## Archive for the ‘Julia’ Category

A question I get asked a lot is ‘How can I do nonlinear least squares curve fitting in X?’ where X might be MATLAB, Mathematica or a whole host of alternatives. Since this is such a common query, I thought I’d write up how to do it for a very simple problem in several systems that I’m interested in

This is the Julia version. For other versions,see the list below

- Simple nonlinear least squares curve fitting in Maple
- Simple nonlinear least squares curve fitting in Mathematica
- Simple nonlinear least squares curve fitting in MATLAB
- Simple nonlinear least squares curve fitting in Python
- Simple nonlinear least squares curve fitting in R

**The problem**

xdata = -2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9 ydata = 0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001

and you’d like to fit the function

using nonlinear least squares. You’re starting guesses for the parameters are p1=1 and P2=0.2

For now, we are primarily interested in the following results:

- The fit parameters
- Sum of squared residuals

Future updates of these posts will show how to get other results such as confidence intervals. Let me know what you are most interested in.

**Julia solution using Optim.jl**

Optim.jl is a free Julia package that contains a suite of optimisation routines written in pure Julia. If you haven’t done so already, you’ll need to install the Optim package

Pkg.add("Optim")

The solution is almost identical to the example given in the curve fitting demo of the Optim.jl readme file:

using Optim model(xdata,p) = p[1]*cos(p[2]*xdata)+p[2]*sin(p[1]*xdata) xdata = [-2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9] ydata = [0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001] beta, r, J = curve_fit(model, xdata, ydata, [1.0, 0.2]) # beta = best fit parameters # r = vector of residuals # J = estimated Jacobian at solution @printf("Best fit parameters are: %f and %f",beta[1],beta[2]) @printf("The sum of squares of residuals is %f",sum(r.^2.0))

The result is

Best fit parameters are: 1.881851 and 0.700230 @printf("The sum of squares of residuals is %f",sum(r.^2.0))

**Notes**

I used Julia version 0.2 on 64bit Windows 7 to run the code in this post

From the wikipedia page on Division by Zero: *“The IEEE 754 standard specifies that every floating point arithmetic operation, including division by zero, has a well-defined result”.*

MATLAB supports this fully:

>> 1/0 ans = Inf >> 1/(-0) ans = -Inf >> 0/0 ans = NaN

Julia is almost there, but doesn’t handled the signed 0 correctly (This is using Version 0.2.0-prerelease+3768 on Windows)

julia> 1/0 Inf julia> 1/(-0) Inf julia> 0/0 NaN

Python throws an exception. (Python 2.7.5 using IPython shell)

In [4]: 1.0/0.0 --------------------------------------------------------------------------- ZeroDivisionError Traceback (most recent call last) in () ----> 1 1.0/0.0 ZeroDivisionError: float division by zero In [5]: 1.0/(-0.0) --------------------------------------------------------------------------- ZeroDivisionError Traceback (most recent call last) in () ----> 1 1.0/(-0.0) ZeroDivisionError: float division by zero In [6]: 0.0/0.0 --------------------------------------------------------------------------- ZeroDivisionError Traceback (most recent call last) in () ----> 1 0.0/0.0 ZeroDivisionError: float division by zero

**Update:**

Julia **does** do things correctly, provided I make it clear that I am working with floating point numbers:

julia> 1.0/0.0 Inf julia> 1.0/(-0.0) -Inf julia> 0.0/0.0 NaN

I support scientific applications at The University of Manchester (see my LinkedIn profile if you’re interested in the details) and part of my job involves working on code written by researchers in a variety of languages. When I say ‘variety’ I really mean it – MATLAB, Mathematica, Python, C, Fortran, Julia, Maple, Visual Basic and PowerShell are some languages I’ve worked with this month for instance.

Having to juggle the semantics of so many languages in my head sometimes leads to momentary confusion when working on someone’s program. For example, I’ve been doing a lot of Python work recently but this morning I was hacking away on someone’s MATLAB code. Buried deep within the program, it would have been very sensible to be able to do the equivalent of this:

a=rand(3,3) a = 0.8147 0.9134 0.2785 0.9058 0.6324 0.5469 0.1270 0.0975 0.9575 >> [x,y,z]=a(:,1) Indexing cannot yield multiple results.

That is, I want to be able to take the first column of the matrix a and broadcast it out to the variables x,y and z. The code I’m working on uses MUCH bigger matrices and this kind of assignment is occasionally useful since the variable names x,y,z have slightly more meaning than a(1,3), a(2,3), a(3,3).

The only concise way I’ve been able to do something like this using native MATLAB commands is to first convert to a cell. In MATLAB 2013a for instance:

>> temp=num2cell(a(:,1)); >> [x y z] = temp{:} x = 0.8147 y = 0.9058 z = 0.1270

This works but I think it looks ugly and introduces conversion overheads. The problem I had for a short time is that I subconsciously expected multiple assignment to ‘Just Work’ in MATLAB since the concept makes sense in several other languages I use regularly.

from pylab import rand a=rand(3,3) [a,b,c]=a[:,0]

a = RandomReal[1, {3, 3}] {x,y,z}=a[[All,1]]

a=rand(3,3); (x,y,z)=a[:,1]

I’ll admit that I don’t often need this construct in MATLAB but it would definitely be occasionally useful. I wonder what other opinions there are out there? Do you think multiple assignment is useful (in any language)?

