def _not_divisible(n): return lambda x: x % n > 0

标签：for   lam   lambda   while   filter   round   lse   ext   prime   

如下代码中如何理解：传入的实际参数，替换def中的x还是n？

def _not_divisible(n):
    return lambda x: x % n > 0

def _odd_iter():
    n = 1
    while True:
        n = n + 2
        yield n


def _not_divisible(n):
    return lambda x: x % n > 0


def primes(n):
    return lambda x: x % n > 0


def primes():
    yield 2
    it = _odd_iter()
    while True:
        n = next(it)
        yield n
        it = filter(_not_divisible(n), it)


for n in primes():
    if n < 1000:
        print(n)
    else:
        break

def _not_divisible(n): return lambda x: x % n > 0

标签：for   lam   lambda   while   filter   round   lse   ext   prime   

原文地址：https://www.cnblogs.com/LewisAAA/p/9270972.html