I am a beginner exploring the world of Mathematica. I expected the following code
T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] -> Length[List[b]]}
should return the value 2, rather than 1. Anyone could explain where I was wrong?
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2 Answers
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This is because the code Length[List[b]] is evaluated before the rule is applied. Using RuleDelayed rather than Rule would fix it:
T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] :> Length[{b}]}
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The function Trace can be helpful in diagnosing the problem. The documentation says:
Trace returns a set of nested lists. Each individual list corresponds to a single evaluation chain, which contains the sequence of forms found for a particular expression. The list has sublists which give the histories of subsidiary evaluations.
With your input:
Trace[
T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] -> Length[List[b]]}
]
{{{{Length[{b}],1},T[a___,1,b___]->1,T[a___,1,b___]->1}, {T[a___,1,b___]->1}},T[6,5,4,1,2,3]/.{T[a___,1,b___]->1},1}
Observe that the first evaluation chain is {Length[{b}],1} which clearly shows the problem. Now compare the Trace when correctly using RuleDelayed:
Trace[
T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] :> Length[List[b]]}
]
{{{T[a___,1,b___]:>Length[{b}],T[a___,1,b___]:>Length[{b}]}, {T[a___,1,b___]:>Length[{b}]}},T[6,5,4,1,2,3]/. {T[a___,1,b___]:>Length[{b}]},Length[{2,3}],2}
answered Apr 8, 2013 at 13:17
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$\begingroup$ Thank you. This would be very helpful to understand the behavior of Mathematica. $\endgroup$ Commented Apr 9, 2013 at 23:23
Tis a bad choice of variable/function name. I suggest using names starting in lowercase letters, especially when using single letters. This avoids conflicts with built-ins (e.g.N,D...) $\endgroup$