1、列表list：a=[value1,value2,value3,value4,…]

方法論methods：list.append(x) #列表追加，等同于a[len(a):] = [x]list.extend(L) #列表加長，等同于a[len(a):] = Llist.insert(i, x) #列表插入，a.insert(len(a), x)等同于a.append(x)list.remove(x) #列表刪除，從first查找value=xlist.pop([i]) #列表剔除并返回值，默認a.pop()是remove the last onelist.index(x) #返回value=x的索引號list.count(x) #返回value=x出現的次數list.sort(cmp=None, key=None, reverse=False) #等同于sorted(iterable[, cmp[, key[, reverse]]])函數list.reverse() #列表反向

應用論using：1、堆棧Stacks>>> stack = [3, 4, 5]>>> stack.append(6)>>> stack.append(7)>>> stack

[3, 4, 5, 6, 7]>>> stack.pop()7

>>> stack

[3, 4, 5, 6]>>> stack.pop()6

>>> stack.pop()5

>>> stack

[3, 4]2、隊列Queues>>> from collections import 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'])

工具論tools：filter(function, sequence) #返回sequence中符合function的值>>> def f(x): return x % 3 == 0 or x % 5 == 0

...>>> filter(f, range(2, 25))

[3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]map(function, sequence) #返回一個由function生成的列表>>> def cube(x): return x*x*x

...>>> map(cube, range(1, 11))

[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]reduce(function, sequence) #返回sequence前兩個items在function中值>>> def add(x,y): return x+y

...>>> reduce(add, range(1, 11))55

##如果sequence中只有一個值，返回之>>> def sum(seq):

... def add(x,y): return x+y

... return reduce(add, seq, 0)

...>>> sum(range(1, 11))55

>>> sum([])

0

另類函數del a[index1:index2]>>> a = [-1, 1, 66.25, 333, 333, 1234.5]>>> del a[0]>>> a

[1, 66.25, 333, 333, 1234.5]>>> del a[2:4]>>> a

[1, 66.25, 1234.5]>>> del a[:]>>> a

[]>>> del a>>> a

Traceback (most recent call last):

File "", line 1, in NameError: name 'a' is not defined#### 列表a已被清除

2、元組（Tuples）和序列（Sequences ）>>> t = 12345, 54321, 'hello!'>>> t[0]12345

>>> t

(12345, 54321, 'hello!')>>> # Tuples may be nested:... u = t, (1, 2, 3, 4, 5)>>> u

((12345, 54321, 'hello!'), (1, 2, 3, 4, 5))###元組在輸出時總有括號，輸入時可有可無。元組可用于表示坐標點，員工記錄；元組就像字符串，不可更改，不可單獨賦值##包含0個或一個元素的元組>>> empty = ()>>> singleton = 'hello', # <-- note trailing comma>>> len(empty)

0>>> len(singleton)1

>>> singleton

('hello',)

3、字符集sets

無序的，沒有重復的字符集合。創建字符集用set()函數，而不是set{}>>> 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 testingTrue>>> 'crabgrass' in fruit

False>>> # 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'])

4、終于輪到字典了>>> tel = {'jack': 4098, 'sape': 4139}>>> tel['guido'] = 4127

>>> tel

{'sape': 4139, 'guido': 4127, 'jack': 4098}>>> tel['jack']4098

>>> del tel['sape']>>> tel['irv'] = 4127

>>> tel

{'guido': 4127, 'irv': 4127, 'jack': 4098}>>> tel.keys()

['guido', 'irv', 'jack']>>> tel.has_key('guido')

True###鏈表中存儲關鍵字-值對元組的話，字典可以從中直接構造。關鍵字-值對來自一個模式時，可以用鏈表推導式簡單的表達關鍵字-值鏈表。>>> dict([('sape', 4139), ('guido', 4127), ('jack', 4098)])

{'sape': 4139, 'jack': 4098, 'guido': 4127}>>> dict([(x, x**2) for x in vec]) # use a list comprehension{2: 4, 4: 16, 6: 36}

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