Matlab -- random walk with boundaries, vectorized - matlab

Suppose I have a vector J of jump sizes and an initial starting point X_0. Also I have boundaries 0, B (assume 0 < X_0 < B). I want to do a random walk where X_i = [min(X_{i-1} + J_i,B)]^+. (positive part). Basically if it goes over a boundary, it is made equal to the boundary. Anyone know a vectorized way to do this? The current way I am doing it consists of doing cumsums and then finding places where it violates a condition, and then starting from there and repeating the cumsum calculation, etc until I find that I stop violating the boundaries. It works when the boundaries are rarely hit, but if they are hit all the time, it basically becomes a for loop.
In the code below, I am doing this across many samples. To 'fix' the ones that go out of the boundary, I have to loop through the samples to check...(don't think there is a vectorized 'find')
% X_init is a row vector describing initial resource values to use for
% each sample
% J is matrix where each col is a sequence of Jumps (columns = sample #)
% In this code the jumps are subtracted, but same thing
X_intvl = repmat(X_init,NumJumps,1) - cumsum(J);
X = [X_init; X_intvl];
for sample = 1:NumSamples
k = find(or(X_intvl(:,sample) > B, X_intvl(:,sample) < 0),1);
while(~isempty(k))
change = X_intvl(k-1,sample) - X_intvl(k,sample);
X_intvl(k:end,sample) = X_intvl(k:end,sample)+change;
k = find(or(X_intvl(:,sample) > B, X_intvl(:,sample) < 0),1);
end
end

Interesting question (+1).
I faced a similar problem a while back, although slightly more complex as my lower and upper bound depended on t. I never did work out a fully-vectorized solution. In the end, the fastest solution I found was a single loop which incorporates the constraints at each step. Adapting the code to your situation yields the following:
%# Set the parameters
LB = 0; %# Lower bound
UB = 5; %# Upper bound
T = 100; %# Number of observations
N = 3; %# Number of samples
X0 = (1/2) * (LB + UB); %# Arbitrary start point halfway between LB and UB
%# Generate the jumps
Jump = randn(N, T-1);
%# Build the constrained random walk
X = X0 * ones(N, T);
for t = 2:T
X(:, t) = max(min(X(:, t-1) + Jump(:, t-1), UB), 0);
end
X = X';
I would be interested in hearing if this method proves faster than what you are currently doing. I suspect it will be for cases where the constraint is binding in more than one or two places. I can't test it myself as the code you provided is not a "working" example, ie I can't just copy and paste it into Matlab and run it, as it depends on several variables for which example (or simulated) values are not provided. I tried adapting it myself, but couldn't get it to work properly?
UPDATE: I just switched the code around so that observations are indexed on columns and samples are indexed on rows, and then I transpose X in the last step. This will make the routine more efficient, since Matlab allocates memory for numeric arrays column-wise - hence it is faster when performing operations down the columns of an array (as opposed to across the rows). Note, you will only notice the speed-up for large N.
FINAL THOUGHT: These days, the JIT accelerator is very good at making single loops in Matlab efficient (double loops are still pretty slow). Therefore personally I'm of the opinion that every time you try and obtain a fully-vectorized solution in Matlab, ie no loops, you should weigh up whether the effort involved in finding a clever solution is worth the slight gains in efficiency to be made over an easier-to-obtain method that utilizes a single loop. And it is important to remember that fully-vectorized solutions are sometimes slower than solutions involving single loops when T and N are small!

I'd like to propose another vectorized solution.
So, first we should set the parameters and generate random Jumpls. I used the same set of parameters as Colin T Bowers:
% Set the parameters
LB = 0; % Lower bound
UB = 20; % Upper bound
T = 1000; % Number of observations
N = 3; % Number of samples
X0 = (1/2) * (UB + LB); % Arbitrary start point halfway between LB and UB
% Generate the jumps
Jump = randn(N, T-1);
But I changed generation code:
% Generate initial data without bounds
X = cumsum(Jump, 2);
% Apply bounds
Amplitude = UB - LB;
nsteps = ceil( max(abs(X(:))) / Amplitude - 0.5 );
for ii = 1:nsteps
ind = abs(X) > (1/2) * Amplitude;
X(ind) = Amplitude * sign(X(ind)) - X(ind);
end
% Shifting X
X = X0 + X;
So, instead of for loop I'm using cumsum function with smart post-processing.
N.B. This solution works significantly slower than Colin T Bowers's one for tight bounds (Amplitude < 5), but for loose bounds (Amplitude > 20) it works much faster.

