I am running a factor analysis via psych package.
cormat <- polychoric(na.omit(mself1))
##Warning message:
In cor.smooth(mat) : Matrix was not positive definite, smoothing was done
fa(cormat,nfactors=3,n.obs = 183 ,n.iter=1, rotate="oblimin",fm="ml")
## Error in if (dim(r)[1] == dim(r)[2]) { : argument is of length zero
I just wondered why the argument is of length zero which means the file is empty, when I type cormat the Polychoric matrix came out.
How could I fix this problem, or it has anything with the "not positive definite"?
Thanks!
The psych package stores the actual matrix in $rho in your cormat object. In your case, fa(cormat$rho) will fix that problem. Bill also includes the fa.poly() function, which will compute the polychoric correlations from a raw data matrix before running a factor analysis on that.
Be a bit cautious with your results though. Your resulting matrix was non-Gamian (i.e., had a negative eigenvalue), which can happen when they're estimated pairwise or you have large sampling errors (see Garrido, Abad, & Ponsosa, 2013).
Related
The error description is as follows:
Error using full request 68813x68813 (35.3GB) array to exceed the preset maximum array size Creating an array larger than this limit can take a long time and result in no response from MATLAB for more information, see Array Size Limits or Default Items panel.
Error schurmat_sblk (line 35)
if issparse(schur); schur = full(schur); end;
The function file schurmat_sblk is a file in cvx\sdpt3\Solver,
How can I do to avoid this error?
My cvx codes are as follows:
The value you may need are: n=8; d=2^n;m=(d^2)*0.05;the size of Pauli is m*d^2, it's a sparse matrix. The size of y is m*1;
function [rhoE] = test_compressed_cc(n,~,m,Pauli,y)
d = 2^n;
cvx_begin sdp quiet
% how to define the variable ?
variable rhoE(d,d) hermitian;
rhoE == hermitian_semidefinite(d);
% ||x||_tr=tr(sqrt(x^\daggerx))=tr(sqrt(x^2))=Tr(x)
minimize(trace(rhoE));
subject to
(d/m)*(Pauli * vec(rhoE)) == y;
rhoE >= 0;
cvx_end
end
On the other hand, maybe CVX can't solve the 8 qubit case, does anyone know how SVT should be used to solve this convex program.
Paper link: https://arxiv.org/pdf/0909.3304.pdf
Welcome any comment : )
I am trying to convert .m script to C++ using MATLAB Coder.
function P=r_p(1,var1,var3)
p=[[3,7]
[10,15]
[6,19]
[21,19]
[43,11]
[969,2]
[113,9]
[43,59]
[21,15]
[6,15]
[10,18]
[3,15]];
tmax=sum(p(:,1))+41;
coder.varsize('x');
x=ones(9,11).*[0:10:100]; % getting error in this line: [9x11]~=[1x11]. Since size of x is varying in for loop, so i should tell coder that it is variable size, So I used Varsize
for t=11:tmax
a1=(rand-0.5)*1;
a9=(rand-0.5)*1.25;
a2=(rand-0.5)*1.5;
a8=(rand-0.5)*1.75;
a3=(rand-0.5)*2.0;
a7=(rand-0.5)*2.25;
a4=(rand-0.5)*2.5;
a6=(rand-0.5)*2.75;
a5=(rand-0.5)*3;
x(1,t+1)=x(1,t)+a1;
if x(1,t+1)<(100-var1) || x(1,t+1)>(100+var1) % loop 1: x(1,11)+a1 value is is writing to x(1,12) So coder gives error "Index exceeds array dimensions. Index value 12 exceeds valid range [1-11] of array x".
x(1,t+1)=x(1,t); % In matlab it works fine, but coder throws error.
end
end
My question is Let say loop 1,
x(1,12)= x(1,11)+a1 In matlab this assignment works fine, but when converting it is throwing error " Index exceeds array dimensions. Index value 12 exceeds valid range [1-11] of array x" As I declared x as variable size coder should assign x(1,11)+a1 value to x(1,12) but it is not doing, instead throwing error. Why?
Since t is looping for 1289, if I specify bounds for x like
coder.varsize('x',[1290,1290],[0,0]) then Coder gives error in other part of the code i.e dimensions doesn't match. Ofcourse it should because dimension of x doesn't match with [ones(12,9)p(1,2)/9;(P_1s+var3/100P_1s.*randn(size(P_1s))/2)/9;zeros(30,9)].
is declaring x as variable size is correct step or not? if yes then what should be the work around for "index exceeds array dimensions error"
Please Let me know, what am I missing to convert it to C++ code
MATLAB Coder doesn't support 2 things you're using here: implicit expansion and growing arrays by assigning past the end of a dimension.
