[Haskell-cafe] AND/OR Perceptron
hectorg87 at gmail.com
Thu Oct 29 16:32:44 EDT 2009
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 <inforichland at gmail.com>wrote:
> 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.
> - Tim
> On Thu, Oct 29, 2009 at 2:27 PM, Hector Guilarte <hectorg87 at 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=
>> 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:
>> Thanks a lot,
>> Hector Guilarte
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
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