Related

How to setup equation that involves a sum from x=1 to infinity and loops?

I am getting confused on how to properly set up this equation. To find a value of V(i,j). The end result would be plotting V over time. I understand that there needs to be loops to allow this equation to work, however I am lost when it comes to setting it up. Basically I am trying to take the sum from n=1 to infinity of (1-(-1)^n)/(n^4 *pi^4)*sin((n*pi*c*j)/L)*sin((n*pi*i)/L)
I originally thought that I should make it a while loop to increment n by 1 until I reach say 10 or so just to get an idea of what the output would look like. All of the variables were unknown and values were added again to see what the plot would look like.
I have down another code where the equation is just dependent on i and j. However with this n term, I am thrown off. Any advice would be great as to setting up the equation. Thank you.
L=10;
x=linspace(0,L,30);
t1= 50;
X=30;
p=1
c=t1/1000;
V=zeros(X,t1);
V(1,:)=0;
V(30,:)=0;
R=((4*p*L^3)/c);
n=1;
t=1:50;
while n < 10
for i=1:31
for j=1:50
V(i,j)=R*sum((1-(-1)^n)/(n^4 *pi^4)*sin((n*pi*c*j)/L)*sin((n*pi*i)/L));
end
end
n=n+1;
end
figure(1)
plot(V(i,j),t)
Various ways of doing so:
1) Computing the sum up to one Nmax in one shot:
Nmax = 30;
Vijn = #(i,j,n) R*((1-(-1)^n)/(n^4 *pi^4)*sin((n*pi*c*j)/L)*sin((n*pi*i)/L));
i = 1:31;
j = 1:50;
n = 1:Nmax;
[I,J,N] = ndgrid(i,j,n);
V = arrayfun(Vijn,I,J,N);
Vc = cumsum(V,3);
% now Vc(:,:,k) is sum_n=1^{k+1} V(i,j,n)
figure(1);clf;imagesc(Vc(:,:,end));
2) Looping indefinitely
n = 1;
V = 0;
i = 1:31;
j = 1:50;
[I,J] = meshgrid(i,j);
while true
V = V + R*((1-(-1)^n)/(n^4 *pi^4)*sin((n*pi*c*J)/L).*sin((n*pi*I)/L));
n = n + 1;
figure(1);clf;
imagesc(V);
title(sprintf('N = %d',n))
drawnow;
pause(0.25);
end
Note that in your example you won't need many terms, since:
Every second term is zero (for even n, the term 1-(-1)^n is zero).
The terms decay with 1/n^4. In norms: n=1 contributes ~2e4, n=3 contributes ~4e2, n=5 contributes 5e1, n=7 contributes ~14, etc. Visually, there is a small difference between n=1 and n=1+n=3 but barely a noticeable one for n=1+n=3+n=5.
Given that so few terms are needed, the first approach is probably the better one. Also, skip the even indices, as you don't need them.

