I have an array in Matlab
A = [1 2 3 4 5 6 7 8 9;
67 67 67 86 86 86 86 67 67]';
where every point in the first row of A corresponds to a "code" either 67 or 86. I am trying to extract these blocks of "67s" and "86s" such that every time a block starts the corresponding elements are put into the 3rd dimension of a different array called X, where the .
So for e.g. in A I have 3 different blocks, so I would like to end up with an array X of size 1x9x3. And for e.g. the first 67 block I would like to have X
X(1,:,1) = [1 2 3];
I understand that I would "fill up" this vector X using a for loop
for i=1:size(A,2)
for j=1:size(A,2) %actually j should be up till the number of blocks present
X(1,i,j) = A(1,i)
end
end
But this isn't correct or complete of course because firstly I'm unsure how to separate out the "blocks" and how to correctly "fill in" the j's in X(1,i,j). Secondly how can I get the code to recognise how many blocks there are?
Can anyone help?
Thanks
One possible approach, based on this answer:
>> B = accumarray([0; cumsum(diff(A(:,2)) ~= 0)] + 1, A(:,1), [], #(x) {x}, [])
Now you have this:
>> B{1}
ans =
1
2
3
>> B{2}
ans =
4
5
6
7
>> B{3}
ans =
8
9
I have the following vector A:
A = [34 35 36 5 6 7 78 79 7 9 10 80 81 82 84 85 86 102 3 4 6 103 104 105 106 8 11 107 201 12 202 203 204];
For n = 2, I counted the elements larger or equal to 15 within A:
D = cellfun(#numel, regexp(char((A>=15)+'0'), [repmat('0',1,n) '+'], 'split'));
The above expression gives the following output as duration values:
D = [3 2 7 4 6] = [A(1:3) **stop** A(7:8) **stop** A(12:18) **stop** A(22:25) **stop** A(28:33)];
The above algorithm computes the duration values by counting the elements larger or equal to 15. The counting also allows less than 2 consecutive elements smaller than 15 (n = 2). The counter stops when there are 2 or more consecutive elements smaller than 15 and starts over at the next substring within A.
Eventually, I want a way to find the median position points of the duration events A(1:3), A(7:8), A(12:18), A(22:25) and A(28:33), which are correctly computed. The result should look like this:
a1 = round(median(A(1:3))) = 2;
a2 = round(median(A(7:8))) = 8;
a3 = round(median(A(12:18))) = 15;
a4 = round(median(A(22:25))) = 24;
a5 = round(median(A(28:33))) = 31;
I edited the question to make it more clear, because the solution that was provided here assigns the last number within the row of 2 or more consecutive numbers smaller than 15 (3 in this case) after A(1:3) to the next substring A(7:8)and the same with the other substrings, therefore generating wrong duration values and in consequence wrong median position points of the duration events when n = 2 or for any given even n.
Anyone has any idea how to achieve this?
I want to pass columns in various matrices to a for loop.
If my two matrices had the same number of columns, I might do something like this:
mat1 = matrix(rep(1:25), 5,5)
mat2 = matrix(rep(26:50), 5,5)
array.mat = array(c(mat1,mat2), dim=c(5,5,2))
mat1.ncol = ncol(mat1)
mat2.ncol = ncol(mat2)
mat.ncol = c(mat1.ncol, mat2.ncol)
mat.ncol
array.mat
for (dimi in 1:2){
dim.col = mat.ncol[dimi]
for (coli in 1:dim.col){
st = shapiro.test(array.mat[,coli,dimi])$p.value
if(st > .001){
array.mat[,coli,dimi] = log(array.mat[,coli,dimi])
}}}
But, my data don't have the same number of columns, so I'd like to use a list of matrices instead.
mat1 = matrix(rep(1:10), 5,2)
mat2 = matrix(rep(26:50), 5,5)
list.mat=list(a=mat1, b=mat2)
list.mat
But I can't figure out how I'd pass the columns of the matrices?
list.mat$a[1:5]
gives the first column of the first matrix, but how would you pass $a and [startindex:endindex] in a loop? All the other answers I see tend to pass the ith element (e.g., column) of both matrices. I need to keep the two matrices (a and b) separate for later computations, but I want them together (the list of the two matrices) for these types of loops.
Once again, I'm probably just thinking about this incorrectly. Thanks for any thoughts.
Can you use numeric indices? e.g., matrix 1, column 1: list.mat[[1]][,1]
In a loop:
for (m in 1:2) {
for (i in 1:ncol(list.mat[[m]])) {
cat('Here is matrix', m, ', columnn', i, '\n')
print(list.mat[[m]][,i])
}
}
result:
Here is matrix 1 , columnn 1
[1] 1 2 3 4 5
Here is matrix 1 , columnn 2
[1] 6 7 8 9 10
Here is matrix 2 , columnn 1
[1] 26 27 28 29 30
Here is matrix 2 , columnn 2
[1] 31 32 33 34 35
Here is matrix 2 , columnn 3
[1] 36 37 38 39 40
Here is matrix 2 , columnn 4
[1] 41 42 43 44 45
Here is matrix 2 , columnn 5
[1] 46 47 48 49 50
I have a 12-D array and am using each dimension as an index value in an optimization problem.
