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
Related
I have 4x4 matrix A
[1 2 3 4;
2 2 2 3;
5 5 5 5;
4 4 4 4]
I know how to locate all values less than 4. A<4. But I'm not sure how to write an 'if' statement for; three or more values, all which are less than 4, contained in the same row. For instance; see above A(1,:) and A(2,:) satisfies my conditions.
You can basically do A<4 to know which ones are smaller. If you want to know which rows contain N values smaller than 4 then you can do
rows=find(sum(A<4,2)>=3)
This basically does:
find smaller than 4
Count how many of them in each row (sum(_,2))
find if they are 3 or more
give the row index of those find()
I have a matrix of growing length for example a 4-by-x matrix A where x is increasing in a loop. I want to find the smallest column c where all columns before that, each, carry one single number. The matrix A can look like:
A = [1 2 3 4;
1 2 3 5;
1 2 3 1;
1 2 3 0];
where c=3, and x=4.
At each iteration of the loop where A grows in length, the value of index c grows as well. Therefore, at each iteration, I want to update the value of c. How efficiently can I code this in Matlab?
Let's say you had the matrix A and you wanted to check a particular column iito see if all its elements are the same. The code would be:
all(A(:, ii)==A(1, ii)) % checks if all elements in column are same as first element
Also, keep in mind that once the condition is broken, x cannot be updated anymore. Therefore, your code should be:
x = 0;
while true
%% expand A by one column
if ~all(A(:, x)==A(1, x)) % true if all elements in column are not the same as first element
break;
end
x = x+1;
end
You could use this:
c = find(arrayfun(#(ind)all(A(1, ind)==A(:, ind)), 1:x), 1, 'first');
This finds the first column where not all values are the same. If you run this in a loop, you can detect when entries in a column start to differ:
for x = 1:maxX
% grow A
c = find(arrayfun(#(ind)~all(A(1, ind)==A(:, ind)), 1:x), 1, 'first');
% If c is empty, all columns have values equal to first row.
% Otherwise, we have to subtract 1 to get the number of columns with equal values
if isempty(c)
c = x;
else
c = c - 1;
end
end
Let me give a try as well:
% Find the columns which's elements are same and sum the logical array up
c = sum(A(1,:) == power(prod(A,1), 1/size(A,1)))
d=size(A,2)
To find the last column such that each column up to that one consists of equal values:
c = find(any(diff(A,1,1),1),1)-1;
or
c = find(any(bsxfun(#ne, A, A(1,:)),1),1)-1;
For example:
>> A = [1 2 3 4 5 6;
1 2 3 5 5 7;
1 2 3 1 5 0;
1 2 3 0 5 8];
>> c = find(any(diff(A,1,1),1),1)-1
c =
3
You can try this (easy and fast):
Equal_test = A(1,:)==A(2,:)& A(2,:)==A(3,:)&A(3,:)==A(4,:);
c=find(Equal_test==false,1,'first')-1;
You can also check the result of find if you want.
I'm suppose to find 10 most repeated elements in a vector with n elements,
(the elements are from 1-100)
does anyone know how to do that?
I know how to find the one that is most repeated element in a vector but I don't know how to find 10 most repeated elements with n being unknown.
a = randi(10,1,100);
y = hist(a,1:max(a));
[~,ind] = sort(y,'descend');
out = ind(1:10);
for number of occurrences use y(ind(1:10)).
I had some doubts so I tested it many times, it seems to work.
You can use unique for that case. In my example, I have 4 numbers and I want to grep the 2 with the most occurances.
A = [1 1 3 3 1 1 2 2 1 1 1 2 3 3 3 4 4 4 4];
B = sort(A); % Required for the usage of unique below
[~,i1] = unique(B,'first');
[val,i2] = unique(B,'last');
[~,pos] = sort(i2-i1,'descend');
val(pos(1:2))
1 3
Replace val(pos(1:2)) by val(pos(1:10)) in your case to get the 10 most values. The get the number of elements you can use i1 and i2.
num = i2-i1+1;
num(1:2)
ans =
7 3
Since you already know how to find the most repeated element, you could use the following algorithm:
Find the most repeated element of the vector
Remove the most repeated element from the vector
Repeat the process on the new vector to find the 2nd most repeated element
Continue until you have the 10 most repeated elements
The code would look something like:
count = 0;
values = [];
while count < 10
r = Mode(Vector);
values = [values r]; % store most repeated values
Vector = Vector(find(Vector~=r));
count = count + 1;
end
Not efficient, but it'll get the job done
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 have quite big array. To make things simple lets simplify it to:
A = [1 1 1 1 2 2 3 3 3 3 4 4 5 5 5 5 5 5 5 5];
So, there is a group of 1's (4 elements), 2's (2 elements), 3's (4 elements), 4's (2 elements) and 5's (8 elements). Now, I want to keep only columns, which belong to group of 3 or more elements. So it will be like:
B = [1 1 1 1 3 3 3 3 5 5 5 5 5 5 5 5];
I was doing it using for loop, scanning separately 1's, 2's, 3's and so on, but its extremely slow with big arrays...
Thanks for any suggestions how to do it in more efficient way :)
Art.
A general approach
If your vector is not necessarily sorted, then you need to run to count the number of occurrences of each element in the vector. You have histc just for that:
elem = unique(A);
counts = histc(A, elem);
B = A;
B(ismember(A, elem(counts < 3))) = []
The last line picks the elements that have less than 3 occurrences and deletes them.
An approach for a grouped vector
If your vector is "semi-sorted", that is if similar elements in the vector are grouped together (as in your example), you can speed things up a little by doing the following:
start_idx = find(diff([0, A]))
counts = diff([start_idx, numel(A) + 1]);
B = A;
B(ismember(A, A(start_idx(counts < 3)))) = []
Again, note that the vector need not to be entirely sorted, just that similar elements are adjacent to each other.
Here is my two-liner
counts = accumarray(A', 1);
B = A(ismember(A, find(counts>=3)));
accumarray is used to count the individual members of A. find extracts the ones that meet your '3 or more elements' criterion. Finally, ismember tells you where they are in A. Note that A needs not be sorted. Of course, accumarray only works for integer values in A.
What you are describing is called run-length encoding.
There is software for this in Matlab on the FileExchange. Or you can do it directly as follows:
len = diff([ 0 find(A(1:end-1) ~= A(2:end)) length(A) ]);
val = A(logical([ A(1:end-1) ~= A(2:end) 1 ]));
Once you have your run-length encoding you can remove elements based on the length. i.e.
idx = (len>=3)
len = len(idx);
val = val(idx);
And then decode to get the array you want:
i = cumsum(len);
j = zeros(1, i(end));
j(i(1:end-1)+1) = 1;
j(1) = 1;
B = val(cumsum(j));
Here's another way to do it using matlab built-ins.
% Set up
A=[1 1 1 1 2 2 3 3 3 3 4 4 5 5 5 5 5];
threshold=2;
% Get the unique elements of the array
uniqueElements=unique(A);
% Count haw many times each unique element occurs
counts=histc(A,uniqueElements);
% Write which elements should be kept
toKeep=uniqueElements(counts>threshold);
% Make a logical index
indexer=false(size(A));
for i=1:length(toKeep)
% For every unique element we want to keep select the indices in A that
% are equal
indexer=indexer|(toKeep(i)==A);
end
% Apply index
B=A(indexer);