I have a 2D matrix with in the 1st dimension different channels, and in the 2nd dimension time samples. I want to rearrange this to a 3D matrix, with in the 1st and 2nd dimension channels, and in the 3rd time samples.
The channels have to mapped according to a certain mapping. Right now I am using a for-loop to do so, but what would be a no-loop solution?
N_samples = 1000;
N_channels = 64;
channel_mapping = reshape(1:64, [8 8]).';
% Results in mapping: (can also be random)
% 1 2 3 4 5 6 7 8
% 9 10 11 12 13 14 15 16
% 17 18 19 20 21 22 23 24
% 25 26 27 28 29 30 31 32
% 33 34 35 36 37 38 39 40
% 41 42 43 44 45 46 47 48
% 49 50 51 52 53 55 55 56
% 57 58 59 60 61 62 63 64
data = rand(N_channels, N_samples);
data_grid = NaN(8,8, N_samples);
for k = 1:N_samples
tmp = data(:, k);
data_grid(:, :, k) = tmp(channel_mapping);
end
You can do it in one go as follows:
data_grid = reshape(data(channel_mapping, :), 8, 8, []);
Related
Say I have an Mx4 array A where the values in the first column are a number 1 to 12. Now I want to gather the rest of the columns in 12 separate Mx3 arrays depending on which number is in column 1.
How would I go about doing that?
You can use unique and splitapply as follows. The result is a cell array of arrays.
M = [2 11 41 51;
1 10 20 30;
1 62 83 22;
4 73 53 53;
2 84 94 14]; % example data
L = 5; % Group labels are 1:L (L=12 in your case)
[u,~,w] = unique(M(:,1));
result = cell(L,1);
result(u) = splitapply(#(x){x}, M(:,2:end), w);
This gives
>> celldisp(result)
result{1} =
10 20 30
62 83 22
result{2} =
11 41 51
84 94 14
result{3} =
[]
result{4} =
73 53 53
result{5} =
[]
I have the following multidimensional array:
a(:,:,1) =
11 12 13
21 22 23
a(:,:,2) =
21 22 23
31 32 33
a(:,:,3) =
31 32 33
41 42 43
The end result looks as follows:
e =
11 21 31
12 22 32
13 23 33
21 31 41
22 32 42
23 33 43
In order to obtain array e, I did the following:
b=permute(a,[2,1,3])
c=reshape(b,1,6,[])
d=permute(c,[2,1,3])
Is there a cleaner or more elegant way of achieving the same result?
Here permute used once:
b = permute(a,[2 1 3]);
result = reshape(b, [], 3)
You can do it like the following:
reshape([a(:,:,1);a(:,:,2);a(:,:,3)].',6,3)
For example
I have one binary array with size of 9 as b = [0 1 0 1 0 1 1 1 1], Then another array 'm' with size of 7 as m = [21 28 36 45 45 66 66]. Here i want to change all the zeros of 'b' by 1st element of m then replace 1's of b by consecutive elements of 'm' so my output 1D array should be like k = [21 28 21 36 21 45 45 66 66].
Below is my code i really don't know where i did mistake please help me to solve this
b= [0 1 0 1 0 1 1 1 1];
b=b(:);
m = [21 28 36 45 45 66 66];
m = m(:);
k=zeros(size(b));
for i=1:length(b)
for j=2:length(m)
if b(i)==0
k(i)=m(1);
else
k(i)=m(j);
end
end
end
am getting output as
k = [21 66 21 66 21 66 66 66 66]
Use logical indexing instead - it is much faster and more readable:
b = [0 1 0 1 0 1 1 1 1];
m = [21 28 36 45 45 66 66];
k = zeros(size(b));
k(b==0) = m(1); % fill values where b=0 with m(1)
k(b==1) = m(2:sum(b)+1); % fill values where b=1 with consecutive m values
Result:
>> k
k =
21 28 21 36 21 45 45 66 66
I need to create a data field that will go through a vector. Data field is constant length, and it is going through the data vector shifting data field with data field length. I need the mean value of that field (A vector) that corresponds to a mean value of another field (B vector).
Example:
A=[1 5 7 8 9 10 11 13 15 18 19 25 28 30 35 40 45 48 50 51];
B=[2 4 8 9 12 15 16 18 19 20 25 27 30 35 39 40 45 48 50 55];
I want to do next:
A=[{1 5 7 8 9} 10 11 13 15 18 19 25 28 30 35 40 45 48 50 51];
B=[{2 4 8 9 12} 15 16 18 19 20 25 27 30 35 39 40 45 48 50 55];
I want to take data from field of 5 points and get mean value. And then shift whole data field with data field length.
A=[1 5 7 8 9 {10 11 13 15 18} 19 25 28 30 35 40 45 48 50 51];
B=[2 4 8 9 12 {15 16 18 19 20} 25 27 30 35 39 40 45 48 50 55];
I need two vectors, C and D with mean values of this method.
C=[6 13.4 27.4 45.2];
D=[7 17.6 31.2 47.6];
I started something with
n = length(A);
for k = 1:n
....
but nothing I tried worked.
reshape the vector into a 5-row matrix and then compute the mean of each column:
C = mean(reshape(A,5,[]),1);
D = mean(reshape(B,5,[]),1)
I have a 3x3x2000 array of rotation matrices that I need to transform into a 2000x9 array.
I think I have to use a combination of permute() and reshape(), but I don't get the correct output order.
This is what I need:
First row of 3x3 array needs to be columns 1:3 in the output
Second row of 3x3 array needs to be columns 4:6 in the output
Third row of 3x3 array needs to be columns 7:9 in the output
I have tried all possible combinations of numbers 1 2 3 in the following code:
out1 = permute(input, [2 3 1]);
out2 = reshape(out1, [2000 9]);
But I always end up with the wrong order. Any tips for a Matlab newbie?
How about a simple for-loop?
for i=1:size(myinput,3)
myoutput(i,:)=[myinput(1,:,i) myinput(2,:,i) myinput(3,:,i)];
% or
% myoutput(i,:)=reshape(myinput(:,:,i),[],9);
end
It's not simple as using permute and reshape, but it is transparent and easier for debugging. Once everything in your program runs perfectly, you can consider to rewrite such for-loops in your code...
You had a mix-up in your permute
a = reshape(1:9*6, 3, 3, []);
a is a 3x3x6 matrix, each
a(:,:,i) = 9*(i-1) + [1 4 7
2 5 8
3 6 9];
So
out1 = permute(a, [3,1,2]);
out2 = reshape(out1, [], 9);
Or in one line
out3 = reshape(permute(a, [3,1,2]), [], 9);
So
out2 = out3 =
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54