How to Implement Nested Dictionaries

How do you create nested dict in Python?

A nested dict is a dictionary within a dictionary. A very simple thing.

>>> d = {}
>>> d['dict1'] = {}
>>> d['dict1']['innerkey'] = 'value'
>>> d['dict1']['innerkey2'] = 'value2'
>>> d
{'dict1': {'innerkey': 'value', 'innerkey2': 'value2'}}

You can also use a defaultdict from the collections package to facilitate creating nested dictionaries.

>>> import collections
>>> d = collections.defaultdict(dict)
>>> d['dict1']['innerkey'] = 'value'
>>> d  # currently a defaultdict type
defaultdict(<type 'dict'>, {'dict1': {'innerkey': 'value'}})
>>> dict(d)  # but is exactly like a normal dictionary.
{'dict1': {'innerkey': 'value'}}

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You can populate that however you want.

I would recommend in your code something like the following:

d = {}  # can use defaultdict(dict) instead

for row in file_map:
    # derive row key from something 
    # when using defaultdict, we can skip the next step creating a dictionary on row_key
    d[row_key] = {} 
    for idx, col in enumerate(row):
        d[row_key][idx] = col

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According to your comment:

may be above code is confusing the question. My problem in nutshell: I
have 2 files a.csv b.csv, a.csv has 4 columns i j k l, b.csv also has
these columns. i is kind of key columns for these csvs'. j k l column
is empty in a.csv but populated in b.csv. I want to map values of j k
l columns using 'i` as key column from b.csv to a.csv file

My suggestion would be something like this (without using defaultdict):

a_file = "path/to/a.csv"
b_file = "path/to/b.csv"

# read from file a.csv
with open(a_file) as f:
    # skip headers
    f.next()
    # get first colum as keys
    keys = (line.split(',')[0] for line in f) 

# create empty dictionary:
d = {}

# read from file b.csv
with open(b_file) as f:
    # gather headers except first key header
    headers = f.next().split(',')[1:]
    # iterate lines
    for line in f:
        # gather the colums
        cols = line.strip().split(',')
        # check to make sure this key should be mapped.
        if cols[0] not in keys:
            continue
        # add key to dict
        d[cols[0]] = dict(
            # inner keys are the header names, values are columns
            (headers[idx], v) for idx, v in enumerate(cols[1:]))

Please note though, that for parsing csv files there is a csv module.

What is the best way to implement nested dictionaries?

What is the best way to implement nested dictionaries in Python?

This is a bad idea, don't do it. Instead, use a regular dictionary and use dict.setdefault where apropos, so when keys are missing under normal usage you get the expected KeyError. If you insist on getting this behavior, here's how to shoot yourself in the foot:

Implement __missing__ on a dict subclass to set and return a new instance.

This approach has been available (and documented) since Python 2.5, and (particularly valuable to me) it pretty prints just like a normal dict, instead of the ugly printing of an autovivified defaultdict:

class Vividict(dict):
    def __missing__(self, key):
        value = self[key] = type(self)() # retain local pointer to value
        return value                     # faster to return than dict lookup

(Note self[key] is on the left-hand side of assignment, so there's no recursion here.)

and say you have some data:

data = {('new jersey', 'mercer county', 'plumbers'): 3,
        ('new jersey', 'mercer county', 'programmers'): 81,
        ('new jersey', 'middlesex county', 'programmers'): 81,
        ('new jersey', 'middlesex county', 'salesmen'): 62,
        ('new york', 'queens county', 'plumbers'): 9,
        ('new york', 'queens county', 'salesmen'): 36}

Here's our usage code:

vividict = Vividict()
for (state, county, occupation), number in data.items():
    vividict[state][county][occupation] = number

And now:

>>> import pprint
>>> pprint.pprint(vividict, width=40)
{'new jersey': {'mercer county': {'plumbers': 3,
                                  'programmers': 81},
                'middlesex county': {'programmers': 81,
                                     'salesmen': 62}},
 'new york': {'queens county': {'plumbers': 9,
                                'salesmen': 36}}}

Criticism

A criticism of this type of container is that if the user misspells a key, our code could fail silently:

>>> vividict['new york']['queens counyt']
{}

And additionally now we'd have a misspelled county in our data:

>>> pprint.pprint(vividict, width=40)
{'new jersey': {'mercer county': {'plumbers': 3,
                                  'programmers': 81},
                'middlesex county': {'programmers': 81,
                                     'salesmen': 62}},
 'new york': {'queens county': {'plumbers': 9,
                                'salesmen': 36},
              'queens counyt': {}}}

Explanation:

We're just providing another nested instance of our class Vividict whenever a key is accessed but missing. (Returning the value assignment is useful because it avoids us additionally calling the getter on the dict, and unfortunately, we can't return it as it is being set.)

