How can I use replacement rules to match and work with a list. For example,
{1, {2, 3}} /. {{a_, b_} -> ((a + # &) /@ b)}
outputs
{2,3}
but I expected
{3,4}
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2 Answers
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Lists have nothing to do with the problem. The reason why you obtained the result above is that -> (Rule) evaluates its arguments before going further. So
(a + # &) /@ b
inside your rule first evaluated to
b
resulting in the rule
{a_, b_} -> b
and only after that the replacement
{1, {2, 3}} /. {{a_, b_} -> b}
evaluated to {2, 3}. You can use :> (RuleDelayed) to prevent the second argument of :> from evaluation:
{1, {2, 3}} /. {{a_, b_} :> ((a + # &) /@ b)}
(*{3, 4}*)
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Specifying b with a head List works
{1, {2, 3}} /. {a_, b_List} -> a + b
But why not just use Plus:
Plus @@ {1, {2, 3}}
answered Dec 21, 2018 at 17:13
->to:>as in/. {{a_, b_} :> ((a + # &) /@ b)}$\endgroup$