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# 不可不知的Python数据结构

1.1. 关于列表更多的内容

Python 的列表数据类型包含更多的方法。这里是所有的列表对象方法：

list.append(x)

list.extend(L)

list.insert(i, x)

list.remove(x)

list.pop([i])

list.clear()

list.index(x)

list.count(x)

list.sort()

list.reverse()

list.copy()

>>> a = [66.25, 333, 333, 1, 1234.5]

>>> print(a.count(333), a.count(66.25), a.count('x'))

2 1 0

>>> a.insert(2, -1)

>>> a.append(333)

>>> a

[66.25, 333, -1, 333, 1, 1234.5, 333]

>>> a.index(333)

1

>>> a.remove(333)

>>> a

[66.25, -1, 333, 1, 1234.5, 333]

>>> a.reverse()

>>> a

[333, 1234.5, 1, 333, -1, 66.25]

>>> a.sort()

>>> a

[-1, 1, 66.25, 333, 333, 1234.5]

>>> a.pop()

1234.5

>>> a

[-1, 1, 66.25, 333, 333]

1.1.1. 把列表当作堆栈使用

>>> 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]

1.1.2. 把列表当作队列使用

>>> 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 arrival

deque(['Michael', 'Terry', 'Graham'])

1.1.3. 列表推导式

>>> squares = []

>>> for x in range(10):

...     squares.append(x**2)

...

>>> squares

[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

squares = list(map(lambda x: x**2, range(10)))

squares = [x**2 for x in range(10)]

>>> [(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)]

>>> combs = []

>>> for x in [1,2,3]:

...     for y in [3,1,4]:

...         if x != y:

...             combs.append((x, y))

...

>>> combs

[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

>>> vec = [-4, -2, 0, 2, 4]

>>> # create a new list with the values doubled

>>> [x*2 for x in vec]

[-8, -4, 0, 4, 8]

>>> # filter the list to exclude negative numbers

>>> [x for x in vec if x >= 0]

[0, 2, 4]

>>> # apply a function to all the elements

>>> [abs(x) for x in vec]

[4, 2, 0, 2, 4]

>>> # call a method on each element

>>> freshfruit = ['  banana', '  loganberry ', 'passion fruit  ']

>>> [weapon.strip() for weapon in freshfruit]

['banana', 'loganberry', 'passion fruit']

>>> # create a list of 2-tuples like (number, square)

>>> [(x, x**2) for x in range(6)]

[(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25)]

>>> # the tuple must be parenthesized, otherwise an error is raised

>>> [x, x**2 for x in range(6)]

File "", line 1, in ?

[x, x**2 for x in range(6)]

^

SyntaxError: invalid syntax

>>> # flatten a list using a listcomp with two 'for'

>>> vec = [[1,2,3], [4,5,6], [7,8,9]]

>>> [num for elem in vec for num in elem]

[1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> from math import pi

>>> [str(round(pi, i)) for i in range(1, 6)]

['3.1', '3.14', '3.142', '3.1416', '3.14159']

1.1.4. 嵌套的列表推导式

>>> matrix = [

...     [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]]

>>> transposed = []

>>> for i in range(4):

...     transposed.append([row[i] for row in matrix])

...

>>> transposed

[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

>>> transposed = []

>>> for i in range(4):

...     # the following 3 lines implement the nested listcomp

...     transposed_row = []

...     for row in matrix:

...         transposed_row.append(row[i])

...     transposed.append(transposed_row)

...

