python - 找到圆形素数 [英] python - finding circular prime number

问题描述

我正在尝试从给定的限制中找到圆形素数的数量.prime(x) 将返回一个数字是否为素数.rotations() 将返回一个旋转数字列表.最后， prime_count() 将根据给定的限制输出循环素数的总数.prime() 和 rotations() 都给了我正确的输出；然而， prime_count() 并没有像它应该的那样递增.关于我做错了什么的任何想法?

I am trying trying to find the number of circular primes from a given limit. The prime(x) will return whether a number is a prime or not. The rotations() will return a list of rotated numbers. Lastly, prime_count() will output the total amount of circular primes based on the given limit. Both prime() and rotations() gave me the correct output; however, prime_count() is not incrementing like it should. Any ideas on what i did wrong?

def prime(number): #return true or false
    return all(number% i for i in range(2,number))

def rotations(num): #rotating number and return list
    list = []
    m = str(num)
    counter = 0 
    while counter < len(str(num)):
        m=m[1:] + m[0]
        list.append(int(m))
        counter+=1
    list1=sorted(list,key=int)
    return list1

def prime_count(limit): #return numbers of circular primes from given limit
    counter = 0 

    for i in range(1,limit+1):
        a=rotations(i)
        for j in a:
            if j == prime(j): 
                counter+=1 

    return counter

print(prime_count(100))

推荐答案

您的代码存在一些问题:

There are a few problems with your code:

1. 你的 prime 函数有一个错误:

In [8]: prime(1)
Out[8]: True

由于 range(2, n) 为空且 any([]) == True<，它错误地为任何小于 2 的数字返回 True/code>.

It erroneously returns True for any number less than 2 due to range(2, n) being empty and any([]) == True.

prime_count 应该计算低于 limit 的循环素数的总数.prime(j) 返回一个布尔值，但您检查 j == prime(j)，只有当 j 为零或一，这绝对不是你想要的.尝试创建一个 is_circular_prime 函数，它接受一个整数 n 并返回素数是否是循环的.然后，prime_count 变得容易编写.

prime_count should be counting the total number of circular primes below limit. prime(j) returns a boolean, but you check j == prime(j), which can only be true if j is zero or one, which definitely isn't what you want. Try creating an is_circular_prime function that takes in an integer n and returns whether or not the prime is circular. Then, prime_count becomes easy to write.

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