在Python所有的數(shù)據(jù)結(jié)構(gòu)中，list具有重要地位，并且非常的方便，這篇文章主要是講解list列表的高級(jí)應(yīng)用，基礎(chǔ)知識(shí)可以查看博客。
此文章為python英文文檔的翻譯版本，你也可以查看英文版：https://docs.python.org/2/tutorial/datastructures.html
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(gè)對(duì)象 7 stack [3, 4, 5, 6] stack.pop() 6 stack.pop() 5 stack [3, 4]
use a list as a queue: #像隊(duì)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 arrivaldeque(['Michael', 'Terry', 'Graham'])three built-in functions： 三個(gè)重要的內(nèi)建函數(shù)
filter(), map(), and reduce().
1）、filter(function, sequence):：
按照f(shuō)unction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(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]
2）、map(function, sequence):
map函數(shù)實(shí)現(xiàn)按照f(shuō)unction函數(shù)的規(guī)則對(duì)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(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]
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(gè)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(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(zhǎng)度不一致會(huì)出現(xiàn)錯(cuò)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f(shuō)unction函數(shù)操作，如 將列表第一個(gè)數(shù)與第二個(gè)數(shù)進(jìn)行function操作，得到的結(jié)果和列表中下一個(gè)數(shù)據(jù)進(jìn)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(gè)應(yīng)用：
squares = [x**2 for x in range(10)]
#生成一個(gè)列表，列表是由列表range(10)生成的列表經(jīng)過(guò)平方計(jì)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(gè)列表，列表的每一項(xiàng)為元組，每個(gè)元組是由x和y組成，x是由列表[1,2,3]提供，y來(lái)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(shuō)明一下吧：
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]]
這里兩層嵌套比較麻煩，簡(jiǎn)單講解一下：對(duì)矩陣matrix，for row in matrix來(lái)取出矩陣的每一行，row[i]為取出每行列表中的第i個(gè)（下標(biāo)），生成一個(gè)列表，然后i又是來(lái)源于for i in range(4) 這樣就生成了一個(gè)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>>> fruitset(['orange', 'pear', 'apple', 'banana'])>>> 'orange' in fruit # fast membership testingTrue>>> 'crabgrass' in fruitFalse>>> # Demonstrate set operations on unique letters from two words...>>> a = set('abracadabra')>>> b = set('alacazam')>>> a # unique letters in aset(['a', 'r', 'b', 'c', 'd'])>>> a - b # letters in a but not in bset(['r', 'd', 'b'])>>> a | b # letters in either a or bset(['a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'])>>> a & b # letters in both a and bset(['a', 'c'])>>> a ^ b # letters in a or b but not bothset(['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 tic1 tac2 toe
zip():
zip()是Python的一個(gè)內(nèi)建函數(shù)，它接受一系列可迭代的對(duì)象作為參數(shù)，將對(duì)象中對(duì)應(yīng)的元素打包成一個(gè)個(gè)tuple（元組），然后返回由這些tuples組成的list（列表）。若傳入?yún)?shù)的長(zhǎng)度不等，則返回list的長(zhǎng)度和參數(shù)中長(zhǎng)度最短的對(duì)象相同。利用*號(hào)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(gè)簡(jiǎn)單點(diǎn)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...applebananaorangepear
python的set和其他語(yǔ)言類似, 是一個(gè) 基本功能包括關(guān)系測(cè)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']
以上就是本文的全部?jī)?nèi)容，希望對(duì)大家的學(xué)習(xí)有所幫助。
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