Choosing a max size

Unlike Python’s set and dict, Hirola does not manage its size automatically by default (although it can be reconfigured to). To prevent wasted resizing (which is what Python does under the hood), you have full control of and responsibility for how much space the table uses. Obviously the table has to be large enough to fit all the keys in it. Additionally, when a hash table gets to close to full it becomes much slower. Depending on how much you favour speed over memory you should add 20-50% extra headroom. If you intend to a lot of looking up of the same small set of values then it can continue to run faster if you increase max to 2-3x its minimal size.

Structured key data types

To indicate that an array axis should be considered as a single key, use NumPy’s structured dtypes. In the following example, the data type (points.dtype, 3) indicates that a 3D point - a triplet of floats - should be considered as one object. See help(HashTable.dtype) for more information of specifying dtypes. Only the last axis or last axes may be thought of as single keys. For other setups, first convert with numpy.transpose().

import numpy as np
from hirola import HashTable

# Create a cloud of 3D points with duplicates. This is 3000 points in total,
# with up to 1000 unique points.
points = np.random.uniform(-30, 30, (1000, 3))[np.random.choice(1000, 3000)]

# Create an empty hash table.
# In practice, you generally don't know how many unique elements there are
# so we'll pretend we don't either an assume the worst case of all 3000 are
# unique. We'll also give 25% padding for speed.
table = HashTable(len(points) * 1.25, (points.dtype, 3))

# Add all points to the table.
ids = table.add(points)

Duplicate-free contents can be accessed from table.keys:

>>> table.keys  # <--- These are `points` but with no duplicates.
array([[  3.47736554, -15.17112511,  -9.51454466],
       [ -6.46948046,  23.64504329, -16.25743105],
       [-27.02527253, -16.1967225 , -10.11544157],
       ...,
       [  3.75972597,   1.24130412,  -8.14337206],
       [-13.62256791,  11.76551455, -13.31312988],
       [  0.19851678,   4.06221179, -22.69006592]])
>>> table.keys.shape
(954, 3)

Each point’s location in table.keys is returned by table.add(), similarly to numpy.unique(..., return_args=True).

>>> ids  # <--- These are the indices in `table.keys` of each point in `points`.
array([  0,   1,   2, ..., 290, 242, 669])
>>> np.array_equal(table.keys[ids], points)
True

Lookup the indices of points without adding them using table.get().

Handling of nearly full hash tables

HashTables become very slow when almost full. As of v0.3.0, an efficiency warning will notify you if a table exceeds 90% full. This warning can be reconfigured into an error, silenced or set to resize the table automatically to make room. These are demonstrated in the example constructors below:

# The default: Issue a warning when the table is 90% full.
HashTable(..., almost_full=(0.9, "warn"))

# Disable all "almost full" behaviours.
HashTable(..., almost_full=None)

# To consider a table exceeding 80% full as an error use:
HashTable(..., almost_full=(0.8, "raise"))

# To automatically triple in size whenever the table exceeds 80% full use:
HashTable(..., almost_full=(0.8, 3.0))

Resizing tables is slow which is why it’s not enabled by default. It should be avoided unless you really have no idea how big your table will need to be.

Using a HashTable as a dict

A dict requires a second array for values. The output of HashTable.add() and HashTable.get() should be used as indices of values:

import numpy as np
from hirola import HashTable

# The `keys` - will be populated with names of African countries.
countries = HashTable(40, (str, 20))
# The `values` - will be populated with the names of each country's capital city.
capitals = np.empty(countries.max, (str, 20))

Add or set items using the pattern values[table.add(key)] = value:

capitals[countries.add("Algeria")] = "Al Jaza'ir"

Or in bulk:

new_keys = ["Angola", "Botswana", "Burkina Faso"]
new_values = ["Luanda", "Gaborone", "Ouagadougou"]
capitals[countries.add(new_keys)] = new_values

Like Python dicts, overwriting values is exactly the same as writing them.

Retrieve values with values[table[key]]:

>>> capitals[countries["Botswana"]]
'Gaborone'
>>> capitals[countries["Botswana", "Algeria"]]
array(['Gaborone', "Al Jaza'ir"], dtype='<U20')

View all keys and values with table.keys and values[:len(table)]. A HashTable remembers the order keys were first added so this dict is automatically a sorted dict.

