无法解释python中打印语句的输出

Question

我在下面的代码中使用集合理解来计算 2 到 n 之间的素数，n=132。

只要变量 n 的值 <= 131，生成的素数就会按适当的升序打印，即 {2,3,5,7,11,...}。

每当 n > 131 时，打印顺序就会出现偏差，例如 {2,3,131,5,7,...}。

无论 'n' 的值如何，变量 'noPrimes' 的值总是以正确的顺序打印。

我不太明白为什么？

Environment: Python: 3.7.2, macOS: Mojave 10.14.4, IDE: WingPro version 7.0.1.2
Code: 

    from math import sqrt
    n = 132
    sqrt_n = int (sqrt(n))
    noPrimes = {j for i in range (2, (sqrt_n + 1)) for j in range (i*2, n, i)}
    primes = {x for x in range (2,n) if x not in noPrimes}
    print ("Printing 'noPrimes':")
    print (noPrimes)
    print ("Printing 'Primes':")
    print (primes)

Answer 1

from math import sqrt
n = 132
sqrt_n = int (sqrt(n))
noPrimes = {j for i in range (2, (sqrt_n + 1)) for j in range (i*2, n, i)}
primes = [x for x in range (2,n) if x not in noPrimes]
print ("Printing 'noPrimes':")
print (sorted(noPrimes))
print ("Printing 'Primes':")
print (primes)

Answer 2

在您的示例中， noPrimes和primes都被设置，并且 set 是无序的，从文档中可以明显看出： https : noPrimes

集合对象是不同的可散列对象的无序集合。

from math import sqrt
n = 132
sqrt_n = int (sqrt(n))
noPrimes = {j for i in range (2, (sqrt_n + 1)) for j in range (i*2, n, i)}
primes = {x for x in range (2,n) if x not in noPrimes}
print ("Printing 'noPrimes':")
print (noPrimes)
print ("Printing 'Primes':")
print (primes)
print(type(noPrimes))
print(type(primes))

所以原始情况下的输出将是 .

Printing 'noPrimes':
{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 128, 129, 130}
Printing 'Primes':
{2, 3, 131, 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}
<class 'set'>
<class 'set'>

所以你得到的答案是有效的，只是他们没有顺序

如果你想要订单，你想要做的是像这样的列表理解

from math import sqrt
n = 132
sqrt_n = int (sqrt(n))
#noPrimes is a list
noPrimes = [j for i in range (2, (sqrt_n + 1)) for j in range (i*2, n, i)]
#primes is a list 
primes = [x for x in range (2,n) if x not in noPrimes]
print ("Printing 'noPrimes':")
print (noPrimes)
print ("Printing 'Primes':")
print (primes)
print(type(noPrimes))
print(type(primes))

然后输出将是

Printing 'noPrimes':
[4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128, 130, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121]
Printing 'Primes':
[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]
<class 'list'>
<class 'list'>

Answer 3

@101arrowz 在他的评论中提供了正确答案

集合是无序的。 你不会总是得到序列化的输出。 如果您希望它们按顺序排列，您应该使用列表理解

为了扩展这个答案，您可能会看到在一定时间内排序的一个原因是一些整数在运行前预先填充，给它们一个确定性的 id 顺序（对于我来说，2.7 和 3.6 cPython 在下面是 -5,256）。

def collapse(items):
    ret = []
    if not items:
        return ret
    first = last = items[0]
    for i in items[1:]:
        if i == last + 1:
            last = i
        else:
            ret.append((first, last))
            first = last = i
    ret.append((first, last))
    return ret


def pretty_collapse(items):
    vals = collapse(items)
    return ', '.join('%d..%d' % (a, b) if a != b else '%d' % a
                     for a, b in vals)


int_by_id = list(sorted(range(-50, 300), key=id))
print(pretty_collapse(int_by_id))

给

-5..256, -48, -49, -50, -32, -47, -36, -37, -42..-41, -43, -33, -34, -35, -44, -45, -46, -38, -39, -40, -24, -25, -26, -27, -28, -29, -30, -31, -23..-6, 257..299

请注意在 Python 中使用“is”还是“==”进行数字比较更好？