Adding a constraint to my if else if statement.

So the follwoing for loop will generate 8133 random values for 'astrosRuns' between the possible range of (0:16). The random values for 'astrosRuns' are generated depending upon the random values of either 'j' or 'k'.

There are 8133 steps in this for loop. I want to add the constraint that the average of every group of 10 'astroRuns' values equals 4. (that's the average in values not the sum) Meaning that a group of ten values could possibly be: (1,4,6,6,2,2,9,4,2,4). These ten steps have an average in values of 4. And then the next group of 10 values must also have an average in values of 4 however, it could be a different combination in values like: (8,5,7,4,0,0,1,4,5,6).

Here is the example code:

for i=1:8133;
    %    Let j = the probability for the score value
    %     Let k = the probability that the adjacent values will be equal
    %
    j =rand;
    k =rand;

      if 0<=k<=.0063
          astrosRuns(i+1)=astrosRuns(i);

      elseif 0<=j<=.0687
          astrosRuns(i+1)=0;

      elseif .0688<=j<=.1880
          astrosRuns(i+1)=1;

      elseif .1881<=j<=.3286
          astrosRuns(i+1)=2;

      elseif .3287<=j<=.4801
          astrosRuns(i+1)=3;

      elseif .4802<=j<=.6110
          astrosRuns(i+1)=4;

      elseif .06111<=j<=.7223
          astrosRuns(i+1)=5;

      elseif .7224<=j<=.8050
          astrosRuns(i+1)=6;

      elseif .8051<=j<=.8696
          astrosRuns(i+1)=7;

      elseif .8697<=j<=.9156
          astrosRuns(i+1)=8;

      elseif .9157<=j<=.9455
          astrosRuns(i+1)=9;

      elseif .9456<=j<=.9671
          astrosRuns(i+1)=10;

      elseif .9672<=j<=.9797
          astrosRuns(i+1)=11;

      elseif .9798<=j<=.9874
          astrosRuns(i+1)=12;

      elseif .9875<=j<=.9931
          astrosRuns(i+1)=13;

      elseif .9932<=j<=.9962
          astrosRuns(i+1)=14;

      elseif .9963<=j<=.9984
          astrosRuns(i+1)=15;

      else
          .9985<=j<=1;
          astrosRuns(i+1)=16;
      end
  end

So basically, I want to have the values for 'astrosRuns' randomly created but also follow a law of average of 4.

I hope this is clearer now. Thanks for all the help!

[Oleg Komarov]

I don't understand how do you want the average rule combined with the binning. Can you elaborate on that? Because the binning takes into consideration 1 value per time while the average rule 10 values.

For example, do you want to discard the past 10 astroRuns if their average is above 4?

Answer 1

[Sean de Wolski]

This could be written without the for-loop or the if-statement by using histc.

x = rand(8033,1);
edges = [0,.0687,.1880] %etc. for all of yoru values

[~, astroRuns] = histc(x,edges); %edited!

And

doc histc

for how this works. Also, to find out why what you are doing now does not work:

Loren's Blog: 3<x<7

[Clifford Shelton]

Could you be a little more detailed with what you are trying to help me with?

I don't understand how what your example shown addresses my need to have a constraint on the average values of 10 steps in the loop.

I also don't understand how your example will generate my desired values with desired probabilities between (0:16).

Thanks for the help. But I'm a beginner and need things really explained in detail!

[Sean de Wolski]

Okay. First I had a small mistake that is fixed now.

What histc does is take a vector of numbers, x, and divides them into bins, defined by their edges. This is the same thing you are trying to do with your if/else statements. I.e. if it's greater than this edge and less than the next edge, increment this value.

So let's picture it with bigger integers:

x = [2.5 3.5 3.5 7.7];
edges = 1:10;
[bins,astroRuns] = histc(x,edges)

bins = 0 1 2 0 0 0 1 0 0 0 astroRuns = 2 3 3 7

[Sean de Wolski]

Now the second bin ranging from 2 to 3 has one value in it (2.5). This shows up in bins. We are not quite as worried about bins here as: "where did each value go? I.e. you increment by setting 2,3...16 to where that bin is true. I used the second output of histc to identify the bin corresponding to each x-value.

Welcome to MATLAB and Answers!

[Clifford Shelton]

Ok..I think I understand what you are saying. But I have no clue how to write the code.

Do I want my edges variable to be the range of values (0:16) or do I want my edges variable to be the ranges for the probabilities of 'k' in my above code (0,.0687,.1880,.3286,...etc.)

Furthermore, I'm still clueless about how to do an average for each group of 10values.

Answer 2

[Daniel Shub]

This has been said many times before on this site:

0<=k<=.0063

does not do what you think it does. For all k greater than or equal to 0, this statement is false, even if k is less than or equal to .0063.

[Clifford Shelton]

umm..ok...so then what should I say to have it do what I want?

[Oleg Komarov]

Just <= 0.0063 since 0 is ensured by rand.

[Daniel Shub]

Loren's blog deals with the issues nicely.

[Daniel Shub]

@Oleg, good point, but .4802<=j<=.6110 will cause problems.

[Oleg Komarov]

Yes but k has only one condition.