Haskell中的木薯解析错误

时间:2018-07-02 18:58:29

标签: csv parsing haskell

我正在尝试使用木薯将csv转换为向量。尝试转换的csv Im是用于机器学习的菲舍尔虹膜数据集。它由四个双打和一个字符串组成。 我的代码如下:

{-# LANGUAGE OverloadedStrings #-}

module Main where
import Data.Csv
import qualified Data.ByteString.Lazy as BS
import qualified Data.Vector as V

data Iris = Iris
  { sepal_length  :: !Double
  , sepal_width   :: !Double
  , petal_length  :: !Double
  , petal_width   :: !Double
  , iris_type     :: !String
 } deriving (Show, Eq, Read)

instance FromNamedRecord Iris where
  parseNamedRecord r =
    Iris
      <$> r .: "sepal_length"
      <*> r .: "sepal_width"
      <*> r .: "petal_length"
      <*> r .: "petal_width"
      <*> r .: "iris_type"

printIris :: Iris -> IO ()
printIris r  = putStrLn $  show (sepal_length r) ++ show (sepal_width r)
   ++ show(petal_length r) ++ show(petal_length r) ++ "hola"

main :: IO ()
main = do
  csvData <- BS.readFile "./iris/test-iris"
  print csvData
  case decodeByName csvData of
    Left err -> putStrLn err
    -- forM : O(n) Apply the monadic action to all elements of the vector,
    -- yielding a vector of results.
    Right (h, v) -> V.forM_ v $ printIris

运行此命令时,似乎csvData的格式正确,打印csvData的第一行返回以下内容:

"5.1,3.5,1.4,0.2,Iris-setosa\n4.9,3.0,1.4,0.2,Iris- setosa\n4.7,3.2,1.3,0.2,Iris-setosa\n4.6,3.1,1.5,0.2,Iris-setosa\n5.0,3.6,1.4,0.2,Iris-setosa\n5.4,3.9,1.7,0.4,Iris-setosa\n4.6,3.4,1.4,0.3,Iris-setosa\n5.0,3.4,1.5,0.2,Iris-setosa\n4.4,2.9,1.4,0.2,Iris-setosa\n4.9,3.1,1.5,0.1,Iris-setosa\n5.4,3.7,1.5,0.2,Iris-setosa\n4.8,3.4,1.6,0.2,Iris-setosa\n4.8,3.0,1.4,0.1,Iris-setosa\n4.3,3.0,1.1,0.1,Iris-setosa\n5.8,4.0,1.2,0.2,Iris-setosa\n5.7,4.4,1.5,0.4,Iris-set

但是出现以下错误:

parse error (Failed reading: conversion error: no field named "sepal_length")  at 
4.7,3.2,1.3,0.2,Iris-setosa
4.6,3.1,1.5,0.2,Iris-setosa
5.0,3.6,1.4,0.2,Iris-setosa
5.4,3.9,1.7,0.4 (truncated)

有人对我为什么会收到此错误有任何想法吗? csv没有遗漏任何值,如果我替换另一行产生错误的行,则会遇到相同的错误。

1 个答案:

答案 0 :(得分:3)

您的数据似乎没有标头,由decodeByName

假定
  

假定数据以标头开头。

添加标题,或使用decode NoHeaderFromRecord类型的类。