7
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Is there any way to get different values for different blanks with ReplaceAll[]? For example, I would like to do something like:

{_,_,_}/. _Blank -> RandomInteger[{1, 10}]

But, the output comes out as all the same random integer for each blank, such as:

{4, 4, 4}

Is there an efficient way to get an output that is more random, like:

{4, 7, 1}

,as shown above?

Thanks!

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    $\begingroup$ I suppose something like Replace[{_, _, _}, _ :> RandomInteger[{1, 10}], {1}] would do what you ask for, but why not just use RandomInteger[{1,10}, 3] instead? $\endgroup$
    – Carl Woll
    Commented Sep 13, 2018 at 0:06
  • $\begingroup$ Thanks, that's a good idea - I should have been more specific though, because it wasn't just for {,,}, but also for variable lengths, such as {,}, or even {,{,}} and such $\endgroup$
    – Jmeeks29ig
    Commented Sep 13, 2018 at 15:50

2 Answers 2

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You can simply change the Rule -> to RuleDelayed :> to give

{_, _, _} /. _Blank :> RandomInteger[{1, 10}]

(*{5, 1, 3}*)
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  • $\begingroup$ Awesome, thanks! $\endgroup$
    – Jmeeks29ig
    Commented Sep 13, 2018 at 14:34
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Just for variety: you can also temporarily redefine Blank as RandomInteger[{1,10}]& using Block:

Block[{Blank = RandomInteger[{1, 10}] &}, {_, _, x, y, _, z}]

{5, 7, x, y, 5, z}

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  • $\begingroup$ Thanks, I appreciate it! $\endgroup$
    – Jmeeks29ig
    Commented Sep 13, 2018 at 14:34

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