3.11.2 in操作符和index()
先看一段代码
>>> 'record' in music_mediaFalse>>> 'H' in music_mediaTrue>>> music_media.index('Hello')1>>> music_media.index('e')2>>> music_media.index('world')Traceback (most recent call last): File " ", line 1, in ValueError: 'world' is not in list 这里我们发现，index()调用时若给定的对象不在列表中Python会报错，所以检查一个元素是否在列表中应该用in操作符而不是index()，index()仅适用于已经确定某元素确实在列表中获取其索引的情况。
针对上述代码，可以做出如下修改
for eachthing in ('record', 'H', 'Hello', 'e', 'world'): if eachthing in music_media: print music_media.index(eachthing) else: print '%s is not in %s!!' % (eachthing, music_media) 3.11.3 关于可变对象和不可变对象的方法的返回值问题
3.12 列表的特殊特性
3.12.1 用列表构建堆栈
# -*- coding:utf-8 -*-stack = []def pushit(): stack.append(raw_input('Enter New string: ').strip()) def popit(): if len(stack) == 0: print "Cannot pop from an empty stack!" else: print 'Removed [', `stack.pop()`, ']' def viewstack(): print stack # calls str() internally CMDs = {'u': pushit, 'o': popit, 'v': viewstack}def showmenu(): pr = ''' p(U)sh p(o)p (V)iew (Q)uit Enter choice: ''' while True: while True: try: choice = raw_input(pr).strip()[0].lower() except (EOFError, KeyboardInterrupt, IndexError): choice = 'q' print '\nYou picked: [%s]' % choice if choice not in 'uovq': print 'Invalid option, try again' else: break if choice == 'q': break CMDs[choice]() if __name__ == '__main__': showmenu() 3.12.2 用列表构建队列
# -*- coding:utf-8 -*-queue = []def enQ(): queue.append(raw_input('Enter New string: ').strip()) def deQ(): if len(queue) == 0: print 'Cannot pop from an empty queue!' else: print 'Removed [', `queue.pop(0)`, ']' def viewQ(): print queue # calls str internally CMDs = {'e': enQ, 'd': deQ, 'v': viewQ}def showmenu(): pr = ''' (E)nqueue (D)equeue (V)iew (Q)uit Enter choice: ''' while True: while True: try: choice = raw_input(pr).strip()[0].lower() except (EOFError, KeyboardInterrupt, IndexError): choice = 'q' print '\nYou picked: [%s]' % choice if choice not in 'edvq': print 'Invalid option, try again' else: break if choice == 'q': break CMDs[choice]() if __name__ == '__main__': showmenu() 3.13 元组
①元组用圆括号而列表用方括号
②元组是不可变