For implicit expansion, you can use:
x=bsxfun(#times,ones(9,11),[0:10:100]);
Assigning past the end of an array in MATLAB will grow the array. That's an error in Coder. There are 2 ways to overcome this:
Allocate your array to have the right number of elements up front
Use concatenation to grow an array: x = [x, newColumn]
In this example, you know tmax so I'd suggest just changing the allocation of x to have the right number of columns up front:
% Current initial value
x=bsxfun(#times,ones(9,11),[0:10:100]);
% Extra columns - please check my upper bound value
x=[x, zeros(9,tmax)];
I have a simple task of expanding the string FX according to the following rules:
X -> X+YF+
Y-> -FX-Y
In OpenCL, string manipulation is not supported but the use of an array of characters is. How would a kernel program that expands this string in parallel look like in openCL?
More details:
Consider the expansion of 'FX' in the python code below.
axiom = "FX"
def expand(s):
switch = {
"X": "X+YF+",
"Y": "-FX-Y",
}
return switch.get(s, s)
def expand_once(string):
return [expand(c) for c in string]
def expand_n(s, n):
for i in range(n):
s = ''.join(expand_once(s))
return s
expanded = expand_n(axiom, 200)
The result expanded will be a result of expanding the axiom 'FX' 200 times. This is a rather slow process thus the need to do it on openCL for parallelization.
This process results in an array of strings which I will then use to draw a dragon curve.
below is an example of how I would come up with such a dragon curve: This part is not of much importance. The expansion on OpenCL is the crucial part.
import turtles
from PIL import Image
turtles.setposition(5000, 5000)
turtles.left(90) # Go up to start.
for c in expanded:
if c == "F":
turtles.forward(10)
elif c == "-":
turtles.left(90)
elif c == "+":
turtles.right(90)
# Write out the image.
im = Image.fromarray(turtles.canvas)
im.save("dragon_curve.jpg")
Recursive algorithms like this don't especially lend themselves to GPU acceleration, especially as the data set changes its size on each iteration.
If you do really need to do this iteratively, the challenge is for each work-item to know where in the output string to place its result. One way to do this would be to assign work groups a specific substring of the input, and on every iteration, keep count of the total number of Xs and Ys in each workgroup-sized substring of the output. From this you can calculate how much that substring will expand in one iteration, and if you accumulate those values, you'll know the offset of the output of each substring expansion. Whether this is efficient is another question. :-)
However, your algorithm is actually fairly predictable: you can calculate precisely how large the final string will be given the initial string and number of iterations. The best way to generate this string with OpenCL would be to come up with a non-recursive function which analytically calculates the character at position N given M iterations, and then call that function once per work-item, with the (known!) final length of the string as the work size. I don't know if it's possible to come up with such a function, but it seems like it might be, and if it is possible, this is probably the most efficient way to do it on a GPU.
It seems like this might be possible: as far as I can tell, the result will be highly periodic:
FX
FX+YF+
FX+YF++-FX-YF+
FX+YF++-FX-YF++-FX+YF+--FX-YF+
FX+YF++-FX-YF++-FX+YF+--FX-YF++-FX+YF++-FX-YF+--FX+YF+--FX-YF+
^^^^^^ ^^^^^^^ ^^^^^^^ ^^^^^^^ ^^^^^^^ ^^^^^^^ ^^^^^^^ ^^^^^^^
A* B A B A B A B
As far as I can see, those A blocks are all identical, and so are the Bs. (apart from the first A which is effectively at position -1) You can therefore determine the characters at 14 positions out of every 16 completely deterministically. I strongly suspect it's possible to work out the pattern of +s and -s that connects them too. If you figure that out, the solution becomes pretty easy.
Note though that when you have that function, you probably don't even need to put the result in a giant string: you can just feed your drawing algorithm with that function directly.
This is an interview question. Say you have an array of four ints named A, and also this function:
int check(int x, int y){
if (x<=y) return 1;
return 0;
}
Now, you want to create a function that will sort A, and you can use only the function checkfor comparisons. How many calls for check do you need?
(It is ok to return a new array for result).
I found that I can do this in 5 calls. Is it possible to do it with less calls (on worst case)?
