
X = horzcat(X,b(i:len+(i-1)));
end
You could probably remove the for loop with some further thought.
I want to write a Matlab script.
In my example I have a vector A=[1 3 4 5 7 8 9 10 11 13 14 15 16 17 19 20 21]
Now I want to cut the vector automatically at the points where a number is missing(here the numbers 2, 6, 12, 18 are missing).
As a result I want to have the vectors [1] and [3 4 5] and [7 8 9 10 11] and [13 14 15 16 17] and [19 20 21]. So as you can see the new vectors have different lenghts.
I thought about using a for loop, but I am not sure how to write these new vectors.
Thank you for your help :)
One liner with diff, cumsum & accumarray -
out = accumarray(cumsum([0 ; diff(A(:))~=1])+1,A(:),[],#(x) {x})
Sample run -
>> A
A =
1 3 4 5 7 8 9 10 ...
11 13 14 15 16 17 19 20 21
>> celldisp(out)
out{1} =
1
out{2} =
3
4
5
out{3} =
7
8
9
10
11
out{4} =
13
14
15
16
17
out{5} =
19
20
21
This is one approach:
s = [find(diff(A(:).')>1) numel(A)]; %'// detect where consecutive difference exceeds 1
s = [s(1) diff(s)]; %// sizes of groups
result = mat2cell(A(:).', 1, s); %'// split into cells according to those sizes
In your example, this gives
>> celldisp(result)
result{1} =
1
result{2} =
3 4 5
result{3} =
7 8 9 10 11
result{4} =
13 14 15 16 17
result{5} =
19 20 21
Another approach (computes group sizes differently):
s = diff([0 sum(bsxfun(#lt, A(:), setdiff(1:max(A(:).'), A(:).')), 1) numel(A)]);
result = mat2cell(A(:).', 1, s);
I made a matrix in Matlab, say,
A = magic(5);
A =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Now I found the indices I want using the find function as:
ind = find(A(:,5)>3 & A(:,4)>= 8);
ind =
1
2
3
Now if I want to get a subset of matrix A for those indices using B = A(ind) function, I only get the first column of the matrix:
B = A(ind)
B =
17
23
4
How can I get all the columns as subset??
Oops ... I got it
B = A(ind,:);