I am trying to create two vectors
first:
[ 10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10 ]
second:
[ 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 ]
I had almost no problems with the first one:
i = 1:10;
for t = 1: 2*length(i)-1
y1(t) = abs(length(x)-t)+1;
end
But there are some problems with the second...
Does anyone have any idea how I can create it using the same for loop?Thanks in advance
If you want to do it with a loop:
N = 10;
for t = 1:2*N-1
y2(t) = -abs(t-N)+N;
end
But its probably easier to use the following, first make an array 1:N, and then concatenate the array N-1:-1:1:
y2 = [1:N, N-1:-1:1]
Related
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])]);
By using ndgrid, we can obtain the matrices representing the grid:
[Y, X, Z]=ndgrid(1:2,3:4,5:6)
Y(:,:,1) =
1 1
2 2
Y(:,:,2) =
1 1
2 2
X(:,:,1) =
3 4
3 4
X(:,:,2) =
3 4
3 4
Z(:,:,1) =
5 5
5 5
Z(:,:,2) =
6 6
6 6
However, there are actually 8 grid "points"
(3,1,5), (3,1,6), (3,2,5), (3,2,6), (4,1,5), (4,1,6), (4,2,5), (4,2,6)
How can I create a matrix of these 8 vectors (using ndgrid or not in the process)? That is,
3 1 5
3 1 6
3 2 5
3 2 6
4 1 5
4 1 6
4 2 5
4 2 6
I've seen this related question, but it uses meshgrid, which only works for two dimensions.
Easy. Just linearize the output from ndgrid:
[Y, X, Z]=ndgrid(1:2,3:4,5:6);
out = [X(:) Y(:) Z(:)]
If you want the same ordering as in your question, use sortrows:
out = sortrows([X(:) Y(:) Z(:)])
You just need to straighten these 3D vectors:
>> vertices = [X(:),Y(:),Z(:)]
vertices =
3 1 5
3 2 5
4 1 5
4 2 5
3 1 6
3 2 6
4 1 6
4 2 6
I have a matrix
q = [1 2 3 4 5 6;
7 8 9 10 11 12];
and I want to create an array d such that
d(:,:,1) = 1 2
7 8
d(:,:,2) = 3 4
9 10
d(:,:,3) = 5 6
11 12
I know how to do that using loops but i prefer not to use loops.
With reshape
out = reshape(q,size(q,1),2,[])
I have a variable like this that is all one row:
1 2 3 4 5 6 7 8 9 2 4 5 6 5
I want to write a for loop that will find where a number is less than the previous one and put the rest of the numbers in a new row, like this
1 2 3 4 5 6 7 8 9
2 4 5 6
5
I have tried this:
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
m = zeros(size(test));
for i=1:numel(test)-1;
for rows=1:size(m,1)
if test(i) > test(i+1);
m(i+1, rows+1) = test(i+1:end)
end % for rows
end % for
But it's clearly not right and just hangs.
Let x be your data vector. What you want can be done quite simply as follows:
ind = [find(diff(x)<0) numel(x)]; %// find ends of increasing subsequences
ind(2:end) = diff(ind); %// compute lengths of those subsequences
y = mat2cell(x, 1, ind); %// split data vector according to those lenghts
This produces the desired result in cell array y. A cell array is used so that each "row" can have a different number of columns.
Example:
x = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
gives
y{1} =
1 2 3 4 5 6 7 8 9
y{2} =
2 4 5 6
y{3} =
5
If you are looking for a numeric array output, you would need to fill the "gaps" with something and filling with zeros seem like a good option as you seem to be doing in your code as well.
So, here's a bsxfun based approach to achieve the same -
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5] %// Input
idx = [find(diff(test)<0) numel(test)] %// positions of row shifts
lens = [idx(1) diff(idx)] %// lengths of each row in the proposed output
m = zeros(max(lens),numel(lens)) %// setup output matrix
m(bsxfun(#le,[1:max(lens)]',lens)) = test; %//'# put values from input array
m = m.' %//'# Output that is a transposed version after putting the values
Output -
m =
1 2 3 4 5 6 7 8 9
2 4 5 6 0 0 0 0 0
5 0 0 0 0 0 0 0 0
Hi may I know how i can perform this using matlab?
I tried nchoosek but only works on 1 type of combination. I would like to output all together in array
Let S={a,b,c,d,e}
I would like to get combinations as such which start from 3 combinations:
the 3-combinations : {a,b,c} , {a,b,d} , {a,b,e} , {a,c,d} , {a,c,e} , {a,d,e}
the 4-combinations : {a,b,c,d} , {a,b,c,e} , {a,c,d,e}
the 5-combinations : {a,b,c,d,e}
So the output would be like this:
{a,b,c} {a,b,d} {a,b,e} {a,c,d} {a,c,e} {a,d,e}{a,b,c,d} {a,b,c,e} {a,c,d,e}{a,b,c,d,e}
Thanks
You can use a loop there or arrayfun which is just a compact way to express such a loopy approach and not a vectorized approach -
combs = arrayfun(#(x) nchoosek(S,x),3:numel(S),'Uniform',0)
The output would be a cell array with each cell representing values for each combination. So, when you run the code, you would get -
>> combs{1}
ans =
2 7 4
2 7 1
2 7 9
2 4 1
2 4 9
2 1 9
7 4 1
7 4 9
7 1 9
4 1 9
which would be your 3-combinations set.
>> combs{2}
ans =
2 7 4 1
2 7 4 9
2 7 1 9
2 4 1 9
7 4 1 9
would be your 4-combinations set and so on.