I noticed that SciPy has an implementation of the Discrete Sine Transform, and I was comparing it to the one that's in MATLAB. The MATLAB documentation notes that for best performance, the size of the inputs should be 2^p -1, presumably for a divide and conquer strategy. Is this also true for the SciPy implementation?
Although this question is old, I happen to have just ran some tests and then stumbled upon this question.
The answer is yes. Internally, scipy seems to converts the array to size M = 2*(N+1).
Ideally, M = 2^i, for some integer i. Therefore, N should follow N = 2^i - 1. The following picture shows how timings scale with fft-size. Note that the orange line is much smoother, indicating no unexpected memory overhead.
Green line: N = 2^i
Blue line: N = 2^i + 1
Orange line: N = 2^i - 1
UPDATE
After digging some more into the documentation of scipy.fftpack, I found that the above answer is only partly true. According to the documentation, "SciPy’s FFTPACK has efficient functions for radix {2, 3, 4, 5}". This means that instead of efficiently doing arrays of size M = 2^i, it can handle any M = 2^i * 3^j * 5^k (4 is not a prime). The optimum for scipy.fftpack.dst (or dct) is then M - 1. Finding those numbers can be a little awkward, but luckily there's a function for that, too!
Please note that the above graph is log-log scale, so speedups of 40 or so are not uncommon. Thus, choosing a fast size can make you calculations orders of magnitudes faster! (I found this out the hard way).
Related
To apply the combination of SVD perturbation:
I = imread('image.jpg');
Ibw = single(im2double(I));
[U S V] = svd(Ibw);
% calculate derviced image
P = U * power(S, i) * V'; % where i is between 1 and 2
%To compute the combined image of SVD perturbations:
J = (single(I) + (alpha*P))/(1+alpha); % where alpha is between 0 and 1
I applied this method to a specific face recognition model and I noticed the accuracy was highly increased!! So it is very efficient!. Interestingly, I used the value i=3/4 and alpha=0.25 according to a paper that was published in a journal in 2012 in which the authors used i=3/4 and alpha=0.25. But I didn't make attention that i must be between 1 and 2! (I don't know if the authors make an error of dictation or they in fact used the value 3/4). So I tried to change the value of i to a value greater than 1, the accuracy decreased!!. So can I use the value 3/4 ? If yes, how can I argument therefore my approach?
The paper that I read is entitled "Enhanced SVD based face recognition". In page 3, they used the value i=3/4.
(http://www.oalib.com/paper/2050079)
Kindly I need your help and opinions. Any help will be very appreciated!
The idea to have the value between one and two is to magnify the singular values to make them invariant to illumination changes.
Refer to this paper: A New Face Recognition Method based on SVD Perturbation for Single Example Image per Person: Daoqiang Zhang,Songcan Chen,and Zhi-Hua Zhou
Note that when n equals to 1, the derived image P is equivalent to the original image I . If we
choose n>1, then the singular values satisfying s_i > 1 will be magnified. Thus the reconstructed
image P emphasizes the contribution of the large singular values, while restraining that of the
small ones. So by integrating P into I , we get a combined image J which keeps the main
information of the original image and is expected to work better against minor changes of
expression, illumination and occlusions.
My take:
When you scale the singular values in the exponent, you are basically introducing a non-linearity, so its possible that for a specific dataset, scaling down the singular values may be beneficial. Its like adjusting the gamma correction factor in a monitor.
I have got a problem to remove highly correlated components. Can I ask how to do this?
For example, I have got 40 instances with 20 features (random created). Feature 2 and 18 is highly correlated with feature 4. And feature 6 is highly correlated with feature 10. Then how to remove the highly correlated (redundant) features such as 2, 18 and 10? Essentially, I need the index of remaining features 1, 3, 4, 5, 6, ..., 9, 11, ..., 17, 19, 20.
Matlab codes:
x = randn(40,20);
x(:,2) = 2.*x(:,4);
x(:,18) = 3.*x(:,4);
x(:,6) = 100.*x(:,10);
x_corr = corr(x);
size(x_corr)
figure, imagesc(x_corr),colorbar
Correlation matrix x_corr looks like
edit:
I worked out a way:
x_corr = x_corr - diag(diag(x_corr));
[x_corrX, x_corrY] = find(x_corr>0.8);
for i = 1:size(x_corrX,1)
xx = find(x_corrY == x_corrX(i));
x_corrX(xx,:) = 0;
x_corrY(xx,:) = 0;
end
x_corrX = unique(x_corrX);
x_corrX = x_corrX(2:end);
im = setxor(x_corrX, (1:20)');
Am I right? Or you have a better idea please post. Thanks.
edit2: Is this method the same as using PCA?
