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Python函数式编程之map/reduce/filter进阶

Python函数式编程之map/reduce/filter进阶

作者: QuantumEnergy | 来源:发表于2017-10-01 11:14 被阅读0次

    说明

    本文重点在于示例代码,在熟悉基本概念(map/reduce/filter/lambda)的基础上阅读最好

    背景

    函数式编程是一种编程范式,我们常见的是命令式编程,首先大概了解下概念:

    命令式:冯诺依曼机的序列

    函数式:基于λ演算

    函数式编程一般有如下特点:

    • Referential transparency
    • No Side Effect
    • Currying
    • Closure
    • Higher-order function
    • Lazy evaluation
    • Lambda

    进入正题,主要看看python函数式编程,几个高阶函数map/reduce/filter的使用

    定义(python2)

    sequence是一种泛型,包括list,tuple,string

    map(function, sequence[, sequence, ...]) -> list
    1. map将传入的函数依次作用到序列的每个元素,并把结果作为新的list返回

    reduce(function, sequence[, initial]) -> value

    1. 把一个函数作用在一个序列上,reduce把结果继续和序列的下一个元素做累积计算
    2. function接受两个参数
    3. 返回值类型取决于function return

    filter(function or None, sequence) -> list, tuple, or string

    1. 如果是None,返回是True的元素
    2. 返回和sequence相同的类型

    示例

    map/reduce/filter其实很简单,要熟练掌握,最好的办法就是code,示例代码包括map/reduce/filter使用的各个方面,重点地方已注释,认真看完code,一定可以掌握map/reduce/filter的

    # coding:utf-8
    from operator import add
    
    
    def foo():
        a = [i for i in range(1, 10)]
        b = map(lambda x: x**2, a)
        c = reduce(lambda x, y: x + y, a)
        d = filter(lambda x: not x % 3, a)
        print b
        print c
        print d
    '''
    [1, 4, 9, 16, 25, 36, 49, 64, 81]
    45
    [3, 6, 9]
    '''
    
    
    def foo1():
        a = [i for i in range(1, 10)]
        b = map(lambda x: x**2, a)
        b = reduce(lambda x, y, z='hi': z, a)
        # z始终是hi
        b1 = reduce(lambda x, y='hi': y, a)
        # y始终是a中的元素,结束时是最后一个元素
        c = map(lambda x, y='hi': y, a)
        c1 = map(lambda x, y: x - y, a, a)
        d = filter(lambda x, y='hi': y, a)
        # y始终是hi,为True,所以a中没有元素被过滤掉
        print b
        print b1
        print c
        print c1
        print d
    '''
    hi
    9
    ['hi', 'hi', 'hi', 'hi', 'hi', 'hi', 'hi', 'hi', 'hi']
    [0, 0, 0, 0, 0, 0, 0, 0, 0]
    [1, 2, 3, 4, 5, 6, 7, 8, 9]
    '''
    
    
    def foo2():
        # sequence:list tuple str
        s = "h x010y i"
        # s中每个字符作为参数单独调用
        b = map(lambda x: x, s)
        c = reduce(lambda x, y: x, s)
        # x始终是s的第一个字符(h)
        d = reduce(lambda x, y: y, s)
        e = reduce(lambda x, y: (x,y), s)
        # reduce调用的过程
        f = filter(lambda x: x.isalpha() or x.isspace(), s)
        # 保留字母和空格
        print b
        print c
        print d
        print e
        print f
    '''
    ['h', ' ', 'x', '0', '1', '0', 'y', ' ', 'i']
    h
    i
    (((((((('h', ' '), 'x'), '0'), '1'), '0'), 'y'), ' '), 'i')
    h xy i
    '''
    
    
    def foo3():
        a = "a01bcdfalsetrueFalseTrue0None1"
        b = [0, None, False,True, 1, '', 'a']
        c = (0, None, False,True, 1, '', 'a')
        # filter可以接受None作为第一个参数,此时由sequence中
        # 元素本身的真假值进行过滤,返回值保持sequence本身的类型
        print filter(None, a)
        # a中每个字符都是真(0是字符,也为真),返回值是str
        print filter(None, b)
        # 返回值是list
        print filter(None, c)
        # 返回值是tuple
    '''
    a01bcdfalsetrueFalseTrue0None1
    [True, 1, 'a']
    (True, 1, 'a')
    '''
    
    
    def foo4():
        bar = [[1, 2, 4, 5], [3, 5, 7, 2, 6]]
        bar1 = ['1', '2', '3']
        b = reduce(add, bar)
        c = map(sum, zip(*bar))
        # 二维数组反转求和
        d = map(int, bar1)
        # 字符串转int
        e = sorted(set(b), key = b.index)
        # 对list去重后,保持原有的元素顺序
        print b
        print c
        print d
        print e
    '''
    [1, 2, 4, 5, 3, 5, 7, 2, 6]
    [4, 7, 11, 7]
    [1, 2, 3]
    [1, 2, 4, 5, 3, 7, 6]
    '''
    
    
    def foo5():
        a = [i for i in range(1, 4)]
        b = [lambda y: y*x for x in a] # late binding (x延迟绑定)
        c = [lambda y, x=x: y*x for x in a] # x是local变量
        d = (lambda y: y*x for x in a) # lazy evaluation(生成器惰性求值)
        e = map(lambda x: lambda y: y*x, a) # closure(闭包)
        for bar in b:
            print bar(2)
        print '--------'
        for bar in c:
            print bar(2)
        print '--------'
        for bar in d:
            print bar(2)
        print '--------'
        for bar in e:
            print bar(2)
        print '--------'
    
        for t in e:
            for j in t.__closure__:
                print j.cell_contents
    
    '''
    6
    6
    6
    --------
    2
    4
    6
    --------
    2
    4
    6
    --------
    2
    4
    6
    --------
    1
    2
    3
    '''
    
    if __name__ == '__main__':
        foo()
        foo1()
        foo2()
        foo3()
        foo4()
        foo5()
        pass
    

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