Erlang (programming language)/Tutorials/Folding: Difference between revisions
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Fun with folding | Fun with folding | ||
Revision as of 17:44, 2 February 2009
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Fun with folding
Fold is a powerful tool, when you get to know it. lists:foldr(F,S,L) takes three arguments: F is some folding function, S is some starting value, and L is some List that needs folding. Let us do some simple folding. The following fold calculates the lenght of a list. Here the current list item, _C is ignored, and the accumulator, A, counts the number of elements in the list.
1> lists:foldr( fun(_C,A)->A+1 end, 0, [1,2,3]). 3
We could reverse a list with fold if we like.
2> lists:foldr(fun(C,A)->A++[C] end, [], [a,b,c]). [c,b,a]
Or to get fancy we could try a finite difference.
3> lists:foldr( fun(C,{P,A})->{C,[P-C]++A} end, {0,[]}, [1,4,9,16]).
{1,[3,5,7,-16]}
In example 3, we used a tuple to remember the previous value so we could use it for next difference. Fold is a function with a limited attention span. We might like to delete the last item because is has a non-intuitive relation to the source list.
Calculating a continuted calculation with fold seems natural
lists:foldr(fun(C,A)->(A+1)/C end, 1, lists:seq(1,1000)). 1.718281828459045
The answer appears to be the value e-1, which we can verify with:
math:exp(1) - 1. 1.718281828459045