Question

我正在尝试压缩两个[[Float]]数组，然后对每个元素执行计算。我认为生成的zip-ed数组将具有类型[([Float], [Float])]。

但是我继续在Haskell告诉我的地方收到错误：

[1 of 1] Compiling Mod               ( Mod.hs, interpreted )

NN.hs:38:42:
    Couldn't match type ‘[Float]’ with ‘Float’
    Expected type: [Float]
      Actual type: [[Float]]
    In the second argument of ‘zip’, namely ‘bs’
    In the second argument of ‘map’, namely ‘(zip ws bs)’
Failed, modules loaded: none.

我似乎无法弄清楚MCVE，因为这就是（kinda / sorta）试图做的事情：

feedforward :: ([[Float]], [[Float]]) -> Float -> [Float]
feedforward (ws, bs) a = map (equation a) (zip ws bs)
  where equation a (w, b) = (sum $ zipWith (*) w b) + a

但实际上编译了！而我的真实代码看起来像这样：

data NNetwork = NNetwork { nlayers :: Int
                         , sizes   :: [Int]
                         , biases  :: [[Float]]
                         , weights :: [[Float]] }
                deriving (Show, Ord, Eq)

toFloat x = fromIntegral x :: Float

sig :: Float -> Float
sig a = 1 / (1 + exp (-a))


feedforward :: NNetwork -> Float -> [Float]
feedforward net a = map (eq22 a) (zip ws bs)
  where bs = biases net
        ws = weights net
        dot a b = sum $ zipWith (*) a b
        rep a b = replicate (length a) b
        eq22 a (w, b) = sig $ (dot w (rep w a)) + b

然而无效！

- 提前致谢

Answer 1

问题是：

eq22 :: Float -> ([Float], [Float]) ->  Float
eq22 a (w, b) = sig $ (dot w (rep w a)) + b

除此之外，b是[Float]并用作Float（请参阅+ b）。你的意思是b以上的某种功能吗？

此外，这些不是数组，而是列表。

Haskell奇怪的`zip`错误

1 个答案: