Python 实现一个函数 return 一个列表，其中 y[i] 是素数 <= 到 i

Question

c) 使用您在 (b) 部分中的函数，实现函数 primesCount(n)，它给出一个整数 n，returns 长度为 n 的列表 y，由 y[i] = 素数的数量给出小于或等于 i，因为 i = 0, 1, . . . ，n-1。示例 Input/Output：输入数字：6 [0, 0, 1, 2, 2, 3]

第 (b) 部分功能：

def generatePrimes(n):

    prime = [False for i in range(n)]
    I = []
    for i in range(2, n):
       if i > 1:
           for j in range(2, int(math.sqrt(i))+1):
               if i % j == 0:
                   break
           else:
               I.append(i)
               prime[i] = True
    return prime

全部代码（b部分+c部分函数）：

import math

def generatePrimes(n):
    prime = [False for i in range(n)]  # creates a list of n-length all consisting of F elem
    I = []
    for i in range(2, n):
       if i > 1:
           for j in range(2, int(math.sqrt(i))+1):
               if i % j == 0:
                   break
           else:
               I.append(i)   # unecessary
               prime[i] = True #change False to True if number is prime
    return prime   # return the list (indexes with False are composite numbers while indexes 
                    with True are prime number

def primesCount(n):
    I = []
    for i in range(n):
        I.append(generatePrimes(i))
    return I
   
n = int(input("Enter a number: "))
print(primesCount(n))

预计input/output:

input: Enter a number: 6 

output: [0, 0, 1, 2, 2, 3]

实际input/output:

input: Enter a number: 6 

output: [[], [False], [False, False], [False, False, True],[False, False, True, True]]

我需要的是将 False 和 True 转换为整数，例如 False + False + False + True + True = 0 + 0 + 1 + 1 = 2

我想将上面的输出转换为 [0, 0, 1, 2, 2, 3] ( [假] = 0, [假 + 假] = 0, [假, 假, 真] = 1...)

我为此绞尽脑汁好几个小时似乎无法正常工作

Answer 1

def generatePrimes(n):
    prime = [False for i in range(n)]  # creates a list of n-length all consisting of F elem
    I = []
    for i in range(2, n):
       if i > 1:
           for j in range(2, int(math.sqrt(i))+1):
               if i % j == 0:
                   break
           else:
               I.append(i)   # unecessary
               prime[i] = True #change False to True if number is prime


    final_output = []
    counter = 0
    for value in prime:
        if prime:
            counter += 1
    final_output.append(counter)

    return final_output

只需遍历每个值并在值为真时迭代一个计数器。把这个放在 return 素数之前。

Answer 2

我已经更新了你的代码。请看一下

import math

def generatePrimes(n):
    prime = [False for i in range(n)]  # creates a list of n-length all consisting of F elem
    I = []
     
    for i in range(2, n):
       if i > 1:
           for j in range(2, int(math.sqrt(i))+1):
               if i % j == 0:
                   break
           else:
               I.append(i)   # unecessary
               prime[i] = True #change False to True if number is prime
    return prime   # return the list (indexes with False are composite numbers while indexes 
                    #with True are prime number

def primesCount(n):
    I = []
    for i in range(n+1):
        if len(generatePrimes(i)):
            I.append(sum(generatePrimes(i)))
    return I
   
n = int(input("Enter a number: "))
print(primesCount(n))

Python 实现一个函数 return 一个列表，其中 y[i] 是素数 <= 到 i

Python implementing a function to return a list where y[i] is the number of prime numbers <= to i

python

primes

function

python-3.x