This is how I thought of doing it (pseudo code):
int[4] B=new int[4];
/*
The idea: put minimum values in even cells and maximum values in odd cells using check.
Then swap (if needed) between minimum values and also between maximum values.
And finally, swap the second element (max of minimums)
and the third element (min of maximums) if needed.
*/
if (check(A[0],A[1])==1){ //A[0]<=A[1]
B[0]=A[0];
B[2]=A[1];
}
else{
B[0]=A[1];
B[2]=A[0];
}
if (check(A[2],A[3])==1){ //A[2]<=A[3]
B[1]=A[2];
B[3]=A[3];
}
else{
B[1]=A[3];
B[3]=A[2];
}
if (check(B[0],B[1])==0){ //B[0]>B[1]
swap(B[0],B[1]);
}
if (check(B[2],B[3])==0){ //B[2]>B[3]
swap(B[2],B[3]);
}
if (check(B[1],B[2])==0){ // B[1]>B[2]
swap(B[1],B[2]);
}
There are 24 possible orderings of a 4-element list. (4 factorial) If you do only 4 comparisons, then you can get only 4 bits of information, which is enough to distinguish between 16 different cases, which isn't enough to cover all the possible output cases. Therefore, 5 comparisons is the optimal worst case.
In The Art of Computer Programming, p. 183 (Section 3.5.1), Donald Knuth has the following table of lower and upper bounds on the minimum numbers of comparisons:
The ceil(ln n!) is the "information theoretic" lower bound, whereas B(n) is the maximum number of comparisons in an insertion binary sort. Since the lower and upper bounds are equal for n=4, 5 comparisons are needed.
The information theoretic bound is derived by recognizing that there are n! possible orderings of n unique items. We distinguish these cases by asking S yes-no questions in the form of is X<Y?. These questions form a tree which has at most 2^S tips. We need n!<=2^S; solving for S gives ceil(lg(n!)).
Incidentally, you can use Stirling's approximation to show that this implies that sorting requires O(n log n) time.
The rest of the section goes on to describe a number of approaches to creating these bounds and studying this question, though work is on-going (see, for instance Peczarski (2011)).
I am familiar with iterative methods on paper, but MATLAB coding is relatively new to me and I cannot seem to find a way to code this.
In code language...
This is essentially what I have:
A = { [1;1] [2;1] [3;1] ... [33;1]
[1;2] [2;2] [3;2] ... [33;2]
... ... ... ... ....
[1;29] [2;29] [3;29] ... [33;29] }
... a 29x33 cell array of 2x1 column vectors, which I got from:
[X,Y] = meshgrid([1:33],[1:29])
A = squeeze(num2cell(permute(cat(3,X,Y),[3,1,2]),1))
[ Thanks to members of stackOverflow who helped me do this ]
I have a function that calls each of these column vectors and returns a single value. I want to institute a 2-D 5-point stencil method that evaluates a column vector and its 4 neighbors and finds the maximum value attained through the function out of those 5 column vectors.
i.e. if I was starting from the middle, the points evaluated would be :
1.
A{15,17}(1)
A{15,17}(2)
2.
A{14,17}(1)
A{14,17}(2)
3.
A{15,16}(1)
A{15,16}(2)
4.
A{16,17}(1)
A{16,17}(2)
5.
A{15,18}(1)
A{15,18}(2)
Out of these 5 points, the method would choose the one with the largest returned value from the function, move to that point, and rerun the method. This would continue on until a global maximum is reached. It's basically an iterative optimization method (albeit a primitive one). Note: I don't have access to the optimization toolbox.
Thanks a lot guys.
EDIT: sorry I didn't read the iterative part of your Q properly. Maybe someone else wants to use this as a template for a real answer, I'm too busy to do so now.
One solution using for loops (there might be a more elegant one):
overallmax=0;
for v=2:size(A,1)-1
for w=2:size(A,2)-1
% temp is the horizontal part of the "plus" stencil
temp=A((v-1):(v+1),w);
tmpmax=max(cat(1,temp{:}));
temp2=A(v,(w-1):(w+1));
% temp2 is the vertical part of the "plus" stencil
tmpmax2=max(cat(1,temp2{:}));
mxmx=max(tmpmax,tmpmax2);
if mxmx>overallmax
overallmax=mxmx;
end
end
end
But if you're just looking for max value, this is equivalent to:
maxoverall=max(cat(1,A{:}));