我如何获得这样的嵌套列表?

时间:2016-07-09 07:28:08

标签: python recursion nested-lists

variable tree structure

- nestedList1 variable

aa3
 |
aa1      aa2      bb1
   \    /        /
     aa       bb
       \     /
         root

- nestedList2 variable

              bb4
               |
aa3           bb2     bb3
 |              \     /
aa1      aa2      bb1    cc1
   \    /        /        |
     aa       bb         cc
       \       |        /
              root


nestedList1 = ['root', ['aa', ['aa1', ['aa3'], 'aa2'], 'bb', ['bb1']]]
nestedList2 = ['root', ['aa', ['aa1', ['aa3'], 'aa2'], 'bb', ['bb1', ['bb2', ['bb4'], 'bb3']], 'cc', ['cc1']]]

def ConvertTraverse(nlist, depth=0):
    convertlist = []
    for leaf in nlist:
        if isinstance(leaf, list):
            tmplist = ConvertTraverse(leaf, depth+1)
            convertlist.insert(0, tmplist)
        else:
            convertlist += [leaf]
    return convertlist

print ConvertTraverse(nestedList1)
print ConvertTraverse(nestedList2)
  • 结果
    nestedList1:[[['bb1'], [['aa3'], 'aa1', 'aa2'], 'aa', 'bb'], 'root']
    nestedList2:[[['cc1'], [[['bb4'], 'bb2', 'bb3'], 'bb1'], [['aa3'], 'aa1', 'aa2'], 'aa', 'bb', 'cc'], 'root']

我想要的只是下面的结果。

  • 结果
    nestedList1:[[[['aa3'], 'aa1', 'aa2'], 'aa', ['bb1'], 'bb'], 'root']
    nestedList2:[[[['aa3'], 'aa1', 'aa2'], 'aa', [[['bb4'], 'bb2', 'bb3'], 'bb1'], 'bb', ['cc1'], 'cc'], 'root']

如何获得这样的嵌套列表? 我想要一个嵌套列表,命令下订单遍历。

2 个答案:

答案 0 :(得分:2)

基本上,为了对列表重新排序,您需要做什么:每当n元素是标签,n+1元素是子列表时,交换这两个元素。您可以在以下几行就地执行此操作:

def reorder(lst):
    for i, (cur, nxt) in enumerate(zip(lst, lst[1:])):
        if isinstance(cur, str) and isinstance(nxt, list):
            reorder(nxt)
            lst[i:i+2] = [nxt, cur]

对于 not-in-place 解决方案,您只需创建列表的深层副本,然后在副本上使用该副本。

答案 1 :(得分:0)

我可能在这里不合适,或者完全忽略了这一点,但是如果你将每个分支完全收集到括号中,我会冒险声称我会更容易。即将每个分支写为独特的[root,[branch1],[branch2],...]

nestedList1 = ['root', ['aa', ['aa1', ['aa3']], ['aa2']], ['bb', ['bb1']]]
nestedList2 = ['root', ['aa', ['aa1', ['aa3']], ['aa2']], ['bb', ['bb1', ['bb2', ['bb4']], ['bb3']]], ['cc', ['cc1']]]

然后你可以递归地改变顺序,使每个分支离开 - 1st,trunk-2nd。

def recursivereverese(l):
    if len(l)<=1 or type(l) is not list:
        return l
    else:
        new = []
        for k in l[::-1]:
            new.append(recursivereverese(k))
        return new

修改后的嵌套列表的结果:

In [127]: recursivereverese(nestedList1)
Out[127]: [[['bb1'], 'bb'], [['aa2'], [['aa3'], 'aa1'], 'aa'], 'root']

In [128]: recursivereverese(nestedList2)
Out[128]: 
[[['cc1'], 'cc'],
 [[['bb3'], [['bb4'], 'bb2'], 'bb1'], 'bb'],
 [['aa2'], [['aa3'], 'aa1'], 'aa'],
 'root']

这是你追求的吗?

找出更好的绘图分支是一个不同的主题。