我现在正在开发一个程序，其目标是以某种方式分析孪生素数(孪生素数是 (p,p+2) 形式的素数)。现在我有一个代码可以计算 %10 形式的孪生素数余数。
这是代码:
def twin_prime_counter_type10(n):
not_prime = []
prime = []
A = range(n + 1)
B = range(n + 1)
for i in xrange(2, n+1):
if i not in not_prime:
prime.append(i)
for j in xrange(i*i, n+1, i):
not_prime.append(j)
for n,i in enumerate(prime):
if not A[n] == prime[n]:
A[i] = 1
count1_3 = 0
count7_9 = 0
count9_1 = 0
for i in B:
if B[i] % 10 == 3 and B[i - 2] % 10 == 1:
if A[i] * A[i-2] == 1:
count1_3 += 1
elif B[i] % 10 == 9 and B[i - 2] % 10 == 7:
if A[i] * A[i-2] == 1:
count7_9 += 1
elif B[i] % 10 == 1 and B[i - 2] % 10 == 9:
if A[i] * A[i-2] == 1:
count9_1 += 1
print count1_3
print count7_9
print count9_1
打印筛(10000)
这部分代码工作正常，但我想知道是否有人知道一种方法，当我找到 (1) 对(孪生素数)时，我还可以记录它们在列表中出现的顺序。我不需要任何人实际编写执行此操作的代码，我只是想问问是否有人知道我可以用来执行此任务的 Python 内置工具。
感谢您的帮助。
请您参考如下方法:
假设 sieve 是 Sieve of Eratosthenes , 或 similar , 然后你可以使用 pairwise itertools 中的食谱仅获取带有生成器表达式的孪生素数的模块
from itertools import tee #, izip # <-- for python 2
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return zip(a, b) #izip(a, b) # <-- for python 2
def twin_primes(n):
return ( (p1,p2) for p1,p2 in pairwise(sieve(n)) if p1+2 == p2 )
# or use list comprehension, but I prefer generator as they use
# less memory and can be infinite, as sieve can be made to be
# infinite as well
例如
>>> print( list(twin_primes(100)) )
[(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)]
>>>
同样，您可以将它与枚举一起使用，以获取与素数列表相关的对中第一个素数的索引，例如
>>> primes=list(sieve(100))
>>> 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]
>>> twins=[ (i,pair) for i,pair in enumerate( pairwise(primes) ) if pair[0]+2 == pair[1] ]
>>> twins
[(1, (3, 5)), (2, (5, 7)), (4, (11, 13)), (6, (17, 19)), (9, (29, 31)), (12, (41, 43)), (16, (59, 61)), (19, (71, 73))]
>>> primes[1]
3
>>> primes[4]
11
>>> primes[9]
29
>>>
