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)
Related
I'd like to take a single array lets say 3x5 as follows
1 3 5
1 3 5
2 8 6
4 5 7
4 5 8
and write a code that will create a new array that adds the third column together with the previous number if the numbers in the first and second columns equal the numbers in the row below it.
since the first two values in row 1 and 2, then add the third elements in row 1 and 2 together
so the output from the array above should look like this
1 3 10
2 8 6
4 5 15
The function accumarray(subs,val) accumulate elements of vector val using the subscripts subs. So we can use this function to sum the elements in the third column having the same value in the first and second column. We can use unique(...,'rows') to determine which pairs of value are unique.
%Example data
A = [1 3 5,
1 3 5,
2 3 6,
4 5 7,
4 5 7]
%Get the unique pair of value based on the first two column (uni) and the associated index.
[uni,~,sub] = unique(A(:,1:2),'rows');
%Create the result using accumarray.
R = [uni, accumarray(sub,A(:,3))]
If the orders matters the script would be a little bit more complex:
%Get the unique pair of value based on the first two column (uni) and the associated index.
[uni,~,sub] = unique(A(:,1:2),'rows');
%Locate the consecutive similar row with this small tricks
dsub = diff([0;sub])~=0;
%Create the adjusted index
subo = cumsum(dsub);
%Create the new array
R = [uni(sub(dsub),:), accumarray(subo,A(:,3))]
Or you can get an identical result with a for loop:
R = A(1,:)
for ii = 2:length(A)
if all(A(ii-1,1:2)==A(ii,1:2))
R(end,3) = R(end,3)+A(ii,3)
else
R(end+1,:) = A(ii,:)
end
end
Benchmark:
With an array A of size 100000x3 on the mathworks live editor:
The for loop take about 5.5s (no pre-allocation, so it's pretty slow)
The vectorized method take about 0.012s
Given a 2-column matrix, for instance. The input is:
[ 1,2;
3,4;
5,5]
The expected output is:
[1,2;
3,4;]
Does anyone know how do accomplish this? Many thanks for your time and attention.
You could use logical indexing:
A = [1 2;3 4;5 5];
match = A(:,1) == A(:,2); // 1 where row has the same elements in both columns
A(match,:) = []; // make the match columns empty
You would need to make this more generic for another case, but for two columns and your example this will work.
Your question suggests your matrix may have an arbitrary number of columns. In that case you may want to delete a row if it has (a) any two elements equal, or (b) all elements equal.
One possible approach is:
Apply sort along each row;
Use diff to compute differences between consecutive elements;
Generate a logical index with all to (a) keep rows for which all such differences are non-zero, or with any to (b) keep rows for which any such difference is non-zero:
So:
X = [1 2 3;
3 4 3;
5 5 5];
Y = X(all(diff(sort(X,2),[],2),2),:);
Z = X(any(diff(sort(X,2),[],2),2),:);
gives
Y =
1 2 3
Z =
1 2 3
3 4 3
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
I have a 3X3 cell array and each element store a (x,y) point.
The point are generate by random number from [0,1].
What I want to do is sort the cell array so that it looks like following
ex: 9 points
each circle is one 2D point
index:(1,1) in the left top corner and (3,3) to the right bottom corner as the usual array index
that is to ensure the topological order.
How do I do it?
Thank in advance.
for the example
pairs = [4 9 2 6 5 1 7 8 3; 9 6 2 1 3 8 7 4 5] (row 1 = x-values, row 2 = y-values))
what I want to do is put them in the cell array so that they can be connected by read lines like the image's topology.
The number of permutations is factorial(9), which is not terribly large. So a brute-froce approach is feasible: test all permutations for your desired conditions, and pick the first that is valid.
In the following I'm using a 2x3x3 array, instead of a 3x3 cell array containing length-2 vectors, because it's much easier that way.
N = 3;
data = rand(2,N,N);
permutations = perms(1:N^2); %// generate all permutations
for k = 1:numel(permutations)
dx = reshape(data(1,permutations(k,:)),N,N); %// permuted x data
dy = reshape(data(2,permutations(k,:)),N,N); %// permuted y data
if all(all(diff(dy,[],1)<0)) && all(all(diff(dx,[],2)>0)) %// solution found
disp(dx) %// display solution: x values
disp(dy) %// y values
break %// we only want one solution
end
end
Note that for some choices of data there may not be a solution.
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{:}