If I had a matrix A such as:
63 55 85 21 71
80 65 85 48 53
55 60 93 71 66
21 65 40 33 21
61 90 80 48 50
... and so on how would I find the minimum values of each column and remove those numbers from the matrix completely, meaning essentially I would have one less row overall.
I though about using:
[C,I] = min(A);
A(I) = [];
but that wouldn't remove the necessary numbers, and also reshape would not work either. I would like for this to work with an arbitrary number of rows and columns.
A = [
63 55 85 21 71
80 65 85 48 53
55 60 93 71 66
21 65 40 33 21
61 90 80 48 50
];
B = zeros( size(A,1)-1, size(A,2));
for i=1:size(A,2)
x = A(:,i);
maxIndex = find(x==min(x(:)),1,'first');
x(maxIndex) = [];
B(:,i) = x;
end
disp(B);
Another vectorized solution:
M = mat2cell(A,5,ones(1,size(A,2)));
z = cellfun(#RemoveMin,M);
B = cell2mat(z);
disp(B);
function x = RemoveMin(x)
minIndex = find(x==min(x(:)),1,'first');
x(minIndex) = [];
x = {x};
end
Another solution:
[~,I] = min(A);
indexes = sub2ind(size(A),I,1:size(A,2));
B = A;
B(indexes) = [];
out = reshape(B,size(A)-[1 0]);
disp(out);
Personally I prefer the first because:
For loops aren't evil - many times they are actually faster (By using JIT optimizer)
The algorithm is clearer to the developer who reads your code.
But of course, its up to you.
Your original approach works if you convert the row indices resulting from min into linear indices:
[m, n] = size(A);
[~, row] = min(A,[],1);
A(row + (0:n-1)*m) = [];
A = reshape(A, m-1, n);
Related
How to improve the distance calculation on the 2 separated datasets?
This is the code:
X = [ 3.6 79
1.8 54
3.333 74
2.283 62
4.533 85
2.883 55
4.7 88
3.6 85
1.95 51
4.35 85
1.833 54
3.917 84
4.2 78
1.75 47
4.7 83
2.167 52
1.75 62
4.8 84
1.6 52
4.25 79
1.8 51
1.75 47
3.45 78
3.067 69
4.533 74
3.6 83
1.967 55
4.083 76
3.85 78
4.433 79
4.3 73
4.467 77
3.367 66
4.033 80
3.833 74
2.017 52
1.867 48
4.833 80
1.833 59
4.783 90 ]
clc;
close all;
figure;
h(1) = plot(X(:,1),X(:,2),'bx');
hold on;
X1 = X(1:3,:);
X2 = X(4:40,:);
h(2) = plot(X1(1:3,1), X1(1:3,2),'rs','MarkerSize',10);
k=5;
[D2 ind] = sort(squeeze(sqrt(sum(bsxfun(#minus,X2,permute(X1,[3 2 1])).^2,2))))
ind_closest = ind(1:k,:)
x_closest = X(ind_closest,:)
for j = 1:length(x_closest);
h(3) =plot(x_closest(j,1),x_closest(j,2),'ko','MarkerSize',10);
end
The output is shown as in the picture below:
The problem is, the code does not pick the closest data points of red squared data points. I also tried to use pdist2 function from statistical toolbox,the result yields similar with the bsxfun function that i applied in my code.
I'm not sure which part in the code need to improve so that i can pick the data points that closest to the target.
