I have the following matrix:
[ 2 5 7 8 1 3 4 6 5 7 3 1;
1 1 1 1 2 2 2 2 3 3 3 3;]
The first row represents values and the second characteristic
I want to get the max value if the value in the second row is the same, i.e. their characteristic is the same. So, what I would like to have is:
[ 8 6 7], since 8 is the highest value when the second row is 1, 6 when the second row is is 2, and 7 when the second row is 3. I can do it with a loop, but I would like vectorized solution, and if possible of course, in one line.
accumarray does exactly what you want
x=[ 2 5 7 8 1 3 4 6 5 7 3 1; 1 1 1 1 2 2 2 2 3 3 3 3;]
accumarray(x(2,:)',x(1,:)',[],#max)
Related
I'm looking for a quick way in MATLAB to do the following:
Given a permutation matrix of a vector, say [1, 2, 3], I would like to remove all duplicate reverse rows.
So the matrix P = perms([1, 2, 3])
3 2 1
3 1 2
2 3 1
2 1 3
1 3 2
1 2 3
becomes
3 2 1
3 1 2
2 3 1
You can noticed that, symetrically, the first element of each rows have to be bigger than the last one:
n = 4; %row size
x = perms(1:n) %all perms
p = x(x(:,1)>x(:,n),:) %non symetrical perms
Or you can noticed that the number of rows contained by the p matrix follows this OEIS sequence for each n and correspond to size(x,1)/2 so since perms output the permutation in reverse lexicographic order:
n = 4; %row size
x = perms(1:n) %all perms
p = x(1:size(x,1)/2,:) %non symetrical perms
You can use MATLAB's fliplr method to flip your array left to right, and then use ismember to find rows of P in the flipped version. At last, iterate all locations and select already found rows.
Here's some code (tested with Octave 5.2.0 and MATLAB Online):
a = [1, 2, 3];
P = perms(a)
% Where can row x be found in the left right flipped version of row x?
[~, Locb] = ismember(P, fliplr(P), 'rows');
% Set up logical vector to store indices to take from P.
n = length(Locb);
idx = true(n, 1);
% Iterate all locations and set already found row to false.
for I = 1:n
if (idx(I))
idx(Locb(I)) = false;
end
end
% Generate result matrix.
P_star = P(idx, :)
Your example:
P =
3 2 1
3 1 2
2 3 1
2 1 3
1 3 2
1 2 3
P_star =
3 2 1
3 1 2
2 3 1
Added 4 to the example:
P =
4 3 2 1
4 3 1 2
4 2 3 1
4 2 1 3
4 1 3 2
4 1 2 3
3 4 2 1
3 4 1 2
3 2 4 1
3 2 1 4
3 1 4 2
3 1 2 4
2 4 3 1
2 4 1 3
2 3 4 1
2 3 1 4
2 1 4 3
2 1 3 4
1 4 3 2
1 4 2 3
1 3 4 2
1 3 2 4
1 2 4 3
1 2 3 4
P_star =
4 3 2 1
4 3 1 2
4 2 3 1
4 2 1 3
4 1 3 2
4 1 2 3
3 4 2 1
3 4 1 2
3 2 4 1
3 1 4 2
2 4 3 1
2 3 4 1
As demanded in your question (at least from my understanding), rows are taken from top to bottom.
Here's another approach:
result = P(all(~triu(~pdist2(P,P(:,end:-1:1)))),:);
pdist computes the distance between rows of P and rows of P(:,end:-1:1).
~ negates the result, so that true corresponds to coincident pairs.
triu keeps only the upper triangular part of the matrix, so that only one of the two rows of the coincident pair will be removed.
~ negates back, so that true corresponds to non-coincident pairs.
all gives a row vector with true for rows that should be kept (because they do not coincide with any previous row).
This is used as a logical index to select rows of P.
In each operation we can either push an element to the end of the array or at the beginning of it
for example an array 3 2 5 1 4 6 would take 4 steps.
after first operation 2 3 5 1 4 6
after second operation 1 2 3 5 4 6
after third operation 2 3 1 4 6 5
after fourth operation 1 2 3 4 5 6
I think in the best case, the array is already sorted - 0 operations needed.
In the worst case, its sorted already, but in the opposite order (eg 6 5 4 3 2 1), you gonna need number of elements-1 operations.
I have a matrix like following,
A =
1 2 3
4 5 6
7 8 9
10 11 12
4 5 6
7 8 9
4 5 6
1 2 3
I could extract unique rows in this matrix using command A_unique = unique(A,'rows') and result as follows
A_unique =
1 2 3
4 5 6
7 8 9
10 11 12
I need to find number of times each rows exists in the main matrix A
Some thing like following
A_unique_count =
2
3
2
1
How can I find count of unique rows? Can anyone help? Thanks in Advance
Manu
The third output of unique gives you the index of the unique row in the original array. You can use this with accumarray to count the number of occurrences.
