Python Generate a dynamic dictionary from the list of keys

Question

I do have a list as given below -

keyList1 = ["Person", "Male", "Boy", "Student", "id_123", "Name"]
value1 = "Roger"

How can I generate dynamic dictionary which can be retrieved as below -

mydict["Person"]["Male"]["Boy"]["Student"]["id_123"]["Name"] = value

The list could be anything; Variable Length or consisting of "N" number of elements unknown to me...

Now I do have another list, so that My dictionary should be updated accordingly

keyList2 = ["Person", "Male", "Boy", "Student", "id_123", "Age"]
value2 = 25

i.e. If Keys "Person", "Male", "Boy", "Student", "id_123" already exists, the new key "age" should be appended ...

I would suggest to append all items in the list and use the resultant string as key. It would be lot easier. — rajpy
– rajpy, Commented Jul 4, 2013 at 4:54
The answers to this question may help: stackoverflow.com/questions/16384174/… — DJG
– DJG, Commented Jul 4, 2013 at 4:54

roman · Accepted Answer · 2013-07-04 05:28:53Z

8

I'm just learning python, so my code could be not very pythonic, but here's my code

d = {}

keyList1 = ["Person", "Male", "Boy", "Student", "id_123", "Name"]
keyList2 = ["Person", "Male", "Boy", "Student", "id_123", "Age"]
value1 = "Roger"
value2 = 3

def insert(cur, list, value):
    if len(list) == 1:
        cur[list[0]] = value
        return
    if not cur.has_key(list[0]):
        cur[list[0]] = {}
    insert(cur[list[0]], list[1:], value)

insert(d, keyList1, value1)
insert(d, keyList2, value2)

{'Person': {'Male': {'Boy': {'Student': {'id_123': {'Age': 3, 'Name': 'Roger'}}}}}}

edited Jul 4, 2013 at 5:28

answered Jul 4, 2013 at 4:55

roman

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4 Comments

nneonneo Over a year ago

This ends up being pretty similar to my solution. One simple suggestion: avoid .has_key whenever possible; prefer the quicker and more understandable if list[0] not in cur.

roman Over a year ago

thanks, I agree that it's more readable and readability counts, but why it's quicker?

nneonneo Over a year ago

Try it using timeit. On my computer, in is twice as fast as .has_key (0.0672µs vs. 0.117µs). .has_key is a method lookup, whereas in is a builtin which is dispatched much quicker.

John La Rooy Over a year ago

.has_key is deprecated and doesn't exist in Python3

nneonneo · Accepted Answer · 2013-07-04 05:18:47Z

You can do this by making nested defaultdicts:

from collections import defaultdict

def recursive_defaultdict():
    return defaultdict(recursive_defaultdict)

def setpath(d, p, k):
    if len(p) == 1:
        d[p[0]] = k
    else:
        setpath(d[p[0]], p[1:], k)

mydict = recursive_defaultdict()

setpath(mydict, ["Person", "Male", "Boy", "Student", "id_123", "Name"], 'Roger')

print mydict["Person"]["Male"]["Boy"]["Student"]["id_123"]["Name"]
# prints 'Roger'

This has the nice advantage of being able to write

mydict['a']['b'] = 4

without necessarily having to use the setpath helper.

You can do it without recursive defaultdicts too:

def setpath(d, p, k):
    if len(p) == 1:
        d[p[0]] = k
    else:
        setpath(d.setdefault(p[0], {}), p[1:], k)

Rao · Accepted Answer · 2013-07-04 05:50:16Z

3

Perhaps you could subclass dict:

class ChainDict(dict):
    def set_key_chain(self, keyList, value):
        t = self
        for k in keyList[:-1]:
            t = t.setdefault(k, {})
        t.setdefault(keyList[-1], value)

c = ChainDict()
c.set_key_chain(['Person', 'Male', 'Boy', 'Student', 'id_123', 'Name'], 'Roger')
print c
>>{'Person': {'Male': {'Boy': {'Student': {'id_123': {'Name': 'Roger'}}}}}}

c.set_key_chain(['Person', 'Male', 'Boy', 'Student', 'id_123', 'Age'], 25)
print c
>>{'Person': {'Male': {'Boy': {'Student': {'id_123': {'Age': 25,
      'Name': 'Roger'}}}}}}

answered Jul 4, 2013 at 5:50

Rao

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Comments

Steve Barnes · Accepted Answer · 2013-07-04 05:09:58Z

1

Create your own class derived from dict where the init method takes a list and a single value as inputs and iterate through the list setting the keys to value, define an update method that takes a list and a new value and for each item that is not already a key set it to the new value, (assuming that is what you need).

Forget the idea of

mydict["Person"]["Male"]["Boy"]["Student"]["id_123"]["Name"] = value1`

as it is confusing with subindexes.

answered Jul 4, 2013 at 5:09

Steve Barnes

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Comments

Shruti Joshi · Accepted Answer · 2013-07-04 05:20:28Z

0

I am trying to dealing with similar stuff, so I can suggest some guidelines, but again I am naive in Python, so this is just a guideline...

you have a list of keys , so you can definitely start with a loop iterating for each value and then assign the value

like

for i in keylist:
if type(keylist[i]) == dict:
        do something
    else:
        keylist[i] = {}

in the do something, you need to increment i and change index to [i][i+1] and then follow the same untill i+n = len(keylist)

answered Jul 4, 2013 at 5:20

Shruti Joshi

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Comments

Torben Klein · Accepted Answer · 2013-07-04 05:35:37Z

0

Use tuple(keyList1) as key. (tuples are immutable and therefore can be dict keys).

You will have a world of headache with the nested dict approach. (nested loops for enumeration, legacy data when the hierarchy needs to change, etc.).

On a second thought, maybe you should define a person class

class Person(object):
    gender = "Male"
    group = "Student"
    id = 123
    Name = "John Doe"

then use a list of all persons and filter with e.g.

male_students = [s for s in  ALL_PERSONS where s.gender=="Male" and s.group="Student"]

... for <= 10000 students you should be fine performancewise.

answered Jul 4, 2013 at 5:35

Torben Klein

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Comments

John La Rooy · Accepted Answer · 2013-07-04 05:21:24Z

-2

>>> mydict = {}
>>> keyList1 = ["Person", "Male", "Boy", "Student", "id_123", "Name"]
>>> value1 = "Roger"
>>> reduce(lambda x, y: x.setdefault(y, {}), keyList1, mydict)
{}
>>> mydict["Person"]["Male"]["Boy"]["Student"]["id_123"]["Name"] = value1

You can also do it in one step like this

>>> keyList2 = ["Person", "Male", "Boy", "Student", "id_123", "Age"]
>>> value2 = 25
>>> reduce(lambda x,y: x.setdefault(y,{}), keyList2[:-1], mydict).update({keyList2[-1]: value2})

edited Jul 4, 2013 at 5:21

answered Jul 4, 2013 at 4:54

John La Rooy

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2 Comments

Steve Barnes Over a year ago

>>> mydict {'Person': {'Male': {'Boy': {'Student': {'id_123': {'Name': 'Roger'}}}}}}

John La Rooy Over a year ago

@AbhishekKulkarni, er.. where does reduce(lambda x, y: x.setdefault(y, {}), keyList1, mydict) depend on the number of elements or the specific keys?

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