在这个earlier question中，我问过如何编写一个对非二进制整数树求和的函数，并且出现了几个答案。
@Sibi说：data Tree a = Empty | Node a [Tree a] deriving (Eq, Show)
addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n []) = n
addNums (Node n (x:xs)) = n + (addNums x) + addNums (Node 0 xs)
@ user3237465说：
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show, Foldable)
myNums :: (Num a) => Tree a
myNums = ...
main = print $ sum myNums
和@chi说：
addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n xs) = n + sum (map addNums xs)
如何找到最有效的解决方案？ Haskell中是否有原生的基准测试工具？
答案 0 :(得分：2)
虽然so.com不是建议网站，但我建议您查看标准https://hackage.haskell.org/package/criterion
我可能会举一些明天使用它的例子
如果你真的想深入研究这个问题，你可以通过添加编译器选项--ddump-llvm来分析生成的llvm汇编程序，尽管这是一个相当高级的主题，仅为了完整性而包含在内。
criterion 首先，我将使用haskell堆栈工具解释这一点，所有代码都可以在github/epsilonhalbe找到
首先，我们创建一个项目，并将每个相关定义拆分为一个单独的模块（否则我们需要data Tree，data Tree'和data Tree''）。请参阅Chi.hs作为示例：
module Chi where
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show)
addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n xs) = n + sum (map addNums xs)
myInts :: Tree Int
myInts =
Node 1 [
Node 2 [
Node 4 [Empty], Node 5 [Empty]
],
Node 3 [
Node 6 [Empty], Node 7 [Empty], Node 8 [Empty]
]
]
myDouble :: Tree Double
myDouble =
Node 1 [
Node 2 [
Node 4 [Empty], Node 5 [Empty]
],
Node 3 [
Node 6 [Empty], Node 7 [Empty], Node 8 [Empty]
]
]
注意：对于User3237465.hs，我们需要语言实用主义
{-# LANGUAGE DeriveFoldable #-}
module User3237465 where
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show, Foldable)
addNums :: Num a => Tree a -> a
addNums = sum
myInts ..
myDouble ..
我们构建一个像下面这样的文件夹/文件结构（这是我们用stack new critExample得到的并且有点复制/重命名/删除）
../haskell/critExample/
▾ src/
Chi.hs
Sibi.hs
User3237465.hs
▾ bench/
Benchmarks.hs
critExample.cabal
LICENSE
Setup.hs
stack.yaml
critExample.cabal的内容也需要一些调整，
name: critExample
[... non-important stuff ...]
library
hs-source-dirs: src
-- don't forget to adjust the exposed modules
exposed-modules: Chi
, Sibi
, User3237465
build-depends: base >= 4.7 && < 5
default-language: Haskell2010
-- and add the following benchmark part
benchmark addNums
type: exitcode-stdio-1.0
hs-source-dirs: bench
main-is: Benchmarks.hs
build-depends: base
, critExample
, criterion
default-language: Haskell2010
[...]
Benchmarks.hs
module Main where
import Criterion
import Criterion.Main
import qualified Chi
import qualified Sibi
import qualified User3237465
main :: IO ()
main = defaultMain [
bgroup "myInts" [ bench "Sibi" $ whnf Sibi.addNums Sibi.myInts
, bench "Chi" $ whnf Chi.addNums Chi.myInts
, bench "User3237465" $ whnf User3237465.addNums User3237465.myInts
],
bgroup "myDouble" [ bench "Sibi" $ whnf Sibi.addNums Sibi.myDouble
, bench "Chi" $ whnf Chi.addNums Chi.myDouble
, bench "User3237465" $ whnf User3237465.addNums User3237465.myDouble ]
]
请注意whnf仅评估弱头普通格式，即它看到的第一个构造函数 - 对于列表，它将在第一个元素看到{{1对于元组运算符它不会评估一个东西，但对于(:)或Int它会完全评估东西。如果您需要“深度”评估，请使用Double代替nf - 如果您不确定需要什么，请尝试whnf通常不合理的快速（如超长列表的纳秒数 - 因为它只检查该列表的头部。）
您可以使用whnf构建项目，然后使用stack build（触发所有可用的基准测试）或stack bench调用基准测试（如果您有多个基准测试套件且仅需要，则非常有用）要运行特定的一个），使用总是stack bench critExample:addNums。
如果你想要花哨的html输出（ - 并且相信我你想要它，因为bryan o'sullivan为了让它变得性感而付出了很多努力）你将不得不：
projectname:name of benchmarks given in cabal-file
当然，如果您不使用Linux操作系统，这条路径可能会有所不同。
基准测试的结果 - 我不知道它们有多么有代表性 - 我在虚拟化的linux中运行它们！
./.stack-work/dist/x86_64-linux/Cabal-1.22.4.0/build/addNums/addNums --output index.html正如评论中所述 - 使用Running 1 benchmarks...
