[英]How to transform a list into a hierarchy dict
How would I turn a list like ["one","two","three","four"]
into something like {"one": {"two": {"three":{"four"}}}}
where each item in the list would be a descendant of the other element in the dictionary? 如何将像
["one","two","three","four"]
这样的列表变成{"one": {"two": {"three":{"four"}}}}
列表中的每个项目都是字典中其他元素的后代? I think it could be done in a recursive function, but I'm not sure how. 我认为它可以在递归函数中完成,但我不确定如何。
This is what I tried: 这是我试过的:
l = ["one","two","three","four"]
d = {}
for v in l[:-1]:
d[v] = {l}
d = d[v]
print(d)
Thanks! 谢谢!
A recursive solution 递归解决方案
def dictify(d):
if len(d) == 1:
return {d[0]}
else:
return {d[0]: dictify(d[1:])}
For example 例如
>>> dictify(["one","two","three","four"])
{'one': {'two': {'three': {'four'}}}}
Note that the in above solution, the inner-most object is actually a set
, not a dict
. 注意在上面的解决方案中,最里面的对象实际上是一个
set
,而不是一个dict
。 If you want all objects to be dict
then you can modify the solution to 如果您希望所有对象都是
dict
那么您可以将解决方案修改为
def dictify(d):
if len(d) == 1:
return {d[0]: None}
else:
return {d[0]: dictify(d[1:])}
Resulting in 导致
>>> dictify(["one","two","three","four"])
{'one': {'two': {'three': {'four': None}}}}
If you wanted the structure to look like the following 如果您希望结构如下所示
{'one': {'two': {'three': {'four': None}}}}
You could generate that with something like this. 你可以用这样的东西生成它。 This solution uses recursion.
此解决方案使用递归。
arr = ["one", "two", "three", "four"]
def gen(arr):
if len(arr) == 0:
return None
else:
key = arr.pop(0)
val = gen(arr)
dict = {}
dict[key] = val
return dict
print gen(arr)
If you'd prefer a non-recursive solution: 如果您更喜欢非递归解决方案:
def endeepen(lst):
old = None
for v in lst[::-1]:
ret = {}
ret[v] = old
old = ret
return ret
Just iterates the list in reverse and buries the each dct as the previous elements value: 只需反向迭代列表并将每个dct作为前面的元素值隐藏:
>>> endeepen(l)
{'one': {'two': {'three': {'four': None}}}}
If you really want the last element to be a set, you can do so with a minor correction: 如果你真的想要最后一个元素是一个集合,你可以通过一个小的修正来做到这一点:
def endeepen(lst):
old = {lst[-1]}
for v in lst[len(lst) - 2::-1]:
ret = {}
ret[v] = old
old = ret
return ret
which then gives: 然后给出:
>>> endeepen(l)
{'one': {'two': {'three': set(['four'])}}}
Note: In both cases, I haven't covered edge conditions, so empty or very short lists len(1) <= 1
may misbehave. 注意:在这两种情况下,我都没有覆盖边缘条件,所以空(或)非常短的列表
len(1) <= 1
可能行为不端。
l = ["one","two","three","four"]
d = {}
d2 = d
for v in l[:-1]:
d2[v] = {}
d2 = d2[v]
d2[l[-2]] = {l[-1]}
print(d)
>>> {'one': {'two': {'three': {'three': {'four'}}}}}
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