Pyhton数据结构列表是什么-创新互联

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Pyhton数据结构列表是什么？很多新手对此不是很清楚，为了帮助大家解决这个难题，下面小编将为大家详细讲解，有这方面需求的人可以来学习下，希望你能有所收获。

1、列表常用操作方法

list.append(x)

list末尾追加元素 x，等价于 a[len(a):]=[x] 

 list.extend(iterable)

list末尾追加可迭代类型元素（如添加[1,2])等价于 a[len(a):]=iterable 

list.insert(i,x)

list指定位置i添加元素x

list.remove(x)

list中删除元素

list.pop([i])

从指定位置[i]处删除元素,未指定位置时，默认从末尾元素删除。

list.clear()

清空list数据

list.index(x[, start[, end]])

返回x在list中首次出现的位置, start，end 指定查找的范围。

list.count(x)

返回x在list中的个数

list.sort(key=None, reverse=False)

对list进行排序

list.reverse()

list元素倒序

list.copy()

返回list的复制数据，等价于a[:]

>>> fruits = ['orange', 'apple', 'pear', 'banana', 'kiwi', 'apple', 'banana']

>>> fruits.count('apple')

2

>>> fruits.count('tangerine')

0

>>> fruits.index('banana')

3

>>> fruits.index('banana', 4)  # Find next banana starting a position 4

6

>>> fruits.reverse()

>>> fruits

['banana', 'apple', 'kiwi', 'banana', 'pear', 'apple', 'orange']

>>> fruits.append('grape')

>>> fruits

['banana', 'apple', 'kiwi', 'banana', 'pear', 'apple', 'orange', 'grape']

>>> fruits.sort()

>>> fruits

['apple', 'apple', 'banana', 'banana', 'grape', 'kiwi', 'orange', 'pear']

>>> fruits.pop()

'pear'

2、列表作为堆栈

堆栈的原则是数据 先进后出 

>>> 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]

3、列表作为队列

队列的原则是数据 先进先出

>>> 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'])

4、列表推导

根据List提供的相关方法，我们可以自己根据需求创建list

>>> squares = []

>>> for x in range(10):

...     squares.append(x**2)

...

>>> squares

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

也可以使用lambda表达式来创建

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

或者更直接

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

使用多个for循环或者if 组合语句也可以创建list

>>> [(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)]

等价于

>>> 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)]

如果表达式为元组类型，使用时必须用 () 

>>> 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]

lists推导也可以用混合表达式和内置函数

>>> from math import pi

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

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

5、列表内置推导

下面是一个 3* 4的矩阵

>>> matrix = [

...     [1, 2, 3, 4],

...     [5, 6, 7, 8],

...     [9, 10, 11, 12],

... ]

下面方法可以将matrix数据转换为 4*3的矩阵。

>>> [[row[i] for row in matrix] for i in range(4)]

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

等价于

>>> 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]]

或者是这样

>>> 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]]

上面操作完全可以使用list的内置函数 zip() 来实现

>>> list(zip(*matrix))

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

6、del表达式

使用可以直接删除List中的某个数值

>>> 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

[]

也可以删除整个 list

>>> del a

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