# keys
>>> countries.keys
array(['Algeria', 'Angola', 'Botswana', 'Burkina Faso'], dtype='<U20')
# values
>>> capitals[:len(countries)]
array(["Al Jaza'ir", 'Luanda', 'Gaborone', 'Ouagadougou'], dtype='<U20')

Depending on the usage scenario, it may or may not make sense to want an equivalent to dict.items(). If you do want an equivalent, use numpy.rec.fromarrays([table.keys, values[:len(table)]]), possibly adding a names= option:

>>> np.rec.fromarrays([countries.keys, capitals[:len(countries)]],
...                   names="countries,capitals")
rec.array([('Algeria', "Al Jaza'ir"), ('Angola', 'Luanda'),
           ('Botswana', 'Gaborone'), ('Burkina Faso', 'Ouagadougou')],
          dtype=[('countries', '<U20'), ('capitals', '<U20')])

If the keys and values have the same dtype then numpy.c_ works too.

>>> np.c_[countries.keys, capitals[:len(countries)]]
array([['Algeria', "Al Jaza'ir"],
       ['Angola', 'Luanda'],
       ['Botswana', 'Gaborone'],
       ['Burkina Faso', 'Ouagadougou']], dtype='<U20')

Using a HashTable as a set

To get set-like capabilities from a HashTable, leverage the contains() method. For these examples we will experiment with integer multiples of 3 and 7.

import numpy as np

of_3s = np.arange(0, 100, 3)
of_7s = np.arange(0, 100, 7)

We’ll only require one array to be converted into a hash table. The other can remain as an array. If both are hash tables, simply use one table’s keys attribute as the array.

from hirola import HashTable

table_of_3s = HashTable(len(of_3s) * 1.25, of_3s.dtype)
table_of_3s.add(of_3s)

Use table.contains() as a vectorised version of in.

>>> table_of_3s.contains(of_7s)
array([ True, False, False,  True, False, False,  True, False, False,
        True, False, False,  True, False, False])

From the above, the common set operations can be derived with following:

• set.intersection() - Values in the array and in the set:

>>> of_7s[table_of_3s.contains(of_7s)]
array([ 0, 21, 42, 63, 84])

• Set subtraction - Values in the array which are not in the set:

>>> of_7s[~table_of_3s.contains(of_7s)]
array([ 7, 14, 28, 35, 49, 56, 70, 77, 91, 98])

• set.union() - Values in either the table or in the tested array (with no duplicates):

>>> np.concatenate([table_of_3s.keys, of_7s[~table_of_3s.contains(of_7s)]], axis=0)
array([ 0,  3,  6,  9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48,
       51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99,
        7, 14, 28, 35, 49, 56, 70, 77, 91, 98])

Using a HashTable as a collections.Counter

For this example, let’s give ourselves something a bit more substantial to work on. Counting word frequencies in Shakespeare’s Hamlet play is the trendy example for collections.Counter and it’s what we’ll use too.

from urllib.request import urlopen
import re
import numpy as np

hamlet = urlopen("https://gist.githubusercontent.com/provpup/2fc41686eab7400b796b/raw/b575bd01a58494dfddc1d6429ef0167e709abf9b/hamlet.txt").read()
words = np.array(re.findall(rb"([\w']+)", hamlet))

A counter is just a dict with integer values and a dict is just a hash table with a separate array for values.

from hirola import HashTable

word_table = HashTable(len(words), words.dtype)
counts = np.zeros(word_table.max, dtype=int)

The only new functionality that is not defined in using a hash table as a dict is the ability to count keys as they are added. To count new elements use the rather odd line np.add(counts, table.add(keys), 1).

np.add.at(counts, word_table.add(words), 1)

This line does what you might expect counts[word_table.add(words)] += 1 to do but, due to the way NumPy works, the latter form fails to increment each count more than once if words contains duplicates.

Use NumPy’s indirect sorting functions to get most or least common keys.

# Get the most common word.
>>> word_table.keys[counts[:len(word_table)].argmax()]
b'the'

# Get the top 10 most common words. Note that these are unsorted.
>>> word_table.keys[counts[:len(word_table)].argpartition(-10)[-10:]]
array([b'it', b'and', b'my', b'of', b'in', b'a', b'to', b'the', b'I',
       b'you'], dtype='|S14')

# Get all words in ascending order of commonness.
>>> word_table.keys[counts[:len(word_table)].argsort()]
array([b'END', b'whereat', b"griev'd", ..., b'to', b'and', b'the'],
      dtype='|S14')