I want to define a recursive function that sorts an input vector and uses a sequence of secondary vectors to break any ties (or randomises them if it runs out of tiebreak vectors)
Given some input vector I and some tiebreaker matrix T, the pseudocode for the algorithm is as follows:
check if T is empty, if so, we reached stopping condition, therefore randomise input
get order of indices for sorted I, using matlab's standard sort function
find indices of duplicate values
for each duplicate value,
call function recursively on T(:,1) with rows corresponding to the indices of that duplicate value, with T(:,2:end)(with appropriate rows) as the new tiebreaker matrix - if empty then this call will just return random indices
fix the order of the sorted indices in the original sorted I
return the sorted I and corresponding indices
Here is what I have so far:
function [vals,idxs] = tiebreak_sort(input, ties)
% if the tiebreak matrix is empty, then return random
if isempty(ties)
idxs = randperm(size(input,1));
vals = input(idxs);
return
end
% sort the input
[vals,idxs] = sort(input);
% check for duplicates
[~,unique_idx] = unique(vals);
dup_idx = setdiff(1:size(vals,1),unique_idx);
% iterate over each duplicate index
for i = 1:numel(dup_idx)
% resolve tiebreak for duplicates
[~,d_order] = tiebreak_sort(ties(input==input(i),1),...
ties(input==input(i),2:end));
% fix the order of sorted indices (THIS IS WHERE I AM STUCK)
idxs(vals==input(i)) = ...?
end
return
I want to find a way to map the output of the recursive call, to the indices in idxs, to fix their order based on the (possibly recursive) tie breaks, but my brain is getting twisted in knots thinking about it..
Can I just use the fact that Matlabs sort function is stable and preserves the original order, and do it like this?
% find indices of duplicate values
dups = find(input==input(i));
% fix the order of sorted indices
idxs(vals==input(i)) = dups(d_order);
Or will that not work? is there another way of doing what I am trying to do, in general?
Just to give a concrete example, this would be a sample input:
I = [1 2 2 1 2 2]'
T = [4 1 ;
3 7 ;
3 4 ;
2 2 ;
1 8 ;
5 3 ]
and the output would be:
vals = [1 1 2 2 2 2]'
idxs = [4 1 5 3 2 6]'
Here, there are clearly duplicates in the input, so the function is called recursively on the first column of the tiebreaker matrix, which was able to fix the 1s but it needed a second recursive call on the 3s of the first column to break those ties.
No need to define a function, sortrows does that:
[S idxs] = sortrows([I T]);
vals = S(:,1);
Related
I have a vector of numbers (temperatures), and I am using the MATLAB function mink to extract the 5 smallest numbers from the vector to form a new variable. However, the numbers extracted using mink are automatically ordered from lowest to largest (of those 5 numbers). Ideally, I would like to retain the sequence of the numbers as they are arranged in the original vector. I hope my problem is easy to understand. I appreciate any advice.
The function mink that you use was introduced in MATLAB 2017b. It has (as Andras Deak mentioned) two output arguments:
[B,I] = mink(A,k);
The second output argument are the indices, such that B == A(I).
To obtain the set B but sorted as they appear in A, simply sort the vector of indices I:
B = A(sort(I));
For example:
>> A = [5,7,3,1,9,4,6];
>> [~,I] = mink(A,3);
>> A(sort(I))
ans =
3 1 4
For older versions of MATLAB, it is possible to reproduce mink using sort:
function [B,I] = mink(A,k)
[B,I] = sort(A);
B = B(1:k);
I = I(1:k);
Note that, in the above, you don't need the B output, your ordered_mink can be written as follows
function B = ordered_mink(A,k)
[~,I] = sort(A);
B = A(sort(I(1:k)));
Note: This solution assumes A is a vector. For matrix A, see Andras' answer, which he wrote up at the same time as this one.
First you'll need the corresponding indices for the extracted values from mink using its two-output form:
[vals, inds] = mink(array);
Then you only need to order the items in val according to increasing indices in inds. There are multiple ways to do this, but they all revolve around sorting inds and using the corresponding order on vals. The simplest way is to put these vectors into a matrix and sort the rows:
sorted_rows = sortrows([inds, vals]); % sort on indices
and then just extract the corresponding column
reordered_vals = sorted_rows(:,2); % items now ordered as they appear in "array"
A less straightforward possibility for doing the sorting after the above call to mink is to take the sorting order of inds and use its inverse to reverse-sort vals:
reverse_inds = inds; % just allocation, really
reverse_inds(inds) = 1:numel(inds); % contruct reverse permutation
reordered_vals = vals(reverse_inds); % should be the same as previously
I have a m-by-1 matrix, A. I want to find out which elements are duplicates and get their row values. (Just the row values because the matrix is m-by-1.)
