Python array from set of keys

Question

I need to create array/dictionaries in python based on an array which contains keys. I found an equivalent solution in PHP. Unfortunately I don't know how I can implement this in Python. Can somebdoy give me any tips?

a = ['one', 'two', 'three']
b = ['one', 'four', 'six']

I want to get the following result:

c = {'one': {'two': 'three', 'four': 'six}}

The PHP solution uses references for that. Perhaps this is a better example:

ar[0] = ['box0', 'border0', 'name']
var[1] = ['box0', 'border0', 'type']
var[2] = ['box0', 'border1', 'name']
var[3] = ['box1', 'border2', 'name']
var[4] = ['box1', 'border0', 'color']

$val = 'value'

In PHP the result would look like:

$result = array(
    'box0' => array(
      'border0' => array('name' => $val, 'type' => $val, 'color' => $val), 
      'border1' => array('name' => $val),
    ),
    'box1' => array(
      'border0' => array('color' => $val),
      'border2' => array('name' => $val)
    )
) );

Is this strictly for the case of 3-tuples of key, subkey, and value, or are you trying to generalize this to lists of length N? — Joel Cornett
– Joel Cornett, Commented Mar 10, 2013 at 16:47
related: Python: Change values in dict of nested dicts using items in a list — jfs
– jfs, Commented Mar 10, 2013 at 17:14

nneonneo · Accepted Answer · 2013-03-10 16:48:26Z

2

The PHP answer constructs a dictionary from a path of keys. So here's an equivalent in Python:

from collections import defaultdict
def set_with_path(d, path, val):
    end = path.pop()
    for k in path:
        d = d.setdefault(k, {})
    d[end] = val

Example:

>>> d = {}
>>> set_with_path(d, ['one', 'two', 'three'], 'val')
>>> d
{'one': {'two': {'three': 'val'}}}
>>> set_with_path(d, ['one', 'four', 'six'], 'val2')
>>> d
{'one': {'four': {'six': 'val2'}, 'two': {'three': 'val'}}}

answered Mar 10, 2013 at 16:48

nneonneo

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Comments

Eric Urban · Accepted Answer · 2013-03-10 16:36:46Z

0

x = dict()
for list in (a,b):
    if not x.has_key(list[0]):
        x[list[0]] = []
    x[list[0]] += list[1:]

answered Mar 10, 2013 at 16:36

Eric Urban

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2 Comments

DSM Over a year ago

That will produce {'one': ['two', 'three', 'four', 'six']} instead.

Eric Urban Over a year ago

The phrasing of the question let me to believe he wanted a dictionary of lists.

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