python中的埃拉托色尼问题的值错误

发布于 2024-10-21 08:59:39 字数 573 浏览 2 评论 0原文

这是我的埃拉托色尼筛法的版本，用于查找 n 以内的素数。我觉得它应该可以工作，但我收到一个错误，我不太明白为什么：

    mylist.remove(i)
ValueError: list.remove(x): x not in list

为什么它提到 x？

这是代码：

def number_list(n):

    mylist = []
    for i in range(2,n+1):
        mylist.append(i)
    return mylist


def sieve():

    n = eval(input("What is the maximum limit? "))
    mylist = number_list(n)

    primes = []

    for i in mylist:
        p = mylist.pop(0)
        primes.append(p)
        if i % p == 0:
            mylist.remove(i)
    return primes

原文

Here's my version of the Sieve of Eratosthenes for finding prime numbers up to a limit n. I feel like it should work but I get an error and I don't really understand why:

    mylist.remove(i)
ValueError: list.remove(x): x not in list

Why does it mention x?

Here's the code:

def number_list(n):

    mylist = []
    for i in range(2,n+1):
        mylist.append(i)
    return mylist


def sieve():

    n = eval(input("What is the maximum limit? "))
    mylist = number_list(n)

    primes = []

    for i in mylist:
        p = mylist.pop(0)
        primes.append(p)
        if i % p == 0:
            mylist.remove(i)
    return primes

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惯饮孤独 2024-10-28 08:59:39

您在迭代时修改 mylist 。这会产生奇怪的结果。在第一次运行循环期间，i 将被分配列表的第一项，即 2。然后，该语句

p = mylist.pop(0)

从列表的开头删除此项，并且 p 也设置为 2。接下来，您的检查 i % p == 0 将产生 True，并且您尝试从列表中删除 i，但它已经消失了，导致您引用的错误消息。

这个循环的完整逻辑似乎被打破了。对于遇到的每个素数，您需要从列表中删除该素数的所有倍数。您通常需要两个嵌套循环来实现此目的。

此外，您的函数

def number_list(n):
    mylist = []
    for i in range(2,n+1):
        mylist.append(i)
    return mylist

相当于

def number_list(n):
    return range(2, n + 1)

，即您根本不需要此函数，只需使用 range(2, n + 1) 即可。

You are modifying mylist while iterating over it. This will yield strange results. During the first run of your loop, i will be assigned the first item of your list, i.e. 2. Then, the statement

p = mylist.pop(0)

removes this very item from the beginning of the list, and p is also set to 2. Next, your check i % p == 0 will yield True, and you try to remove i from the list, but it is already gone, resulting in the error message you cited.

The complete logic of this loop seems to be broken. For each prime you encounter, you need to remove all multiples of that prime from the list. You usually need two nested loops to achieve this.

Furthermore, your function

def number_list(n):
    mylist = []
    for i in range(2,n+1):
        mylist.append(i)
    return mylist

is equivalent to

def number_list(n):
    return range(2, n + 1)

i.e. you would not need this function at all, just use range(2, n + 1) instead.

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简单 2024-10-28 08:59:39

这是埃拉托斯特尼筛法的工作代码。范围对象必须创建为列表，因为我们需要删除项目。

def all_primes_to(n):
    numbers = list(range(2,n+1))
    primes = []
    while numbers:
        prime = numbers.pop(0)
        primes.append(prime)
        remove_multiples(numbers, prime)
    return primes

def remove_multiples(lst, x):
    length = len(lst)
    i = 0
    while i < length:
        if lst[i] % x == 0:
            del lst[i]
            length -= 1
        i += 1

Here is a working code for the Sieve of Eratosthenes. The range object must be made to a list because we need to delete items.

def all_primes_to(n):
    numbers = list(range(2,n+1))
    primes = []
    while numbers:
        prime = numbers.pop(0)
        primes.append(prime)
        remove_multiples(numbers, prime)
    return primes

def remove_multiples(lst, x):
    length = len(lst)
    i = 0
    while i < length:
        if lst[i] % x == 0:
            del lst[i]
            length -= 1
        i += 1

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