I want to ask a question in some matrix operations in MATLAB.
Assume we have this matrix:
A = [1 1 17
1 1 14
1 2 10
1 2 11
2 1 9
2 1 9
2 2 13
2 2 12
3 1 18
3 1 15]
I want the first column, say M and the second column, say D to control the entire matrix to result to one row matrix depending on the following condition:
the program will ask the user to enter the values of M then D as follows:
M = input(' ENTER M VALUE = ') ;
D = input(' ENTER D VALUE = ') ;
Now, the output will be the corresponding 2 values to M and D, and these two values will be taken from the third column,
for example:
if M = 1 and D = 2 , the output is B = 10 ; 11
another example:
if M = 3 and D = 1 , the output is B = 18 ; 15
and so on.
Actually, I know how to solve this using if statement but I have large data and this will take very long time. I am sure that there is a short way to do that.
Thanks.
The short way to do it is
B = A(A(:,1)==M & A(:,2)==D, 3);
Related
I have 3 datasets in one 40×7 double data set and I want to multiply these and plot the result over a time period. My question is how to do that.
My attempt has been to divide the datasets into 3 individual cells of data and call them A, B and C (each with 40×3) by using the operator a=[{A}] and so on.
Next, I multiplied them with each other by using .*a.*b.*c and with the time t=[0:40] (which is the x-axis).
I get the following error and I don't know what to do.
Index exceeds matrix dimensions
Anyone that can help me?
Code as provided by OP in a comment:
a = ans(:,1:3);
b = ans(:,4:6);
c = ans(:,7:9);
A[{xyz}];
B=[{a}];
C=[{c}];
t=[0:41];
D=(A.*B.*C);
plot(D,t);
One way to do this is to combine reshape and permute, and take the product along the third dimension.
Assume the following is your data. This will be 40x9 in your case.
rows = 4; cols = 6; N = 2;
x = ceil(10*rand(rows,cols))
x =
9 6 1 6 5 8
6 9 9 5 10 6
10 9 7 9 6 3
7 10 10 3 9 7
r = permute(reshape(x, rows, N, []), [3,1,2])
result = prod(r,3).'
result =
54 6 40
54 45 60
90 63 18
70 30 63
So looking at your code, it seems unnecessary to create a cell from it:
you can for example:
ans=rand(40,9);
a = ans(:,1:3); b = ans(:,4:6); c = ans(:,7:9);
D=a.*b.*c;
t=[1:40];
plot(D,t)
or
ans=rand(40,9);
a = ans(:,1:3); b = ans(:,4:6); c = ans(:,7:9);
D=a.*b.*c;
t=[0:39];
plot(D,t)
Your t vector has to have the same length as your inital matrix - your matrix has 40 rows but the vector you create by t=[0:40] is 41 entries long so you either do t=[1:40] or t=[0:39]
and if t is your x-axis you should plot it in the right order so plot(t,D) instead of plot(D,t)
N.B: This question is more complex than my previous question: Matlab: How I can make this transformation on the matrix A?
I have a matrix A 4x10000, I want to use it to find another matrix C, based on a predefined vector U.
I'll simplify my problem with a simple example:
from a matrix A
20 4 4 74 20 20 4
36 1 1 11 36 36 1
77 1 1 15 77 77 1
3 4 2 6 7 8 15
and
U=[2 3 4 6 7 8 2&4&15 7&8 4|6].
& : AND
| : OR
I want, first, to find an intermediate entity B:
2 3 4 6 7 8 2&4&15 7&8 4|6
[20 36 77] 0 1 0 0 1 1 0 1 0 4
[4 1 1] 1 0 1 0 0 0 1 0 1 4
[74 11 15] 0 0 0 1 0 0 0 0 1 2
we put 1 if the corresponding value of the first line and the vector on the left, made a column in the matrix A.
the last column of the entity B is the sum of 1 of each line.
at the end I want a matrix C, consisting of vectors which are left in the entity B, but only if the sum of 1 is greater than or equal to 3.
