问题
I would like to create a d-dimensional tensor using d as an input and without the if statement as below:
if d == 2
B = zeros(r,r);
for i = 1:r
B(i,i) = 1;
end
elseif d == 3
B = zeros(r,r,r);
for i = 1:r
B(i,i,i) = 1;
end
end
Is there a more efficient way?
回答1:
You can use accumarray:
f = @(d,r)accumarray(repmat((1:r).',1 , d), 1);
> f(2,5)
=
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
Here is the basic signature of accumarray:
accumarray( subs , val )
Using accumarray we can create an n-dimensional array where subs represents the position of points that will be filled in the array and val represents the value of them.
If subs provided as a matrix , its number of columns determines the number of dimensions of the resultant array and each row represents position of each point.
For example for d = 2 and r = 5 we want to create a (5 x 5) array that has 1s in the following positions: [1 1],[2 2],[3 3],[4 4],[5 5].
Using repmat we can create subs:
subs = repmat ((1:5).' , 1, 2)
=
1 1
2 2
3 3
4 4
5 5
val is set to 1 so all specified positions will be filled by 1.
.
回答2:
This does what you want for arbitrary r and d:
B = zeros(repmat(r,1,d)); % initiallize as d×d×···×d array filled with 0
B(linspace(1,end,r)) = 1; % write 1 in the "diagonal". end can be replaced by r^d
Some notes on how it works:
- zeros can take a vector as input, which allows the number of dimensions to be specified dynamically.
- Linear indexing is being used to specify the positions containing
1. - end is being used within a function call. This is a not very well known feature.
Equivalently, you can first build the array in linear form and then reshape it:
B = reshape(mod(1:r^d, (r^d-1)/(r-1))==1, repmat(r,1,d));
Notes:
- The step between entries containing
1, in linear indexing, is(r^d-1)/(r-1). - reshape allows a vector as input to specify dimensions, similar to
zeros.
来源:https://stackoverflow.com/questions/48251186/create-a-d-dimensional-tensor-dynamically