Find n minimum values in an array?

Question

ME 2015-4-15

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评论： Chris Volpe 2023-2-1

I have an array, I need to be able to select 2, or 4 or so on 'n' minimum (smallest) values from the specific array? I know i can use 'min' function but this only gives one smallest value. I do not want to sort the array in order and pick values. is there another way ?????

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Margarida Costa 2017-1-22

And is there an easy way to get back the indexes of those values?

Nicholas Ayres 2020-8-21

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The "optional second output" of sort is the index order.

[sortedVals,indexes] = sort(myvals);

So

myVals = [1 0.1 7 10 4];
[sortedVals,indexes] = sort(myVals);

gives

sortedVals = [0.1 1 4 7 10]
indexes = [2 1 5 3 4]

Figured I'd pop this here in case anyone else comes along with the same question. :)

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Answer 1

Titus Edelhofer 2015-4-15

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Hi,

if you don't want to sort, and n is not too large, you could remove the index, something like

x = rand(1000, 1);
n = 5;
val = zeros(n,1);
for i=1:n
  [val(i),idx] = min(x);
  % remove for the next iteration the last smallest value:
  x(idx) = [];
end

Titus

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Rik 2019-7-8

I know this is an old thread, but still:

Even replacing x(idx) with NaN instead of removing the entry doesn't really improve the speed. This might be dependent on the release and of the size of the input array.

zaid tahir 2021-3-5

@Titus Edelhoferthanks for the code. Helped make a very complex problem, very simple.

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Answer 2

Jan 2017-1-23

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While sorting is the best approach to get the data, you can easily obtain an unsorted result:

n = 2;
[xs, index] = sort(x);
result      = x(sort(index(1:n)))

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Chris Volpe 2023-2-1

Seems to me you want to leave out the "sort" call in line 3:

result = x(index(1:n))

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Answer 3

Salam Ismaeel 2018-8-30

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use: mink() command for R2018a

https://www.mathworks.com/help/matlab/ref/mink.html

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Bill Tubbs 2020-6-22

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This is the correct answer. Thanks!

Example:

>> X = [4 1 2 1 5 4 2];
>> [minValues, idx] = mink(X,3)
minValues =
     1     1     2
idx =
     2     4     3

Martin Vankát 2020-10-25

Thank you! This one helped.

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Answer 4

Titus Edelhofer 2015-4-15

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编辑：Titus Edelhofer 2015-4-15

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Hi,

if it's not floating point but integers (so no problem with roundoff),you could do the following:

x = [1 3 2 4 1 3 5 4 1 3 1]
% find the first 2 min occurrences:
idxMin = find(x==min(x), 2, 'first')

This tells that the smallest value 1 is found at positions 1 and 5. Or are you looking for the n smallest values in the sense, you are looking for the smallest, the second smallest etc.? In this case the answer is simply

n = 2;
xs = sort(x);
% pick the n smallest:
xs(1:n)

Titus

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Titus Edelhofer 2015-4-15

If this was what you were looking for, then please mark the question as answered ... Thanks, Titus

ME 2015-4-15

not entirely, because i don't want to sort the array.

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Answer 5

Luis Gomez 2022-8-30

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编辑：Luis Gomez 2022-8-30

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I do it like this,

% k miminum values of a matrix (return the position, not the value)
% not optimized: it just worked for my application.
a = magic(5)
a = 5×5
    17    24     1     8    15
    23     5     7    14    16
     4     6    13    20    22
    10    12    19    21     3
    11    18    25     2     9
%Copy the original data
aa = a;
%Size of the matrix
[rows columns] = size(a);
%Select the k mininum values
k = 3;
k_mins = zeros(k,1); %to store solution (not really needed)
for i = 1: k
    [m, I] = min(a(:));
    a(I) = inf;
    k_mins(i) = I;
end
k_mins
k_mins = 3×1
    11
    20
    24
% to get the [i, j] usual matrix locations for ecah value within k_mins
% for instance, to get the "i,j" indices por first element of k_mins
% vector,
[i j] = ind2sub([rows, columns],k_mins(1))
i = 1
j = 3
%the value
aa(i,j)
ans = 1

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Rik 2022-8-30

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I'm of the opinion you shouldn't wait for performance to become an issue before you look for optimizations. Good should not be the enemy of better. If you make them a habit, you will have a lot less work once performance does become important. One important part is why you serialize inside the loop. For large k, you will do this operation many times, and it is really not necessary.

It is also a good idea to look at what mlint is suggesting. It is suggesting you put a comma between m and I. I see no reason why you wouldn't.

I also don't see where you are actually returning the lowest values. They aren't in k_mins (despite its name), because that only stores the indices. The actual minimum values require an extra step, which is made impossible due the overwriting, as you can see below.

So, putting your code in a function with the two adjustments:

% k miminum values of a matrix
a = magic(5)
a = 5×5
    17    24     1     8    15
    23     5     7    14    16
     4     6    13    20    22
    10    12    19    21     3
    11    18    25     2     9
%Select the k mininum values
k = 3;
find_k_lowest_values(a,k)
ans = 3×1
   Inf
   Inf
   Inf
% show time spent in microseconds on this example
1e6*timeit(@() find_k_lowest_values(a,k))
ans = 14.8830
1e6*timeit(@() find_k_lowest_values__alternative(a,k))
ans = 6.3830
function k_lowest_values=find_k_lowest_values(a,k)
% Write single line purpose here
%
% Explain the function and the available syntax options.
% This help text should ensure you never *have* to look at the code itself.
% A comment should explain what is stored in k_mins
k_mins = zeros(k,1);
for i = 1: k
    % A comment should explain why we serialize inside the loop
    [~,I] = min(a(:));
    % A comment should explain why we set this value to inf
    a(I) = inf;
    k_mins(i) = I;
end
% A comment should explain why we do this
k_lowest_values=a(k_mins);
end

If you insist on a loop, I would suggest this:

function k_lowest_values=find_k_lowest_values__alternative(a,k)
% Write single line purpose here
%
% Explain the function and the available syntax options.
% This help text should ensure you never *have* to look at the code itself.
% We only need the data inside a, the shape is not important. By converting
% to a vector we can use min() inside the loop.
a=a(:);
% Pre-allocate the output.
k_lowest_values = zeros(k,1);
for n = 1:k
    % Store the lowest value in the output and get the index.
    [k_lowest_values(n),ind] = min(a);
    % Replace the position of the lowest value with inf to ignore it on the
    % next iteration.
    a(ind) = inf;
end
end

Walter Roberson 2022-8-30

Why not mink()?

Rik 2022-8-30

@Walter, to be honest I always forget that function exists ever since it was introduced in R2017b.

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