【Python】【问题】关于lambda函数

m0_52606838 2021-02-16 01:19:58

新人小白，最近在用Python实现筛法求质数的时候遇到一个奇怪的情况：

def odd(): #生成一个无穷奇数序列

    n = 1

    while 1:

        yield 2*n+1

        n += 1



def nd(n): #筛选函数

    return lambda x: x%n>0



def prime(seq):

    yield 2

    while 1:

        p = next(seq)

        yield(p)

        seq = filter(nd(p), seq) #筛法更新质数序列



oseq = odd()

pr = prime(oseq)

print([next(pr) for i in range(100)])

正确输出：

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541]

筛选函数只包含一个lambda表达式，但是如果不定义筛选函数，直接在筛选时使用lambda表达式，就会出现错误结果（输出了所有奇数）：

def odd(): #生成一个无穷奇数序列

    n = 1

    while 1:

        yield 2*n+1

        n += 1



def prime(seq):

    yield 2

    while 1:

        p = next(seq)

        yield(p)

        seq = filter(lambda x: x%p>0, seq) #筛法更新质数序列



oseq = odd()

pr = prime(oseq)

print([next(pr) for i in range(100)])

输出：

[2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151, 153, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181, 183, 185, 187, 189, 191, 193, 195, 197, 199]

另外在尝试lambda+filter的性质时也出现了奇怪的结果：