Note, these are the same semantics as the most upvoted answer but in half the lines of code - nosklo's implementation:

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

Demonstration of Usage

Below is just an example of how this dict could be easily used to create a nested dict structure on the fly. This can quickly create a hierarchical tree structure as deeply as you might want to go.

import pprint

class Vividict(dict):
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

d = Vividict()

d['foo']['bar']
d['foo']['baz']
d['fizz']['buzz']
d['primary']['secondary']['tertiary']['quaternary']
pprint.pprint(d)

Which outputs:

{'fizz': {'buzz': {}},
 'foo': {'bar': {}, 'baz': {}},
 'primary': {'secondary': {'tertiary': {'quaternary': {}}}}}

And as the last line shows, it pretty prints beautifully and in order for manual inspection. But if you want to visually inspect your data, implementing __missing__ to set a new instance of its class to the key and return it is a far better solution.

Other alternatives, for contrast:

dict.setdefault

Although the asker thinks this isn't clean, I find it preferable to the Vividict myself.

d = {} # or dict()
for (state, county, occupation), number in data.items():
    d.setdefault(state, {}).setdefault(county, {})[occupation] = number

and now:

>>> pprint.pprint(d, width=40)
{'new jersey': {'mercer county': {'plumbers': 3,
                                  'programmers': 81},
                'middlesex county': {'programmers': 81,
                                     'salesmen': 62}},
 'new york': {'queens county': {'plumbers': 9,
                                'salesmen': 36}}}

A misspelling would fail noisily, and not clutter our data with bad information:

>>> d['new york']['queens counyt']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'queens counyt'

Additionally, I think setdefault works great when used in loops and you don't know what you're going to get for keys, but repetitive usage becomes quite burdensome, and I don't think anyone would want to keep up the following:

d = dict()

d.setdefault('foo', {}).setdefault('bar', {})
d.setdefault('foo', {}).setdefault('baz', {})
d.setdefault('fizz', {}).setdefault('buzz', {})
d.setdefault('primary', {}).setdefault('secondary', {}).setdefault('tertiary', {}).setdefault('quaternary', {})

Another criticism is that setdefault requires a new instance whether it is used or not. However, Python (or at least CPython) is rather smart about handling unused and unreferenced new instances, for example, it reuses the location in memory:

>>> id({}), id({}), id({})
(523575344, 523575344, 523575344)

An auto-vivified defaultdict

This is a neat looking implementation, and usage in a script that you're not inspecting the data on would be as useful as implementing __missing__:

from collections import defaultdict

def vivdict():
    return defaultdict(vivdict)

But if you need to inspect your data, the results of an auto-vivified defaultdict populated with data in the same way looks like this:

>>> d = vivdict(); d['foo']['bar']; d['foo']['baz']; d['fizz']['buzz']; d['primary']['secondary']['tertiary']['quaternary']; import pprint; 
>>> pprint.pprint(d)
defaultdict(<function vivdict at 0x17B01870>, {'foo': defaultdict(<function vivdict 
at 0x17B01870>, {'baz': defaultdict(<function vivdict at 0x17B01870>, {}), 'bar': 
defaultdict(<function vivdict at 0x17B01870>, {})}), 'primary': defaultdict(<function 
vivdict at 0x17B01870>, {'secondary': defaultdict(<function vivdict at 0x17B01870>, 
{'tertiary': defaultdict(<function vivdict at 0x17B01870>, {'quaternary': defaultdict(
<function vivdict at 0x17B01870>, {})})})}), 'fizz': defaultdict(<function vivdict at 
0x17B01870>, {'buzz': defaultdict(<function vivdict at 0x17B01870>, {})})})

This output is quite inelegant, and the results are quite unreadable. The solution typically given is to recursively convert back to a dict for manual inspection. This non-trivial solution is left as an exercise for the reader.

