I am trying to export XY data objects from sets of the size of 20-40k elements, but Abaqus is slowing down considerably, and even crashing. In fact, when I create the xy data, Abaqus gives me a warning saying that "the number of xyDataObjects is very large, and might cause performance issues". And so it does.
My usual procedure to is to create the xy data and then export in rpt format. Can someone suggest another method less prone to crashing? Would it be more efficient to divide the output element set into two or more subsets, and concatenate them after exporting?
The method recommended by #hgazibara in the comments is certainly sufficient, but it is laborious.
An easier method, I found, is a package called Abaqus2Matlab, which scrapes any variable you want from the odb. See here: http://www.abaqus2matlab.com/
I'm working with openCV and I'm a newbie in this field. I'm researching about Camshift. I want to extend this method by using multiple histograms. It means when tracking an object has many than one apperance (ex: rubik cube with six apperance), if we use only one histogram, Camshift will most likely fail.
I know calcHist function in openCV (http://docs.opencv.org/modules/imgproc/doc/histograms.html#calchist) has a parameter is "accumulate", but I don't know how to use and when to use (apply for camshiftdemo.cpp in opencv samples folder). This function can help me solve this problem? Or I have to use difference solution?
I have an idea, that is: create an array histogram for object, for every appearance condition that strongly varies in color, we pre-compute and store all to this array. But when we compute new histogram? It means that the pre-condition to start compute new histogram is what?
And what happend if I have to track multiple object has same color?
Everybody please help me. Thank you so much!
I want to get the properly rendered projection result from a Stage3D framework that presents something of a 'gray box' interface via its API. It is gray rather than black because I can see this critical snippet of source code:
matrix3D.copyFrom (renderable.getRenderSceneTransform (camera));
matrix3D.append (viewProjection);
The projection rendering technique that perfectly suits my needs comes from a helpful tutorial that works directly with AGAL rather than any particular framework. Its comparable rendering logic snippet looks like this:
cube.mat.copyToMatrix3D (drawMatrix);
drawMatrix.prepend (worldToClip);
So, I believe the correct, general summary of what is going on here is that both pieces of code are setting up the proper combined matrix to be sent to the Vertex Shader where that matrix will be a parameter to the m44 AGAL operation. The general description is that the combined matrix will take us from Object Local Space through Camera View Space to Screen or Clipping Space.
My problem can be summarized as arising from my ignorance of proper matrix operations. I believe my failed attempt to merge the two environments arises precisely because the semantics of prepending one matrix to another is not, and is never intended to be, equivalent to appending that matrix to the other. My request, then, can be summarized in this way. Because I have no control over the calling sequence that the framework will issue, e.g., I must live with an append operation, I can only try to fix things on the side where I prepare the matrix which is to be appended. That code is not black-boxed, but it is too complex for me to know how to change it so that it would meet the interface requirements posed by the framework.
Is there some sequence of inversions, transformations or other manuevers which would let me modify a viewProjection matrix that was designed to be prepended, so that it will turn out right when it is, instead, appended to the Object's World Space coordinates?
I am providing an answer more out of desperation than sure understanding, and still hope I will receive a better answer from those more knowledgeable. From Dunn and Parberry's "3D Math Primer" I learned that "transposing the product of two matrices is the same as taking the product of their transposes in reverse order."
Without being able to understand how to enter text involving superscripts, I am not sure if I can reduce my approach to a helpful mathematical formulation, so I will invent a syntax using functional notation. The equivalency noted by Dunn and Parberry would be something like:
AB = transpose (B) x transpose (A)
That comes close to solving my problem, which problem, to restate, is really just a problem arising out of the fact that I cannot control the behavior of the internal matrix operations in the framework package. I can, however, perform appropriate matrix operations on either side of the workflow from local object coordinates to those required by the GPU Vertex Shader.
I have not completed the test of my solution, which requires the final step to be taken in the AGAL shader, but I have been able to confirm in AS3 that the last 'un-transform' does yield exactly the same combined raw data as the example from the author of the camera with the desired lens properties whose implementation involves prepending rather than appending.
BA = transpose (transpose (A) x transpose (B))
I have also not yet tested to see if these extra calculations are so processing intensive as to reduce my application frame rate beyond what is acceptable, but am pleased at least to be able to confirm that the computations yield the same result.
I am trying to run this code but it keeps crashing:
log10(x):=log(x)/log(10);
char(x):=floor(log10(x))+1;
mantissa(x):=x/10**char(x);
chop(x,d):=(10**char(x))*(floor(mantissa(x)*(10**d))/(10**d));
rnd(x,d):=chop(x+5*10**(char(x)-d-1),d);
d:5;
a:10;
Ibwd:[[30,rnd(integrate((x**60)/(1+10*x^2),x,0,1),d)]];
for n from 30 thru 1 step -1 do Ibwd:append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd);
Maxima crashes when it evaluates the last line. Any ideas why it may happen?