It seems quite clear that this idea of yours, to simply remove highly correlated variables from the analysis is NOT the same as PCA. PCA is a good way to do rank reduction of what seems to be a complicated problem, into one that turns out to have only a few independent things happening. PCA uses an eigenvalue (or svd) decomposition to achieve that goal.
Anyway, you might have a problem. For example, suppose that A is highly correlated to B, and B is highly correlated to C. However, it need not be true that A and C are highly correlated. Since correlation can be viewed as a measure of the angle between those vectors in their corresponding high dimensional vector space, this can be easily made to happen.
As a trivial example, I'll create two variables, A and B, that are correlated at a "moderate" level.
n = 50;
A = rand(n,1);
B = A + randn(n,1)/2;
corr([A,B])
ans =
1 0.55443
0.55443 1
So here 0.55 is the correlation. I'll create C to be virtually the average of A and B. It will be highly correlated by your definition.
C = [A + B]/2 + randn(n,1)/100;
corr([A,B,C])
ans =
1 0.55443 0.80119
0.55443 1 0.94168
0.80119 0.94168 1
Clearly C is the bad guy here. But if one were to simply look at the pair [A,C] and remove A from the analysis, then do the same with the pair [B,C] and then remove B, we would have made the wrong choices. And this was a trivially constructed example.
In fact, it is true that the eigenvalues of the correlation matrix might be of interest.
[V,D] = eig(corr([A,B,C]))
V =
-0.53056 -0.78854 -0.311
-0.57245 0.60391 -0.55462
-0.62515 0.11622 0.7718
D =
2.5422 0 0
0 0.45729 0
0 0 0.00046204
The fact that D has two significant diagonal elements, and a tiny one tells us that really, this is a two variable problem. What PCA will not easily tell us is which vector to simply remove though, and the problem would only be less clear with more variables, with many interactions between all of them.
I think the answer of woodchips is quite good. But when you're using eigenvalues, you can run into some trouble. If the dataset is large enough, there will always be some small eigenvalues, but you won't be sure what they tell you.
Instead, consider grouping your data by a simple clustering method. It's easy to implement in Matlab.
http://www.mathworks.de/de/help/stats/cluster-analysis-1-1.html
edit:
If you disregard the points that woodchips made, you're solution is okay, as an algorithm.
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.
I have a Problem. I have a Matrix A with integer values between 0 and 5.
for example like:
x=randi(5,10,10)
Now I want to call a filter, size 3x3, which gives me the the most common value
I have tried 2 solutions:
fun = #(z) mode(z(:));
y1 = nlfilter(x,[3 3],fun);
which takes very long...
and
y2 = colfilt(x,[3 3],'sliding',#mode);
which also takes long.
I have some really big matrices and both solutions take a long time.
Is there any faster way?
+1 to #Floris for the excellent suggestion to use hist. It's very fast. You can do a bit better though. hist is based on histc, which can be used instead. histc is a compiled function, i.e., not written in Matlab, which is why the solution is much faster.
Here's a small function that attempts to generalize what #Floris did (also that solution returns a vector rather than the desired matrix) and achieve what you're doing with nlfilter and colfilt. It doesn't require that the input have particular dimensions and uses im2col to efficiently rearrange the data. In fact, the the first three lines and the call to im2col are virtually identical to what colfit does in your case.
function a=intmodefilt(a,nhood)
[ma,na] = size(a);
aa(ma+nhood(1)-1,na+nhood(2)-1) = 0;
aa(floor((nhood(1)-1)/2)+(1:ma),floor((nhood(2)-1)/2)+(1:na)) = a;
[~,a(:)] = max(histc(im2col(aa,nhood,'sliding'),min(a(:))-1:max(a(:))));
a = a-1;
Usage:
x = randi(5,10,10);
y3 = intmodefilt(x,[3 3]);
For large arrays, this is over 75 times faster than colfilt on my machine. Replacing hist with histc is responsible for a factor of two speedup. There is of course no input checking so the function assumes that a is all integers, etc.
