python中l(wèi)ist列表高級(jí)函數(shù)

stack = [3, 4, 5] 
stack.append(6) 
stack.append(7) 
stack 
[3, 4, 5, 6, 7] 
stack.pop() #刪除最后一個(gè)對(duì)象 
7 
stack 
[3, 4, 5, 6] 
stack.pop() 
6 
stack.pop() 
5 
stack 
[3, 4]

> from collections import deque #這里需要使用模塊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函數(shù)為定義整數(shù)對(duì)象x,x性質(zhì)為是3或5的倍數(shù)
> filter(f, range(2, 25)) #篩選
[3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]

> def cube(x): return x*x*x #這里是立方計(jì)算 還可以使用 x**3的方法
...
> map(cube, range(1, 11)) #對(duì)列表的每個(gè)對(duì)象進(jìn)行立方計(jì)算
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

 seq = range(8) #定義一個(gè)列表
> def add(x, y): return x+y #自定義函數(shù)，有兩個(gè)形參
...
> map(add, seq, seq) #使用map函數(shù)，后兩個(gè)參數(shù)為函數(shù)add對(duì)應(yīng)的操作數(shù)，如果列表長度不一致會(huì)出現(xiàn)錯(cuò)誤
[0, 2, 4, 6, 8, 10, 12, 14]

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

matrix = [ #此處定義一個(gè)矩陣
... [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] #刪除下標(biāo)為0的元素
>a
[1, 66.25, 333, 333, 1234.5]
>del a[2:4] #從列表中刪除下標(biāo)為2，3的元素
>a
[1, 66.25, 1234.5]
>del a[:] #全部刪除 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 #相當(dāng)于向字典中添加數(shù)據(jù)
>>> tel
{'sape': 4139, 'guido': 4127, 'jack': 4098}
>>> tel['jack'] #取數(shù)據(jù)
4098
>>> del tel['sape'] #刪除數(shù)據(jù)
>>> tel['irv'] = 4127 #修改數(shù)據(jù)
>>> tel
{'guido': 4127, 'irv': 4127, 'jack': 4098}
>>> tel.keys() #取字典的所有key值
['guido', 'irv', 'jack']
>>> 'guido' in tel #判斷元素的key是否在字典中
True
>>> tel.get('irv') #取數(shù)據(jù)
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.

>>> 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)]

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

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

>>> 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']