stack = [3, 4, 5] 
stack.append(6) 
stack.append(7) 
stack 
[3, 4, 5, 6, 7] 
stack.pop() #删除最后一个对象 
7 
stack 
[3, 4, 5, 6] 
stack.pop() 
6 
stack.pop() 
5 
stack 
[3, 4]

> from collections import deque #这里需要使用模块deque 
> queue = deque(["Eric", "John", "Michael"])
> queue.append("Terry")      # Terry arrives
> queue.append("Graham")     # Graham arrives
> queue.popleft()         # The first to arrive now leaves
'Eric'
> queue.popleft()         # The second to arrive now leaves
'John'
> queue              # Remaining queue in order of arrival
deque(['Michael', 'Terry', 'Graham'])

> def f(x): return x % 3 == 0 or x % 5 == 0
... #f函数为定义整数对象x,x性质为是3或5的倍数
> filter(f, range(2, 25)) #筛选
[3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]

> def cube(x): return x*x*x #这里是立方计算 还可以使用 x**3的方法
...
> map(cube, range(1, 11)) #对列表的每个对象进行立方计算
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

 seq = range(8)  #定义一个列表
> def add(x, y): return x+y #自定义函数，有两个形参
...
> map(add, seq, seq) #使用map函数，后两个参数为函数add对应的操作数，如果列表长度不一致会出现错误
[0, 2, 4, 6, 8, 10, 12, 14]

def add(x,y): return x+y
...
reduce(add, range(1, 11))
55

matrix = [          #此处定义一个矩阵
...   [1, 2, 3, 4],
...   [5, 6, 7, 8],
...   [9, 10, 11, 12],
... ]
[[row[i] for row in matrix] for i in range(4)]
#[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

> a = [-1, 1, 66.25, 333, 333, 1234.5]
>del a[0] #删除下标为0的元素
>a
[1, 66.25, 333, 333, 1234.5]
>del a[2:4] #从列表中删除下标为2，3的元素
>a
[1, 66.25, 1234.5]
>del a[:] #全部删除 效果同 del a
>a
[]

> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']
> fruit = set(basket)        # create a set without duplicates
> fruit
set(['orange', 'pear', 'apple', 'banana'])
> 'orange' in fruit         # fast membership testing
True
> 'crabgrass' in fruit
False

> # Demonstrate set operations on unique letters from two words
...
> a = set('abracadabra')
> b = set('alacazam')
> a                 # unique letters in a
set(['a', 'r', 'b', 'c', 'd'])
> a - b               # letters in a but not in b
set(['r', 'd', 'b'])
> a | b               # letters in either a or b
set(['a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'])
> a & b               # letters in both a and b
set(['a', 'c'])
> a ^ b               # letters in a or b but not both
set(['r', 'd', 'b', 'm', 'z', 'l'])

> tel = {'jack': 4098, 'sape': 4139}
> tel['guido'] = 4127 #相当于向字典中添加数据
> tel
{'sape': 4139, 'guido': 4127, 'jack': 4098}
> tel['jack'] #取数据
4098
> del tel['sape'] #删除数据
> tel['irv'] = 4127   #修改数据
> tel
{'guido': 4127, 'irv': 4127, 'jack': 4098}
> tel.keys()    #取字典的所有key值
['guido', 'irv', 'jack']
> 'guido' in tel #判断元素的key是否在字典中
True
> tel.get('irv') #取数据
4127

> {x: x**2 for x in (2, 4, 6)}
{2: 4, 4: 16, 6: 36}

> for i, v in enumerate(['tic', 'tac', 'toe']):
...   print i, v
...
0 tic
1 tac
2 toe

> questions = ['name', 'quest', 'favorite color']
> answers = ['lancelot', 'the holy grail', 'blue']
> for q, a in zip(questions, answers):
...   print 'What is your {0}"htmlcode">

> a = [1,2,3]
> b = [4,5,6]
> c = [4,5,6,7,8]
> zipped = zip(a,b)
[(1, 4), (2, 5), (3, 6)]
> zip(a,c)
[(1, 4), (2, 5), (3, 6)]
> zip(*zipped)
[(1, 2, 3), (4, 5, 6)]


reversed():反转


> for i in reversed(xrange(1,10,2)):
...   print i
...




sorted(): 排序


> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']
> for f in sorted(set(basket)):       #这里使用了set函数
...   print f
...
apple
banana
orange
pear


python的set和其他语言类似, 是一个 基本功能包括关系测试和消除重复元素.
To change a sequence you are iterating over while inside the loop (for example to duplicate certain items), it is recommended that you first make a copy. Looping over a sequence does not implicitly make a copy. The slice notation makes this especially convenient:


> words = ['cat', 'window', 'defenestrate']
> for w in words[:]: # Loop over a slice copy of the entire list.
...   if len(w) > 6:
...     words.insert(0, w)
...
> words
['defenestrate', 'cat', 'window', 'defenestrate']



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