I've tried
k = find(~unique(A));
but k contains the wrong values.
Here's an example of what I'm trying to do. Consider the array A;
A = [4
5
5
5
7
8
4];
Since 4 and 5 are the repeated elements here, I would like to get the row values of these elements and put them in a new array. The resulting array would be
RowValues= [1
2
3
4
7];
Note: The above is just an example and the actual array I am dealing with contains rational numbers of the type -0.0038, 1.3438 and so on in the array A.
Here is a solution using intersect:
s = sort(A);
c = intersect(s(1:2:end),s(2:2:end));
RowValues = find(ismember(A,c));
I have compared this method with the method proposed by #SardarUsama with a large [1 x 10000000] input in Octave. Here is the result:
=======INTERSECT==========
Elapsed time is 1.94979 seconds.
=======ACCUMARRAY==========
Elapsed time is 2.5205 seconds.
Find the count of each element of A using unique and accumarray, filter out the non-repeating values, use ismember to get the logical indices of repeating values and then use find to convert them to linear indices.
[C, ~, ic] = unique(A);
RowValues = find(ismember(A, C(accumarray(ic,1)>1)));
Why you get the wrong indices with your code?
Applying logical NOT on the vector of unique values would convert them to a logical vector containing true at the index where unique value is zero and false where it is non-zero and hence finding the non-zero (false in this case) elements of such a vector would lead nowhere.
I need to find all possible combinations of numbers 1:8 such that sum of all elements is equal to 8
The combinations need to be arranged in an ascending order.
Eg
1 7
2 2 4
1 3 5
1 2 2 3
1 1 1 1 1 1 1 1
A number can repeat itself. But a combination must not..
i.e 1 2 2 3 and 2 1 2 3
I need the the solution in ascending order So there will be only one possibility of every combination
I tried a few codes online suggested on Find vector elements that sum up to specific number in MATLAB
VEC = [1:8];
NUM = 8;
n = length(VEC);
finans = zeros(2^n-1,NUM);
for i = 1:(2^n - 1)
ndx = dec2bin(i,n) == '1';
if sum(VEC(ndx)) == NUM
l = length(VEC(ndx));
VEC(ndx)
end
end
but they dont include the possibilities where the numbers repeat.
I found a better approach through recursion and it's more elegant (I like elegant) and faster than my previous attempt (0.00399705213 seconds on my computer).
EDIT: You will need my custom function stretchmat.m that stretches a vector to fit the size of another matrix. Kinda like repmat but stretching the first parameter (see help for details). Very useful!
script.m
% Define funciton to prepend a cell x with a variable i
cellprepend = #(x,i) {[i x]};
% Execute and time function
tic;
a = allcomb(cellprepend,1,8); % Solution in a
toc;
allcomb.m
function a = allcomb( cellprepend, m, n )
% Add entire block as a combination
a{1} = n;
% Exit recursion if block size 1
if n == 1
return;
end
% Recurse cutting blocks at different segments
for i = m:n/2
b = allcomb(cellprepend,i,n-i);
a = [a cellfun( cellprepend, b, num2cell( stretchmat( i, b ) ) )];
end
end
So the idea is simple, for solutions that add to 8 is exhaustive. If you look for only valid answers, you can do a depth first search by breaking up the problem into 2 blocks. This can be written recursively as I did above and is kinda similar to Merge Sort. The allcomb call takes the block size (n) and finds all the ways of breaking it up into smaller pieces.
We want non-zero pieces so we loop it from 1:n-1. It then prepends the first block to all the combinations of the second block. By only doing all comb on one of the blocks, we can ensure that all solutions are unique.
As for the sorting, I'm not quite sure what you mean by ascending. From what I see, you appear to be sorting from the last number in ascending order. Can you confirm? Any sort can be appended to the end of script.m.
EDIT 2/3 Notes
For the permutatively unique case, the code can be found here
Thanks to #Simon for helping me QA the code multiple times
EDIT: Look at my second more efficient answer!
