I'm using Matlab, and I'm trying to come up with a vectorized solution for comparing the elements of one array to every element of another array. Specifically I want to find the difference and see if this difference is below a certain threshold.
Ex: a = [1 5 10 15] and b=[12 13 14 15], threshold = 6
so the elements in a that would satisfy the threshold would be 10 and 15 since each value comes within 6 of any of the values in b while 1 and 5 do not. Currently I have a for loop going through the elements of a and subtracting an equivalently sized matrix from b (for 5 it would be a = [5 5 5 5]). This obviously takes a long time so I'm trying to find a vectorized solution. Additionally, the current format I have my data in is actually cells where each cell element has size [1 2], and I have been using the cellfun function to perform my subtraction. I'm not sure if this complicates the solution of each [1 2] block with the [1 2] block of the second cell. A vectorized solution response is fine, there is no need to do the threshold analysis. I just added it in for a little more background.
Thanks in advance,
Manwei Chan
Use bsxfun:
>> ind = any(abs(bsxfun(#minus,a(:).',b(:)))<threshold)
ind =
0 0 1 1
>> a(ind)
ans =
10 15
Related
Suppose I have 3 arrays of different length:
A[1 2 3 4 5 6]; B[1 2 4], C[0 1 5 6]
the MAXIMUM length the three arrays is 6.
So I want B and C to have 6 elements too. Furthermore, the last "filled" elements of new array should be the last element of the original array
At the end, new B should look like [1 2 4 4 4 4], C should look like [0 1 5 6 6 6], how do I implement this?
Thanks a lot!
You can do this using MATLAB's indexing operators. If you assign to elements past the end of an array, MATLAB will grow the array. For vectors (as in this case), you only need to specify a single subscript because the direction to extend the array is not ambiguous.
I'd do it like this - this doesn't assume that you know up-front which of the arrays is the largest. I'm using numel to compute the number of elements in each vector. I'm also relying on the fact that for one of the arrays, end+1:maxLen is an empty range, so no modifications are made in that case.
maxLen = max([numel(A), numel(B), numel(C)]);
A(end+1:maxLen) = A(end)
B(end+1:maxLen) = B(end)
C(end+1:maxLen) = C(end)
I have an array of mxnx4 dimensions and a function with a 4-element vector input. I want to apply my function to give me an mxnx4 output, such that the function is applied mxn times (i.e. the 3rd dimension of my array becomes in the input of my function). Any ideas on how do do this would be massively appreciated. I have looked at arrayfun, but this seems not to address what I want to do. I also want to maintain the formatting of my function as it requires this formatting elsewhere in my code.
I.e.:
F=#(V)[V(1)+V(2);V(2)+V(3); V(3)+V(4); V(4)+V(1)]
to be applied to array M of dimensions 2x3x4 (m=2; n=3)
M = zeros(2,3,4);
M(:,:,1) = [1 2 3;4 5 6];
M(:,:,2) = [7 8 9;10 11 12];
M(:,:,3) = [13 14 15; 16 17 18];
M(:,:,4) = [19 20 21; 22 23 24];
to generate an array of dimensions 2x3x4
C(1,1,:) = F([1 7 13 19])
C(1,2,:) = F([2 8 14 20])
etc.
I can see a for loop would work, applying the following across the mxn matrix
C(m,n,:)=F(M(m,n,:))
However, I need to run this millions of times so would like a faster approach
I have an array
a = [1 2 3 4 5 6 7 8]
I want to get every group of 4
so the result is as such
[1 2 3 4]
[5 6 7 8]
I do not know how many elements there will be but I know it is divisible by 4
so something like a(1:4) and a(5:8) wont work, I can use a loop, but is there a way to not use a loop?
For an unknown number of elements in a you can use reshape you just need to figure out how many rows you will have in the final matrix or (better for your case) the number of columns.
a = 1:4*10;
a2 = reshape(a, 4, []).';
If you went the rows routine you would do this.
a = 1:4*10;
a2 = reshape(a, [], numel(a) / 4).';
You just need to be sure that a has the proper number of elements. numel simply tells you the total element count.
