Question

我一直在学习Python 2.5.4，我有以下问题需要解决：

“编写一个程序，计算从2到某个数n的所有素数的对数之和，并打印出素数的对数，数n和这两个量的比率之和。这适用于n的不同值。“

这是我到目前为止所做的：

from math import *
n = raw_input('This is a logarithm ratio tester. Which number do you want to test? ')
for x in range(2,n):                            #picks numbers to test
    for divisor in range(2, 1+int(sqrt(x+1))):
        if x%divisor == 0:                      #checks if x is prime
            log(x)                              #computes log of prime

不幸的是，我不确定如何实现一个函数来汇总所有日志。我想，一旦我有了这个，我只需要用以下方式结束该程序：

print 'Sum of the logs of the primes is',logsum
print 'n =',n
print 'Ratio of sum to n is',logsum/n

或某些变体。但是什么是获得我标记为logsum的好方法？请注意我已经学习了不超过一周的编程，我很少知道声明/功能，我不是数学家。如果有疑问，假设我是个白痴。谢谢！

Answer 1

我会将一些代码包装在函数中，并修复prime的定义：

def isprime(x):
  for j in range(2,1 + int(sqrt(x - 1))):
    if x%j == 0:
      return False
  return True

def primes(n):
    return [j for j in range(2, n) if isprime(n)]

logsum = 0
for p in primes(n):
    logsum += log(p)

Answer 2

在计算日志之前，您应该检查数字是否为素数，然后将log（x）的值添加到变量logsum中，如下所示：

from math import *
logsum = 0
n = raw_input('This is a logarithm ratio tester. Which number do you want to test? ')
for x in range(2,n):                            #picks numbers to test
    isprime = True
    for divisor in range(2, 1+int(sqrt(x+1))):
        if x%divisor == 0:                      #checks if x is prime
            isprime = False                                  #computes log of prime
            break
    if isprime:
        logsum += log(x)

此外，由于您正在检查大量值，因此实施The Sieve of Eratosthenes可能是继续学习的好方法。

编辑：使用Juri Robl的建议，代码可以简化为：

from math import *
logsum = 0
n = raw_input('This is a logarithm ratio tester. Which number do you want to test? ')
for x in range(2,n):                            #picks numbers to test
    for divisor in range(2, 1+int(sqrt(x+1))):
        if x%divisor == 0:                      #checks if x is prime
            break
    else:
        logsum += log(x)

Answer 3

请注意log（2）+ log（3）+ ... + log（n）= log（2 * 3 * ... * n）

然后你可以使用Eratosthenes的Sieve来获得素数列表，将它们相乘，然后取该产品的对数。

import math
import operator

def filter_primes(alist, newlist):
    for x in alist[1:]:
        if x%alist[0] != 0:
            newlist.append(x)
    return newlist

n = int(raw_input('This is a logarithm ratio tester. Which number do you want to test?'))
alist = range(2, n)
sieve_list = []
primes = [2]

while alist[0] < math.sqrt(n):
    a = filter_primes(alist, sieve_list)
    primes.append(a[0])
    alist = sieve_list
    sieve_list = []
for j in alist[1:]: primes.append(j)

product = reduce(lambda x, y: x*y, primes)
ans = math.log(product)

print 'The sum of the logs is: %d \nn is: %d \nThe ratio of these two quantities is: %s' %   (ans, n, float(ans/n))

Python 2.5.4：如何查找对数值之和

3 个答案: