I'm trying to do the following:

R = {a, b, c, d, e, f, g, h, i};
Table[R[[n]] = RandomInteger[{0, 20}], {n, 1, Length[R]}]

My intention is to assign a random value to each of these variables. But somehow it's not working. What else do I need to do? I've tried to do:

ToExpression[R[[n]] = RandomInteger[{0, 30}]]


Evaluate[R[[n]] = RandomInteger[{0, 30}]]
  • 3
    $\begingroup$ Hello Gustavo, you'll want to avoid assigning things to capital letters. That being said, try Scan[(# = RandomInteger[{0, 30}]) &, R]. $\endgroup$ Commented May 24, 2015 at 22:45

1 Answer 1


Remember that = is really Set, and Set is a pretty ordinary function. Also, RandomInteger can make a list all by itself, so you don't need Table.

Edit 11/15: Jerry Guern pointed out the need to be careful with quoting to have it work repeatedly. Here, I use Unevaluated to quote each variable:

r = Unevaluated /@ {a, b, c, d, e, f, g, h, i};

Now, you may use r repeatedly as input to:

MapThread[Set, {r, RandomInteger[{0, 20}, Length[r]]}];
  • 2
    $\begingroup$ But you can only do this once. If you try to use the 2nd line a second time, it fails. $\endgroup$ Commented Nov 15, 2015 at 2:09

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