A(:,:,i1,i2,i3,i4,i5,i6,i7,i8,i9,i10)
each index value i is a value from 1 to 5.
I want to sort A from greatest to least and keep track of the indices so I know which indices correspond to to what value of A.
So my ideal output would be a 2 column cell/array with one column being the value and the other other column being the index values.
For a simple 3D example: say I have a 3D array: A(:,:,i1).
Where:
A(:,:,1) = 2
A(:,:,2) = 6
A(:,:,3) = 13
A(:,:,4) = 11
A(:,:,5) = 5
I would like my output to be:
13 3
11 4
6 2
5 5
2 1
EDIT:
assume I have 1x1x3x3 sized input such that
A(1,1,1,1) = 3
A(1,1,2,1) = 1
A(1,1,3,1) = 23
A(1,1,1,2) = 12
A(1,1,2,2) = 9
A(1,1,3,2) = 8
A(1,1,1,3) = 33
A(1,1,2,3) = 14
A(1,1,3,3) = 6
the expected output would be:
33 [1,1,1,3]
23 [1,1,3,1]
14 [1,1,2,3]
12 [1,1,1,2]
9 [1,1,2,2]
8 [1,1,3,2]
6 [1,1,3,3]
3 [1,1,1,1]
1 [1,1,2,1]
This should be a generic code for any multi-dimensional input array -
%// Sort A and get the indices
[sorted_vals,sorted_idx] = sort(A(:),'descend');
%// Set storage for indices as a cell array and then store sorted indices into it
c = cell([1 numel(size(A))]);
[c{:}] = ind2sub(size(A),sorted_idx);
%// Convert c to the requested format and concatenate with cell arary version of
%// sorted values for the desired output
out = [num2cell(sorted_vals) mat2cell([c{:}],ones(1,numel(A)),numel(size(A)))];
The generic code owes its gratitude to this fine solution.
I guess this is what you want:
b=A(:);
[sorted_b,ind]=sort(b,'descend');
[dim1,dim2,dim3,dim4]=ind2sub(size(A),ind);
%arranging in the form you want
yourCell=cell(size(b,1),2);
yourCell(:,1)=mat2cell(sorted_b,ones(size(b,1),1),1);
%arranging indices -> maybe vectorized way is there for putting values in "yourCell"
for i=1:size(b,1)
yourCell{i,2}=[dim1(i) dim2(i) dim3(i) dim4(i)];
end
For the array A, given by you, my output looks like:
33 [1,1,1,3]
23 [1,1,3,1]
14 [1,1,2,3]
12 [1,1,1,2]
9 [1,1,2,2]
8 [1,1,3,2]
6 [1,1,3,3]
3 [1,1,1,1]
1 [1,1,2,1]
which matches with your output.
In trying to port an algorithm from C# to Matlab I found that Matlab is inefficient at running for loops. As such I want to vectorize the algorithm.
I have following inputs:
lowrange:
[ 00 10 20 30 40 50 ... ]
highrange:
[ 10 20 30 40 50 60 ... ]
These arrays are equal in length.
I now have a third array Values (which could be any length) and for this array I want to count the occurrences of Values elements between lowerange(i) and highrange(i) (You can see I'm coming from a for loop).
The output should be an array of length lowrange/highrange.
So with the above arrays and input LineData:
[ 1 2 3 4 6 11 12 16 31 34 45 ]
I expect to get:
[ 05 03 00 02 01 00 ... ]
I tried the (for me) obvious thing:
LineData(LineData < PixelEnd & LineData > PixelStart)
But that doesn't work because it just checks LineData on an element by element way. It does not try to apply the comparison over all values in LineData.
Unfortunately, I cannot come up with anything else since I'm not yet used to think in a Matlab 'vector' way, let alone knowing all applicable instructions from memory.
As you are looking to do a basic histogram with given edges, you can use Matlabs built-in function histc:
values = [ 1 2 3 4 6 11 12 16 31 34 45 ];
edges = 0:10:60;
histc(values, edges)
ans =
5 3 0 2 1 0 0
For ranges with identical intervals and starting from 0, here's a bsxfun based counting approach -
LineData = [ 1 2 3 4 6 11 12 16 31 34 45 ] %// Input
interval = 10; %// interval width
num_itervals = 6; %// number of intervals
%// Get matches for each interval and sum them within each interval for the counts
out = sum(bsxfun(#eq,ceil(LineData(:)/interval),1:num_itervals))
Output -
LineData =
1 2 3 4 6 11 12 16 31 34 45
out =
5 3 0 2 1 0
Assuming that the last interval would be the one holding the max of input data, you can try out a diff + indexing based approach too -
LineData = [ 1 2 3 4 6 11 12 16 31 34 45 ] %// Input
interval = 10; %// interval width
labels = ceil(LineData(:)/interval); %// set labels to each input entry
df_labels = diff(labels)~=0; %// mark the change of labels
df_labels_pos = find([df_labels; 1]); %// get the positions of label change
intv_pos= labels([true;df_labels]);%// position of each interval with nonzero counts
%// get counts from interval between label position change and put at right places
out(intv_pos) = [ df_labels_pos(1) ; diff(df_labels_pos)];