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.
Related
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).
I have checked other questions. I didn't find my answer. I have a matrix of n * 2 size. I want to compare the 1st and 2nd column and based on which is greater I want to assign 0/1 to the respective index. Suppose I want an output as
a = 1 2
4 3
7 8
I want the output like this
out = 0 1
1 0
0 1
I did this :
o1 = a(:,1) > a (:,2)
o2 = not(o1)
out = [o1, o2]
This does the job but I am sure there's a better way to do this. Need suggestions on that/.
Forgot to mention, the datatype is float in the matrix.
A more generic solution that can handle matrices with more than two columns:
out = bsxfun(#eq, a, max(a,[],2));
What you did is good. The number of lines doesn't really matter, what matters is the complexity of the operation in each line. Following the comments, I think you could gain some time as well by avoiding copy and multiple allocations:
out = false(size(a)); out(:,1) = (a(:,1) > a(:,2)); out(:,2) = ~out(:,1);
It is good practice to preallocate in Matlab, and in general to avoid copies in any programming language.
Optimizing further the runtime of this by using different operations is pointless IMO. If you really need speed you could Mex it to spare one iteration through the rows (second assignment), it's literally a dozen C lines, although you'd have to be careful about how you write the loop (the naive way would cause cache-miss at each iteration).
I have 12 sets of vectors (about 10-20 vectors each) and i want to pick one vector of each set so that a function f that takes the sum of these vectors as argument is maximized. In addition i have constraints for some components of that sum.
Example:
a_1 = [3 2 0 5], a_2 = [3 0 0 2], a_3 = [6 0 1 1], ... , a_20 = [2 12 4 3]
b_1 = [4 0 4 -2], b_2 = [0 0 1 0], b_3 = [2 0 0 4], ... , b_16 = [0 9 2 3]
...
l_1 = [4 0 2 0], l_2 = [0 1 -2 0], l_3 = [4 4 0 1], ... , l_19 = [3 0 9 0]
s = [s_1 s_2 s_3 s_4] = a_x + b_y + ... + l_z
Constraints:
s_1 > 40
s_2 < 100
s_4 > -20
Target: Chose x, y, ... , z to maximize f(s):
f(s) -> max
Where f is a nonlinear function that takes the vector s and returns a scalar.
Bruteforcing takes too long because there are about 5.9 trillion combinations, and since i need the maximum (or even better the top 10 combinations) i can not use any of the greedy algorithms that came to my mind.
The vectors are quite sparse, about 70-90% are zeros. If that is helping somehow ...?
The Matlab Optimization toolbox didnt help either since it doesnt much support for discrete optimization.
Basically this is a lock-picking problem, where the lock's pins have 20 distinct positions, and there are 12 pins. Also:
some of the pin's positions will be blocked, depending on the positions of all the other pins.
Depending on the specifics of the lock, there may be multiple keys that fit
...interesting!
Based on Rasman's approach and Phpdna's comment, and the assumption that you are using int8 as data type, under the given constraints there are
>> d = double(intmax('int8'));
>> (d-40) * (d+100) * (d+20) * 2*d
ans =
737388162
possible vectors s (give or take a few, haven't thought about +1's etc.). ~740 million evaluations of your relatively simple f(s) shouldn't take more than 2 seconds, and having found all s that maximize f(s), you are left with the problem of finding linear combinations in your vector set that add up to one of those solutions s.
Of course, this finding of combinations is no easy feat, and the whole method breaks down anyway if you are dealing with
int16: ans = 2.311325368800510e+018
int32: ans = 4.253529737045237e+037
int64: ans = 1.447401115466452e+076
So, I'll discuss a more direct and more general approach here.
Since we're talking integers and a fairly large search space, I'd suggest using a branch-and-bound algorithm. But unlike the bintprog algorithm, you'd have to use different branching strategies, and of course, these should be based on a non-linear objective function.
