Question

#!/usr/bin/python
#
# Description: bitwise factorization and then trying to find
# an elegant way to print numbers

# Source: http://forums.xkcd.com/viewtopic.php?f=11&t=61300#p2195422
# bug with large numbers such as 99, but main point in simplifying it
#
def primes(n):
    # all even numbers greater than 2 are not prime.
    s = [False]*2 + [True]*2 + [False,True]*((n-4)//2) + [False]*(n%2)
    i = 3;
    while i*i < n:
        # get rid of ** and skip even numbers.
        s[i*i : n : i*2] = [False]*(1+(n-i*i)//(i*2))
        i += 2
        # skip non-primes
        while not s[i]: i += 2
    return s


# TRIAL: can you find a simpler way to print them?
# feeling the overuse of assignments but cannot see a way to get it simpler
#
p = 49
boolPrimes = primes(p)
numbs = range(len(boolPrimes))
mydict = dict(zip(numbs, boolPrimes))

print([numb for numb in numbs if mydict[numb]])

我正在寻找的东西，你能让TRIAL变得极为简单吗？有这样的方法吗？

a=[True, False, True]
b=[1,2,3]
b_a                    # any such simple way to get it evaluated to [1,3]
                       # above a crude way to do it in TRIAL

Answer 1

对于python2.7 +，您可以使用itertools.compress

itertools.compress(b,a)

例如

>>> from itertools import compress
>>> a=[True, False, True]
>>> b=[1,2,3]
>>> list(compress(b,a))
[1, 3]

否则你可以使用列表理解

>>> [j for i,j in zip(a,b) if i]
[1, 3]

如果要在素数列表中执行此操作，使用枚举可能更简单

>>> primes = [False, False, True, True, False, True]
>>> list(compress(*zip(*enumerate(primes))))
[2, 3, 5]

Answer 2

您可以使用enumerate以及列表推导来大大简化这一点：

p = 49
print([num for num, isprime in enumerate(primes(p)) if isprime])

Answer 3

在这里制作一个词典并不是你想要的，因为词汇是无序的。将对保持为对实际上简化了逻辑，因为您可以迭代对，给出两个元素名称，并在理解中使用这些名称。

使用并行索引列表'压缩'列表的Pythonic方法是使用enumerate。

因此：

print ([x for (x, y) in enumerate(primes(49)) if y])

解释范围（n）和布尔列表，一对一地图，更简单？

3 个答案: