Several comments here about the need to de-vectorize code and use for-loops 
instead. However, vectorized code is a lot more compact and generally 
easier to read than lots of for-loops. I seem to recall that there was 
discussion in the past about speeding up vectorized code in Julia so that 
it could be on par with the vectorized code performance - is this still 
something being worked on for future versions?

Otherwise, if we keep telling people that they need to convert their code 
to use for-loops, I think Julia isn't going to seem very compelling for 
people looking for alternatives to Matlab, R, etc.

On Sunday, October 18, 2015 at 6:41:54 AM UTC-7, Daniel Carrera wrote:
>
> Hello,
>
> Other people have already given advice on how to speed up the code. I just 
> want to comment that Julia really is faster than Matlab, but the way that 
> you make code faster in Julia is almost the opposite of how you do it in 
> Matlab. Specifically, in Matlab the advice is that if you want the code to 
> be fast, you need to eliminate every loop you can and write vectorized code 
> instead. This is because Matlab loops are slow. But Julia loops are fast, 
> and vectorized code creates a lot of overhead in the form of temporary 
> variables, garbage collection, and extra loops. So in Julia you optimize 
> code by putting everything into loops. The upshot is that if you take a 
> Matlab-optimized program and just do a direct line-by-line conversion to 
> Julia, the Julia version can easily be slower. But by the same token, if 
> you took a Julia-optimized program and converted it line-by-line to Matlab, 
> the Matlab version would be ridiculously slow.
>
> Oh, and in Julia you also care about types. If the compiler can infer 
> correctly the types of your variables it will write more optimal code.
>
> Cheers,
> Daniel.
>
>
> On Sunday, 18 October 2015 13:17:50 UTC+2, Vishnu Raj wrote:
>>
>> Although Julia homepage shows using Julia over Matlab gains more in 
>> performance, my experience is quite opposite.
>> I was trying to simulate channel evolution using Jakes Model for wireless 
>> communication system.
>>
>> Matlab code is:
>> function [ h, tf ] = Jakes_Flat( fd, Ts, Ns, t0, E0, phi_N )
>> %JAKES_FLAT 
>> %   Inputs:
>> %       fd, Ts, Ns  : Doppler frequency, sampling time, number of samples
>> %       t0, E0      : initial time, channel power
>> %       phi_N       : initial phase of the maximum Doppler frequeny
>> %       sinusoid
>> %
>> %   Outputs:
>> %       h, tf       : complex fading vector, current time
>>
>>     if nargin < 6,  phi_N = 0;  end
>>     if nargin < 5,  E0 = 1;     end
>>     if nargin < 4,  t0 = 0;     end
>>     
>>     N0 = 8;         % As suggested by Jakes
>>     N  = 4*N0 + 2;  % an accurate approximation
>>     wd = 2*pi*fd;   % Maximum Doppler frequency[rad]
>>     t  = t0 + [0:Ns-1]*Ts;  % Time vector
>>     tf = t(end) + Ts;       % Final time
>>     coswt = [ sqrt(2)*cos(wd*t); 2*cos(wd*cos(2*pi/N*[1:N0]')*t) ];
>>     h  = E0/sqrt(2*N0+1)*exp(j*[phi_N pi/(N0+1)*[1:N0]])*coswt;
>> end
>> Enter code here...
>>
>> My call results in :
>> >> tic; Jakes_Flat( 926, 1E-6, 50000, 0, 1, 0 ); toc
>> Elapsed time is 0.008357 seconds.
>>
>>
>> My corresponding Julia code is
>> function Jakes_Flat( fd, Ts, Ns, t0 = 0, E0 = 1, phi_N = 0 )
>> # Inputs:
>> #
>> # Outputs:
>>   N0  = 8;                  # As suggested by Jakes
>>   N   = 4*N0+2;             # An accurate approximation
>>   wd  = 2*pi*fd;            # Maximum Doppler frequency
>>   t   = t0 + [0:Ns-1]*Ts;
>>   tf  = t[end] + Ts;
>>   coswt = [ sqrt(2)*cos(wd*t'); 2*cos(wd*cos(2*pi/N*[1:N0])*t') ]
>>   h = E0/sqrt(2*N0+1)*exp(im*[ phi_N pi/(N0+1)*[1:N0]']) * coswt
>>   return h, tf;
>> end
>> # Saved this as "jakes_model.jl"
>>
>>
>> My first call results in 
>> julia> include( "jakes_model.jl" )
>> Jakes_Flat (generic function with 4 methods)
>>
>> julia> @time Jakes_Flat( 926, 1e-6, 50000, 0, 1, 0 )
>> elapsed time: 0.65922234 seconds (61018916 bytes allocated)
>>
>> julia> @time Jakes_Flat( 926, 1e-6, 50000, 0, 1, 0 )
>> elapsed time: 0.042468906 seconds (17204712 bytes allocated, 63.06% gc 
>> time)
>>
>> For first execution, Julia is taking huge amount of time. On second call, 
>> even though Julia take considerably less(0.042468906 sec) than first(
>> 0.65922234 sec), it's still much higher to Matlab(0.008357 sec).
>> I'm using Matlab R2014b and Julia v0.3.10 on Mac OSX10.10.
>>
>> - vish
>>
>

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