I'm new to Arm and Neon.
I wanna divide int32x4x2_t A into two int32x4_t B1, int32x4_t B2.
So, if A = [1 2 3 4; 5 6 7 8], I want to make B1 to [1 2 3 4] and B2 to [5 6 7 8]
I tried
B1 = vld1q_s32(A);
B2 = vld1q_s32(A+4);
But it does not work.
How can I fix it?
int32x4x2_t is really just a struct that looks like
typedef struct int32x4x2_t {
int32x4_t val[2];
} int32x4x2_t;
So all you have to do is
B1 = A.val[0];
B2 = A.val[1];
Related
I have two vectors, and I'm trying to find ALL coincidences of one on the other within a certain tolerance without using a for loop.
By tolerance I mean for example if I have the number 3, with tolerance 2, I will want to keep values within 3±2, so (1,2,3,4,5).
A = [5 3 4 2]; B = [2 4 4 4 6 8];
I want to obtain a cell array containing on each cell the numbers of all the coincidences with a tolerance of 1 (or more) units. (A = B +- 1)
I have a solution with zero units (A = B), which would look something like this:
tol = 0;
[tf, ia] = ismembertol(B,A,tol,'DataScale',1); % For tol = 0, this is equivalent to using ismember
idx = 1:numel(B);
ib = accumarray(nonzeros(ia), idx(tf), [], #(x){x}) % This gives the cell array
The output is:
ib =
[]
[]
[2 3 4]
[1]
Which is as desired.
If I change the tolerance to 1, the code doesn't work as intended. It outputs instead:
tol = 1
[tf, ia] = ismembertol(B,A,tol,'DataScale',1); % For tolerance = 1, this is equivalent to using ismember
idx = 1:numel(B);
ib = accumarray(nonzeros(ia), idx(tf), [], #(x){x}) % This gives the cell array
ib =
[5]
[2 3 4]
[]
[1]
When I would expect to obtain:
ib =
[2 3 4 5]
[1 2 3 4]
[2 3 4]
[1]
What am I doing wrong? Is there an alternative solution?
Your problem is that, in the current state of your code, ismembertol only outputs 1 index per element of B found in A, so you lose information in the cases where an element can be found several times within tolerance.
As per the documentation You can use the 'OutputAllIndices',true value pair argument syntax, to output what you want in ia with just a call to ismembertol:
A = [5 3 4 2]; B = [2 4 4 4 6 8];
tol = 0;
[tf, ia] = ismembertol(A,B,tol,'DataScale',1,'OutputAllIndices',true);
celldisp(ia) % tol = 0
ia{1} =
0
ia{2} =
0
ia{3} =
2
3
4
ia{4} =
1
celldisp(ia) % tol = 1
ia{1} =
2
3
4
5
ia{2} =
1
2
3
4
ia{3} =
2
3
4
ia{4} =
1
Here is a manual approach, just to provide another method. It computes an intermediate matrix of all absolute differences (using implicit expansion), and from the row and column indices of the entries that are less than the tolerance it builds the result:
A = [5 3 4 2];
B = [2 4 4 4 6 8];
tol = 1;
[ii, jj] = find(abs(A(:).'-B(:))<=tol);
ib = accumarray(jj, ii, [numel(A) 1], #(x){x});
Note that this approach
may be memory-intensive, because of the intermediate matrix;
can be made to work in old Matlab versions, because it doesn't use ismembertol; but then implicit expansion has to be replaced by explicitly calling bsxfun:
[ii, jj] = find(abs(bsxfun(#minus, A(:).', B(:)))<=tol);
ib = accumarray(jj, ii, [numel(A) 1], #(x){x});
I have a question that seems like it should have a simple answer that can avoid for loops.
Suppose I have an N x 4 array defined in MATLAB:
A = [1 2 3 4; 1 2 3 4; 1 2 3 4; 1 2 3 4; 1 2 3 4; 1 2 3 4];
In this example, N = 6, but it is arbitrary. I want to re-arrange A into a new array, B, which is 2 x 2 x N array of the form:
B(:,:,1) = [1 2; 3 4];
B(:,:,2) = [1 2; 3 4];
...
B(:,:,N) = [1 2; 3 4];
This seems like a simple problem and I tried a variety of things such as:
B = reshape(A',2,2,N);
However, this results in
B(:,:,1) = [1 3; 2 4];
B(:,:,2) = [1 3; 2 4];
...
B(:,:,N) = [1 3; 2 4];
I feel like there must be a simple way of doing this in one line using some combination of "reshape", "permute" and/or "transpose" that I am missing. Any suggestions are appreciated.
