列表推导：
生成一个列表：[0,2,4,6,8]
>>> [i for i in range(10) if i % 2 == 0] [0,2,4,6,8]
enumerate
seq = ["one","two","three"] for i,element in enumerate(seq): seq[i] = '%d:%s' % (i,seq[i])
上面的代码将生成下面的列表： ['0:one','1:two','2:three']
迭代器：
>>> i = iter('abc') >>> i.next() 'a' >>> i.next() 'b' >>> i.next() 'c' >>> i.next() Traceback (most recent call last): File "<stdin>", line 1, in <module> StopIteration >>>
遍历完毕将产生 StopIteration 异常
生成器
>>> def fibonacci(): ... a, b = 0, 1 ... while True: ... yield b ... a, b = b, a+b ... >>> fib = fibonacci() >>> fib.next() 1 >>> fib.next() 1 >>> fib.next() 2 >>> [fib.next() for i in range(10)] [3, 5, 8, 13, 21, 34, 55, 89, 144, 233] >>>
协同程序
协同不同于多线程，线程是抢占式的，下面的例子可以说明这点：
#!/usr/bin/env python #coding=utf-8 #协同 import multitask from threading import Thread import time def conroutine_1(): for i in range(3): print 'c1' yield i def conroutine_2(): for i in range(3): print 'c2' yield i def conroutine_3(): for i in range(3): print 'c3' def conroutine_4(): for i in range(3): print 'c4' print "==========协同========" multitask.add(conroutine_1()) multitask.add(conroutine_2()) multitask.run() print "==========多线程========" a = Thread(target=conroutine_3,args=()) b = Thread(target=conroutine_4,args=()) a.start() b.start()
上面的程序运行的结果可能是这样的：
==========协同======== c1 c2 c1 c2 c1 c2 ==========多线程======== c3 c3c4 c3 c4 c4
从结果来看，协同程序c1和c2依次执行，多线程的执行结果就不好说了，有多种可能
下面的程序是用生成器形成的echo服务器
#!/usr/bin/env python #coding=utf-8 from __future__ import with_statement from contextlib import closing import socket import multitask def client_handler(sock): with closing(socket): while True: data = (yield multitask.recv(sock,1024)) if not data: break yield multitask.send(sock,data) def echo_server(hostname,port): addrinfo = socket.getaddrinfo(hostname,port, socket.AF_UNSPEC, socket.SOCK_STREAM) (family,socktype,proto, canonname,sockaddr)=addrinfo[0] with closing(socket.socket(family,socktype,proto)) as sock: sock.setsockopt(socket.SOL_SOCKET,socket.SO_REUSEADDR,1) sock.bind(sockaddr) sock.listen(5) while True: multitask.add(client_handler((yield multitask.accept(sock))[0])) if __name__=='__main__': hostname = 'localhost' port = 1111 multitask.add(echo_server(hostname,port)) try: multitask.run() except KeyboardInterrupt: pass
上面的代码理解起来有点难度，我也有些地方不太懂，测试的话可以用telnet,方法是：
telnet localhost 1111
然后再输入一些信息，回车以后会显示你发送的信息。
itertools模块
islice：窗口迭代器
#!/usr/bin/env python #coding=utf-8 import itertools def starting_at_five(): value = raw_input("input1").strip() while value != '': for el in itertools.islice(value.split(),4,None): yield el value = raw_input("input2").strip() iter = starting_at_five() while True: print iter.next()
下面是一些测试输入输出：
input1: 1 2 3 4 5 6 5 6 input2: 1 2 input2: 1 2 3 4 5 6 7 8 5 6 7 8 input2:
从上面的例子可以看出，上面的代码是输出第4个之后的元素，利用这个功能我们可以输出特定位置的元素
tee
#!/usr/bin/env python #coding=utf-8 import itertools def with_head(iterable,headsize=1): a, b = itertools.tee(iterable) print list(itertools.islice(a,headsize)),b seq = [1] with_head(seq) seq = [1,2,3,4] with_head(seq,4)
上面的代码我也没看懂，具体怎么用还需要进一步学习代码运行结果是：
[1] <itertools.tee object at 0xb71f866c>
[1, 2, 3, 4] <itertools.tee object at 0xb71f862c>
uniq迭代器：
使用行程长度编码来压缩数据：将字符串中每组相邻的重复字符替换成字符本身的重复字数，没重复则为1,代码实现：
#!/usr/bin/env python #coding=utf-8 import itertools def compress(data): return ((len(list(group)),name) for name,group in itertools.groupby(data)) def decompress(data): return (car * size for size,car in data) print list(compress('aaasdffffffffffffffffffffff')) compressed = list(compress('aaasdffffffffffffffffffffff')) print "".join(decompress(compressed))
运行结果：
[(3, 'a'), (1, 's'), (1, 'd'), (22, 'f')] aaasdffffffffffffffffffffff
本文转自ma6174博客园博客，原文链接：http://www.cnblogs.com/ma6174/archive/2013/01/06/2847563.html，如需转载请自行联系原作者