for my example:
20 4
C = 36 1
77 1
This was a complex one indeed and because of the many restrictions and labeling processes involved, it won't be as efficient as the solution to the previous problem. Here's the code to solve the posted problem -
find_labels1 = 2:8; %// Labels to be detected - main block
find_labels2 = {[2 4 15],[7 8],[4 6]}; %// ... side block
A1 = A(1:end-1,:); %// all of A except the last row
A2 = A(end,:); %// last row of A
%// Find unique columns and their labels for all of A execpt the last row
[unqmat_notinorder,row_ind,inv_labels] = unique(A1.','rows'); %//'
[tmp_sortedval,ordered_ind] = sort(row_ind);
unqcols = unqmat_notinorder(ordered_ind,:);
[tmp_matches,labels] = ismember(inv_labels,ordered_ind);
%// Assign labels to each group
ctl = numel(unique(labels));
labelgrp = arrayfun(#(x) find(labels==x),1:ctl,'un',0);
%// Work for the main comparisons
matches = bsxfun(#eq,A2,find_labels1'); %//'
maincols = zeros(ctl,numel(find_labels1));
for k = 1:ctl
maincols(k,:) = any(matches(:,labelgrp{k}),2);
end
%// Work for the extra comparisons added that made this problem extra-complex
lens = cellfun('length',find_labels2);
lens(end) = 1;
extcols = nan(ctl,numel(find_labels2));
for k = 1:numel(find_labels2)
idx = find(ismember(A2,find_labels2{k}));
extcols(:,k)=arrayfun(#(n) sum(ismember(labelgrp{n},idx))>=lens(k),1:ctl).'; %//'
end
C = unqcols(sum([maincols extcols],2)>=3,:).' %//'# Finally the output
I will give you a partial answer. I think you can take from here. Idea is to concatenate first 3 rows of A with each element of U replicated as last column. After you get the 3D matrix, replicate your original A and then just compare the rows. The rows which are equal, that is equivalent to putting one in your table.
B=(A(1:3,:).';
B1=repmat(B,[1 1 length(U)]);
C=permute(U,[3 1 2]);
D=repmat(C,[size(B1,1),1,1]);
E=[B1 D];
F=repmat(A',[1 1 size(E,3)]);
Now compare F and E, row-wise. If the rows are equal, then you put 1 in your table. For replicating & and |, you can form some kind of indicator vector.
Say,
indU=[1 2 3 4 5 6 7 7 7 8 8 -9 -9];
Same positive value indicates &, same negative value indicates |. Different value indicate different entries.
I hope you can take from here.
suppose that we are creating following matrix from given signal
function [ x ]=create_matrix1(b,l)
n = length(b);
m = n-l+1;
x = zeros(m,l);
for i=1:m
x(i,:)=b(i:i+l-1);
end;
end
with some window length,for example
X=[2;1;3;4;5;7]
X =
2
1
3
4
5
7
>> B=create_matrix1(X,3)
B =
2 1 3
1 3 4
3 4 5
4 5 7
if we have given matrix and windows length ,how can i reconstruct original signal?let say i know that windows length is 3,thanks in advance,i think i should sum elements on anti diagonal and divide by number of elements in this anti diagonal ,but how can i do it by code?thanks in advance
Your original vector is located along the top and right edge of your matrix B and can be reconstructed like so:
>> X_reconstructed = [B(1,1:end-1).'; B(:,end)]
X_reconstructed =
2
1
3
4
5
7
In case the matrix B is some noisy matrix and you actually want to do the averages along the diagonals:
>> BB = fliplr(B);
>> X_mean = arrayfun(#(i) mean(diag(BB,i)), size(B,2)-1:-1:-size(B,1)+1).'
X_mean =
2
1
3
4
5
7
just lets make it simple, assume that I have a 10x3 matrix in matlab. The numbers in the first two columns in each row represent the x and y (position) and the number in 3rd columns show the corresponding value. For instance, [1 4 12] shows that the value of function in x=1 and y=4 is equal to 12. I also have same x, and y in different rows, and I want to average the values with same x,y. and replace all of them with averaged one.
