I have an array A, I want to arrange each row in descending order to get a new array B. How could I do this ?
E.g.
Array A (original array):
11 9 13 10
12 4 1 6
13 5 12 11
Array B (rearranged array):
13 11 10 9
12 6 4 1
13 12 11 5
>> A=[11 9 13 10;12 4 1 6;13 5 12 11]
A =
11 9 13 10
12 4 1 6
13 5 12 11
>> sort(A,2,'descend')
ans =
13 11 10 9
12 6 4 1
13 12 11 5
For details type: help sort at Matlab command window
Related
I would like to explore a better way to apply binary function which iterate via each element of the two argument. Let make the question simpler by using below function as an example:
func:{x+y}
a:til 10
q)a
0 1 2 3 4 5 6 7 8 9
b:a
q)b:a
q)b
0 1 2 3 4 5 6 7 8 9
What I what to get is the cross production such that each element of the argument will cross each other and apply the function. My expected result is
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 7 8 9 10 11 12 13 14 15 16 8 9 10 11 12 13 14 15 16 17 9 10 11 12 13 14 15 16 17 18
My current solution is crossing the the list of argument first:
(func/) each (a cross b)
I wonder is there a better way to doing that? simply using func'[a;b] will just get a pairwise result which not what I want.
The following should be what you are looking for:
a +/:\: b
The same can apply for other defined functions too, for example:
a {x mod y}/:\: b
You do not need cross for this just each-right or each-left. Because '+' is a vector function you can just iterate over one list and use other list as full vector.
q) a+/:b
I need to find the minimum values in each column of matrix "A", and then replace those min values with the values in last row of matrix "B" (which has same number of columns). Like I have these:
>> A = randi(10,10,5)
A =
3 5 9 5 8
7 6 4 10 2
8 4 1 7 4
4 7 2 8 2
7 5 8 7 5
3 7 10 10 1
5 7 8 5 7
8 3 8 2 3
6 10 2 1 10
3 7 6 7 2
>> B = randi(100,3,5)
B =
10 34 66 18 62
99 95 49 54 81
52 1 52 9 95
>> [M,I] = min(A)
M =
3 3 1 1 1
I =
1 8 3 9 6
And I want to replace the values of "M" with "B(end,:), so that:
A(1,1) = B(end,1);
A(8,2) = B(end,2);
A(3,3) = B(end,3);
A(9,4) = B(end,4);
A(6,5) = B(end,5);
I try "A(I) = B(end,:)" and "A(I(1,:)) = B(end,:)" but they do not work! Any ideas how I could do that? My real matrices are huge (1200x100000) so no way to do it by hand!
try this to replace the first min value:
A = [ 3 5 9 5 8;
7 6 4 10 2;
8 4 1 7 4;
4 7 2 8 2;
7 5 8 7 5;
3 7 10 10 1;
5 7 8 5 7;
8 3 8 2 3;
6 10 2 1 10;
3 7 6 7 2];
B =[ 10 34 66 18 62;
99 95 49 54 81;
52 1 52 9 95];
[M,I] = min(A)
A(sub2ind(size(A),I,1:size(A,2)))=B(end,:)
the output will be:
A =
52 5 9 5 8
7 6 4 10 2
8 4 52 7 4
4 7 2 8 2
7 5 8 7 5
3 7 10 10 95
5 7 8 5 7
8 1 8 2 3
6 10 2 9 10
3 7 6 7 2
However, when you have to replace all of the min values, use the code below instead
A = [ 3 5 9 5 8;
7 6 4 10 2;
8 4 1 7 4;
4 7 2 8 2;
7 5 8 7 5;
3 7 10 10 1;
5 7 8 5 7;
8 3 8 2 3;
6 10 2 1 10;
3 7 6 7 2];
B =[ 10 34 66 18 62;
99 95 49 54 81;
52 1 52 9 95];
M = min(A);
for i=1:size(A,2)
A(find(A(:,i) == M(i)),i)=B(end,i);
end;
A
the output is:
A =
52 5 9 5 8
7 6 4 10 2
8 4 52 7 4
4 7 2 8 2
7 5 8 7 5
52 7 10 10 95
5 7 8 5 7
8 1 8 2 3
6 10 2 9 10
52 7 6 7 2
You can acces you matrix by a single index, which looks like this:
Indeces =
1 6 11 16
2 7 12 17
3 8 13 18
4 9 14 19
5 10 15 20
Since you get the indeces for each individual column, you just need to increase it by the column number times the height of the matrix.
