Python 素数、对数、求和和和循环

我正试图制作一个程序来计算所有小于给定数的素数，以及所有这些素数乘积的自然对数。因为我们使用对数，我也可以把每个小于给定数的素数的自然对数相加。所以我创建了这个程序，但是我觉得对数有点奇怪？我试着检查我的程序给出的总和是否正确，但首先我的程序只给出整数作为答案，其次我的计算器给出不同的答案。我的代码有问题吗？素数工作正常，但如前所述，对数是错误的。请参阅下面的代码

def prime_program():
while True:

    answer = raw_input("Do you want to know the primes smaller than a number n and the natural logarithm of the product of those primes? Please answer yes or no    ")

    if answer == "yes":
        n = float(raw_input("Please enter a whole number bigger than 1 ")) 
        counter = 1
        prime = 3.0
        log_of_prime_sum = float(log(2.0)) #I'm starting out with log(2.0) cause 2 is the first prime not counted later



        if n > 2.0:
            print "these are the primes smaller than %d" % n
            print 2.0 #cause I want to see all the primes but only want to check the odd numbers

            while prime < n:

                for x in range(2,int(sqrt(prime))+1): #range only works with integers right?

                    if prime%x == 0.0:       
                        break#because if it isn't a prime I don't want to do anything with it
                else:
                    print(prime) #cause I want to know the primes
                    log_of_prime_sum += float(log(prime)) #cause I want to add all the logs of the primes

                prime += 2.0 #we're starting with three, which is an odd, 
                             #and I'm adding two because I only want to check the odd numbers,
                             #because no even number other than two is prime

            print "the natural logarithm of the product of all the primes smaller than %d is %d"  % (n, log_of_prime_sum)
            ratio = float(float(n)/float((log_of_prime_sum)))
            print "The ratio of %d to the sum of the natural logarithms smaller than %d is %d" % (n, n, ratio) #here I want to print the ratio of n and the log_of_prime_sum
        else:
            print "There is no prime smaller than 2"
    elif answer == "no": #like what why would anyone open this file if he wouldn't want to do this lol
        print "well ok then"
    else:
        print "I'm sorry that was not a valid answer, you can only answer yes or no"
        continue #jumps back to while True:
    return #finished; exit the function
        #here I want to go back to answer = raw_input

prime_program()

---

虽然您的程序非常混乱（并且确定数字是否为素数的方法远远不是最优的，但可能有一个错误）

+=

表示使用操作数“增加”变量，因此可以声明：

log_of_prime_sum = log_of_prime_sum+log_of_prime_sum+float(log(2.0))

防止此类错误的一种方法是，例如，初始化（在程序顶部，

log\u of_prime\u sum

已经使用了

float（log（2.0））

，它甚至会稍微提高算法的性能

换言之，你将素数和的对数加倍

但我建议您使代码更具可读性，并使用更快的素数检查。例如，您可以停止在日志（n）中使用您希望检查的数字n枚举潜在的除法器，以便读取：

for x in range(2,int(sqrt(prime))) :
    if prime%x == 0.0:
        break

可能建议使用

int

进行素数计算，如果希望将其添加到素数和的

日志中，则仅将其转换为
float

如果在终端中运行此程序：

import math

prime = 3
log_of_prime_sum = math.log(2.0) #I'm starting out with 0.0 cause the sum of the log of the primes should begin here
n = 8

print("these are the primes smaller than {}".format(n))
print(2)

while prime < n:
    for x in range(2,int(math.sqrt(prime))+1): #range only works with integers right?
        if prime%x == 0.0:       
            break
    else:
        print(prime)
        log_of_prime_sum += math.log(prime)

    prime += 2

print("the natural logarithm of the product of all the primes smaller than {} is {}".format(n, log_of_prime_sum))
ratio = (n)/(log_of_prime_sum)
print("the ratio is {}".format(ratio))

这段代码缩进正确吗？我想是的。当我运行我的程序时，它运行得很好，除了日志的总和，洛利修正了这一点，使质数检查器更快，但在运行这段代码时，我仍然得到整数：（比如当我想要素数的自然对数之和小于8时，我得到的是5，而不是5.347….而且和n之间的比率（因此n/对数素数之和）被舍入为1，而不是1.4961….我尝试过使用float，但没有任何帮助：（@pokemonfan:你确定这不是一个数字格式错误。我在终端上测试了它。）（参见答案）并得到了预期的结果…嗯，这听起来可能很愚蠢，但数字格式错误是什么？您可以使用
%d
，正如您在中所看到的，这意味着您将数字打印为整数，通过使用
%f
您将打印带有小数点等的浮点。
for x in range(2,int(sqrt(prime))) :
    if prime%x == 0.0:
        break

import math

prime = 3
log_of_prime_sum = math.log(2.0) #I'm starting out with 0.0 cause the sum of the log of the primes should begin here
n = 8

print("these are the primes smaller than {}".format(n))
print(2)

while prime < n:
    for x in range(2,int(math.sqrt(prime))+1): #range only works with integers right?
        if prime%x == 0.0:       
            break
    else:
        print(prime)
        log_of_prime_sum += math.log(prime)

    prime += 2

print("the natural logarithm of the product of all the primes smaller than {} is {}".format(n, log_of_prime_sum))
ratio = (n)/(log_of_prime_sum)
print("the ratio is {}".format(ratio))

$ python3 primefilter.py 
these are the primes smaller than 8
2
3
5
7
the natural logarithm of the product of all the primes smaller than 8 is 5.3471075307174685
the ratio is 1.4961359864267718