MatLab using Fixed Point method to find a root

I wanna find a root for the following function with an error less than 0.05%
f= 3*x*tan(x)=1
In the MatLab i've wrote that code to do so:
clc,close all
syms x;
x0 = 3.5
f= 3*x*tan(x)-1;
df = diff(f,x);
while (1)
x1 = 1 / 3*tan(x0)
%DIRV.. z= tan(x0)^2/3 + 1/3
er = (abs((x1 - x0)/x1))*100
if ( er <= 0.05)
break;
end
x0 = x1;
pause(1)
end
But It keeps running an infinite loop with error 200.00 I dunno why.
Don't use while true, as that's usually uncalled for and prone to getting stuck in infinite loops, like here. Simply set a limit on the while instead:
while er > 0.05
%//your code
end
Additionally, to prevent getting stuck in an infinite loop you can use an iteration counter and set a maximum number of iterations:
ItCount = 0;
MaxIt = 1e5; %// maximum 10,000 iterations
while er > 0.05 & ItCount<MaxIt
%//your code
ItCount=ItCount+1;
end
I see four points of discussion that I'll address separately:
Why does the error seemingly saturate at 200.0 and the loop continue infinitely?
The fixed-point iterator, as written in your code, is finding the root of f(x) = x - tan(x)/3; in other words, find a value of x at which the graphs of x and tan(x)/3 cross. The only point where this is true is 0. And, if you look at the value of the iterants, the value of x1 is approaching 0. Good.
The bad news is that you are also dividing by that value converging toward 0. While the value of x1 remains finite, in a floating point arithmetic sense, the division works but may become inaccurate, and er actually goes NaN after enough iterations because x1 underflowed below the smallest denormalized number in the IEEE-754 standard.
Why is er 200 before then? It is approximately 200 because the value of x1 is approximately 1/3 of the value of x0 since tan(x)/3 locally behaves as x/3 a la its Taylor Expansion about 0. And abs(1 - 3)*100 == 200.
Divisions-by-zero and relative orders-of-magnitude are why it is sometimes best to look at the absolute and relative error measures for both the values of the independent variable and function value. If need be, even putting an extremely (relatively) small finite, constant value in the denominator of the relative calculation isn't entirely a bad thing in my mind (I remember seeing it in some numerical recipe books), but that's just a band-aid for robustness's sake that typically hides a more serious error.
This convergence is far different compared to the Newton-Raphson iterations because it has absolutely no knowledge of slope and the fixed-point iteration will converge to wherever the fixed-point is (forgive the minor tautology), assuming it does converge. Unfortunately, if I remember correctly, fixed-point convergence is only guaranteed if the function is continuous in some measure, and tan(x) is not; therefore, convergence is not guaranteed since those pesky poles get in the way.
The function it appears you want to find the root of is f(x) = 3*x*tan(x)-1. A fixed-point iterator of that function would be x = 1/(3*tan(x)) or x = 1/3*cot(x), which is looking for the intersection of 3*tan(x) and 1/x. However, due to point number (2), those iterators still behave badly since they are discontinuous.
A slightly different iterator x = atan(1/(3*x)) should behave a lot better since small values of x will produce a finite value because atan(x) is continuous along the whole real line. The only drawback is that the domain of x is limited to the interval (-pi/2,pi/2), but if it converges, I think the restriction is worth it.
Lastly, for any similar future coding endeavors, I do highly recommend #Adriaan's advice. If would like a sort of compromise between the styles, most of my iterative functions are written with a semantic variable notDone like this:
iter = 0;
iterMax = 1E4;
tol = 0.05;
notDone = 0.05 < er & iter < iterMax;
while notDone
%//your code
iter = iter + 1;
notDone = 0.05 < er & iter < iterMax;
end
You can add flags and all that jazz, but that format is what I frequently use.
I believe that the code below achieves what you are after using Newton's method for the convergence. Please leave a comment if I have missed something.
% find x: 3*x*tan(x) = 1
f = #(x) 3*x*tan(x)-1;
dfdx = #(x) 3*tan(x)+3*x*sec(x)^2;
tolerance = 0.05; % your value?
perturbation = 1e-2;
converged = 1;
x = 3.5;
f_x = f(x);
% Use Newton s method to find the root
count = 0;
err = 10*tolerance; % something bigger than tolerance to start
while (err >= tolerance)
count = count + 1;
if (count > 1e3)
converged = 0;
disp('Did not converge.');
break;
end
x0 = x;
dfdx_x = dfdx(x);
if (dfdx_x ~= 0)
% Avoid division by zero
f_x = f(x);
x = x - f_x/dfdx_x;
else
% Perturb x and go back to top of while loop
x = x + perturbation;
continue;
end
err = (abs((x - x0)/x))*100;
end
if (converged)
disp(['Converged to ' num2str(x,'%10.8e') ' in ' num2str(count) ...
' iterations.']);
end

