循环列表切片+元素分配Python

Posted 2023-02-23

【中文标题】循环列表切片+元素分配Python

【英文标题】：loops list slicing + elements allocation Python

【发布时间】：2020-08-08 19:02:44

【问题描述】：

我是 Python 的初学者，正在尝试执行以下操作：

main_list[80,80,30,30,30,30,20,10,5,4,3,2,1] #list of integers - 在多个列表中切片 main_list，例如 list1,2,3,..,n，子列表的总和 for i in range of n: print(list(i)) list1[80,20], list2[80,10,5,4,1], list3[30,30,30], listn[30,3,2]

谢谢！

【问题讨论】：

这能回答你的问题吗？ Understanding slice notation

你真正想要的是什么？也许需要代码格式化。

我看到了 sline 符号，但这并不是我所需要的。我是 Python 的新手，我试图在帖子的另一个 cmets 中更详细地解释

【参考方案1】：

我不太清楚你认为什么是可接受的输出，所以我假设它是其元素总和小于 100 的任何列表。

我找到的解决方案是使用递归。对于列表 [a, b, c, d]，我们将检查此子列表的条件：

[a]
[a, b] (if the condition for [a] is met)
[a, b, c] (if the condition for [a, b] is met)
[a, b, c, d] (if the condition for [a, b, c] is met)
[a, c] (if the condition for [a] is met)
[a, c, d] (if the condition for [a, c] is met)
[a, d] (if the condition for [a] is met)

[b]
[b, c] (if the condition for [b] is met)
[b, c, d] (if the condition for [b, c] is met)
[b, d] (if the condition for [b] is met)

[c]
[c, d] (if the condition for [c] is met)

[d]

概念是，对于列表中的“n”元素，我们将寻找大小为“n - 1”到 0（即元素本身）的满足要求的子列表。子列表由每次迭代的研究元素右侧的元素形成，因此对于前 30 个，要使用的子列表将是 [30, 30, 30, 20, 10, 5, 4, 3, 2, 1 ]

为每个元素查找子列表的过程是使用递归的过程。它为子列表的每个元素调用自身，检查它是否满足条件。对于上面的例子，如果 [a, b] 满足条件，那么它也会尝试 [a, b, c] 和 [a, b, d] （通过使用 (a, b) 的总和调用自身和子列表 [c, d]。

我添加了一些打印件，以便您研究它是如何工作的，但您应该只使用脚本末尾的 results 变量来获取结果。

main_list = [80,80,30,30,30,30,20,10,5,4,3,2,1] 

def less_than_hundred(input) -> bool:
    return input < 100

def sublists_meet_condition(condition, input):
    """
    This function is used to call the sublists_for_element function with every element in the original list and its sublist:
    - For the first element (80) it calls the second function with the sublist [80,30,30,30,30,20,10,5,4,3,2,1]
    - For the fifth element (30) it calls the second function with the sublist [30,20,10,5,4,3,2,1]
    Its purpose is to collect all the sublists that meet the requirements for each element
    """
    results = []
    for index, element in enumerate(input):
        print('Iteration  - Element '.format(index, element))
        if condition(element):
            results.append([element])
            print(' = '.format([element], element))
            num_elements = len(input) - index
            main_element = element
            sublist = input[index+1:]
            for result in sublists_for_element(condition, main_element, sublist):
                new_result = [element] + result
                sum_new_result = sum(new_result)
                results.append(new_result)
                print(' = '.format([element] + result, sum_new_result))
    return results

def sublists_for_element(condition, sum_main_elements, sublist):
    """
    This function is used to check every sublist with the given condition.
    The variable sum_main_elements allows the function to call itself and check if for a given list of numbers that meet the conditions [30, 30, 4] for example, any of the elements of the remaining sublists also meets the condition for example adding the number 3 still meets the condition.
    Its purpose is to return all the sublists that meet the requirements for the given sum of main elements and remaining sublist
    """
    num_elements = ''.format('0' if len(sublist) + 1 < 10 else '',len(sublist) + 1)
    #print('Elements:  | Main element:  | Sublist: '.format(num_elements, sum_main_elements, sublist))
    result = []
    for index, element in enumerate(sublist):
        if condition(sum_main_elements + element):
            result.append([element])
            sublist_results = sublists_for_element(condition, sum_main_elements + element, sublist[index+1:])
            for sublist_result in sublist_results:
                result.append([element] + sublist_result)
    return result

results = sublists_meet_condition(less_than_hundred, main_list)

【讨论】：

感谢您的回答！我需要澄清一些事情。我的想法是将上面示例中列表的所有元素相加为 325。这意味着 325/100 它是 3.25 => 这意味着 4 个子列表。现在在这个子列表中，我想添加每个元素 a,b,c,d 但要让 list1 [a,h,i,j] (80,10,5,4) 最接近的总和