Unfortunately, there is nothing like this in the optimization toolbox (or the File Exchange as far as I could find). fmincon is a no-go, since it uses gradient and Hessian information (which will usually be all-zero for integers), and fminsearch is a no-go, since you'll need a really good initial estimate, and the rate of convergence is (roughly) O(N), meaning, for this 20-dimensional problem you'll have to wait quite long before convergence, without the guarantee of having found the global solution.
An interval method could be a possibility, however, I personally have very little experience with this. There is no native interval-related stuff in MATLAB or any of its toolboxes, but there's the freely available INTLAB.
So, if you're not feeling like implementing your own non-linear binary integer programming algorithm, or are not in the mood for an adventure with INTLAB, there's really only one thing left: heuristic methods. In this link there is a similar situation, with an outline of the solution: use the genetic algorithm (ga) from the Global Optimization toolbox.
I would implement the problem roughly like so:
function [sol, fval, exitflag] = bintprog_nonlinear()
%// insert your data here
%// Any sparsity you may have here will only make this more
%// *memory* efficient, not *computationally*
data = [...
... %// this will be an array with size 4-by-20-by-12
... %// (or some permutation of that you find more intuitive)
];
%// offsets into the 3D array to facilitate indexing a bit
offsets = bsxfun(#plus, ...
repmat(1:size(data,1), size(data,3),1), ...
(0:size(data,3)-1)' * size(data,1)*size(data,2)); %//'
%// your objective function
function val = obj(X)
%// limit "X" to integers in [1 20]
X = min(max(round(X),1),size(data,3));
%// "X" will be a collection of 12 integers between 0 and 20, which are
%// indices into the data matrix
%// form "s" from "X"
s = sum(bsxfun(#plus, offsets, X*size(data,1) - size(data,1)));
%// XxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxX
%// Compute the NEGATIVE VALUE of your function here
%// XxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxXxX
end
%// your "non-linear" constraint function
function [C, Ceq] = nonlcon(X)
%// limit "X" to integers in [1 20]
X = min(max(round(X),1),size(data,3));
%// form "s" from "X"
s = sum(bsxfun(#plus, offsets, X(:)*size(data,1) - size(data,1)));
%// we have no equality constraints
Ceq = [];
%// Compute inequality constraints
%// NOTE: solver is trying to solve C <= 0, so:
C = [...
40 - s(1)
s(2) - 100
-20 - s(4)
];
end
%// useful GA options
options = gaoptimset(...
'UseParallel', 'always'...
...
);
%// The rest really depends on the specifics of the problem.
%// Useful to look at will be at least 'TolCon', 'Vectorized', and of course,
%// 'PopulationType', 'Generations', etc.
%// THE OPTIMZIATION
[sol, fval, exitflag] = ga(...
#obj, size(data,3), ... %// objective function, taking a vector of 20 values
[],[], [],[], ... %// no linear (in)equality constraints
1,size(data,2), ... %// lower and upper limits
#nonlcon, options); %// your "nonlinear" constraints
end
Note that even though your constraints are essentially linear, the way by which you must compute the value for your s necessitates the use of a custom constraint function (nonlcon).
Especially note that this is currently (probably) a sub-optimal way to use ga -- I don't know the specifics of your objective function, so a lot more may be possible. For instance, I currently use a simple round() to convert the input X to integers, but using 'PopulationType', 'custom' (with a custom 'CreationFcn', 'MutationFcn' etc.) might produce better results. Also, 'Vectorized' will likely speed things up a lot, but I don't know whether your function is easily vectorized.
And yes, I use nested functions (I just love those things!); it prevents these huge, usually identical lists of input arguments if you use sub-functions or stand-alone functions, and they can really be a performance boost because there is little copying of data. But, I realize that their scoping rules make them somewhat akin to goto constructs, and so they are -ahum- "not everyone's cup of tea"...you might want to convert them to sub-functions to prevent long and useless discussions with your co-workers :)
Anyway, this should be a good place to start. Let me know if this is useful at all.
Unless you define some intelligence on how the vector sets are organized, there will be no intelligent way of solving your problem other then pure brute force.