I felt like playing with Julia and MATLAB this Sunday morning. I found some code that prices European Options in MATLAB using Monte Carlo simulations over at computeraidedfinance.com and thought that I’d port this over to Julia. Here’s the original MATLAB code

function V = bench_CPU_European(numPaths) %Simple European steps = 250; r = (0.05); sigma = (0.4); T = (1); dt = T/(steps); K = (100); S = 100 * ones(numPaths,1); for i=1:steps rnd = randn(numPaths,1); S = S .* exp((r-0.5*sigma.^2)*dt + sigma*sqrt(dt)*rnd); end V = mean( exp(-r*T)*max(K-S,0) )

I ran this a couple of times to see what results I should be getting and how long it would take for 1 million paths:

tic;bench_CPU_European(1000000);toc V = 13.1596 Elapsed time is 6.035635 seconds. >> tic;bench_CPU_European(1000000);toc V = 13.1258 Elapsed time is 5.924104 seconds. >> tic;bench_CPU_European(1000000);toc V = 13.1479 Elapsed time is 5.936475 seconds.

The result varies because this is a stochastic process but we can see that it should be around 13.1 or so and takes around 6 seconds on my laptop. Since it’s Sunday morning, I am feeling lazy and have no intention of considering if this code is optimal or not right now. I’m just going to copy and paste it into a julia file and hack at the syntax until it becomes valid Julia code. The following seems to work

function bench_CPU_European(numPaths) steps = 250 r = 0.05 sigma = .4; T = 1; dt = T/(steps) K = 100; S = 100 * ones(numPaths,1); for i=1:steps rnd = randn(numPaths,1) S = S .* exp((r-0.5*sigma.^2)*dt + sigma*sqrt(dt)*rnd) end V = mean( exp(-r*T)*max(K-S,0) ) end

I ran this on Julia and got the following

julia> tic();bench_CPU_European(1000000);toc() elapsed time: 36.259000062942505 seconds 36.259000062942505 julia> bench_CPU_European(1000000) 13.114855104505445

The Julia code appears to be valid, it gives the correct result of 13.1 ish but at 36.25 seconds is around **6 times slower** than the MATLAB version. The dog needs walking so I’m going to think about this another time but comments are welcome.

**Update (9pm 7th October 2012):** I’ve just tried this Julia code on the Linux partition of the same laptop and 1 million paths took 14 seconds or so:

tic();bench_CPU_European(1000000);toc() elapsed time: 14.146281957626343 seconds

I built this version of Julia from source and so it’s at the current bleeding edge (version 0.0.0+98589672.r65a1 Commit 65a1f3dedc (2012-10-07 06:40:18). The code is still slower than the MATLAB version but better than the older Windows build

**Update: 13th October 2012**

Over on the Julia mailing list, someone posted a faster version of this simulation in Julia

function bench_eu(numPaths) steps = 250 r = 0.05 sigma = .4; T = 1; dt = T/(steps) K = 100; S = 100 * ones(numPaths,1); t1 = (r-0.5*sigma.^2)*dt t2 = sigma*sqrt(dt) for i=1:steps for j=1:numPaths S[j] .*= exp(t1 + t2*randn()) end end V = mean( exp(-r*T)*max(K-S,0) ) end

On the Linux partition of my test machine, this got through 1000000 paths in 8.53 seconds, very close to the MATLAB speed:

julia> tic();bench_eu(1000000);toc() elapsed time: 8.534484148025513 seconds

It seems that, when using Julia, one needs to unlearn everything you’ve ever learned about vectorisation in MATLAB.

**Update: 28th October 2012**

Members of the Julia team have been improving the performance of the randn() function used in the above code (see here and here for details). Using the de-vectorised code above, execution time for 1 million paths in Julia is now down to 7.2 seconds on my machine on Linux. Still slower than the MATLAB 2012a implementation but it’s getting there. This was using Julia version 0.0.0+100403134.r0999 Commit 099936aec6 (2012-10-28 05:24:40)

tic();bench_eu(1000000);toc() elapsed time: 7.223690032958984 seconds 7.223690032958984

- Laptop model: Dell XPS L702X
- CPU:Intel Core i7-2630QM @2Ghz software overclockable to 2.9Ghz. 4 physical cores but total 8 virtual cores due to Hyperthreading.
- RAM: 8 Gb
- OS: Windows 7 Home Premium 64 bit and Ubuntu 12.04
- MATLAB: 2012a
- Julia: Original windows version was Version 0.0.0+94063912.r17f5, Commit 17f50ea4e0 (2012-08-15 22:30:58). Several versions used on Linux since, see text for details.

I first mentioned Julia, a new language for high performance scientific computing, back in the February edition of a Month of Math software and it certainly hasn’t stood still since then. A WalkingRandomly reader, Ohad, recently contacted me to tell me about a forum post announcing some Julia speed improvements.

Julia has a set of micro-benchmarks and the **slowest of them is now only two times slower than the equivalent in C**. That’s compiled language performance with an easy to use scripting language. Astonishingly, Julia is faster than gfortran in a couple of instances. Nice work!

Comparison times between Julia and other scientific scripting languages (MATLAB, Python and R for instance) for these micro-benchmarks are posted on Julia’s website. The Julia team have included the full benchmark source code used for all languages so if you are an expert in one of them, why not take a look at how they have represented your language and see what you think?

Let me know if you’ve used Julia at all, I’m interested in what people think of it.