python函数式编程与应用

拼搏现实的明天。 2022-12-11 09:15 114阅读 0赞

# map 映射 #

*  源码中的注释  
    “”"  
    map(func, \*iterables) --> map object

Make an iterator that computes the function using arguments from  
each of the iterables. Stops when the shortest iterable is exhausted.  
“”"

list1 = [1,2,3,4,5,6,7,8]
    list2 = [1,2,3,4,5,6,7,8]
    a = map(lambda x, y:x*x + y,list1,list2)
    print(list(a))

# reduce 归约 会做一个连续的计算 连续的调用lambda #

*  源码中的注释  
    “”"  
    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.  
“”"

from functools import reduce
    list1 = [1,2,3,4,5,6,7,8]
    list2 = ['1','2','3','4','5','6','7','8']
    # reduce(function, sequence[, initial]) -> value
    a = reduce(lambda x, y:x + y, list1) # list1相加
    a = reduce(lambda x, y:x + y, list1, 10) # 初始值为10 list1相加
    a = reduce(lambda x, y:x + y, list2, 'sum') # 初始值为10 list1相加
    print(a)

# filter 集合的过滤 #

*  源码中的注释

“”"  
filter(function or None, iterable) --> filter object

Return an iterator yielding those items of iterable for which function(item)  
is true. If function is None, return the items that are true.  
“”"

list1 = [1,0,1,0,1,0,1]
    res = filter(lambda x: True if x==1 else False, list1)
    print(list(res))

# 命令式编程 #

# def
        # if else
        # for

# 函数式编程 #

# map reduce filter
        # lambda 算子

# 理论上可以使用后者替换前者 #

# def: lambda
        # if else: 三元表达式
        # for: map reduce

# 三元表达式 #

a = 1
    b = 2
    
    # 第一种写法：
    erroStr = "More" if a > b else "Less"
    print(erroStr) # 运行结果为：Less
    # 第二种写法：
    print({ True: "More", False: "Less"}[a > b]) # 运行结果为：Less
    # 第三种写法：
    print(("FalseValue", "TrueValue")[a > b]) # 运行结果为：FalseValue
    print(("Less", "More")[a > b]) # 运行结果为：Less

# 应用 #

L = []
    x = 2
    L.append(100 if x > 1 else 0)
    
    
    def Max(x, y):
       return x if x > y else y
    
    def fib(n):
       return 1 if n == 1 else fib(n-1) + n
    fib(2)