Thank you so much.
The problem is that the difference becomes negative and your rounding function dies horribly with a negative argument. To find this out, I changed your loop to:
for n from 30 thru 1 step -1 do
block([],
print (1/(2*n-1)-a*last(first(Ibwd))),
print (a*last(first(Ibwd))),
Ibwd: append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd),
print (Ibwd));
The last difference printed before everything fails miserably is -316539/6125000. So now try
rnd(-1,3)
and see the same problem. This all stems from the fact that you're taking the log of a negative number, which Maxima interprets as a complex number by analytic continuation. Maxima doesn't evaluate this until it absolutely has to and, somewhere in the evaluation code, something's dying horribly.
I don't know the "fix" for your specific example, since I'm not exactly sure what you're trying to do, but hopefully this gives you enough info to find it yourself.
If you want to deconstruct a floating point number, let's first make sure that it is a bigfloat.
say z: 34.1
You can access the parts of a bigfloat by using lisp, and you can also access the mantissa length in bits by ?fpprec.
Thus ?second(z)*2^(?third(z)-?fpprec) gives you :
4799148352916685/140737488355328
and bfloat(%) gives you :
3.41b1.
If you want the mantissa of z as an integer, look at ?second(z)
Now I am not sure what it is that you are trying to accomplish in base 10, but Maxima
does not do internal arithmetic in base 10.
If you want more bits or fewer, you can set fpprec,
which is linked to ?fpprec. fpprec is the "approximate base 10" precision.
Thus fpprec is initially 16
?fpprec is correspondingly 56.
You can easily change them both, e.g. fpprec:100
corresponds to ?fpprec of 335.
If you are diddling around with float representations, you might benefit from knowing
that you can look at any of the lisp by typing, for example,
?print(z)
which prints the internal form using the Lisp print function.
You can also trace any function, your own or system function, by trace.
For example you could consider doing this:
trace(append,rnd,integrate);
If you want to use machine floats, I suggest you use, for the last line,
for n from 30 thru 1 step -1 do :
Ibwd:append([[n-1,rnd(1/(2.0*n- 1.0)-a*last(first(Ibwd)),d)]],Ibwd);
Note the decimal points. But even that is not quite enough, because integration
inserts exact structures like atan(10). Trying to round these things, or compute log
of them is probably not what you want to do. I suspect that Maxima is unhappy because log is given some messy expression that turns out to be negative, even though it initially thought otherwise. It hands the number to the lisp log program which is perfectly happy to return an appropriate common-lisp complex number object. Unfortunately, most of Maxima was written BEFORE LISP HAD COMPLEX NUMBERS.
Thus the result (log -0.5)= #C(-0.6931472 3.1415927) is entirely unexpected to the rest of Maxima. Maxima has its own form for complex numbers, e.g. 3+4*%i.
In particular, the Maxima display program predates the common lisp complex number format and does not know what to do with it.
The error (stack overflow !!!) is from the display program trying to display a common lisp complex number.
How to fix all this? Well, you could try changing your program so it computes what you really want, in which case it probably won't trigger this error. Maxima's display program should be fixed, too. Also, I suspect there is something unfortunate in simplification of logs of numbers that are negative but not obviously so.
This is probably waaay too much information for the original poster, but maybe the paragraph above will help out and also possibly improve Maxima in one or more places.
It appears that your program triggers an error in Maxima's simplification (algebraic identities) code. We are investigating and I hope we have a bug fix soon.
In the meantime, here is an idea. Looks like the bug is triggered by rnd(x, d) when x < 0. I guess rnd is supposed to round x to d digits. To handle x < 0, try this:
rnd(x, d) := if x < 0 then -rnd1(-x, d) else rnd1(x, d);
rnd1(x, d) := (... put the present definition of rnd here ...);
When I do that, the loop runs to completion and Ibwd is a list of values, but I don't know what values to expect.
I am in kind of newbies in matlab. I am trying to write a code which divide the image in nonoverlaping blocks of size 3*3 and I am supposed to do an operation of the specific block like getting the value of the center pixel of block and do some operations. But I don't know where to start from. Using command like blockproc won't help. Can anyone suggest me where to start from?
You could easily use blockproc for this:
http://www.mathworks.com/help/toolbox/images/ref/blockproc.html
But if that isn't working for you, what errors do you get?
If you want to do it manually (like extracting the value of the center pixel of each block) you could simply use two loops for this.. but be aware, this is rather an unelegant and not really fast way to do it...
image = imread('image.png');
s = size(image);
for i=2:3:s(1)-1
for j=2:3:s(2)-1
%% here you have the midpoint of each 3x3 block...
%% you could then easily crop the image around it if you
%% really need separated blocks...
end
end
This isn't a really fast way though... but it works...
Hope that helps...