Really appreciate if anyone can help me to improve my code
If the closest point means closest to X, line 19 & line 20 should be replaced as
[D2 ind] = sort(squeeze(sqrt(sum(bsxfun(#minus,X,permute(X1,[3 2 1])).^2,2))))
ind_closest = ind(2:k+1,:)
If the closest point means closest to X2, then try this:
x_closest = X2(ind_closest,:)
In the meanwhile, I modified your code a little bit, since your h(3) could be optimized.
clc; clear; close all;
%load fisheriris
%X=meas(:,3:4);
load X
X=unique(X,'rows');
figure;
h(1) = plot(X(:,1),X(:,2),'bx');
hold on;
X1 = X([5 15 30],:);
h(2) = plot(X1(:,1), X1(:,2),'rs','MarkerSize',10);
[D2,ind] = sort(squeeze(sqrt(sum(bsxfun(#minus,X,permute(X1,[3 2 1])).^2,2))));
k=3;
ind_closest = unique(ind(2:k+1,:));
x_closest = X(ind_closest,:);
h(3) =plot(x_closest(:,1),x_closest(:,2),'ko','MarkerSize',10);
axis equal
It seems to be working fine.
I would like to remove the diagonal of the following matrix;
[0 1 1
0 0 0
0 1 0]
and put this in a vector as such
[1 1 0 0 0 1]
Is there a one-way function to do this?
Most other solutions I found on Stack Overflow delete all zeros.
If two lines are fine...
x = x.'; %'// transpose because you want to work along 2nd dimension first
result = x(~eye(size(x))).'; %'// index with logical mask to remove diagonal
Here's an almost one-liner -
[m,n] = size(x);
x(setdiff(reshape(reshape(1:numel(x),m,n).',1,[]),1:m+1:numel(x),'stable'))
And I will put up my fav bsxfun here -
xt = x.'; %//'
[m,n] = size(x);
out = xt(bsxfun(#ne,(1:n)',1:m)).'
Sample run -
>> x
x =
52 62 37 88
23 68 98 91
49 40 4 79
>> [m,n] = size(x);
>> x(setdiff(reshape(reshape(1:numel(x),m,n).',1,[]),1:m+1:numel(x),'stable'))
ans =
62 37 88 23 98 91 49 40 79
>> xt = x.';
>> xt(bsxfun(#ne,(1:n)',1:m)).'
ans =
62 37 88 23 98 91 49 40 79
Say I have the following columns vector Z
1 53 55 57 60 64 68 70 71 72 74 76 77 78 79 80 255
I want to use it to create a matrix such that each row would contain all the number between (and including) 2 adjacent elements in Z
So the output matrix should be something like this:
1 2 3 .... 53
53 54 55
55 56 57
57 58 60
....
80 81 ... 255
I've been searching for something similar but couldn't find it.
Thanks
See if this works for you -
lens = diff(Z)+1;
mask1 = bsxfun(#le,[1:max(lens)]',lens); %//'
array1 = zeros(size(mask1));
array1(mask1) = sort([1:255 Z(2:end-1)]);
out = array1.'; %//'# out is the desired output
Try this to break the monotony of bsxfun :) :
d = diff(Z);
N = max(d)+1;
R = zeros(length(Z)-1,N);
for i = 1:length(Z)-1
R(i,1:1+d(i)) = Z(i):Z(i+1);
end
EDIT:
I know that the general consensus is that one always should try to avoid loops in Matlab, but is this valid for this example? I know that this is a broad question, so lets focus on this particular problem and compare bsxfun to JIT loop. Comparing the two proposed solutions:
the code used for testing:
Z = [1 53 55 57 60 64 68 70 71 72 74 76 77 78 79 80 255];
%[1 3 4, 6];
nn = round(logspace(1,4,10));
tm1_nn = zeros(length(nn),1);
tm2_nn = zeros(length(nn),1);
for o = 1:length(nn)
tm1 = zeros(nn(o),1);
tm2 = zeros(nn(o),1);
% approach1
for k = 1:nn(o)+1
tic
d = diff(Z);
N = max(d)+1;
R = zeros(length(Z)-1,N);
for i = 1:length(Z)-1
R(i,1:1+d(i)) = Z(i):Z(i+1);
end
tm1(k) = toc;
end
%approach 2
for k = 1:nn(o)+1
tic
lens = diff(Z)+1;
mask1 = bsxfun(#le,[1:max(lens)]',lens); %//'
array1 = zeros(size(mask1));
array1(mask1) = sort([1:255 Z(2:end-1)]);
out = array1.';
tm2(k) = toc;
end
tm1_nn(o) = mean(tm1);%sum(tm1);%mean(tm1);%
tm2_nn(o) = mean(tm2);%sum(tm2);%mean(tm2);%
end
semilogx(nn,tm1_nn, '-ro', nn,tm2_nn, '-bo')
legend('JIT loop', 'bsxfun')
xlabel('log_1_0(Number of runs)')
%ylabel('Sum execution time')
ylabel('Mean execution time')
grid on
I encountered other tasks previously where the loop was faster. (or I mess up the comparison?)
