Pythonの高度な関数をリストする

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}? It is {1}.'.format(q, a)
...
What is your name? It is lancelot.
What is your quest? It is the holy grail.
What is your favorite color? It is blue.