在不完全展平的情況下擴展 Python 中的嵌套列表

Question

假設我有一個包含嵌套列表的列表，例如字符串：

items = ['Hello', ['Ben', 'Chris', 'Linda'], '! The things you can buy today are', ['Apples', 'Oranges']]

我想要一個字符串列表，將嵌套列表組合並展平為所有可能性，這樣的結果是：

new_list = ['Hello Ben ! The things you can buy today are Apples',
            'Hello Ben ! The things you can buy today are Oranges',
            'Hello Chris ! The things you can buy today are Apples',
            'Hello Chris ! The things you can buy today are Oranges',
            'Hello Linda ! The things you can buy today are Apples',
            'Hello Linda ! The things you can buy today are Oranges',]

我一直在查看 itertools 文檔，但沒有按預期正常工作。 我不想對迭代進行硬編碼，因為這個項目列表可以在項目數量和嵌套列表數量之間變化。

例如：

list(itertools.chain(*items))

將展平列表，但它會拆分字符串項中的各個字符。 部分挑戰在於列表中的某些項目是字符串，而其他項目是附加列表。 將不勝感激任何幫助。 謝謝

Answer 1

你需要itertools.product() 。

這是實際操作：

>>> items = [['Hello'], ['Ben', 'Chris', 'Linda']]
>>> list(itertools.product(*items))
[('Hello', 'Ben'), ('Hello', 'Chris'), ('Hello', 'Linda')]

由於itertools.product()將列表列表作為輸入，因此您的代碼中需要進行一些轉換以將'Hello'轉換為['Hello'] -

import itertools
items = ['Hello', ['Ben', 'Chris', 'Linda'], '! The things you can buy today are', ['Apples', 'Oranges']]
new_items = itertools.product(*[item if isinstance(item, list) else [item] for item in items])
new_list = [' '.join(x) for x in new_items]

new_list ：

['Hello Ben ! The things you can buy today are Apples',
 'Hello Ben ! The things you can buy today are Oranges',
 'Hello Chris ! The things you can buy today are Apples',
 'Hello Chris ! The things you can buy today are Oranges',
 'Hello Linda ! The things you can buy today are Apples',
 'Hello Linda ! The things you can buy today are Oranges']

Answer 2

你可以用回溯來解決：

def test(itm):
    n = len(itm)
    results = []
    def dfs(idx, res):
        if idx == n:
            results.append(res.copy())
            return
        if isinstance(itm[idx], list):
            for d in itm[idx]:
                res.append(d)
                dfs(idx+1, res)
                res.pop()
        else:
                res.append(itm[idx])
                dfs(idx+1, res)
                res.pop()

    dfs(0, [])
    return results

output：

test(items)

[['Hello', 'Ben', '! The things you can buy today are', 'Apples'],
 ['Hello', 'Ben', '! The things you can buy today are', 'Oranges'],
 ['Hello', 'Chris', '! The things you can buy today are', 'Apples'],
 ['Hello', 'Chris', '! The things you can buy today are', 'Oranges'],
 ['Hello', 'Linda', '! The things you can buy today are', 'Apples'],
 ['Hello', 'Linda', '! The things you can buy today are', 'Oranges']]