The Naive approach! Where the cartprod.m function can be found here.
% Create all permutations
p(1:8) = {0:8};
M = fliplr( cartprod( p{:} ) );
% Check sums
r = sum( M, 2 ) == 8;
M = M(sum( M, 2 ) == 8,:); % Solution here
There are definitely more efficient solutions than this but if you just need a quick and dirty solution for small permutations, this will work. Please note that this made Matlab take 3.5 GB of RAM to temporarily store the permutations.
First save all combinations with repetitions in a cell array. In order to do that, just use nmultichoosek.
v = 1 : 8;
combs = cell(length(v),0);
for i = v
combs{i} = nmultichoosek(v,i);
end
In this way, each element of combs contains a matrix where each row is a combination. For instance, the i-th row of combs{4} is a combination of four numbers.
Now you need to check the sum. In order to do that to all the combinations, use cellfun
sums = cellfun(#(x)sum(x,2),combs,'UniformOutput',false);
sums contains the vectors with the sum of all combinations. For
instance, sums{4} has the sum of the number in combination combs{4}.
The next step is check for the fixed sum.
fixed_sum = 10;
indices = cellfun(#(x)x==fixed_sum,sums,'UniformOutput',false);
indices contains arrays of logical values, telling if the combination satisfies the fixed sum. For instance, indices{4}(1) tells you if the first combination with 4 numbers sums to fixed_sum.
Finally, retrieve all valid combinations in a new cell array, sorting them at the same time.
valid_combs = cell(length(v),0);
for i = v
idx = indices{i};
c = combs{i};
valid_combs{i} = sortrows(c(idx,:));
end
valid_combs is a cell similar to combs, but with only combinations that sum up to your desired value, and sorted by the number of numbers used: valid_combs{1} has all valid combinations with 1 number, valid_combs{2} with 2 numbers, and so on. Also, thanks to sortrows, combinations with the same amount of numbers are also sorted. For instance, if fixed_sum = 10 then valid_combs{8} is
1 1 1 1 1 1 1 3
1 1 1 1 1 1 2 2
This code is quite efficient, on my very old laptop I am able to run it in 0.016947 seconds.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
I have two vectors with the same elements but their order is not same. For eg
A
10
9
8
B
8
9
10
I want to find the mapping between the two
B2A
3
2
1
How can I do this in matlab efficiently?
I think the Matlab sort is efficient. So:
[~,I]=sort(A); %sort A; we want the indices, not the values
[~,J]=sort(B); %same with B
%I(1) and J(1) both point to the smallest value, and a similar statement is true
%for other pairs, even with repeated values.
%Now, find the index vector that sorts I
[~,K]=sort(I);
%if K(1) is k, then A(k) is the kth smallest entry in A, and the kth smallest
%entry in B is J(k)
%so B2A(1)=J(k)=J(K(1)), where BSA is the desired permutation vector
% A similar statement holds for the other entries
%so finally
B2A=J(K);
if the above were in script "findB2A" the following should be a check for it
N=1e4;
M=100;
A=floor(M*rand(1,N));
[~,I]=sort(rand(1,N));
B=A(I);
findB2A;
all(A==B(B2A))
There are a couple of ways of doing this. The most efficient in terms of lines of code is probably using ismember(). The return values are [Lia,Locb] = ismember(A,B), where Locb are the indices in B which correspond to the elements of A. You can do [~, B2A] = ismember(A, B) to get the result you want. If your version of MATLAB does not allow ~, supply a throwaway argument for the first output.
You must ensure that there is a 1-to-1 mapping to get meaningful results, otherwise the index will always point to the first matching element.
Here a solution :
arrayfun(#(x)find(x == B), A)
I tried with bigger arrays :
A = [ 7 5 2 9 1];
B = [ 1 9 7 5 2];
It gives the following result :
ans =
3 4 5 2 1
Edit
Because arrayfun is usually slower than the equivalent loop, here a solution with a loop:
T = length(A);
B2A = zeros(1, length(A));
for tt = 1:T
B2A(1, tt) = find(A(tt) == B);
end
I would go for Joe Serrano's answer using three chained sort's.
Another approach is to test all combinations for equality with bsxfun:
[~, B2A] = max(bsxfun(#eq, B(:), A(:).'));
This gives B2A such that B(B2A) equals A. If you want it the other way around (not clear from your example), simply reverse A and B within bsxfun.