Regarding loop structure in Matlab for an iterative procedure

I'm trying to code a loop in Matlab that iteratively solves for an optimal vector s of zeros and ones. This is my code
N = 150;
s = ones(N,1);
for i = 1:N
if s(i) == 0
i = i + 1;
else
i = i;
end
select = s;
HI = (item_c' * (weights.*s)) * (1/(weights'*s));
s(i) = 0;
CI = (item_c' * (weights.*s)) * (1/(weights'*s));
standarderror_afterex = sqrt(var(CI - CM));
standarderror_priorex = sqrt(var(HI - CM));
ratio = (standarderror_afterex - standarderror_priorex)/(abs(mean(weights.*s) - weights'*select));
ratios(i) = ratio;
s(i) = 1;
end
[M,I] = min(ratios);
s(I) = 0;
This code sets the element to zero in s, which has the lowest ratio. But I need this procedure to start all over again, using the new s with one zero, to find the ratios and exclude the element in s that has the lowest ratio. I need that over and over until no ratios are negative.
Do I need another loop, or do I miss something?
I hope that my question is clear enough, just tell me if you need me to explain more.
Thank you in advance, for helping out a newbie programmer.
Edit
I think that I need to add some form of while loop as well. But I can't see how to structure this. This is the flow that I want
With all items included (s(i) = 1 for all i), calculate HI, CI and the standard errors and list the ratios, exclude item i (s(I) = 0) which corresponds to the lowest negative ratio.
With the new s, including all ones but one zero, calculate HI, CI and the standard errors and list the ratios, exclude item i, which corresponds to the lowest negative ratio.
With the new s, now including all ones but two zeros, repeat the process.
Do this until there is no negative element in ratios to exclude.
Hope that it got more clear now.
Ok. I want to go through a few things before I list my code. These are just how I would try to do it. Not necessarily the best way, or fastest way even (though I'd think it'd be pretty quick). I tried to keep the structure as you had in your code, so you could follow it nicely (even though I'd probably meld all the calculations down into a single function or line).
Some features that I'm using in my code:
bsxfun: Learn this! It is amazing how it works and can speed up code, and makes some things easier.
v = rand(n,1);
A = rand(n,4);
% The two lines below compute the same value:
W = bsxfun(#(x,y)x.*y,v,A);
W_= repmat(v,1,4).*A;
bsxfun dot multiplies the v vector with each column of A.
Both W and W_ are matrices the same size as A, but the first will be much faster (usually).
Precalculating dropouts: I made select a matrix, where before it was a vector. This allows me to then form a variable included using logical constructs. The ~(eye(N)) produces an identity matrix and negates it. By logically "and"ing it with select, then the $i$th column is now select, with the $i$th element dropped out.
You were explicitly calculating weights'*s as the denominator in each for-loop. By using the above matrix to calculate this, we can now do a sum(W), where the W is essentially weights.*s in each column.
Take advantage of column-wise operations: the var() and the sqrt() functions are both coded to work along the columns of a matrix, outputting the action for a matrix in the form of a row vector.
Ok. the full thing. Any questions let me know:
% Start with everything selected:
select = true(N);
stop = false; % Stopping flag:
while (~stop)
% Each column leaves a variable out...
included = ~eye(N) & select;
% This calculates the weights with leave-one-out:
W = bsxfun(#(x,y)x.*y,weights,included);
% You can comment out the line below, if you'd like...
W_= repmat(weights,1,N).*included; % This is the same as previous line.
% This calculates the weights before dropping the variables:
V = bsxfun(#(x,y)x.*y,weights,select);
% There's different syntax, depending on whether item_c is a
% vector or a matrix...
if(isvector(item_c))
HI = (item_c' * V)./(sum(V));
CI = (item_c' * W)./(sum(W));
else
% For example: item_c is a matrix...
% We have to use bsxfun() again
HI = bsxfun(#rdivide, (item_c' * V),sum(V));
CI = bsxfun(#rdivide, (item_c' * W),sum(W));
end
standarderror_afterex = sqrt(var(bsxfun(#minus,HI,CM)));
standarderror_priorex = sqrt(var(bsxfun(#minus,CI,CM)));
% or:
%
% standarderror_afterex = sqrt(var(HI - repmat(CM,1,size(HI,2))));
% standarderror_priorex = sqrt(var(CI - repmat(CM,1,size(CI,2))));
ratios = (standarderror_afterex - standarderror_priorex)./(abs(mean(W) - sum(V)));
% Identify the negative ratios:
negratios = ratios < 0;
if ~any(negratios)
% Drop out of the while-loop:
stop = true;
else
% Find the most negative ratio:
neginds = find(negratios);
[mn, mnind] = min(ratios(negratios));
% Drop out the most negative one...
select(neginds(mnind),:) = false;
end
end % end while(~stop)
% Your output:
s = select(:,1);
If for some reason it doesn't work, please let me know.