Lastly, note that randi(IMAX,N,N) returns values in the range 1:IMAX, not 0:IMAX as you seem to state.
One suggestion would be to reshape your array so each 3x3 block becomes a column vector. If your initial array dimensions are divisible by 3, this is simple. If they don't, you need to work a little bit harder. And you need to repeat this nine times, starting at different offsets into the matrix - I will leave that as an exercise.
Here is some code that shows the basic idea (using only functions available in FreeMat - I don't have Matlab on my machine at home...):
N = 100;
A = randi(0,5*ones(3*N,3*N));
B = reshape(permute(reshape(A,[3 N 3 N]),[1 3 2 4]), [ 9 N*N]);
hh = hist(B, 0:5); % histogram of each 3x3 block: bin with largest value is the mode
[mm mi] = max(hh); % mi will contain bin with largest value
figure; hist(B(:),0:5); title 'histogram of B'; % flat, as expected
figure; hist(mi-1, 0:5); title 'histogram of mi' % not flat?...
Here are the plots:
The strange thing, when you run this code, is that the distribution of mi is not flat, but skewed towards smaller values. When you inspect the histograms, you will see that is because you will frequently have more than one bin with the "max" value in it. In that case, you get the first bin with the max number. This is obviously going to skew your results badly; something to think about. A much better filter might be a median filter - the one that has equal numbers of neighboring pixels above and below. That has a unique solution (while mode can have up to four values, for nine pixels - namely, four bins with two values each).
Something to think about.
Can't show you a mex example today (wrong computer); but there are ample good examples on the Mathworks website (and all over the web) that are quite easy to follow. See for example http://www.shawnlankton.com/2008/03/getting-started-with-mex-a-short-tutorial/
i am having problem with writing equations.
r = 25, k= 2, R = 50:25:600, DR = 0.5:0.5:4.0
h= r*[1-cos(asin((sqrt(2*R*DR+DR^2))+r*sin(acos(r-k)/r)/r))]-k
but as a resault i get this: h = 1.9118e+001 +1.7545e+002i.
I just start with Matlab. Thanks
What I get from what you've written is actually
??? Error using ==> mtimes
Inner matrix dimensions must agree.
which is correct because you're trying to multiply two row vectors by one another. Could you please show us the actual code you used?
Anyway, supposing that's dealt with somehow, it looks to me as if you're feeding something to asin that's much bigger than 1. That'll give you complex results. Is the thing you're passing to asin perhaps meant to be divided by R^2 or DR^2 or something of the kind? You have a similar issue a bit later with the argument to acos.
I also suspect that some of your * and ^ and / operators should actually be elementwise ones .*, .^, ./.
If you're trying to do as you said:
so in first equation i used R= 50, DR
= 0.5, r= 25, k=2 and i need to get h. In second equation i used R=75,
DR=1.0, r=25, k=2...for a last
equation i used
R=600,DR=4.0,r=25,k=2.
DR and R need to be the same length... so if R goes between 50 and 600 in increments of 25, DR should go from 0.5 to 12.5 in increments of 0.5, or 0.5 to 4.0 in increments of 0.1522...
once you figure that out, be sure the add a period before every matrix multiplication operation (e.g. * or ^)
EDIT: formula adjusted slightly (bracketing) to reflect success in comment.
When you say you want a table, I guess it is to be an R by DR table (since you have to vectors of different length). To do that you need to use R as a column vector (R' below) and multiply with * (not .*). When R doesn't appear in a term multiply by ones(size(R)) (or use repmat) to get DR into the right shape. To square DR by element, you need DR.^2. There seems to be a misplaced bracket for the acos, surely you divide by r before taking the acos. There must be a division by something like r in the asin (not r^2 because you've taken the sqrt). Finally, the last division by r is redundant as written, since you multiply by r at the same level just before. Anyway, if I do the following:
h= r*(1-cos(asin((sqrt(2*R'*DR+ones(size(R))'*DR.^2)/r)+sin(acos((r-k)/r)))))-k
I get an R by DR table. Results for small R,DR are real; higher R,DR are complex due to the argument of the first asin being >1. The first entry in the table is 4.56, as you require.