我正在尝试在Haskell中绘制随机游走路线。 我使用getStdGen随机数。 我正在使用Chart.Easy进行绘图。 我在haskell上大吃一惊,我无法理解我得到的编译器错误。
import Graphics.Rendering.Chart.Easy
import Graphics.Rendering.Chart.Backend.Cairo
import System.Random
main = toFile def "random_walk.png" $ do
layout_title .= "Random Walk"
setColors [opaque blue, opaque red]
g <- getStdGen
plot (line "random walk" [ take 100 $ walk (0,0) ( randoms g :: [Float])])
walk :: (Float, Float) -> [Float] -> [(Float, Float)]
walk (x,y) (dx:dy:list) = (x+dx,y+dy):(walk (x+dx,y+dy) list)
错误是:
[1 of 1] Compiling Main ( simple.hs, interpreted )
simple.hs:8:7: error:
* Couldn't match type `IO'
with `Control.Monad.Trans.State.Lazy.StateT
(Layout Float Float) (Control.Monad.Trans.State.Lazy.State CState)'
Expected type: Control.Monad.Trans.State.Lazy.StateT
(Layout Float Float)
(Control.Monad.Trans.State.Lazy.State CState)
StdGen
Actual type: IO StdGen
* In a stmt of a 'do' block: g <- getStdGen
In the second argument of `($)', namely
`do layout_title .= "Random Walk"
setColors [opaque blue, opaque red]
g <- getStdGen
plot
(line
"random walk" [take 100 $ walk (0, 0) (randoms g :: [Float])])'
In the expression:
toFile def "random_walk.png"
$ do layout_title .= "Random Walk"
setColors [opaque blue, opaque red]
g <- getStdGen
plot
(line
"random walk" [take 100 $ walk (0, 0) (randoms g :: [Float])])
|
8 | g <- getStdGen
| ^^^^^^^^^
Failed, no modules loaded.
有人知道吗?谁能帮我?还有其他评论吗?
修复代码后,重新设置为 this image.
答案 0 :(得分:3)
问题在于toFile
的第三个参数是特殊EC
单子中的一个动作。此monad无法执行IO,因此您不能在其do-block中包含诸如getStdGen
之类的IO操作。
相反,使用单独的外部IO do-block来获取所需的随机数流作为纯值,然后可以在内部EC do-block中自由使用它,如下所示:
main :: IO ()
main = do
-- this is do-block for the IO monad
nums <- randomRs (-1,1) <$> getStdGen
toFile def "random_walk.png" $ do
-- this is the do-block for the EC monad
layout_title .= "Random Walk"
setColors [opaque blue, opaque red]
plot (line "random walk" [ take 100 $ walk (0,0) nums ])
walk :: (Float, Float) -> [Float] -> [(Float, Float)]
walk (x,y) (dx:dy:list) = (x+dx,y+dy):(walk (x+dx,y+dy) list)