Say you find s s.t. f(s) is max given constraints of s, you still need to figure out how to build s with twelve 4-element vectors (an overdetermined system if there ever was one), where each vector has 20 possible values. Sparsity may help, although I'm not sure how it is possible to have a vector with four elements be 70-90% zero, and sparsity would only be useful if there was some yet to be described methodology in how the vector are organized
So I'm not saying you can't solve the problem, I'm saying you need to rethink how the problem is set-up.
I know, this answer is reaching you really late.
Unfortunately, the problem, as is, show not many patterns to be exploited, besides of brute force -Branch&Bound, Master& Slave, etc.- Trying a Master Slave approach -i.e. solving first the function continuous nonlinear problem as master, and solving the discrete selection as slave could help, but with as many combinations, and without any more information over the vectors, there is not too much space for work.
But based on the given continuous almost everywhere functions, based on combinations of sums and multiplication operators and their inverses, the sparsity is a clear point to be exploited here. If 70-90% of vectors are zero, almost a good part of the solution space will be close to zero, or close to infinite. Hence a 80-20 pseudo solution would discard easily the 'zero' combinations, and use only the 'infinite' ones.
This way, the brute-force could be guided.
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/
Is there a way to extend a vector by making it repeat itself?
>v = [1 2];
>v10 = v x 5; %x represents some function. Something like "1 2" x 5 in perl
Then v10 would be:
>v10
1 2 1 2 1 2 1 2 1 2
This should work for the general case, not just for [1 2]
The function you're looking for is repmat().
v10 = repmat(v, 1, 5)
Obviously repmat is the way to go if you know in which direction you want to expand the vector.
However, if you want a general solution that always repeats the vector in the longest direction, this combination of repmat and indexing should do the trick:
v10=v(repmat(1:length(v),1,5))
Although late, I am posting this because this turned out to be the most popular answer to a similar question here.
This is a Faster Method Than repmat or reshape by an Order of Magnitude
One of the best methods for doing such things is Using Tony's Trick. I came across this trick in one of the Electrical Engineering course lectures notes of Columbia University. Repmat and Reshape are usually found to be slower than Tony's trick as it directly uses Matlabs inherent indexing. To answer you question,
Lets say, you want to tile the row vector r=[1 2 3] N times like r=[1 2 3 1 2 3 1 2 3...], then,
c=r'
cc=c(:,ones(N,1));
r_tiled = cc(:)';
This method has significant time savings against reshape or repmat for large N's.
I conducted a small Matlab test to check the speed differential between repmat and tony's trick. Using the code mentioned below, I calculated the times for constructing the same tiled vector from a base vector A=[1:N]. The results show that YES, Tony's-Trick is FASTER BY AN ORDER of MAGNITUDE, especially for larger N. People are welcome to try it themselves. This much time differential can be critical if such an operation has to be performed in loops. Here is the small script I used;
N= 10 ;% ASLO Try for values N= 10, 100, 1000, 10000
% time for tony_trick
tic;
A=(1:N)';
B=A(:,ones(N,1));
C=B(:)';
t_tony=toc;
clearvars -except t_tony N
% time for repmat
tic;
A=(1:N);
B=repmat(A,1,N);
t_repmat=toc;
clearvars -except t_tony t_repmat N
The Times (in seconds) for both methods are given below;
N=10, time_repmat = 8e-5 , time_tony = 3e-5
N=100, time_repmat = 2.9e-4 , time_tony = 6e-5
N=1000, time_repmat = 0.0302 , time_tony = 0.0058
N=10000, time_repmat = 2.9199 , time_tony = 0.5292
My RAM didn't permit me to go beyond N=10000. I am sure, the time difference between the two methods will be even more significant for N=100000. I know, these times might be different for different machines, but the relative difference in order-of-magnitude of times will stand. Also, I know, the avg of times could have been a better metric, but I just wanted to show the order of magnitude difference in time consumption between the two approaches. My machine/os details are given below :
Relevant Machine/OS/Matlab Details : Athlon i686 Arch, Ubuntu 11.04 32 bit, 3gb ram, Matlab 2011b