how to resize my large vector to a smaller one for plotting?

hey guys, I have a large vector with length of 185437, i want to make a new one, reshaped vector with mean values of this one but I need new length to be 180*delay because on x axis i must have like this, for example:

       delay=0.1
       x_axis=0:delay:180

i need y axis (this new vector) to be same size as this x_axis so I can plot it. I tried something like:

       values_delay_sum_plot=mean(reshape(values_delay_sum,1/delay,length(values_delay_sum)*delay));

but it is not working. First errors was for integer values, later that I can't change size of vector. Note that, values_delay_sum is my large vector, a from it I wanna make a new one for my plot. thanks in advance.

Answer 1

[José-Luis]

 doc resample

[Bojan]

      resample(values_delay_sum,100,round(length(values_delay_sum)/100))

When I try it i got this errors.....

   Error using  * 
   BLAS loading error:
    dlopen: cannot load any more object with static TLS

    Error in rat (line 103)
                C = [C*[d;1] C(:,1)];

    Error in resample (line 70)
     [p,q] = rat( p/q, 1e-12 );  %--- reduce to lowest terms

[José-Luis]

 values_delay_sum = rand(185437,1);
 resample(values_delay_sum,100,round(length(values_delay_sum)/100));

It works for me. What do you have in values_delay_sum?

[Bojan]

values_delay_sum is vector that is random generated and aproxx has between 180500 and 200100 values in it.

[José-Luis]

What does

 class(values_delay_sum) 
 size(values_delay_sum)

return?

[Bojan]

It returns

>> class(values_delay_sum)

    ans =

   double

   >> size(values_delay_sum)

   ans =

           1       16019

[José-Luis]

What's the time stamp of that vector?