I will try Hans. Here we go

Do[ m={{a1,a2},{a3,a4}}; (*2x2 matrix*)
a1 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}]; a3 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}];
a2 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}]; a4 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}];
If[a1>a2 && a3>a4,PutAppend[{a1,a2,a3,a4,m},"list"]],{100}]

At this point we stored the values for a1, a2, a3 ,a4 and m that satisfy the condition of the first Do loop. Now we are going to calculate the value for a matrix M, that is,

dn=ReadList["list"]; (*here I'm calling the list because I need the values found previously*)

Do[ M[i_]:= {{b1,b2},{b3.b4}}*dn[[i,3]]+dn[[i,5]]*b1 (*2x2 matrix where I'm using the data from l"ist"*)
b1 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}]; b3 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}];
b2 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}]; b4 = RandomReal[{0, 1.}]*RandomChoice[{-1., 1.}];
If[Eigenvalues[M[i]][[1]]> 0 && b1> dn[[i,1]] && 0<Eigenvectors[M[i]][[1,1]]<0.5,
PutAppend[{M[i],b1,b2,b3,b4,dn[[i,1]],dn[[i,2]],m},"list2"],{i,1,Length[dn]}]

In the second Do loop, I used the values from the first one, this is a simple example. But think about what if in the first Do loop I need to do 100000 iterations in order to get a lot of data and then use them in the second Do loop, It's not good for the memory of the code. How can I make that in every iteration that the first Do loop finds a set of values, go immediately to the second one to evaluate it?

In the case of a bigger matrix, with more conditions, and more restrictive constraints, what would be the better way to optimize it in time and memory?

But while your example is not so simple, it does show the general structure. So here is an alternative you could try to adapt:

fa[x_] := Divisible[x, 7] (* 1st condition *)
fb[x_] := Divisible[x, 3] (* 2nd condition *)

list = {};
Do[
    temp = RandomInteger[10000];
    If[fa[temp] && fb[temp], AppendTo[list, Sqrt@temp]],
    200
]

From this and your earlier post I assume that you have a set of expressions in a file. And you want to take, from that file, those expressions which fulfills both of two conditions. Am I right?

If so it could be convenient to use the function Select:

file = Range@1000;
Select[file, Divisible[#, 3] && Divisible[#, 7] &]

Felipe, could you give a skeleton example of how you now do it with the two Do loops? Numerical and simple.

Paul, what would be the job of the variable "y" here?

Filipe,

There are many ways to do something like this. Please provide a simple, working example. It’s difficult to help you without understanding exactly what you want.

Regards

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