For example:
A = [1 2 3
4 5 6
7 8 9
10 11 12
4 5 6
7 8 9
4 5 6
1 2 3];
[uniquerow, ~, rowidx] = unique(A, 'rows');
noccurrences = accumarray(rowidx, 1)
Returns:
noccurrences =
2
3
2
1
As expected
I would recommend #excaza's approach. But just for variety:
A_unique_count = diff([0; find([any(diff(sortrows(A), [], 1), 2); 1])]);
If an array contains N number of elements (elements can be repeated) and the goal is to make all the elements equal by a +1 on an element and a -1 on another element in each iteration, how can we determine whether it's possible or not to normalize the array? What will be the optimal algorithm to solve the problem?
Ex.
For the array 1 2 3, if I apply +1 on 1 and -1 on 3, the array becomes 2 2 2. That means it's possible in 1 iteration.
For the array 1 2 1, it's not possible to make all the elements equal.
First, since you're not disturbing the sum by each iteration, since you're increasing one number and decreasing another, the optimal target value is going to be the average.
If this average is a whole number, you should be able to achieve it with the iterations, however if the average is a fractional number then you will not be able to achieve it.
The number of steps is going to be the sum of the distances between each number and the target, divided by 2.
Every iteration pick one number above target and one below and apply the operations to them.
PS! As per commented, if all you want is answers to the following two questions:
Can it be done
What will the value be
Then the answers are:
Yes, provided the average number is a whole number
The value repeated in the whole array is the average number
Anyway, if you want the actual operations getting from the input to the target values, here's a longer example:
1 2 3 4 5 6 7 = 28, 28/7 = 4 (optimal target)
+ -
2 2 3 4 5 6 6
+ -
3 2 3 4 5 6 5
+ -
4 2 3 4 5 6 4
+ -
4 3 3 4 5 5 4
+ -
4 4 3 4 5 4 4
+ -
4 4 4 4 4 4 4
6 steps, let's total the distances from the first number:
1 2 3 4 5 6 7
3 2 1 0 1 2 3 = 12, divided by 2 = 6
Here's the example from the comments on the question:
1 9 10 12 3 7 = 42 / 6 = 7 (optimal target)
Distances:
1 9 10 12 3 7
6 2 3 5 4 0 = 20, divided by 2 = 10 (steps)
1 9 10 12 3 7
+ - step 1
2 8 10 12 3 7
+ - step 2
3 7 10 12 3 7
+ - step 3
4 7 9 12 3 7
+ - step 4
5 7 8 12 3 7
+ - step 5
6 7 7 12 3 7
+ - step 6
7 7 7 11 3 7
- + step 7
7 7 7 10 4 7
- + step 8
7 7 7 9 5 7
- + step 9
7 7 7 8 6 7
- + step 10
7 7 7 7 7 7
Here is a more pseudo-code like algorithm description:
Calculate SUM of all the elements
COUNT all the elements
If AVERAGE (SUM/COUNT) is not whole number, solution is not possible to achieve
STEPS = SUM(ABS(numberN - AVERAGE))/2
Each iteration, pick one number below AVERAGE and one above
Apply + operation to number below and - operation to number above
Repeat steps 5 and 6 until target achieved
I have got a question regarding all the combinations of matrix-rows in Matlab.
I currently have a matrix with the following structure:
1 2
1 3
1 4
2 3
2 4
3 4
Now I want to get all the possible combinations of these "pairs" without using a number twice in the same row:
1 2 3 4
1 3 2 4
1 4 2 3
And it must be possible to make it with n-"doublecolumns". Which means, when my pair-matrix goes for example until "5 6", i want to create the matrix with 3 of these doublecolumns:
1 2 3 4 5 6
1 2 3 5 4 6
1 2 3 6 4 5
1 3 2 4 5 6
1 3 2 5 4 6
....
I hope you understand what I mean :)
Any ideas how to solve this?
Thanks and best regard
Jonas
M = [1 2
1 3
1 4
2 3
2 4
3 4]; %// example data
n = floor(max(M(:))/2); %// size of tuples. Compute this way, or set manually
p = nchoosek(1:size(M,1), n).'; %'// generate all n-tuples of row indices
R = reshape(M(p,:).', n*size(M,2), []).'; %// generate result...
R = R(all(diff(sort(R.'))),:); %'//...removing combinations with repeated values