HeadFirstPython-初识python
下面我们来看一下实例： 在书中，作者给我们展示了一个电影清单。在这里我们要学习列表的基本操作方法。
cast = ["Cless","plain","Jones","Idle"]
针对这个列表的实例，我们能做哪些操作？
打印出这个实例：
In [28]: print(cast) [‘Cless‘, ‘plain‘, ‘Jones‘, ‘Idle‘]
向这个实例添加一个新的数据项
In [29]: cast.append("iamappend")
In [30]: print(cast) [‘Cless‘, ‘plain‘, ‘Jones‘, ‘Idle‘, ‘iamappend‘]
从这个实例中弹出最后一个数据项In [31]: cast.pop() Out[31]: ‘iamappend‘
In [32]: print(cast) [‘Cless‘, ‘plain‘, ‘Jones‘, ‘Idle‘]
从这个实例中移除指定的数据项。比如我要移除plain In [33]: cast.remove("plain")
In [34]: print(cast) [‘Cless‘, ‘Jones‘, ‘Idle‘]
现在数据项比以前少了，我再插入到指定的位置上。In [35]: cast.insert(1,"plain")
In [36]: cast Out[36]: [‘Cless‘, ‘plain‘, ‘Jones‘, ‘Idle‘]
如何合并列表？In [37]: newa = ["Gmail","baidu","sina"]
In [38]: print(newa) [‘Gmail‘, ‘baidu‘, ‘sina‘]
In [39]: cast.extend(newa)
In [40]: cast Out[40]: [‘Cless‘, ‘plain‘, ‘Jones‘, ‘Idle‘, ‘Gmail‘, ‘baidu‘, ‘sina‘]
以上只是简单的描述了一个如何对列表做一个操作。
下一节： 向列表增加更多的数据。 在书13页中，第一题： 原题：movies = ["The Holy Grail","The Life of Brain","The Meaning of Life"]将其变成：movies = ["The Holy Grail",1975,"The Life of Brain",1979,"The Meaning of Life",1983]
操作方法1.直接在进行人工插入In [42]: movies = ["The Holy Grail","The Life of Brain","The Meaning of Life"]
In [43]: movies.insert(1,1975)
In [44]: movies.insert(3,1979)
In [45]: movies.insert(5,1983)
这里可以直接使用 movies.append(1983) 因为这是最后一个了
In [46]: print(movies) [‘The Holy Grail‘, 1975, ‘The Life of Brain‘, 1979, ‘The Meaning of Life‘, 1983]
操作方法2.就是直接新建列表In [47]: movies = [‘The Holy Grail‘, 1975, ‘The Life of Brain‘, 1979, ‘The Meaning of Life‘, 1983]
In [48]: print(movies) [‘The Holy Grail‘, 1975, ‘The Life of Brain‘, 1979, ‘The Meaning of Life‘, 1983]
下一节 处理列表数据
简单的迭代fav_movies = ["The Holy Grail","The Life of Brian"]
In [54]: for eachfclick in favmovies: ....: print(each_fclick) ....: The Holy Grail The Life of Brian
基本的迭代都是由for 或者while进行操作。 两者的区别：while必须要使用状态信息，需要使用计数标识。否则有可能死循环下去。下面例子中 count就是作为计数标识符存在In [57]: while count < len(movies): ....: print(movies[count]) ....: count=count+1 ....: The Holy Grail 1975 The Life of Brain 1979 The Meaning of Life 1983
使用for作迭代时，python解释器将你考虑状态信息。
下一节，在列表中存储列表
很长的一个列表了In [62]: movies = ["The Holy Grail",1975,"Terry Jone & Terry Gillam",91, ["Graham chapman",["Michael Palin","Jone Cleese","Terry Gillam","Eric Idle","Terry Jone"]]]
In [63]: print(movies) [‘The Holy Grail‘, 1975, ‘Terry Jone & Terry Gillam‘, 91, [‘Graham chapman‘, [‘Michael Palin‘, ‘Jone Cleese‘, ‘Terry Gillam‘, ‘Eric Idle‘, ‘Terry Jone‘]]]
这么多层的列表肯定不是人工写的，使用Python如何来实现这么多层的列表呢In [74]: a = ["aaa","ccc"]
In [75]: b = ["aaa111","ccc111"]
In [76]: a.append(b) #如果是用a.extend的话，就是把两个列表进行合并了
In [77]: a Out[77]: [‘aaa‘, ‘ccc‘, [‘aaa111‘, ‘ccc111‘]]
多列表嵌套的迭代问题In [65]: for eachitem in movies: ....: print(eachitem) ....: The Holy Grail 1975 Terry Jone & Terry Gillam 91 [‘Graham chapman‘, [‘Michael Palin‘, ‘Jone Cleese‘, ‘Terry Gillam‘, ‘Eric Idle‘, ‘Terry Jone‘]]
使用简单的for循环无法做打打出第一个数据项
做一个迭代并且加上元素的类型判断In [66]: for eachitem in movies: ....: if isinstance(eachitem,list): ....: for each1item in eachitem: ....: if isinstance(each1item,list): ....: for each2item in each1item: ....: print(each2item) ....: else: ....: print(each1item) ....: else: ....: print(eachitem) ....: The Holy Grail 1975 Terry Jone & Terry Gillam 91 Graham chapman Michael Palin Jone Cleese Terry Gillam Eric Idle Terry Jone
上面的一个例子中，代码有得用的存在，随着列表的嵌套越来越多层的时候，这个循环是有问题。 下面使用一个函数进行简化操作：
In [69]: def printlol(thelist): ....: for eachitem in thelist: ....: if isinstance(eachitem,list): ....: printlol(eachitem) #重点在这一行上 ....: else: ....: print(eachitem) ....:
In [73]: print_lol(movies) The Holy Grail 1975 Terry Jone & Terry Gillam 91 Graham chapman Michael Palin Jone Cleese Terry Gillam Eric Idle Terry Jone
这里有一段按照书上指导写的函数
def printlol(thelist,indent=False,level=0): """ 模块参数说明： indent是一个用于控制是否缩进，默认值为False，启用值为Ture. level是一个用于控制缩进的级别，默认为0 """ for eachitem in thelist: if isinstance(eachitem,list): printlol(eachitem,indent,level+1) else: if indent: "这条语句表示一旦indent为Ture的时候，表示缩进已经打开" #for tabstop in range(level): #print "\t", print "\t" *level , #这种方式也是可以的，可以连这个循环都不用做 "在python 2.x里这使用这种方式。在python 3.x里使用其他方式" print(each_item)