For example :
A = [1 4 12
1 4 14
1 4 10
1 5 5
1 5 7];
I want to have
B = [1 4 12
1 5 6]
I really appreciate your help
Thanks
Ali
Like this?
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7];
[x,y] = consolidator(A(:,1:2),A(:,3),#mean);
B = [x,y]
B =
1 4 12
1 5 6
Consolidator is on the File Exchange.
Using built-in functions:
sparsemean = accumarray(A(:,1:2), A(:,3).', [], #mean, 0, true);
[i,j,v] = find(sparsemean);
B = [i.' j.' v.'];
A = [1 4 12;1 4 14;1 4 10; 1 5 5;1 5 7]; %your example data
B = unique(A(:, 1:2), 'rows'); %find the unique xy pairs
C = nan(length(B), 1);
% calculate means
for ii = 1:length(B)
C(ii) = mean(A(A(:, 1) == B(ii, 1) & A(:, 2) == B(ii, 2), 3));
end
C =
12
6
The step inside the for loop uses logical indexing to find the mean of rows that match the current xy pair in the loop.
Use unique to get the unique rows and use the returned indexing array to find the ones that should be averaged and ask accumarray to do the averaging part:
[C,~,J]=unique(A(:,1:2), 'rows');
B=[C, accumarray(J,A(:,3),[],#mean)];
For your example
>> [C,~,J]=unique(A(:,1:2), 'rows')
C =
1 4
1 5
J =
1
1
1
2
2
C contains the unique rows and J shows which rows in the original matrix correspond to the rows in C then
>> accumarray(J,A(:,3),[],#mean)
ans =
12
6
returns the desired averages and
>> B=[C, accumarray(J,A(:,3),[],#mean)]
B =
1 4 12
1 5 6
is the answer.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Sort a matrix with another matrix
The Matrix A ('10 x 1000' all numbers) look like this:
score(1.1) score(1.2) score(1.3)....score(1.1000)
score(2.1) score(2.2) score(2.3)....score(1.1000)
...
The Matrix B ('1 x 1000' all numbers):
Return(1) Return(2) Return(3) .....Return(1.1000)
Every time I sort a row of Matrix A, I want to sort Matrix B based on the order of the sorted row in Matrix A. Because there are 10 rows in Matrix A, Matrix B will be sorted 10 times and generate a new Matrix C ('10 x 1000') like this: (I am looking for a script to generate this Matrix C)
Return(3) Return(25) Return(600) .......Return(1000)
Return(36）Return(123) Return(2)........Return(212)
....
....
This should do what you want:
A = randn(10,1000);
B = randn(1,1000);
C = zeros(size(A));
for i = 1:10
[a idx] = sort(A(1,:));
A(i,:) = a;
C(i,:) = B(idx);
end
Now the rows of A are sorted, and the rows of C contain the corresponding sorted B.
This solution is a bit more compact, and it's also good to get used to doing this kind of solution for efficiency when your matrices get big. You can solve your problem with two ideas:
In [a, ix] = sort(X), a is a column-sorted version of X, and ix stores which rows moved where in each column. Thus if we do [a, ix] = sort(X.').'; (where the dot-apostrophe is the transpose) we can sort the rows.
B(ix) where ix is a bunch of indeces will make a matrix the same size as ix with the i-jth element being B at ix(i,j)
Then you just need to reshape it. So you can do:
A = rand(4,8);
B = rand(1,8);
n = size(A,1);
m = size(A,2);
[~,ix] = sort(A.');
C = reshape(B(ix'),n,m);
If I understand your question correctly the following should work. Using some sample scores:
>> score = [1 4 7 9; 3 5 1 7; 9 3 1 6]
score =
1 4 7 9
3 5 1 7
9 3 1 6
and sample return vector:
>> r = [10 20 30 40]
r =
10 20 30 40
Transpose the scores and sort since the SORT command works on columns of a matrix. We're only interested in the indices of the sorted values:
>> [~, ix] = sort(score')
ix =
1 3 3
2 1 2
3 2 4
4 4 1
Now transpose these indices and use them to reference the return values:
>> answer = r(ix)'
answer =
10 20 30 40
30 10 20 40
30 20 40 10