Benchmark addNums: RUNNING...
benchmarking myInts/Sibi
time 616.7 ns (614.1 ns .. 619.2 ns)
1.000 R² (1.000 R² .. 1.000 R²)
mean 619.1 ns (615.4 ns .. 626.8 ns)
std dev 17.09 ns (9.625 ns .. 31.62 ns)
variance introduced by outliers: 38% (moderately inflated)
benchmarking myInts/Chi
time 582.6 ns (576.5 ns .. 592.1 ns)
0.998 R² (0.996 R² .. 1.000 R²)
mean 586.2 ns (581.5 ns .. 595.5 ns)
std dev 21.14 ns (11.56 ns .. 33.61 ns)
variance introduced by outliers: 52% (severely inflated)
benchmarking myInts/User3237465
time 606.5 ns (604.9 ns .. 608.2 ns)
1.000 R² (1.000 R² .. 1.000 R²)
mean 607.0 ns (605.5 ns .. 609.2 ns)
std dev 5.915 ns (3.992 ns .. 9.798 ns)
benchmarking myInts/User3237465 -- folding variant see comments
time 371.0 ns (370.2 ns .. 371.7 ns)
1.000 R² (1.000 R² .. 1.000 R²)
mean 372.5 ns (370.8 ns .. 375.0 ns)
std dev 6.824 ns (4.076 ns .. 11.19 ns)
variance introduced by outliers: 22% (moderately inflated)
benchmarking myDouble/Sibi
time 678.9 ns (642.3 ns .. 743.8 ns)
0.978 R² (0.958 R² .. 1.000 R²)
mean 649.9 ns (641.1 ns .. 681.6 ns)
std dev 50.99 ns (12.60 ns .. 105.0 ns)
variance introduced by outliers: 84% (severely inflated)
benchmarking myDouble/Chi
time 643.3 ns (617.4 ns .. 673.6 ns)
0.987 R² (0.979 R² .. 0.996 R²)
mean 640.6 ns (626.7 ns .. 665.6 ns)
std dev 58.35 ns (40.63 ns .. 87.82 ns)
variance introduced by outliers: 88% (severely inflated)
benchmarking myDouble/User3237465
time 630.4 ns (622.9 ns .. 638.5 ns)
0.997 R² (0.994 R² .. 0.999 R²)
mean 637.8 ns (625.4 ns .. 659.8 ns)
std dev 53.15 ns (33.46 ns .. 78.36 ns)
variance introduced by outliers: 85% (severely inflated)
benchmarking myDouble/User3237465 -- folding variant see comments
time 398.1 ns (380.7 ns .. 422.0 ns)
0.988 R² (0.980 R² .. 0.996 R²)
mean 400.6 ns (389.1 ns .. 428.6 ns)
std dev 55.83 ns (28.94 ns .. 103.6 ns)
variance introduced by outliers: 94% (severely inflated)
Benchmark addNums: FINISH
Completed all 2 actions.
和import Data.Foldable (foldl')的另一种变体明显更快（感谢@ User3237465 !!）
答案 1 :(得分：1)
实际上，为了提高效率，请更改类型。出于折叠的目的，你无法击败像教会一样的编码。我可能会推荐：
newtype Tree a = Tree {fold :: forall r. r -> (a -> [r] -> r) -> r}
甚至：
newtype Tree a = Tree {fold :: forall r. r -> (a -> ChurchList r -> r) -> r}
还是最好的：
newtype Tree a = Tree {fold :: forall tree list. tree -> (a -> list -> tree) -> list -> (tree -> list -> list) -> tree}
教会编码更有效率，因为您不必遍历任何东西。