How to turn a list into nested dict in Python

Question

Need to turn x:

X = [['A', 'B', 'C'], ['A', 'B', 'D']]

Into Y:

Y = {'A': {'B': {'C','D'}}}

More specifically, I need to create a tree of folders and files from a list of absolute paths, which looks like this:

paths = ['xyz/123/file.txt', 'abc/456/otherfile.txt']

where, each path is split("/"), as per ['A', 'B', 'C'] in the pseudo example.

As this represents files and folders, obviously, on the same level (index of the array) same name strings can't repeat.

Answer 1

Assuming that {'C', 'D'} means set(['C', 'D']) and your Python version supports dict comprehension and set comprehension, here's an ugly but working solution:

>>> tr = [[1, 2, 3], [1, 2, 4], [5, 6, 7]]
>>> {a[0]: {b[1]: {c[2] for c in [y for y in tr if y[1] == b[1]]} for b in [x for x in tr if x[0] == a[0]]} for a in tr}
{1: {2: set([3, 4])}, 5: {6: set([7])}}

As for your example:

>>> X = [['A', 'B', 'C'], ['A', 'B', 'D']]
>>> {a[0]: {b[1]: {c[2] for c in [y for y in X if y[1] == b[1]]} for b in [x for x in X if x[0] == a[0]]} for a in X}
{'A': {'B': set(['C', 'D'])}}

But please don't use it in a real-world application :)

UPDATE: here's one that works with arbitrary depths:

>>> def todict(lst, d=0):
...     print lst, d
...     if d > len(lst):
...         return {}
...     return {a[d]: todict([x for x in X if x[d] == a[d]], d+1) for a in lst}
...
>>> todict(X)
{'A': {'B': {'C': {}, 'D': {}}}}

Answer 2

There is a logical inconsistency in your problem statement. If you really want ['xyz/123/file.txt', 'abc/456/otherfile.txt']

to be changed to {'xyz': {'123': 'file.txt}, 'abc': {'456': 'otherfile.txt'}}

Then you have to answer how a path 'abc.txt' with no leading folder would be inserted into this data structure. Would the top-level dictionary key be the empty string ''?

Answer 3

X = [['A', 'B', 'C'], ['A', 'B', 'D'],['W','X'],['W','Y','Z']]
d = {}

for path in X:
    current_level = d
    for part in path:
        if part not in current_level:
            current_level[part] = {}
        current_level = current_level[part]

This leaves us with d containing {'A': {'B': {'C': {}, 'D': {}}}, 'W': {'Y': {'Z': {}}, 'X': {}}}. Any item containing an empty dictionary is either a file or an empty directory.

Answer 4

This should be pretty close to what you need:

def path_to_dict(path):
    parts = path.split('/')

    def pack(parts):
        if len(parts) == 1:
            return parts
        elif len(parts):
            return {parts[0]: pack(parts[1:])}
        return parts

    return pack(parts)

if __name__ == '__main__':
    paths = ['xyz/123/file.txt', 'abc/456/otherfile.txt']
    for path in paths:
        print '%s -> %s' % (path, path_to_dict(path))

Results in:

xyz/123/file.txt -> {'xyz': {'123': ['file.txt']}}
abc/456/otherfile.txt -> {'abc': {'456': ['otherfile.txt']}}