I have a matrix like:
A=
10 31 32 22
32 35 52 77
68 42 84 32
I need a function like mode but with range, for example mymode(A,10) that return 30, find most frequent number in range 0-10, 10-20, 20-30, .... and return most number in range.
You can use histc to bin your data into the ranges of your desire and then find the bin with the most members using max on the output of histc
ranges = 0:10:50; % your desired ranges
[n, bins] = histc(A(:), ranges); % bin the data
[v,i] = max(n); % find the bin with most occurrences
[ranges(i) ranges(i+1)] % edges of the most frequent bin
For your specific example this returns
ans =
30 40
which matches with your required output, as the most values in A lay between 30 and 40.
[M,F] = mode( A((A>=2) & (A<=5)) ) %//only interested in range 2 to 5
...where M will give you the mode and F will give you frequency of occurence
> A = [10 31 32 22; 32 35 52 77; 68 42 84 32]
A =
10 31 32 22
32 35 52 77
68 42 84 32
> min = 10
min = 10
> max = 40
max = 40
> mode(A(A >= min & A <= max))
ans = 32
>
I guess by the number of different answers that we may be missing your goal. Here is my interpretation.
If you want to have many ranges and you want to output most frequent number for every range, create a cell containing all desired ranges (they could overlap) and use cellfun to run mode() for every range. You can also create a cell with desired ranges using arrayfun in a similar manner:
A = [10 31 32 22; 32 35 52 77; 68 42 84 32];
% create ranges
range_step = 10;
range_start=[0:range_step:40];
range=arrayfun(#(r)([r r+range_step]), range_start, 'UniformOutput', false)
% analyze ranges
o = cellfun(#(r)(mode(A(A>=r(1) & A<=r(2)))), range, 'UniformOutput', false)
o =
[10] [10] [22] [32] [42]
Assume we have the following data:
H_T = [36 66 21 65 52 67 73; 31 23 19 33 36 39 42]
P = [40 38 39 40 35 32 37]
Using MATLAB 7.0, I want to create three new matrices that have the following properties:
The matrix H (the first part in matrix H_T) will be divided to 3 intervals:
Matrix 1: the 1st interval contains the H values between 20 to 40
Matrix 2: the 2nd interval contains the H values between 40 to 60
Matrix 3: the 3rd interval contains the H values between 60 to 80
The important thing is that the corresponding T and P will also be included in their new matrices meaning that H will control the new matrices depending on the specifications defined above.
So, the resultant matrices will be:
H_T_1 = [36 21; 31 19]
P_1 = [40 39]
H_T_2 = [52; 36]
P_2 = [35]
H_T_3 = [66 65 67 73; 23 33 39 42]
P_3 = [38 40 32 37]
Actually, this is a simple example and it is easy by looking to create the new matrices depending on the specifications, BUT in my values I have thousands of numbers which makes it very difficult to do that.
Here's a quick solution
[~,bins] = histc(H_T(1,:), [20 40 60 80]);
outHT = cell(3,1);
outP = cell(3,1);
for i=1:3
idx = (bins == i);
outHT{i} = H_T(:,idx);
outP{i} = P(idx);
end
then you access the matrices as:
>> outHT{3}
ans =
66 65 67 73
23 33 39 42
>> outP{3}
ans =
38 40 32 37