I am limited in the use of cycles in matlab and I have one task. I have the matrix 2xn with numbers and cell array 1xn. Each element in the first row point to a position of the array. I want to add every number from the second row of matrix to cell, which is pointed by corresponding number in the first row. In addition, I want the cells to be strings.
Let me clarify with example:
A = [[1 4 3 3 1 4 2], [7 4 3 5 6 5 4]]
I want to get array of cells: {'76', '4', '35', '45'}
How can I do this without using a for or while loop?
A = [1 4 3 3 1 4 2; 7 4 3 5 6 5 4]; %// data: 2 x n array
[~, ~, ii] = unique(A(1,:));
R = accumarray(ii(:),A(2,:),[],#(v) {fliplr(regexprep(num2str(v(:).'),'\s',''))}).';
The second line (unique) is used to remove possible gaps in first row. Otherwise those gaps would translate to the result of accumarray, which would then take up more memory uselessly.
The third line (accumarray) aggregates all values of second row of A that have the same value in the first row. The aggregation is done by the anonymous function, which converts numbers to string (num2str), removes spaces (regexprep), and changes orientation (fliplr and .') to match the desired output format.
EDIT:
Thanks to #chappjc's suggestion, the third line can be simplified to
R = accumarray(ii(:),A(2,:),[],#(v) {fliplr(num2str(v(:).','%d'))}).';
Alright, by just pointing at cellfun, arrayfun, cell2mat, cell indexing and array indexing I was arrogant. So, I try to make it up by actually using some of those in a more constructive solution:
%// Using array-indexing to find all matches with the current index in the first row,
%// then print their counterpart(s) into strings
fun = #(ii) sprintf('%d', A(2, A(1, :) == ii));
%// The following could also be 1:size(A, 2) or unique(A(1, :)) depending on the
%// general form of your problem.
range = min(A(1, :)) : max(A(1, :));
%// Using arrayfun to loop over all values in the first row
R = arrayfun(fun, range, 'UniformOutput', false)
I want to group my elements using the repeated segments in the array. The breaking is basically depend on where the repeated segments are, in my real data contains ~10000 elements and I want to know if there is a easier way to do that.
Here is a short example to clarify what I want:
Let's say I have an array,
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
What I want is to break A into [1 5 3],[6 9], and [9 5 2];
What is the easiest to code this using matlab??
Thanks.
For a vectorized solution, you can find out the places where either forward or backward differences to the neighbor are zero, and then use bwlabel (from the Image Processing Toolbox) and accumarray to gather the data.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
d = diff(A)==0;
%# combine forward and backward difference
%# and invert to identify non-repeating elments
goodIdx = ~([d,false]|[false,d]);
%# create list of group labels using bwlabel
groupIdx = bwlabel(goodIdx);
%# distribute the data into cell arrays
%# note that the first to inputs should be n-by-1
B = accumarray(groupIdx(goodIdx)',A(goodIdx)',[],#(x){x})
EDIT
Replace the last two lines of code with the following if you want the repeating elements to appear in the cell array as well
groupIdx = cumsum([1,abs(diff(goodIdx))]);
B = accumarray(groupIdx',A',[],#(x){x})
EDIT2
If you want to be able to split consecutive groups of identical numbers as well, you need to calculate groupIdx as follows:
groupIdx = cumsum([1,abs(diff(goodIdx))|~d.*~goodIdx(2:end)])
Here is a solution that works if I understand the question correctly. It can probably be optimised further.
A=[1 5 3 4 4 4 6 9 8 8 9 5 2];
% //First get logical array of non consecutive numbers
x = [1 (diff(A)~=0)];
for nn=1:numel(A)
if ~x(nn)
if x(nn-1)
x(nn-1)=0;
end
end
end
% //Make a cell array using the logical array
y = 1+[0 cumsum(diff(find(x))~=1)];
x(x~=0) = y;
for kk = unique(y)
B{kk} = A(x==kk);
end
B{:}