python3中reduce函数用法

﹏ヽ暗。殇╰゛Y 2023-07-03 08:24 18阅读 0赞

reduce函数用途：对迭代对象中的元素从左至右两两送入参数1传入的function中进行运算

--------------------

导入reduce函数：

from functools import reduce

reduce函数使用方法：help(reduce)

Help on built-in function reduce in module _functools:
    reduce(...)
        reduce(function, sequence[, initial]) -> value
        
        Apply a function of two arguments cumulatively to the items of a sequence,
        from left to right, so as to reduce the sequence to a single value.
        For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates
        ((((1+2)+3)+4)+5).  If initial is present, it is placed before the items
        of the sequence in the calculation, and serves as a default when the
        sequence is empty.

参数1：function句柄，用于完成迭代对象中两个元素之间的操作

参数2：迭代对象

--------------------

使用实例：

1、使用reduce完成列表求和

from functools import reduce
    
    x = list(range(1, 11))
    print(x)
    print(sum(x))
    
    
    def add_op(i, j):
        '''定义列表中两个元素之间的加法操作'''
        return i + j
    
    
    print(reduce(add_op, x))
    
    
    输出结果如下：
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    55
    55

2、使用reduce完成列表的合并

x = [
        [1, 2, 3, 4, 5],
        [6, 7, 8, 9, 10],
        [11, 12, 13, 14, 15],
        [16, 17, 18, 19, 20]
    ]
    
    # 将x中的各个元素合并为一个列表
    # 通常采用的做法如下
    res = []
    for e in x:
        res.extend(e)
    print(res)
    
    
    # 采用reduce合并列表
    def union_list(l1, l2):
        return l1 + l2
    
    
    res = reduce(union_list, x)
    print(res)
    
    输出结果如下：
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]