You are only missing a final permute. This is needed because Matlab is column-major, so it fills the new array down, then across:
B = permute(reshape(A.', 2,2,N), [2 1 3]);
I have vector c:
c = [2 5 3];
I want to generate vectors with their lengths equal to each value in c in a consecutive order. So, I should obtain 3 vectors:
c1 = [1 2];
c2 = [3 4 5 6 7];
c3 = [8 9 10];
Next, I want to align these vectors in a 1x3 cell array:
out = {c1 c2 c3};
This may seem straightforward, but I can't figure how to do it automatically. Any ideas?
You could use mat2cell to accomplish this. We first create an array from 1 to sum(c) and then use mat2cell to group the array into pieces where each piece is the size of each element of c.
out = mat2cell(1:sum(c), 1, c);
This reduces the need for intermediate variables and gives you your cell array directly.
out{1} =
1 2
out{2} =
3 4 5 6 7
out{3} =
8 9 10
I have two vectors a and b as an example:
a = [1 2 3 4];
b = [5 6 7 8];
I want to create strings from the indices of a and b:
c1 = a(1):b(1) = [1 2 3 4 5];
c2 = a(2):b(2) = [2 3 4 5 6];
c3 = a(3):b(3) = [3 4 5 6 7];
c4 = a(4):b(4) = [4 5 6 7 8];
Then I want to concatenate the obtained strings:
C = cat(2, c1, c2, c3, c4) = [1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8];
I would like a general solution to help me automatize this algorithm.
Solution
This should do the trick without using a loop:
>> a = [1 3 4 5];
>> b = [5 6 7 8];
>> resultstr = num2str(cell2mat(arrayfun(#(x,y) x:y,a,b,'UniformOutput',false)))
resultstr =
1 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8
Performance
I tried to do a quick comparison between this method, Luis Mendo's method and the for loop (see e.g. A. Visser's answer). I created two arrays with pseudo-random numbers from 1 to 50 and 501 to 1000 and timed the calculation for array sizes from 1 to 300, disregarding the string conversion.
Luis Mendo's anwer is the clear winner, in terms of time complexity arrayfun seems to be on par with bsxfun. The for loop fares much worse, except for very small array sizes, but I'm not sure the timing can be trusted there.
The code can be found here. I'd be very happy to get some feedback, I'm a bit unsure about those measurements.
This can be done without loops (be they for, arrayfun or cellfun) using bsxfun's masking capability. This works also if the "strings" have different lengths, as in the following example.
a = [1 2 3];
b = [3 5 6];
m = (0:max(b-a)).'; %'
C = bsxfun(#plus, a, m);
mask = bsxfun(#le, m, b-a);
C = C(mask).';
The result in this example is:
C =
1 2 3 2 3 4 5 3 4 5 6
Try something like:
A = [1:5; 5:8];, C = [];
for i = A, C = [C i(1):i(2)]; end
Cstr = num2str(C);
I would do this:
a = [1 3 4 5]; b = [5 6 7 8];
c =[];
for i =1:length(a)
c = [c [a(i):b(i)]];
end
num2str(c)
First of all, don't assign variables as C1, C2, C3 etc, use cells for that.
I'd use a for-loop for this, though there probably is a better alternative.
a = [1 3 4 5]; b = [5 6 7 8];
C = [];
for ii = 1:length(a)
tmp = a(ii):b(ii);
C = [C tmp];
end
This way tmp stores your separate arrays, then the following line adds it to the pre-existing array C.
Just for fun:
a = [1 2 3 4]; b = [5 6 7 8];
A=fliplr(gallery('circul',fliplr([a,b])));
B=A(1:numel(a)+1,1:numel(a));
C=num2str(B(:).')
I have a matrix like this:
A = 1 2 3
4 5 6
7 8 9
My question is how I want to make my matrix to be like this:
A11 = 1
A12 = 2
A13 = 3
A21 = 4
A22 = 5
A23 = 6
A31 = 7
A32 = 8
A33 = 9
Because i have to multiply A21 with A22 which is 4x5=20.
Your question is not clear to me.
To create the matrix, use ',' (or nothing) to delimit columns, ';' to delimit rows.
A = [1 2 3 ; 4 5 6 ; 7 8 9];
To access the matrix, you can use a 1-dimensional index as well as a 2-dimensional index.
E.g. A21 is A(2, 1) as well as A(0*3+2).
If you actually need variables such as 'A11', 'A12' etc. you could do as follows:
A = [1 2 3; 4 5 6; 7 8 9];
for i = 1:size(A,1)
for j = 1:size(A,2)
eval(sprintf('A%d%d = %f;',i,j,A(i,j)));
end
end
A21 * A22
# will result in 20
Maybe not the best way, but it will create the variables for you.