use a list as a stack: #像棧一樣使用列表
stack = [3, 4, 5] stack.append(6) stack.append(7) stack [3, 4, 5, 6, 7] stack.pop() #刪除最后一個對象 7 stack [3, 4, 5, 6] stack.pop() 6 stack.pop() 5 stack [3, 4]
use a list as a queue: #像隊列一樣使用列表
> 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'])
three built-in functions： 三個重要的內(nèi)建函數(shù)
filter(), map(), and reduce().
1）、filter(function, sequence):：
按照function函數(shù)的規(guī)則在列表sequence中篩選數(shù)據(jù)
> def f(x): return x % 3 == 0 or x % 5 == 0 ... #f函數(shù)為定義整數(shù)對象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]
2）、map(function, sequence):
map函數(shù)實現(xiàn)按照function函數(shù)的規(guī)則對列表sequence做同樣的處理，
這里sequence不局限于列表，元組同樣也可。
> def cube(x): return x*x*x #這里是立方計算 還可以使用 x**3的方法 ... > map(cube, range(1, 11)) #對列表的每個對象進行立方計算 [1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
注意：這里的參數(shù)列表不是固定不變的，主要看自定義函數(shù)的參數(shù)個數(shù)，map函數(shù)可以變形為：def func(x,y) map(func,sequence1，sequence2） 舉例：
seq = range(8) #定義一個列表 > def add(x, y): return x+y #自定義函數(shù)，有兩個形參 ... > map(add, seq, seq) #使用map函數(shù)，后兩個參數(shù)為函數(shù)add對應(yīng)的操作數(shù)，如果列表長度不一致會出現(xiàn)錯誤 [0, 2, 4, 6, 8, 10, 12, 14]
3）、reduce(function, sequence):
reduce函數(shù)功能是將sequence中數(shù)據(jù)，按照function函數(shù)操作，如 將列表第一個數(shù)與第二個數(shù)進行function操作，得到的結(jié)果和列表中下一個數(shù)據(jù)進行function操作，一直循環(huán)下去…
舉例:
def add(x,y): return x+y ... reduce(add, range(1, 11)) 55
List comprehensions：
這里將介紹列表的幾個應(yīng)用：
squares = [x**2 for x in range(10)]
#生成一個列表，列表是由列表range(10)生成的列表經(jīng)過平方計算后的結(jié)果。
[(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]
#[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)] 這里是生成了一個列表，列表的每一項為元組，每個元組是由x和y組成，x是由列表[1,2,3]提供，y來源于[3,1,4]，并且滿足法則x!=y。
Nested List Comprehensions：
這里比較難翻譯，就舉例說明一下吧：
matrix = [ #此處定義一個矩陣 ... [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]]
這里兩層嵌套比較麻煩，簡單講解一下：對矩陣matrix，for row in matrix來取出矩陣的每一行，row[i]為取出每行列表中的第i個（下標(biāo)），生成一個列表，然后i又是來源于for i in range(4) 這樣就生成了一個列表的列表。
The del statement:
刪除列表指定數(shù)據(jù)，舉例：
> 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 []
Sets: 集合
> 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'])
Dictionaries：字典
>>> 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
也可以使用規(guī)則生成字典：
>>> {x: x**2 for x in (2, 4, 6)}
{2: 4, 4: 16, 6: 36}enumerate()：遍歷元素及下標(biāo)
enumerate 函數(shù)用于遍歷序列中的元素以及它們的下標(biāo)：
>>> for i, v in enumerate(['tic', 'tac', 'toe']): ... print i, v ... 0 tic 1 tac 2 toe
zip():
zip()是Python的一個內(nèi)建函數(shù)，它接受一系列可迭代的對象作為參數(shù)，將對象中對應(yīng)的元素打包成一個個tuple（元組），然后返回由這些tuples組成的list（列表）。若傳入?yún)?shù)的長度不等，則返回list的長度和參數(shù)中長度最短的對象相同。利用*號操作符，可以將list unzip（解壓）。
>>> 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.
有關(guān)zip舉一個簡單點兒的例子：
>>> 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():反轉(zhuǎn)
>>> 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)): #這里使用了set函數(shù) ... print f ... apple banana orange pear
python的set和其他語言類似, 是一個 基本功能包括關(guān)系測試和消除重復(fù)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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