Thanks Tim! I got it! I have never declared a function before in a let ... in statement, I always do it in a where after I call it...
On Fri, Oct 30, 2009 at 3:03 PM, Tim Wawrzynczak <inforichl...@gmail.com>wrote: > Hector, > > That line is declaring a function named 'f' of two arguments: one is 'w', > and the other is a tuple. The tuple's fst is 'inputs', and its snd is > 'expected.' This function (f) is used in the next line, in the declaration > of the list 'newWeights,' which uses f as the function which does the fold > over the allInputs list. > > Cheers, > - Tim > > On Thu, Oct 29, 2009 at 2:27 PM, Hector Guilarte <hector...@gmail.com>wrote: > >> Hi cafe, >> >> I'm trying to implement a Perceptron in Haskell and I found one in: >> http://jpmoresmau.blogspot.com/2007/05/perceptron-in-haskell.html (Thanks >> JP Moresmau) but there is one line I don't understand, I was wondering if >> someone could explain it to me. I know the theory behind a perceptron, my >> question is more about the Haskell syntax in that line I don't understand. >> >> epoch :: [([Float],Float)] -> -- ^ Test Cases and Expected Values for each >> test case >> [Float] -> -- ^ weights >> ([Float],Float) -- ^ New weights, delta >> epoch allInputs weights= >> let >> f w (inputs,expected) = step inputs w expected -- I don't >> understand this line >> newWeights = foldl f weights allInputs -- Neither this one >> delta = (foldl (+) 0 (map abs (zipWith (-) newWeights weights))) / >> (fromIntegral $ length weights) >> in (newWeights,delta) >> >> What is f and what is w? I really don't get it, Is like it is defining a >> function f which calls step unziping the input, taking one of the elements >> from the fst and it's corresponding snd and invoking step with that, along >> with w (which seems to be a list according to step's signature but I don't >> know where it comes from), and then applying fold to the weights and all the >> Inputs using that f function... But I don't get it! >> >> Maybe if someone could rewrite that redefining f as an separate function >> and calling fold with that function I'll get it. >> >> The input for epoch would be something like this: >> epoch [([0,0],0),([0,1],0),([1,0],0),([1,1],1)] [-0,413,0.135] >> >> and the output for that examples is: >> ([0.0,412.9],3.333537e-2) >> >> >> Thanks a lot, >> >> Hector Guilarte >> >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> >> >
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