Integration via trapezoidal sums in MATLAB

I need help finding an integral of a function using trapezoidal sums.
The program should take successive trapezoidal sums with n = 1, 2, 3, ...
subintervals until there are two neighouring values of n that differ by less than a given tolerance. I want at least one FOR loop within a WHILE loop and I don't want to use the trapz function. The program takes four inputs:
f: A function handle for a function of x.
a: A real number.
b: A real number larger than a.
tolerance: A real number that is positive and very small
The problem I have is trying to implement the formula for trapezoidal sums which is
Δx/2[y0 + 2y1 + 2y2 + … + 2yn-1 + yn]
Here is my code, and the area I'm stuck in is the "sum" part within the FOR loop. I'm trying to sum up 2y2 + 2y3....2yn-1 since I already accounted for 2y1. I get an answer, but it isn't as accurate as it should be. For example, I get 6.071717974723753 instead of 6.101605982576467.
Thanks for any help!
function t=trapintegral(f,a,b,tol)
format compact; format long;
syms x;
oldtrap = ((b-a)/2)*(f(a)+f(b));
n = 2;
h = (b-a)/n;
newtrap = (h/2)*(f(a)+(2*f(a+h))+f(b));
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap;
for i=[3:n]
dx = (b-a)/n;
trapezoidsum = (dx/2)*(f(x) + (2*sum(f(a+(3:n-1))))+f(b));
newtrap = trapezoidsum;
end
end
t = newtrap;
end
The reason why this code isn't working is because there are two slight errors in your summation for the trapezoidal rule. What I am precisely referring to is this statement:
trapezoidsum = (dx/2)*(f(x) + (2*sum(f(a+(3:n-1))))+f(b));
Recall the equation for the trapezoidal integration rule:
Source: Wikipedia
For the first error, f(x) should be f(a) as you are including the starting point, and shouldn't be left as symbolic. In fact, you should simply get rid of the syms x statement as it is not useful in your script. a corresponds to x1 by consulting the above equation.
The next error is the second term. You actually need to multiply your index values (3:n-1) by dx. Also, this should actually go from (1:n-1) and I'll explain later. The equation above goes from 2 to N, but for our purposes, we are going to go from 1 to N-1 as you have your code set up like that.
Remember, in the trapezoidal rule, you are subdividing the finite interval into n pieces. The ith piece is defined as:
x_i = a + dx*i; ,
where i goes from 1 up to N-1. Note that this starts at 1 and not 3. The reason why is because the first piece is already taken into account by f(a), and we only count up to N-1 as piece N is accounted by f(b). For the equation, this goes from 2 to N and by modifying the code this way, this is precisely what we are doing in the end.
Therefore, your statement actually needs to be:
trapezoidsum = (dx/2)*(f(a) + (2*sum(f(a+dx*(1:n-1))))+f(b));
Try this and let me know if you get the right answer. FWIW, MATLAB already implements trapezoidal integration by doing trapz as #ADonda already pointed out. However, you need to properly structure what your x and y values are before you set this up. In other words, you would need to set up your dx before hand, then calculate your x points using the x_i equation that I specified above, then use these to generate your y values. You then use trapz to calculate the area. In other words:
dx = (b-a) / n;
x = a + dx*(0:n);
y = f(x);
trapezoidsum = trapz(x,y);
