Question

我正在尝试在Andrew Ng的ML课程中实现梯度下降算法。在读完数据之后，我尝试实现以下内容，将我的theta值列表更新1000次，期望收敛。

有问题的算法是gradientDescent。我知道通常导致此问题的原因是alpha可能太大，但是当我将alpha更改为n因子时，我的结果会更改n因子。当我将iterations更改为n时，会发生同样的情况。我想说这可能与哈斯克尔的懒惰有关，但我完全不确定。任何帮助将不胜感激。

module LR1V where

import qualified Data.Matrix as M
import System.IO
import Data.List.Split
import qualified Data.Vector as V

main :: IO ()
main = do
    contents <- getContents
    let lns = lines contents :: [String]
        entries = map (splitOn ",") lns :: [[String]]
        mbPoints = mapM readPoints entries :: Maybe [[Double]]
    case mbPoints of 
        Just points -> runData points
        _           -> putStrLn "Error: it is possible the file is incorrectly formatted"

readPoints :: [String] -> Maybe [Double]
readPoints dat@(x:y:_) = return $ map read dat
readPoints _ = Nothing

runData :: [[Double]] -> IO ()
runData pts = do
    let (mxs,ys) = runPoints pts
        c = M.ncols mxs
        m = M.nrows mxs
        thetas = M.zero 1 (M.ncols mxs)
        alpha = 0.01
        iterations = 1000
        results = gradientDescent mxs ys thetas alpha m c iterations
    print results

runPoints :: [[Double]] -> (M.Matrix Double, [Double])
runPoints pts = (xs, ys) where
    xs = M.fromLists $ addX0 $ map init pts
    ys = map last pts

-- X0 will always be 1
addX0 :: [[Double]] -> [[Double]]
addX0 = map (1.0 :)

-- theta is 1xn and x is nx1, where n is the amount of features
-- so it is safe to assume a scalar results from the multiplication
hypothesis :: M.Matrix Double -> M.Matrix Double -> Double
hypothesis thetas x = 
    M.getElem 1 1 (M.multStd thetas x)

gradientDescent :: M.Matrix Double
                   -> [Double] 
                   -> M.Matrix Double
                   -> Double 
                   -> Int 
                   -> Int 
                   -> Int
                   -> [Double]
gradientDescent mxs ys thetas alpha m n it = 
    let x i = M.colVector $ M.getRow i mxs
        y i = ys !! (i-1)
        h i = hypothesis thetas (x i)
        thL = zip [1..] $ M.toList thetas :: [(Int, Double)]
        z i j = ((h i) - (y i))*(M.getElem i j $ mxs)
        sumSquares j = sum [z i j | i <- [1..m]]
        thetaJ t j = t - ((alpha * (1/ (fromIntegral m))) * (sumSquares j))
        result = map snd $ foldl (\ts _ -> [(j,thetaJ t j) | (j,t) <- ts]) thL [1..it] in
    result

和数据......

6.1101,17.592
5.5277,9.1302
8.5186,13.662
7.0032,11.854
5.8598,6.8233
8.3829,11.886
7.4764,4.3483
8.5781,12
6.4862,6.5987
5.0546,3.8166
5.7107,3.2522
14.164,15.505
5.734,3.1551
8.4084,7.2258
5.6407,0.71618
5.3794,3.5129
6.3654,5.3048
5.1301,0.56077
6.4296,3.6518
7.0708,5.3893
6.1891,3.1386
20.27,21.767
5.4901,4.263
6.3261,5.1875
5.5649,3.0825
18.945,22.638
12.828,13.501
10.957,7.0467
13.176,14.692
22.203,24.147
5.2524,-1.22
6.5894,5.9966
9.2482,12.134
5.8918,1.8495
8.2111,6.5426
7.9334,4.5623
8.0959,4.1164
5.6063,3.3928
12.836,10.117
6.3534,5.4974
5.4069,0.55657
6.8825,3.9115
11.708,5.3854
5.7737,2.4406
7.8247,6.7318
7.0931,1.0463
5.0702,5.1337
5.8014,1.844
11.7,8.0043
5.5416,1.0179
7.5402,6.7504
5.3077,1.8396
7.4239,4.2885
7.6031,4.9981
6.3328,1.4233
6.3589,-1.4211
6.2742,2.4756
5.6397,4.6042
9.3102,3.9624
9.4536,5.4141
8.8254,5.1694
5.1793,-0.74279
21.279,17.929
14.908,12.054
18.959,17.054
7.2182,4.8852
8.2951,5.7442
10.236,7.7754
5.4994,1.0173
20.341,20.992
10.136,6.6799
7.3345,4.0259
6.0062,1.2784
7.2259,3.3411
5.0269,-2.6807
6.5479,0.29678
7.5386,3.8845
5.0365,5.7014
10.274,6.7526
5.1077,2.0576
5.7292,0.47953
5.1884,0.20421
6.3557,0.67861
9.7687,7.5435
6.5159,5.3436
8.5172,4.2415
9.1802,6.7981
6.002,0.92695
5.5204,0.152
5.0594,2.8214
5.7077,1.8451
7.6366,4.2959
5.8707,7.2029
5.3054,1.9869
8.2934,0.14454
13.394,9.0551
5.4369,0.61705

当alpha为0.01时，我的thetas评估为[58.39135051546406,653.2884974555699]。当alpha为0.001时，我的值将变为[5.839135051546473,65.32884974555617]。当iterations更改为10,000时，我的值将恢复为之前的值。

Answer 1

似乎每次更新theta值时，近似函数h(x)每次都使用初始theta向量，而不是更新的向量。现在，我得到了我的theta值的近似值。但是，通过较大因子增加迭代次数会以奇怪的方式改变我的结果。

梯度下降算法不在Haskell中收敛

1 个答案: