How can this rule
a -> {1, 2, 3}
be transformed into this
{a -> 1, a -> 2, a -> 3}
by way of a function?
Answers 3175 and 10524 only address extracting values from the right-hand side, so (as far as I can tell) this question is distinct.
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1$\begingroup$ I don't really understand what you mean by "without referring to the symbol/string on the left-hand side explicitly in the code". Can you give an example? It would be best to have an example input and an example output. $\endgroup$– SzabolcsCommented Nov 14, 2014 at 20:30
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1 Answer
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Use Thread
Thread[ a -> {1, 2, 3} ]
and
Join @@ Thread /@ {a -> {1, 2, 3}, b -> {4, 5, 6}}
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answered Nov 14, 2014 at 20:25
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$\begingroup$ How would this be done without knowing that "a" is "a". In your answer, you refer to "a" in your code. I'll update the question to make it clearer. A related question would be: How do you extract the left-hand side of a rule efficiently? $\endgroup$– higgyCommented Nov 14, 2014 at 20:27
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$\begingroup$ @higgy
a -> bis just an alternate notation forRule[a,b], so you can extractausingFirst,Part,Extract, etc. See here: reference.wolfram.com/language/tutorial/PartsOfExpressions.html $\endgroup$– SzabolcsCommented Nov 14, 2014 at 20:29 -
$\begingroup$ If extracting the left-hand side is a necessary step in all possible solutions to this question, then edit your answer I and will accept it. I'm also curious as to whether there are solutions that don't require extracting the left-hand side first, i.e., something more direct. $\endgroup$– higgyCommented Nov 14, 2014 at 20:34
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$\begingroup$ @higgy You don't need to know the first part of the rule to apply
Threadto it. If you havex = a -> {1,2,3}then you can just useThread[x], no need to know what's insidex, other than that it's a rule with a list on the RHS. Can you give an example input and a corresponding example output? I thoughta -> {1, 2, 3}and{a -> 1, a -> 2, a -> 3}could be considered such.Threadtakes you from one to the other. $\endgroup$– SzabolcsCommented Nov 14, 2014 at 20:35 -
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