You can use the above code as a reference to see if you are implementing the trapezoidal rule correctly. Your implementation and using the above code should generate the same results. All you have to do is change the value of n, then run this code to generate the approximation of the area for different subdivisions underneath your curve.
Edit - August 17th, 2014
I figured out why your code isn't working. Here are the reasons why:
The for loop is unnecessary. Take a look at the for loop iteration. You have a loop going from i = [3:n] yet you don't reference the i variable at all in your loop. As such, you don't need this at all.
You are not computing successive intervals properly. What you need to do is when you compute the trapezoidal sum for the nth subinterval, you then increment this value of n, then compute the trapezoidal rule again. This value is not being incremented properly in your while loop, which is why your area is never improving.
You need to save the previous area inside the while loop, then when you compute the next area, that's when you determine whether or not the difference between the areas is less than the tolerance. We can also get rid of that code at the beginning that tries and compute the area for n = 2. That's not needed, as we can place this inside your while loop. As such, this is what your code should look like:
function t=trapintegral(f,a,b,tol)
format long; %// Got rid of format compact. Useless
%// n starts at 2 - Also removed syms x - Useless statement
n = 2;
newtrap = ((b-a)/2)*(f(a) + f(b)); %// Initialize
oldtrap = 0; %// Initialize to 0
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap; %//Save the old area from the previous iteration
dx = (b-a)/n; %//Compute width
%//Determine sum
trapezoidsum = (dx/2)*(f(a) + (2*sum(f(a+dx*(1:n-1))))+f(b));
newtrap = trapezoidsum; % //This is the new sum
n = n + 1; % //Go to the next value of n
end
t = newtrap;
end
By running your code, this is what I get:
trapezoidsum = trapintegral(#(x) (x+x.^2).^(1/3),1,4,0.00001)
trapezoidsum =
6.111776299189033
Caveat
Look at the way I defined your function. You must use element-by-element operations as the sum command inside the loop will be vectorized. Take a look at the ^ operations specifically. You need to prepend a dot to the operations. Once you do this, I get the right answer.
Edit #2 - August 18th, 2014
You said you want at least one for loop. This is highly inefficient, and whoever specified having one for loop in the code really doesn't know how MATLAB works. Nevertheless, you can use the for loop to accumulate the sum term. As such:
function t=trapintegral(f,a,b,tol)
format long; %// Got rid of format compact. Useless
%// n starts at 3 - Also removed syms x - Useless statement
n = 3;
%// Compute for n = 2 first, then proceed if we don't get a better
%// difference tolerance
newtrap = ((b-a)/2)*(f(a) + f(b)); %// Initialize
oldtrap = 0; %// Initialize to 0
while (abs(newtrap-oldtrap)>=tol)
oldtrap = newtrap; %//Save the old area from the previous iteration
dx = (b-a)/n; %//Compute width
%//Determine sum
%// Initialize
trapezoidsum = (dx/2)*(f(a) + f(b));
%// Accumulate sum terms
%// Note that we multiply each term by (dx/2), but because of the
%// factor of 2 for each of these terms, these cancel and we thus have dx
for n2 = 1 : n-1
trapezoidsum = trapezoidsum + dx*f(a + dx*n2);
end
newtrap = trapezoidsum; % //This is the new sum
n = n + 1; % //Go to the next value of n
end
t = newtrap;
end
Good luck!

