python - 创建嵌套/递归列表

Question

如何递归创建列表？

我有这个列表:

l = ['a', 'b', 'new', 'c', 'd', 'new', 'z', 'x', 'c', 'fin', 'f', 'fin', 
     'g', 'l', 'new', 'z', 'x', 'c', 'fin', 'j']

预期的输出是:

r = ['a', 'b', ['c', 'd', ['z', 'x', 'c'] 'f'], 'g', 'l', ['z', 'x', 'c'] 'j']

到目前为止我尝试了什么:

def asd(l, index=0):
    r = []
    for i in l[index:]:
        index += 1
        if i == 'new':
            i, index = asd(l, index)
        r.append(i)
        if i == 'fin':
            return r
    return r, index

r, index = asd(l)

我不明白如何让它发挥作用。谁能帮帮我？

Answer 1

这是一种非递归解决方案，可以创建您的列表，一次解析，无需任何昂贵的 index() 操作:

l = ['a', 'b', 'new', 'c', 'd', 'new', 'f', 'fin', 'g', 'fin', 'j']

rv = []

curr = [rv]  # things are always added to the last element if not 'fin' or 'new'

for elem in l:
    if elem == "new":
        # create a new list, put it at end of curr 
        curr.append([])
        # add that list to the one before
        curr[-2].append(curr[-1])
    elif elem == "fin":
        # done, remove from curr
        curr.pop() 
    else:
        curr[-1].append(elem)

print(rv)

输出:

['a', 'b', ['c', 'd', ['f'], 'g'], 'j']

---

l = ['a', 'b', 'new', '1', '2', '3', 'fin', 'c', 'new', 'x', 'y', 'z', 'fin',]  

导致

['a', 'b', ['1', '2', '3'], 'c', ['x', 'y', 'z']]

你需要针对不平衡/不正确的new/fin

---

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