anonymous function with rng call inside

I would like to have a time domain function that uses a random number, but is stable over an interval. The anonymous function below does this, but it is not elegant.

f = @(t) {rng(ceil(t/10)), rand(1)}; % returns new random numbers each decade of t

f(2)

ans =

1×2 cell array

{1×1 struct} {[0.4170]}

The random number in second cell is what I want, but it can only be accessed through another variable. Is there a way in Matlab to simply return the random number without all the indirection?

Thank you for any help you can offer.

[William Buller]

Sometimes you just need to ask the question and then it comes to you!

f = @(t) isstruct(rng(ceil(t/10)))*rand(1);

[Paul]

Are there bounds on t?

[Matt J]

f = @(t) isstruct(rng(ceil(t/10)))*rand(1);

Not a good idea. You are assuming sub-expressions in the function are evaluated left-to-right. That is not reliable.Why does the function have to be anonymous?

[Walter Roberson]

@Matt J

MATLAB has a well-defined order of operations, that mostly guarantees left-to-right operations (as modified by the operation precedences, that leads to oddities like c^a^b ). The main exception is the vaguely-documented linear algebra exceptions where for example a'*b might be specially evaluated rather than being evaluated as (a')*b

Answer 1

[Matt J]

function y=f(t)

  rng(ceil(t/10)); 
  y=rand(1);

end

Answer 2

[Paul]

t = 2*(rand(1,1000)-0.5)*100;
figure
plot(t,f(t,10),'b.',t,f(t,10),'r.')

function y = f(t,interval)
   tinterval = ceil(t/interval);
   [C,ia,ic] = unique(tinterval);
   yout = rand(size(C));
   y = yout(ic);
end

Answer 3

[Matt J]

rng('default')

T= rand(1,10);

t=1:15;

T(ceil(t/10))

ans =

Columns 1 through 9

0.8147 0.8147 0.8147 0.8147 0.8147 0.8147 0.8147 0.8147 0.8147

Columns 10 through 15

0.8147 0.9058 0.9058 0.9058 0.9058 0.9058

Answer 4

[Walter Roberson]

f = @(t) struct('rng', rng(ceil(t/10)), 'r1', rand(1)).r1

f = function_handle with value:
    @(t)struct('rng',rng(ceil(t/10)),'r1',rand(1)).r1

f(2)

ans = 0.4170