Building a nested python Dictionary from list

Question

I have a string that could be different lengths, and I want to create a nested dictionary. I have this so far, and just cant seem to figure out how to get over the variable depth issue.

    string = "a/b/c/b"
    x = string.split('/')
    y = {}
    for item in x:
      y[item] = dict()
      .............

I have tried many different ways, but just don't know how to build it dynamically. The final result I would like to get to is:

{'a' :{'b' : {'c': {'d': {}}}}

Would love some feedback on design and ideas to get this going.

Thanks,

jamylak · Accepted Answer · 2017-03-17 02:30:00Z

3

Just update the loop as follows:

y = {}
for item in reversed(x):
    y = {item: y}

edited Mar 17, 2017 at 2:30

jamylak

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answered Mar 17, 2017 at 2:21

Ozgur Vatansever

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Comments

Community · Accepted Answer · 2017-05-23 11:54:03Z

2

One Line Reduce version of @ozgur's answer

>>> string = "a/b/c/d"
>>> reduce(lambda x, y: {y: x}, reversed(string.split('/')), {})
{'a': {'b': {'c': {'d': {}}}}}

But I prefer the original answer from @ozgur

edited May 23, 2017 at 11:54

CommunityBot

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answered Mar 17, 2017 at 2:29

jamylak

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2 Comments

AChampion Over a year ago

I preferred your first answer that you deleted. Relying on reversed() can be problematic if you end up having to deal with iterators/generators but that is outside the scope of OPs question.

jamylak Over a year ago

@AChampion The reversed one seems so simple though. But that's a good point, although rare, there could be a generator or iterator that you want to step through

McGrady · Accepted Answer · 2017-03-17 02:17:08Z

0

Try this:

string = "a/b/c/b"
x = string.split('/')
x.reverse()
y = {}
count=0
for item in x:
    if count==0:
        tmp={item:{}}
    else:
        tmp={item: tmp}
    count+=1
print tmp

Output:

{'a': {'b': {'c': {'b': {}}}}}

answered Mar 17, 2017 at 2:17

McGrady

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Comments

jamylak · Accepted Answer · 2017-03-17 02:21:09Z

0

>>> text = 'a/b/c/d'
>>> d = node = {}
>>> for c in text.split('/'):
...   node = node.setdefault(c, {})
... 
>>> d
{'a': {'b': {'c': {'d': {}}}}}

answered Mar 17, 2017 at 2:21

jamylak

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1 Comment

Donald Duck is with Ukraine Over a year ago

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

AChampion · Accepted Answer · 2017-03-17 02:25:51Z

0

One simple way of doing this is recursion:

def fn(s):
    if not s:
        return {}
    x, *y = s   # Python3, for Python2 x, y = s[0], s[1:]
    return {x:fn(y)}

>>> fn("a/b/c/b".split('/'))
{'a': {'b': {'c': {'b': {}}}}}

But if you want to do it iteratively then you are very close, just use a cursor to walk down the structure:

>>> y = {}
>>> c = y
>>> for item in "a/b/c/b".split('/'):
...     c[item] = {}
...     c = c[item]
>>> y
{'a': {'b': {'c': {'b': {}}}}}

edited Mar 17, 2017 at 2:25

answered Mar 17, 2017 at 2:17

AChampion

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