# Hash a Nested Python Dictionary

March 2021 ยท 6 minute read

I came upon a Python situation recently where I had to hash a dictionary of arbitary depth, where:

1. all keys are strings
2. all values are either strings, numbers (numpy or native python ints/floats), basic iterables (non-object dtype numpy ndarrays, numerical pandas Series, or native python lists or tuples of python native numbers/strings), OR…
3. …values are themselves dictionaries subject to these 3 rules.

For example:

some_dictionary = {

"a": 12,
"b": "some_string",
"c": {
"q": np.ndarray([1,2,3...]),
"some_key": pd.Series([19.283, 20.873....]),
"k": 11,
"a": {
"nested_key": [8, 9, 11],
"other nested key": (11, 15),
"c": "eleven",
"d" : {
# of arbitrary depth
...
}
}
}
"d": np.ndarray([0.001, 0.002, 0.003...]),
"something": np.int64(40000),
...
# of arbitrary length
}


How would we possibly hash this?

## A solution

If you’re feeling intimidated, then you know how I felt when there was no answer to be found on StackOverflow or seemingly anywhere in the Googlesphere. So, in the spirit of a frantic coding interview, I wrote a brief (? maybe?) solution that works (pretty well? as far as I know?).

import hashlib
import json

import numpy as np
import pandas as pd

# Collapse the dictionary to a single representation
def immutify_dictionary(d):
d_new = {}
for k, v in d.items():

# convert to python native immutables
if isinstance(v, (np.ndarray, pd.Series)):
d_new[k] = tuple(v.tolist())

# immutify any lists
elif isinstance(v, list):
d_new[k] = tuple(v)

# recursion if nested
elif isinstance(v, dict):
d_new[k] = immutify_dictionary(v)

# ensure numpy "primitives" are casted to json-friendly python natives
else:
# convert numpy types to native
if hasattr(v, "dtype"):
d_new[k] = v.item()
else:
d_new[k] = v

return dict(sorted(d_new.items(), key=lambda item: item[0]))

# Make a json string from the sorted dictionary
# then hash that string
def hash_dictionary(d):
d_hashable = immutify_dictionary(d)
s_hashable = json.dumps(d_hashable).encode("utf-8")
m = hashlib.sha256(s_hashable).hexdigest()
return m


Here’s an example:

d = {
"q": [1, 2, 3],
"b": {
"c": "12",
"d": 15,
"e": np.asarray([4, 5, 6]),
"f": {
"z": 12,
"a": [7, 8]
}
},
"c": np.int64(400)
}

>>> hash_dictionary(d)

'e68f1472030c1dbba09fda5eca116c9e7d5ae37e4742d619102b63972fdf9e50'


Note if we permute the keys (Python 3.6+ dictionaries are insertion-sorted, but we are considering dictionaries unsorted as they can be permuted while holding the same data) we obtain the same result:

d_permuted = {
"b": {
"c": "12",
"f": {
"a": [7, 8],
"z": 12,
},
"e": np.asarray([4, 5, 6]),
"d": 15,
},
"c": np.int64(400),
"q": [1, 2, 3],
}

>>> hash_dictionary(d_permuted)

'e68f1472030c1dbba09fda5eca116c9e7d5ae37e4742d619102b63972fdf9e50'


But if we change any of the keys or the values, we get a different hash:

d_new_value = {
"b": {
"c": "12",
"f": {
"a": [7, 8],
"z": 12,
},
"e": np.asarray([4, 5, 6]),
"d": 15,
},

# changed this value
"c": np.int64(399),
"q": [1, 2, 3],
}

>>> hash_dictionary(d_new_value)

'2e3756de5565262aba8f0f888d2938cde93e5b6bc4735b730f673a798a9d72d5'


If you find a more concise/performant solution, or find an edge case where this does not work, please email me! This solution relies on making a copy of the dictionary which is not efficient in terms of memory, so there is likely a better way to do inplace or without sorting.

## Walkthrough

#### Mutability

In Python, dictionaries are unhashable because they are mutable, meaning they can be changed. So we need a way to get an immutable version of (or ‘immutify’) a dictionary subject to the constraints above. Everything inside the dictionary must also be immutable; for example, Python lists are not hashable because they are mutable.

If you are dealing with the most basic case, where the dictionary is flat:

{
key_1: value_1,
key_2: value_2,
key_3: value_3,
}


Python’s builtin frozenset function makes it trivial to immutify this dictionary. You can think of what frozenset does as similar to creating an immutable set of tuples, which together is immutable and therefore hashable. Provided your values are themselves hashable (e.g., native Python numbers, strings, tuples, etc.) frozenset immediately solves your problem.

However, if we make the dictionary a bit more complicated and include mutable datatypes as values, we will need to immutify them before creating a frozenset:

{
key 1: [item1, item2, item3],
key 2: (item4, item5),
key 3: value_3,
key 4: np.ndarray([1,2,3])
}


The lists can be changed to tuples with the tuple() function; the builtin ndarray.tolist() will take care of changing numpy and pandas objects to native lists, which then can be cast to immutable tuples.

For values which are themselves numpy primitive data types (e.g., int64), these should be cast to python natives.

#### Recursion

We add yet another layer of complexity when we consider that any dictionary value can itself be another dictionary. The only way to deal with this is recursion - or dynamic programming if we’re feeling particularly fancy.

The pseudocode for dealing with this problem is:

function immutify_dictionary(d)
for key, value in d
if type(value) = dictionary
immutify_dictionary(value))
else:
immutify_iterable_or_primitive(value)


The algorithm runs over values until it either runs into a “stop point” (array or primitive) or recursively continues by finding a dictionary value.

#### Regular hash doesn’t cut it

Provided we can now get each part of our dictionary into an immutable form, we still face a problem. The builtin python hash, which can natively hash python objects, does not give reliable digests because it is seeded with a random number from the system as a source of entropy. Instead, we need to use the hashlib module (specifically, SHA256).

However, the hashlib algorithms require strings as input, not any immutable python object.

So as opposed to using hash on a frozenset, where ordering does not matter, we must (1) order the dictionary deterministically and then (2) convert the dictionary to a string. If dictionary A and dictionary B have the same keys and same values but are string-ified with different orderings, they will have different hashes! Essentially…

Essentially, if two dictionaries have the same information contained in them, they must produce the same dictionary string and therefore the same hash.

At first it may seem like our previous work immutifying the dictionary is wasted, since we’ll be immutifying by just converting our dictionaries to a big string. This is not true. Since different formats of the same data must be represented identically, different formats of arrays or numbers must collapse to the same string. In other words, running the code for immutification will collapse the various numpy/pandas/native arrays into the same tuples and the various primitive numerical types to the native python types. Our work was not wasted!

This is where the final line of immutify_dictionary comes in:

return dict(sorted(d_new.items(), key=lambda item: item[0]))


This line simply re-sorts the dictionary by the keys, producing an insertion-sorted dictionary as output.

Next, this sorted, deterministically collapsed dictionary can truly become immutable:

d_hashable = immutify_dictionary(d)
s_hashable = json.dumps(d_hashable).encode("utf-8")


The s_hashable here is just itself a very long “hash” of the input dictionary. The same information, regardless of insertion order or array/primitive format, collapses to the same string.

#### The final step

Much shorter than the string representation of the dictionary is the hexdigest of the SHA256 hash of that string. That is where hashlib finally comes in:

   m = hashlib.sha256(s_hashable).hexdigest()


The output is the 256-bit, or 64-hex character digest of our dictionary.