python巧妙实现数学阶乘n！

# 正确的结果

1*2*3*4*5

120

result = 1  # 给定一个初始值
n = 5

for i in range(1, n+1):
    print("累乘前result: ", result)
    print("循环数i的值: ", i)
    result = result * i  # 不断地累成result
    print("累乘后result: ", result)
    print("------------")

result

累乘前result:  1

循环数i的值:  1

累乘后result:  1

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

累乘前result:  1

循环数i的值:  2

累乘后result:  2

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

累乘前result:  2

循环数i的值:  3

累乘后result:  6

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

累乘前result:  6

循环数i的值:  4

累乘后result:  24

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

累乘前result:  24

循环数i的值:  5

累乘后result:  120

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

120

def recursion(n):
    if n == 0 or n == 1:  # 特殊情况
        return 1
    else:
        return n * recursion(n-1)  # 递归函数

recursion(5)

120

# 在python3中reduce函数被移入到functools中；不再是内置函数

from functools import reduce 

n = 5

reduce(lambda x,y: x*y, range(1,n+1))  

120

reduce(function, iterable[, initializer])

# 使用自定义函数

from functools import reduce 

number = range(1,6)
# number = [1,2,3,4,5]

def add(x,y):
    return x+y

reduce(add, number)  # 1+2+3+4+5

15

# 使用匿名函数

from functools import reduce 

number = range(1,6)

reduce(lambda x,y: x+y, number)  # 1+2+3+4+5

15

# 求出下面的阶乘的累加求和

1 + 1*2 + 1*2*3 + 1*2*3*4 + 1*2*3*4*5

153

# 定义累乘函数

def func(n):
    result = 1
    
    for i in range(1, n+1):
        result = result * i  # 不断地累成re
          
    return result
    
func(5)  # 测试案例

120

# func(1) + func(2) + func(3) +  func(4) + func(5)

# 调用累乘函数
sum(func(i)  for i in range(1,6))

153

# 定义累乘函数

def func(n):
    result = 1  # 定义初始值
    
    for i in range(1, n+1):
        result = result * i  # 不断地累成re
    
    # if result == 1 :  等价于下面的条件
    if n==0 or n==1:
        return 1
    else:
        return result + func(n-1)  #在这里实现递归 func(n-1)
    
func(5)  

153

def recursion(n):
    """
    之前定义的递归函数
    """
    if n == 0 or n == 1:
        return 1
    else:
        return n * recursion(n-1)  

# recursion(1) + recursion(2) + recursion(3) +  recursion(4) + recursion(5)

# 调用定义的递归函数
sum(recursion(i)  for i in range(1,6))

153

from functools import reduce 

n = 5

reduce(lambda x,y: x*y, range(1,n+1))  

120

sum(reduce(lambda x,y: x*y, range(1,n+1)) for n in range(1,6))

153

[reduce(lambda x,y: x*y, range(1,n+1)) for n in range(1,6)]

[1, 2, 6, 24, 120]

reduce(lambda x,y:x+y, [reduce(lambda x,y: x*y, range(1,n+1)) for n in range(1,6)] )

153