python中list列表的高级函数

821次阅读  |  发布于5年以前

在Python所有的数据结构中,list具有重要地位,并且非常的方便,这篇文章主要是讲解list列表的高级应用,基础知识可以查看博客。
此文章为python英文文档的翻译版本,你也可以查看英文版:https://docs.python.org/2/tutorial/datastructures.html

use a list as a stack: #像栈一样使用列表


    stack = [3, 4, 5] 
    stack.append(6) 
    stack.append(7) 
    stack 
    [3, 4, 5, 6, 7] 
    stack.pop() #删除最后一个对象 
    7 
    stack 
    [3, 4, 5, 6] 
    stack.pop() 
    6 
    stack.pop() 
    5 
    stack 
    [3, 4]

use a list as a queue: #像队列一样使用列表


    > from collections import deque #这里需要使用模块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: 三个重要的内建函数

filter(), map(), and reduce().
1)、filter(function, sequence)::
按照function函数的规则在列表sequence中筛选数据


    > def f(x): return x % 3 == 0 or x % 5 == 0
    ... #f函数为定义整数对象x,x性质为是3或5的倍数
    > filter(f, range(2, 25)) #筛选
    [3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]

2)、map(function, sequence):
map函数实现按照function函数的规则对列表sequence做同样的处理,
这里sequence不局限于列表,元组同样也可。


    > def cube(x): return x*x*x #这里是立方计算 还可以使用 x**3的方法
    ...
    > map(cube, range(1, 11)) #对列表的每个对象进行立方计算
    [1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

注意:这里的参数列表不是固定不变的,主要看自定义函数的参数个数,map函数可以变形为:def func(x,y) map(func,sequence1,sequence2) 举例:


     seq = range(8)  #定义一个列表
    > def add(x, y): return x+y #自定义函数,有两个形参
    ...
    > map(add, seq, seq) #使用map函数,后两个参数为函数add对应的操作数,如果列表长度不一致会出现错误
    [0, 2, 4, 6, 8, 10, 12, 14]

3)、reduce(function, sequence):
reduce函数功能是将sequence中数据,按照function函数操作,如 将列表第一个数与第二个数进行function操作,得到的结果和列表中下一个数据进行function操作,一直循环下去…
举例:


    def add(x,y): return x+y
    ...
    reduce(add, range(1, 11))
    55

List comprehensions:
这里将介绍列表的几个应用:
squares = [x2 for x in range(10)]
#生成一个列表,列表是由列表range(10)生成的列表经过平方计算后的结果。
[(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)] 这里是生成了一个列表,列表的每一项为元组,每个元组是由x和y组成,x是由列表[1,2,3]提供,y来源于[3,1,4],并且满足法则x!=y。

Nested List Comprehensions:
这里比较难翻译,就举例说明一下吧:


    matrix = [          #此处定义一个矩阵
    ...   [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]]

这里两层嵌套比较麻烦,简单讲解一下:对矩阵matrix,for row in matrix来取出矩阵的每一行,row[i]为取出每行列表中的第i个(下标),生成一个列表,然后i又是来源于for i in range(4) 这样就生成了一个列表的列表。

The del statement:
删除列表指定数据,举例:


    > a = [-1, 1, 66.25, 333, 333, 1234.5]
    >del a[0] #删除下标为0的元素
    >a
    [1, 66.25, 333, 333, 1234.5]
    >del a[2:4] #从列表中删除下标为2,3的元素
    >a
    [1, 66.25, 1234.5]
    >del a[:] #全部删除 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 #相当于向字典中添加数据
    >>> tel
    {'sape': 4139, 'guido': 4127, 'jack': 4098}
    >>> tel['jack'] #取数据
    4098
    >>> del tel['sape'] #删除数据
    >>> tel['irv'] = 4127   #修改数据
    >>> tel
    {'guido': 4127, 'irv': 4127, 'jack': 4098}
    >>> tel.keys()    #取字典的所有key值
    ['guido', 'irv', 'jack']
    >>> 'guido' in tel #判断元素的key是否在字典中
    True
    >>> tel.get('irv') #取数据
    4127

也可以使用规则生成字典:


    >>> {x: x**2 for x in (2, 4, 6)}
    {2: 4, 4: 16, 6: 36}

enumerate():遍历元素及下标
enumerate 函数用于遍历序列中的元素以及它们的下标:


    >>> for i, v in enumerate(['tic', 'tac', 'toe']):
    ...   print i, v
    ...
    0 tic
    1 tac
    2 toe

zip():
zip()是Python的一个内建函数,它接受一系列可迭代的对象作为参数,将对象中对应的元素打包成一个个tuple(元组),然后返回由这些tuples组成的list(列表)。若传入参数的长度不等,则返回list的长度和参数中长度最短的对象相同。利用*号操作符,可以将list unzip(解压)。


    >>> 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.

有关zip举一个简单点儿的例子:


    >>> 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():反转


    >>> 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)):       #这里使用了set函数
    ...   print f
    ...
    apple
    banana
    orange
    pear

python的set和其他语言类似, 是一个 基本功能包括关系测试和消除重复元素.

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内容,希望对大家的学习有所帮助。

Copyright© 2013-2020

All Rights Reserved 京ICP备2023019179号-8