如何找到一个数组中最接近的素数，在该数组另一个号码？ [英] How to find the nearest prime number in an array, to another number in that array?

问题描述

我想找到最近的素数（即该数组present），任何阵列中的另一个号码？
例如：

I wanted to find out the nearest prime number (that is present in that array), to any another number in the array ?
Example :

list a -> [1,2,4,6,8,12,9,5,0,15,7]

所以，最近的素数为 4 是 2 键，在出现 15 这将是 7 。在这里，我假设列表中的每一个元素是不同的。照片 我花了几个小时就可以了，但解决不了，有没有什么快捷办法来解决这个问题呢？

So the nearest prime number to 4 would be 2 and in case of 15 it would be 7. Here i am assuming that every element in the list is distinct.
I spent hours on it but couldn't solve, is there any efficient way to solve this problem ?

推荐答案

首先，你需要一个良好的质数检查。 维基百科有一个实现 - 这很可能是一个有点进一步优化取决于Python版本，等等。

First, you need a good prime number checker. Wikipedia has an implementation -- It could probably be optimized a bit further depending on python version, etc.

现在，让所有质数的指数列表：

Now, make a list of the indices of all prime numbers:

indices = [i for i, val in enumerate(data) if is_prime(val)]

接下来，选择任意的号码，发现它的索引（或不乱......）。

Next, pick an arbitrary number and find it's index (or not arbitrary ...).

n = random.choice(data)
idx = data.index(n)

我们快到了......平分用自己的方式揣摩出的指数 N 适合的指数列表中。

we're almost there ... bisect your way to figure out where the index of n fits in the indices list.

indices_idx = bisect.bisect_left(indices, idx)

现在，找出更接近数是否是在左边还是右边，我们需要看的值。

Now, to figure out whether the closer number is on the left or the right we need to look at the values.

# Some additional error handling needs to happen here to make sure that the index
# actually exists, but this'll work for stuff in the center of the list...
prime_idx_left = indices[indices_idx - 1]
prime_idx_right = indices[indices_idx]

最后，找出哪些指标更接近并拉出值：

and finally, figure out which index is closer and pull out the value:

if (idx - prime_idx_left) <= (prime_idx_right - idx):
    closest_prime = data[prime_idx_left]
else:
    closest_prime = data[prime_idx_right]

请注意我，你会一遍又一遍地使用同一个列表的假设下熟这件事。如果你没有，你会做的更好，只是：

Note I cooked this up under the assumption that you'll be using the same list over and over. If you're not, you'd do better to just:

• 找到你感兴趣的一些指标。
• 找到的第一个主要的指数权（如果存在的话）
• 找到第一原的索引向左（如果它存在）
• 检查哪一个更接近

例如。

def find_idx_of_prime(lst, start_idx, stop_idx, dir):
    for ix in xrange(start_idx, stop_idx, dir):
        if is_prime(lst[ix]):
            return ix
    return dir*float('inf')

idx = data.index(number)
left_idx = find_idx_of_prime(data, idx, 0, -1)
right_idx = find_idx_of_prime(data, idx, len(data), 1)
prime_idx = left_idx if idx - left_idx < right_idx - idx else right_idx
prime_value = data[prime_idx]  # raises TypeError if no primes are in the list.

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