4行代码求2000000以内素数之和报错求解

cuibaowenown2

1. def Prime(num):
   
2.     prime = []
   
3.     yield [n,prime.append(n)][0] for n in range(2,num) if 0 not in [n%m for m in prime]
   
4. print(sum(Prime(2000000)))

复制代码

cuibaowenown2

自己解决了：

1. def Prime(num):
   
2.     prime = []
   
3.     for n in range(2,num):
   
4.         if 0 not in [n%m for m in prime]:
   
5.             yield [n,prime.append(n)][0]
   
6. print(sum(Prime(2000000)))

复制代码

Ark_king

python很注重缩进的