以递归方式修改列表输出

Question

在Find all possible combinations that overlap by end and start 中，我们得到以下程序，该程序获取仅在结束值和开始值重叠的范围的所有可能组合。输出以字符串格式给出，如下例所示：

def getAllEndOverlappingIndices(lst, i, l):
    r = -1
    if i == len(lst):
        if l:
            print(l)
        return

    n = i + 1
    while n < len(lst) and r > lst[n][0]:
        n += 1
    getAllEndOverlappingIndices(lst, n, l)

    n = i + 1
    r = lst[i][1]
    while n < len(lst) and r > lst[n][0]:
        n += 1
    getAllEndOverlappingIndices(lst, n, l + str(lst[i]))

indices = [(0.0, 2.0), (0.0, 4.0), (2.5, 4.5), (2.0, 5.75), (2.0, 4.0), (6.0, 7.25)]
indices.sort()

print(getAllEndOverlappingIndices(indices, 0, ''))
(6.0, 7.25)
(2.5, 4.5)
(2.5, 4.5)(6.0, 7.25)
(2.0, 5.75)
(2.0, 5.75)(6.0, 7.25)
(2.0, 4.0)
(2.0, 4.0)(6.0, 7.25)
(0.0, 4.0)
(0.0, 4.0)(6.0, 7.25)
(0.0, 2.0)
(0.0, 2.0)(6.0, 7.25)
(0.0, 2.0)(2.5, 4.5)
(0.0, 2.0)(2.5, 4.5)(6.0, 7.25)
(0.0, 2.0)(2.0, 5.75)
(0.0, 2.0)(2.0, 5.75)(6.0, 7.25)
(0.0, 2.0)(2.0, 4.0)
(0.0, 2.0)(2.0, 4.0)(6.0, 7.25)

我希望函数的输出是一个列表列表，每个子列表都包含 分隔的 元组（目前，正在返回元组的字符串）。对于上面给出的示例，输出将改为L = [[(6.0, 7.25)], [(2.5, 4.5)], [(2.5, 4.5), (6.0, 7.25)], ..., [(0.0, 2.0), (2.0, 4.0), (6.0, 7.25)]]。我尝试在函数的开头创建一个列表并附加函数的每个调用，但我相信我正在附加元组的字符串而不是元组本身。是否可以将此函数的输出更改为我想要的描述？我在递归方面没有太多经验，希望有任何提示。

Answer 1

现在，代码所做的是创建一个巨大的字符串，因为您从一个字符串开始并将每个元组转换为一个字符串，然后再将所有内容连接在一起。相反，您可以做的是传入一个空列表并将元组添加到列表中，直到您完成组合。到达组合的末尾后，将其添加到包含所有组合的全局数组中。

# Create a global array to hold all your results.
results = []

def getAllEndOverlappingIndices(lst, i, l):
    r = -1
    if i == len(lst):
        if l:
            # Instead of printing final combination, add the combination to the global list
            results.append(l)
        return

    n = i + 1
    while n < len(lst) and r > lst[n][0]:
        n += 1
    getAllEndOverlappingIndices(lst, n, l)

    n = i + 1
    r = lst[i][1]
    while n < len(lst) and r > lst[n][0]:
        n += 1
    # Wrap the tuple in the list to take advantage of python's list concatenation
    getAllEndOverlappingIndices(lst, n, l + [lst[i]])

indices = [(0.0, 2.0), (0.0, 4.0), (2.5, 4.5), (2.0, 5.75), (2.0, 4.0), (6.0, 7.25)]
indices.sort()
# Pass in an empty list here instead of an empty string
getAllEndOverlappingIndices(indices, 0, [])

输出：

[[(6.0, 7.25)], [(2.5, 4.5)], [(2.5, 4.5), (6.0, 7.25)], [(2.0, 5.75)], [(2.0, 5.75), (6.0, 7.25)], [(2.0, 4.0)], [(2.0, 4.0), (6.0, 7.25)], [(0.0, 4.0)], [(0.0, 4.0), (6.0, 7.25)], [(0.0, 2.0)], [(0.0, 2.0), (6.0, 7.25)], [(0.0, 2.0), (2.5, 4.5)], [(0.0, 2.0), (2.5, 4.5), (6.0, 7.25)], [(0.0, 2.0), (2.0, 5.75)], [(0.0, 2.0), (2.0, 5.75), (6.0, 7.25)], [(0.0, 2.0), (2.0, 4.0)], [(0.0, 2.0), (2.0, 4.0), (6.0, 7.25)]]

为可见性编辑输出：

[[(6.0, 7.25)],
[(2.5, 4.5)],
[(2.5, 4.5), (6.0, 7.25)],
[(2.0, 5.75)],
[(2.0, 5.75), (6.0, 7.25)],
[(2.0, 4.0)],
[(2.0, 4.0), (6.0, 7.25)],
[(0.0, 4.0)],
[(0.0, 4.0), (6.0, 7.25)],
[(0.0, 2.0)],
[(0.0, 2.0), (6.0, 7.25)],
[(0.0, 2.0), (2.5, 4.5)],
[(0.0, 2.0), (2.5, 4.5), (6.0, 7.25)],
[(0.0, 2.0), (2.0, 5.75)],
[(0.0, 2.0), (2.0, 5.75), (6.0, 7.25)],
[(0.0, 2.0), (2.0, 4.0)],
[(0.0, 2.0), (2.0, 4.0), (6.0, 7.25)]]