>>> transposed

[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

>>> list(zip(*matrix))

[(1, 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)]

1.2. del 语句

>>> 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 也可以删除整个变量:

>>> del a

1.3. 元组和序列

>>> 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))

>>> # Tuples are immutable:

... t[0] = 88888

Traceback (most recent call last):

File "", line 1, in

TypeError: 'tuple' object does not support item assignment

>>> # but they can contain mutable objects:

... v = ([1, 2, 3], [3, 2, 1])

>>> v

([1, 2, 3], [3, 2, 1])

>>> empty = ()

>>> singleton = 'hello',    #

>>> len(empty)

>>> len(singleton)

1

>>> singleton

('hello',)

>>> x, y, z = t

1.4. 集合

Python 还包含了一个数据类型 —— set （集合）。集合是一个无序不重复元素的集。基本功能包括关系测试和消除重复元素。集合对象还支持 union（联合），intersection（交），difference（差）和 sysmmetric difference（对称差集）等数学运算。

>>> basket = {'apple', 'orange', 'apple', 'pear', 'orange', 'banana'}

>>> print(basket)                      # show that duplicates have been removed

{'orange', 'banana', 'pear', 'apple'}

>>> 'orange' in basket                 # fast membership testing

True

>>> 'crabgrass' in basket

False

>>> # Demonstrate set operations on unique letters from two words

...

>>> a = set('abracadabra')

>>> b = set('alacazam')

>>> a                                  # unique letters in a

{'a', 'r', 'b', 'c', 'd'}

>>> a - b                              # letters in a but not in b

{'r', 'd', 'b'}

>>> a | b                              # letters in either a or b

{'a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'}

>>> a & b                              # letters in both a and b

{'a', 'c'}

>>> a ^ b                              # letters in a or b but not both

{'r', 'd', 'b', 'm', 'z', 'l'}

>>> a =

>>> a

{'r', 'd'}

1.5. 字典

>>> 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}

>>> list(tel.keys())

['irv', 'guido', 'jack']

>>> sorted(tel.keys())

['guido', 'irv', 'jack']

>>> 'guido' in tel

True

>>> 'jack' not in tel

False

dict() 构造函数可以直接从 key-value 对中创建字典:

>>> dict([('sape', 4139), ('guido', 4127), ('jack', 4098)])

{'sape': 4139, 'jack': 4098, 'guido': 4127}

>>>

>>> dict(sape=4139, guido=4127, jack=4098)

{'sape': 4139, 'jack': 4098, 'guido': 4127}

1.6. 循环技巧

>>> knights = {'gallahad': 'the pure', 'robin': 'the brave'}

>>> for k, v in knights.items():

...     print(k, v)

...

gallahad the pure

robin the brave

>>> 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 ?  It is .'.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.

>>> for i in reversed(range(1, 10, 2)):

...     print(i)

...

9

7

5

3

1

>>> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']

>>> for f in sorted(set(basket)):

...     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']

1.7. 深入条件控制

while 和 if 语句中使用的条件不仅可以使用比较，而且可以包含任意的操作。

>>> string1, string2, string3 = '', 'Trondheim', 'Hammer Dance'

>>> non_null = string1 or string2 or string3

>>> non_null

'Trondheim'

1.8. 比较序列和其它类型

(1, 2, 3)              < (1, 2, 4)

[1, 2, 3]              < [1, 2, 4]

'ABC' < 'C' < 'Pascal' < 'Python'

(1, 2, 3, 4)           < (1, 2, 4)

(1, 2)                 < (1, 2, -1)

(1, 2, 3)             == (1.0, 2.0, 3.0)

(1, 2, ('aa', 'ab'))   < (1, 2, ('abc', 'a'), 4)

Footnotes

[1]别的语言可能会返回一个变化的对象，允许方法连续执行，像 d->insert("a")->remove("b")->sort();。

[2]调用 d.keys() 将会返回一个 dictionary view 对象。它支持支持成员测试以及迭代等操作，但是它的内容不是独立的原始字典 – 它只是一个 视图。

• 发表于:
• 原文链接https://kuaibao.qq.com/s/20200728A0X5GT00?refer=cp_1026
• 腾讯「腾讯云开发者社区」是腾讯内容开放平台帐号（企鹅号）传播渠道之一，根据《腾讯内容开放平台服务协议》转载发布内容。
• 如有侵权，请联系 cloudcommunity@tencent.com 删除。

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