Python 集合理解

作业有两个问题，第二个问题卡住了。

  1. 使用 Python 集合理解(Python 相当于 Set Builder 符号)生成一组所有小于100的素数。回想一下，素数是一个大于1的整数，除了它自己和1之外，不能被其他任何整数整除。将一组素数存储在一个变量中(其他部分需要它)。输出一组素数(例如，使用 print 函数)。

  2. 使用 Python 集合理解生成一组有序对(长度为2的元组) ，其中包含所有由小于100的素数组成的素数对。素数对是一对连续的奇数，它们都是素数。将素数对集存储在一个变量中。你的第一组数字会很有帮助。输出一组素数对。

对于第一种情况，这种方法非常有效:

r= {x for x in range(2, 101)
if not any(x % y == 0 for y in range(2, x))}

然而，我对第二个问题很困惑。我想我可能需要一些笛卡儿积，但是我不确定。

这让我有点接近，但我只是希望连续的对。

cart = { (x, y) for x in r for y in r
if x < y }
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You can generate pairs like this:

{(x, x + 2) for x in r if x + 2 in r}

Then all that is left to do is to get a condition to make them prime, which you have already done in the first example.

A different way of doing it: (Although slower for large sets of primes)

{(x, y) for x in r for y in r if x + 2 == y}
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最佳答案 
primes = {x for x in range(2, 101) if all(x%y for y in range(2, min(x, 11)))}

I simplified the test a bit - if all(x%y instead of if not any(not x%y

I also limited y's range; there is no point in testing for divisors > sqrt(x). So max(x) == 100 implies max(y) == 10. For x <= 10, y must also be < x.

pairs = {(x, x+2) for x in primes if x+2 in primes}

Instead of generating pairs of primes and testing them, get one and see if the corresponding higher prime exists.

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You can get clean and clear solutions by building the appropriate predicates as helper functions. In other words, use the Python set-builder notation the same way you would write the answer with regular mathematics set-notation.

The whole idea behind set comprehensions is to let us write and reason in code the same way we do mathematics by hand.

With an appropriate predicate in hand, problem 1 simplifies to:

 low_primes = {x for x in range(1, 100) if is_prime(x)}

And problem 2 simplifies to:

 low_prime_pairs = {(x, x+2) for x in range(1,100,2) if is_prime(x) and is_prime(x+2)}

Note how this code is a direct translation of the problem specification, "A Prime Pair is a pair of consecutive odd numbers that are both prime."

P.S. I'm trying to give you the correct problem solving technique without actually giving away the answer to the homework problem.


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