This should yield the correct result:
A( I + (0 : size(A,2)-1) * size(A,1) ) = B(end,:)
Is there a MATLAB function to generate this matrix?:
[1 2 3 4 5 6 7 ... n;
2 3 4 5 6 7 8 ... n+1;
3 4 5 6 7 8 9 ... n+2;
...;
n n+1 n+2 ... 2*n-1];
Is there a name for it?
Thanks.
Yes indeed there's a name for that matrix. It's known as the Hankel matrix.
Use the hankel function in MATLAB:
out = hankel(1:n,n:2*n-1);
Example with n=10:
out =
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 11
3 4 5 6 7 8 9 10 11 12
4 5 6 7 8 9 10 11 12 13
5 6 7 8 9 10 11 12 13 14
6 7 8 9 10 11 12 13 14 15
7 8 9 10 11 12 13 14 15 16
8 9 10 11 12 13 14 15 16 17
9 10 11 12 13 14 15 16 17 18
10 11 12 13 14 15 16 17 18 19
Alternatively, you may be inclined to want a bsxfun based approach. That is certainly possible:
out = bsxfun(#plus, (1:n), (0:n-1).');
The reason why I wanted to show you this approach is because in your answer, you used repmat to generate the two matrices to add together to create the right result. You can replace the two repmat calls with bsxfun as it does the replication under the hood.
The above solution is for older MATLAB versions that did not have implicit broadcasting. For recent versions of MATLAB, you can simply do the above by:
out = (1:n) + (0:n-1).';
My standard approach is
repmat(1:n,n,1)+repmat((1:n)',1,n)-1
I need a function that splits a vector in smaller frames with an overlap, like buffer, but instead of column-wise, it should be done row-wise.
This is how buffer works:
x = 1:20
x = buffer(x, 10, 5);
x = 0 1 6 11
0 2 7 12
0 3 8 13
0 4 9 14
0 5 10 15
1 6 11 16
2 7 12 17
3 8 13 18
4 9 14 19
5 10 15 20
What I want would be this though:
x = 0 0 1 2
1 2 3 4
3 4 5 6
5 6 7 8
7 8 9 10
9 10 11 12
11 12 13 14
13 14 15 16
15 16 17 18
17 18 19 20
Is there any function or way to achieve that? Maybe combination of buffer + some rearranging?
First figure out the answer in columns, then transpose the resulting matrix:
buffer(x, 4, 2).'
We have the following case:
Q = [idxcell{:,1}];
Sort = sort(Q,'descend')
Sort =
Columns 1 through 13
23 23 22 22 20 19 18 18 18 18 17 17 17
Columns 14 through 26
15 15 14 14 13 13 13 12 12 12 11 10 9
Columns 27 through 39
9 9 8 8 8 8 8 7 7 7 7 7 7
Columns 40 through 52
7 6 6 6 5 4 4 3 3 3 3 2 2
Columns 53 through 64
2 2 2 2 2 2 2 1 1 1 1 1
How can we sort matrix Sort according to how many times its values are repeated?
Awaiting result should be:
repeatedSort = 2(9) 7(7) 1(5) 8(5) 3(4) 18(4) 6(3) 9(3) 12(3) 13(3) 17(3) 4(2) 14(2) 15(2) 22(2) 23(2) 5(1) 10(1) 11(1) 19(1) 20(1)
or
repeatedSort = 2 7 1 8 3 18 6 9 12 13 17 4 14 15 22 23 5 10 11 19 20
Thank you in advance.
You can use the TABULATE function from the Statistics Toolbox, then call SORTROWS to sort by the frequency.
Example:
x = randi(10, [20 1]); %# random values
t = tabulate(x); %# unique values and counts
t = t(find(t(:,2)),1:2); %# get rid of entries with zero count
t = sortrows(t, -2) %# sort according to frequency
the result, where first column are the unique values, second is their count:
t =
2 4 %# value 2 appeared four times
5 4 %# etc...
1 3
8 3
7 2
9 2
4 1
6 1
Here's one way of doing it:
d = randi(10,1,30); %Some fake data
n = histc(d,1:10);
[y,ii] = sort(n,'descend');
disp(ii) % ii is now sorted according to frequency