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 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 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

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On Sat, Oct 31, 2009 at 8:09 PM, Ryan Ingram wrote:

"where" is just syntactic sugar for "let...in"

That's not perfectly true : "where" and "let...in" don't attach to the same syntactic construction, "where" attaches to a definition and can works for several guard clauses whereas "let ... in expression" is an expression in itself. -- Jedaï