yield 的学习,模拟无限数据 ::-- ZoomQuiet [2007-10-24 04:15:39]
Albert Lee <[email protected]> hide details 12:08 pm (2 minutes ago) reply-to [email protected] to "Python.cn@google" <[email protected]> date Oct 24, 2007 12:08 PM subject [CPyUG:34110] yield 的学习,模拟无限数据 mailed-by googlegroups.com
1. FPy 模拟
1.1. yield 的学习,模拟无限数据
对haskell我最喜爱的特性是惰性计算和无限数据结构,以及函数组合。 这些特性在python中是否可以模拟出来呢? 今天作了一些尝试
python中提供了yield 和 decorators, 应当可以模拟出。
1.1.1. 尝试: 无限数据结构
haskell: [1..] Hugs> take 5 [1..] [1,2,3,4,5]
Python:
>>> def inf_list1(): i = 0 while 1: i += 1 yield i >>> def take(n, it): cnt = 0 for i in it(): cnt += 1 if cnt > n: break yield i >>> take(5, m) >>> [i for i in take(5, inf_list1)] [1, 2, 3, 4, 5]
Python 的 take 也实现为一个 generator ,这样的好处是可以串起来执行,比如,我要实现 haskell中的: Hugs> drop 3 $ take 5 [1..] [4,5]
- 那么再实现一个 drop 函数:
>>> def drop(n, it): t = 0 for i in it(): t += 1 if t > n: yield i >>> def m10(): for i in range(10): yield i >>> [i for i in drop(3, m10)] [3, 4, 5, 6, 7, 8, 9] >>> [i for i in take(5, m10)] [0, 1, 2, 3, 4] >>>
- 不过,在组合 take 和 drop 的时候遇到了麻烦
>>> drop(3, take(5, m10)) >>> [i for i in drop(3, take(5, m10))] Traceback (most recent call last): File "", line 1, in -toplevel- [i for i in drop(3, take(5, m10))] File "", line 3, in drop for i in it(): TypeError: 'generator' object is not callable >>>
- 修改下 drop, take 的定义如下:
>>> def take(n, it): cnt = 0 if callable(it): it=it() for i in it: cnt += 1 if cnt > n: break yield i >>> def drop(n, it): cnt = 0 if callable(it): it = it() for i in it: cnt += 1 if cnt > n: yield i >>> [i for i in take(2, drop(3, take(7, inf_list1)))] [4, 5] >>>
1.1.2. 对应 haskell
Hugs> take 2 $ drop 3 $ take 7 [1..] [4,5] Hugs>
基本达到目的。
1.1.3. 用itertools模块更简单
Qiangning Hong <[email protected]> hide details 12:57 pm (8 minutes ago) reply-to [email protected] to [email protected] date Oct 24, 2007 12:57 PM subject [CPyUG:34115] Re: yield 的学习,模拟无限数据:
!python import sys from itertools import count, islice def inf_list(): return count(1) def take(n, it): return islice(it, n) def drop(n, it): return islice(it, n, sys.maxint) list(take(5, inf_list()) -> [1, 2, 3, 4, 5] list(drop(3, take(5, inf_list()))) -> [4, 5] list(take(2, drop(3, take(7, count(1))))) -> [4, 5]
1.2. 函数的组合
比较简单:
Hugs> ((*2) . (+3)) 5 16
在Python
>>> def m2(x): return x * 2 >>> def add3(x): return x + 3 >>> def compose(f, g): return lambda x:f(g(x)) >>> compose(m2, add3) <function <lambda> at 0x012875F0> >>> compose(m2, add3)(5) 16 >>>
1.2.1. 对compose的装饰:
infix 来自 cookbook
class Infix: def __init__(self, function): self.function = function def __ror__(self, other): return Infix(lambda x, self=self, other=other: self.function(other, x)) def __or__(self, other): return self.function(other) def __rlshift__(self, other): return Infix(lambda x, self=self, other=other: self.function(other, x)) def __rshift__(self, other): return self.function(other) def __call__(self, value1, value2): return self.function(value1, value2) # compose def compose(f, g): return lambda x:f(g(x)) o=Infix(compose) add2 = lambda x:x+2 multi3 = lambda x:x*3 new_f = add2 |o| multi3 print new_f(3) --- >>> (add2 <<o>> multi3) (3) 11
已经比较接近了
1.2.2. curry
一个函数需要两个参数, 只给出一个参数的话,则返回一个函数,再给一个参数才计算(curry) , 用 decorators 貌似可以实现。 对多参数的处理不知道怎么办了。
组合现在可以这样:
>>> f = fun1 <<o>> fun2 <<o>> fun3 >>> f(1)
看来函数重载真是有很多意想不到的作用,可以实现 DSL了。
1 from functools import partial
2
3 def allow_partial_args(func):
4 def _(*args):
5 rturn partial(func, *args)
6 return _
7
8 @allow_partial_args
9 def add(a, b):
10 return a + b
11
12 add(1)(2) -> 3
13 add()(1, 2) -> 3
14 add(1, 2)() -> 3