如何将Haskell转换为f# ?

I'm trying to learn F# by translating some Haskell code I wrote a very long time ago, but I'm stuck!

我想通过翻译我很久以前写的Haskell代码来学习f#，但是我被卡住了!

percent       :: Int -> Int -> Float
percent a b    = (fromInt a / fromInt b) * 100

freqs         :: String -> [Float]
freqs ws       = [percent (count x ws) (lowers ws) | x <- ['a' .. 'z']]

I've managed this:

我管理这个:

let percent a b = (float a / float b) * 100.

although i dont like having to have the . after the 100.

虽然我不喜欢有。在100年。

What is the name of the operation I am performing in freqs, and how do I translate it to F#?

我在freqs中执行的操作的名称是什么，我如何将它转换为f# ?

Edit: count and lowers are Char -> String -> Int and String -> Int respectively, and I have translated these already.

编辑:count和lower分别是Char -> String -> Int和String -> Int，我已经翻译过了。

Thanks

谢谢

2 个解决方案

#1

This is a list comprehension, and in F# it looks like the last two lines below:

这是一个列表理解，在f#中它看起来像下面的最后两行:

// stub out since dunno implementation
let count (c:char) (s:string) = 4
let lowers (s:string) = 10
// your code
let percent a b = (float a / float b) * 100.
let freq ws = [for x in ['a'..'z'] do 
                   yield percent (count x ws) (lowers ws)]

More generally I think Haskell list comprehensions have the form suggested by the example below, and the corresponding F# is shown.

更一般地，我认为Haskell列表的理解具有如下示例所建议的形式，并显示相应的f#。

// Haskell
// [e(x,y) | x <- l1, y <- l2, pred(x,y)]
// F#
[for x in l1 do
    for y in l2 do
        if pred(x,y) then
            yield e(x,y)]

#2

Note that Brian's F# code:

请注意Brian的f#代码:

let freq ws = [for x in ['a'..'z'] do yield percent (count x ws) (lowers ws)]

Can be written more elegantly as:

可以写得更优美:

let freq ws = [for x in 'a'..'z' -> percent (count x ws) (lowers ws)]

#1