Performance

Finally, let's look at performance. I'm subtracting the costs of instantiation.

>>> import timeit
>>> min(timeit.repeat(lambda: {}.setdefault('foo', {}))) - min(timeit.repeat(lambda: {}))
0.13612580299377441
>>> min(timeit.repeat(lambda: vivdict()['foo'])) - min(timeit.repeat(lambda: vivdict()))
0.2936999797821045
>>> min(timeit.repeat(lambda: Vividict()['foo'])) - min(timeit.repeat(lambda: Vividict()))
0.5354437828063965
>>> min(timeit.repeat(lambda: AutoVivification()['foo'])) - min(timeit.repeat(lambda: AutoVivification()))
2.138362169265747

Based on performance, dict.setdefault works the best. I'd highly recommend it for production code, in cases where you care about execution speed.

If you need this for interactive use (in an IPython notebook, perhaps) then performance doesn't really matter - in which case, I'd go with Vividict for readability of the output. Compared to the AutoVivification object (which uses __getitem__ instead of __missing__, which was made for this purpose) it is far superior.

Conclusion

Implementing __missing__ on a subclassed dict to set and return a new instance is slightly more difficult than alternatives but has the benefits of

• easy instantiation
• easy data population
• easy data viewing

and because it is less complicated and more performant than modifying __getitem__, it should be preferred to that method.

Nevertheless, it has drawbacks:

• Bad lookups will fail silently.
• The bad lookup will remain in the dictionary.

Thus I personally prefer setdefault to the other solutions, and have in every situation where I have needed this sort of behavior.

How to create nested dictionaries PYTHON

Use a list comprehension. Iterate over the zipped lists and create the nested dictionaries:

a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]
c = ['N1', 'N2', 'N3', 'N4']

>>> [{k1: {k2: v}} for k1, v, k2 in zip(a, b, c)]
[{'A': {'N1': 1}}, {'B': {'N2': 2}}, {'C': {'N3': 3}}, {'D': {'N4': 4}}]

Create a nested Dictionary with two dictionaries

You can use defaultdict here. multi need not be sorted.

from collections import defaultdict
out=defaultdict(dict)
for v,k,vs in multi:
    out[k]={**out[k],**{v:vs[0]}}

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Output

defaultdict(dict,
            {77766: {14: 2, 15: 1},
             88866: {70: 1, 71: 2, 72: 5, 73: 4, 74: 3},
             99966: {79: 1, 80: 2}})

EDIT:

Sorting the inner dictionaries.

out={k:dict(sorted(v.items(),key=lambda x:x[1])) for k,v in out.items()}

Output:

{77766: {15: 1, 14: 2},
 88866: {70: 1, 71: 2, 74: 3, 73: 4, 72: 5},
 99966: {79: 1, 80: 2}}

How to create a nested dictionary in python with 6 lists

Maybe something like:

out = {k: {k1: v1, k2: v2} for k, k1, v1, k2, v2 in zip(a, b, c, d, e)}

If we want to use excessive number of zips, we could also do:

out = {k: dict(v) for k,v in zip(a, zip(zip(b, c), zip(d, e)))}

Output:

{'A':{1 :9, 0:11} , 'B':{2:8, 3:13}, 'C':{3:7, 5:13} , 'D':{4:6, 7:15}}

How to combine two nested dictionaries with same master keys

Yes:

dict3 = {k: {**v, **dict2[k]} for k, v in dict1.items()}

Firstly, use .items() to iterate over both keys and values at the same time.

Then, for each key k you want the value to be a new dict that is created by dumping — or destructuring — both v and dict2[k] in it.

UPDATE for Python >= 3.9:

Thanks @mwo for mentioning the pipe | operand:

dict3 = {k: v | dict2[k] for k, v in dict1.items()}

How to create dynamic nested dictionaries in Python

from functools import reduce

list_name=["node1","node2","node3","nodeN"]
print(reduce(lambda k, v: {v: k}, reversed(list_name), True))

Output:

{'node1': {'node2': {'node3': {'nodeN': True}}}}

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