How can I speed up this call to quantile in Matlab?

I have a MATLAB routine with one rather obvious bottleneck. I've profiled the function, with the result that 2/3 of the computing time is used in the function levels:
The function levels takes a matrix of floats and splits each column into nLevels buckets, returning a matrix of the same size as the input, with each entry replaced by the number of the bucket it falls into.
To do this I use the quantile function to get the bucket limits, and a loop to assign the entries to buckets. Here's my implementation:
function [Y q] = levels(X,nLevels)
% "Assign each of the elements of X to an integer-valued level"
p = linspace(0, 1.0, nLevels+1);
q = quantile(X,p);
if isvector(q)
q=transpose(q);
end
Y = zeros(size(X));
for i = 1:nLevels
% "The variables g and l indicate the entries that are respectively greater than
% or less than the relevant bucket limits. The line Y(g & l) = i is assigning the
% value i to any element that falls in this bucket."
if i ~= nLevels % "The default; doesnt include upper bound"
g = bsxfun(#ge,X,q(i,:));
l = bsxfun(#lt,X,q(i+1,:));
else % "For the final level we include the upper bound"
g = bsxfun(#ge,X,q(i,:));
l = bsxfun(#le,X,q(i+1,:));
end
Y(g & l) = i;
end
Is there anything I can do to speed this up? Can the code be vectorized?
If I understand correctly, you want to know how many items fell in each bucket.
Use:
n = hist(Y,nbins)
Though I am not sure that it will help in the speedup. It is just cleaner this way.
Edit : Following the comment:
You can use the second output parameter of histc
[n,bin] = histc(...) also returns an index matrix bin. If x is a vector, n(k) = >sum(bin==k). bin is zero for out of range values. If x is an M-by-N matrix, then
How About this
function [Y q] = levels(X,nLevels)
p = linspace(0, 1.0, nLevels+1);
q = quantile(X,p);
Y = zeros(size(X));
for i = 1:numel(q)-1
Y = Y+ X>=q(i);
end
This results in the following:
>>X = [3 1 4 6 7 2];
>>[Y, q] = levels(X,2)
Y =
1 1 2 2 2 1
q =
1 3.5 7
You could also modify the logic line to ensure values are less than the start of the next bin. However, I don't think it is necessary.
I think you shoud use histc
[~,Y] = histc(X,q)
As you can see in matlab's doc:
Description
n = histc(x,edges) counts the number of values in vector x that fall
between the elements in the edges vector (which must contain
monotonically nondecreasing values). n is a length(edges) vector
containing these counts. No elements of x can be complex.
I made a couple of refinements (including one inspired by Aero Engy in another answer) that have resulted in some improvements. To test them out, I created a random matrix of a million rows and 100 columns to run the improved functions on:
>> x = randn(1000000,100);
First, I ran my unmodified code, with the following results:
Note that of the 40 seconds, around 14 of them are spent computing the quantiles - I can't expect to improve this part of the routine (I assume that Mathworks have already optimized it, though I guess that to assume makes an...)
Next, I modified the routine to the following, which should be faster and has the advantage of being fewer lines as well!
function [Y q] = levels(X,nLevels)
p = linspace(0, 1.0, nLevels+1);
q = quantile(X,p);
if isvector(q), q = transpose(q); end
Y = ones(size(X));
for i = 2:nLevels
Y = Y + bsxfun(#ge,X,q(i,:));
end
The profiling results with this code are:
So it is 15 seconds faster, which represents a 150% speedup of the portion of code that is mine, rather than MathWorks.
Finally, following a suggestion of Andrey (again in another answer) I modified the code to use the second output of the histc function, which assigns entries to bins. It doesn't treat the columns independently, so I had to loop over the columns manually, but it seems to be performing really well. Here's the code:
function [Y q] = levels(X,nLevels)
p = linspace(0,1,nLevels+1);
q = quantile(X,p);
if isvector(q), q = transpose(q); end
q(end,:) = 2 * q(end,:);
Y = zeros(size(X));
for k = 1:size(X,2)
[junk Y(:,k)] = histc(X(:,k),q(:,k));
end
And the profiling results:
We now spend only 4.3 seconds in codes outside the quantile function, which is around a 500% speedup over what I wrote originally. I've spent a bit of time writing this answer because I think it's turned into a nice example of how you can use the MATLAB profiler and StackExchange in combination to get much better performance from your code.
I'm happy with this result, although of course I'll continue to be pleased to hear other answers. At this stage the main performance increase will come from increasing the performance of the part of the code that currently calls quantile. I can't see how to do this immediately, but maybe someone else here can. Thanks again!
You can sort the columns and divide+round the inverse indexes:
function Y = levels(X,nLevels)
% "Assign each of the elements of X to an integer-valued level"
[S,IX]=sort(X);
[grid1,grid2]=ndgrid(1:size(IX,1),1:size(IX,2));
invIX=zeros(size(X));
invIX(sub2ind(size(X),IX(:),grid2(:)))=grid1;
Y=ceil(invIX/size(X,1)*nLevels);
Or you can use tiedrank:
function Y = levels(X,nLevels)
% "Assign each of the elements of X to an integer-valued level"
R=tiedrank(X);
Y=ceil(R/size(X,1)*nLevels);
Surprisingly, both these solutions are slightly slower than the quantile+histc solution.