def is_odd(n):return n % 2 == 1newlist = filter(is_odd, range(1,11))print(list(newlist))

2022-06-04

答案：

[1,3,5,7,9]

def is_odd(n):return n%2 == 1lst1 = filter(is_odd,[1,2,3,4,5,6,7,8,9,10])list(lst1)值为？
‌下列函数能够求n的阶乘n!的是 A: f = lambda n: n! B: def f(n): s=1 for i in range(n): s = s * i return s C: def f(n): s=1 for i in range(n,0,-1): s = s * i return s D: def f(n): if n == 1: return 1 else: return f(n-1)*n
给一段代码defis_odd(n):returnn%2==0newlist=filter(is_odd,[1,2,3,4,5,6,7,8,9,10])foriteminnewlist:print(item)其输出结果是? A: [2,4,6,8,10] B: [1,3,5,7,10] C: [1,3,5,7,9] D: [1,3,5,6,9]
函数f打印出1,2,3的6个排列，空缺语句是 def f(n): for a in range(1,n): for b in range(1,n): for c in range(1,n): if ______________: print(a,b,c) _________
智慧职教: def f(n): if n==0: return 0 elif n==1: return 1 return f(n-1)+f(n-2) print(4) 结果是

0
求n!。请补充横线处代码： def fac(n): s=1 for i in range(2,n+1): return s
1
下面哪个是正确的递归函数 A: def fun1(n): if n==1: return 1 else: return n*fun1(n-1) B: def fun2(n): if n==1: return 1 else: return n*fun2(n+1) C: def fun3(n): return n*fun3(n-1) D: def fun4(n): return n*fun4(n+1)
2
在以下函数中，i是（） def fact(n): f=1 for i in range(1,n+1): f=f*i return f print(fact(5))
3
def f(x,n = 2): return（x**n）print(f(5))print(f(3,3)）
4
以下哪个程序段是使用递归函数实现1到100求和 A: def product1(num): product=1 for i in range(1,num+1): product=product *i return productprint(product1(10)) B: def sum1(num): sum2=0 for i in range(1,num+1): sum2+=i return sum2 print(sum1(100)) C: def product2(num): if num==1: return 1 else: return num *product2(num-1)print(product2(10)) D: def sum_a(num1): if num1==1: return 1 else: return num1+sum_a(num1-1) print(sum_a(100))