Question

我编写了这个程序，它会告诉你输入的两个倍数因子。防爆。如果我输入35（一个半英里），程序将打印5和7，这是两个加到35的素数。

但是我想知道是否有更简洁或pythonic的方法来迭代这个元组，所以我不必编写所有这些＆＃34; elif＆＃34;你在下面看到的陈述。

如果我不需要依赖任何外部库，那也很棒。

# multiples of semiprimes 4 - 49
tuple1 = ( 2, 3, 5, 7 )

# tuple 1 calculations
while True:

        try:
                semiprime = int(input('Enter Semiprime: '))

        except ValueError:
                print('INPUT MUST BE AN INTEGER')
                continue

        # index 0 - 3
        if (tuple1[0]) * (tuple1[0]) == semiprime:
                print((tuple1[0]), (tuple1[0]))

        elif (tuple1[0]) * (tuple1[1]) == semiprime:
                print((tuple1[0]), (tuple1[1]))

        elif (tuple1[0]) * (tuple1[2]) == semiprime:
                print((tuple1[0]), (tuple1[2]))

        elif (tuple1[0]) * (tuple1[3]) == semiprime:
                print((tuple1[0]), (tuple1[3]))

        # index 1 - 3
        elif (tuple1[1]) * (tuple1[0]) == semiprime:
                print((tuple1[1]), (tuple1[0]))

        elif (tuple1[1]) * (tuple1[1]) == semiprime:
                print((tuple1[1]), (tuple1[1]))

        elif (tuple1[1]) * (tuple1[2]) == semiprime:
                print((tuple1[1]), (tuple1[2]))

        elif (tuple1[1]) * (tuple1[3]) == semiprime:
                print((tuple1[1]), (tuple1[3]))

        # index 2 - 3
        elif (tuple1[2]) * (tuple1[0]) == semiprime:
                print((tuple1[2]), (tuple1[0]))

        elif (tuple1[2]) * (tuple1[1]) == semiprime:
                print((tuple1[2]), (tuple1[1]))

        elif (tuple1[2]) * (tuple1[2]) == semiprime:
                print((tuple1[2]), (tuple1[2]))

        elif (tuple1[2]) * (tuple1[3]) == semiprime:
                print((tuple1[2]), (tuple1[3]))

        #index 3 - 3
        elif (tuple1[3]) * (tuple1[0]) == semiprime:
                print((tuple1[3]), (tuple1[0]))

        elif (tuple1[3]) * (tuple1[1]) == semiprime:
                print((tuple1[3]), (tuple1[1]))

        elif (tuple1[3]) * (tuple1[2]) == semiprime:
                print((tuple1[3]), (tuple1[2]))

Answer 1

我在评论中暗示了这一点，但意识到功能文档的链接可能还不够。

以下是使用itertools.combinations_with_replacement编写代码的方法：

from itertools import combinations_with_replacement

# multiples of semiprimes 4 - 49
tuple1 = ( 2, 3, 5, 7 )

# tuple 1 calculations
while True:

    try:
        semiprime = int(input('Enter Semiprime: '))

    except ValueError:
        print('INPUT MUST BE AN INTEGER')
        continue

    for (x,y) in combinations_with_replacement(tuple1, 2):
        if x * y == semiprime:
            print(x,y)

好多了，IMO：）

修改：使用itertools.combinations的先前版本不会产生具有相同值的（x，y）对（例如，(x,y) = (2,2)永远不会发生）。 combinations_with_replacement允许重复。感谢@Copperfield指出这一点。

Answer 2

虽然jedwards展示了最恐怖的方法 - 使用你将会了解和喜爱的itertools库 - 这里是更多的经典＆＃34;使用for循环的方法为您想要的模式。我提出来是因为作为一个编程初学者，重要的是要知道这个基本的，命令式的习语：

>>> tuple1 = (2,3,5,7)
>>> for i in range(len(tuple1)):
...   for j in range(i+1, len(tuple1)):
...     print(tuple1[i], tuple1[j])
... 
2 3
2 5
2 7
3 5
3 7
5 7
>>>

因此，您的代码将缩短为：

for i in range(len(tuple1)):
    for j in range(i+1, len(tuple1)):
        if tuple1[i] * tuple1[j] == semiprime
            print(tuple1[i], tuple1[j])

Answer 3

即使@ jedwards 解决方案很棒，（以及简洁/ pythonic ）;另一个可能的解决方案：

def prime_multiples(l,t ):  
    for i in l:  # Iterate over our list.
        for j in t:  # Iterate over the tuple of prime factors.
            #  We check to see that we can divide without a remainder with our factor,
            #  then check to see if that factor exists in our tuple.
            if i%j == 0 and i/j in t:
                print "Prime factors: {} * {} = {}".format(j, i/j, i)
                break  # We could go not break to print out more options.

示例输出：

l = [4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49]
t = ( 2, 3, 5, 7 )
prime_multiples(l, t)
>>> Prime factors: 2 * 2 = 4
... Prime factors: 2 * 3 = 6
... Prime factors: 3 * 3 = 9
... Prime factors: 2 * 5 = 10
... Prime factors: 2 * 7 = 14
... Prime factors: 3 * 5 = 15
... Prime factors: 3 * 7 = 21
... Prime factors: 5 * 5 = 25
... Prime factors: 5 * 7 = 35
... Prime factors: 7 * 7 = 49

在Python 3x中对列表/元组执行计算的最佳方法

3 个答案: