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HeavyHitters

Find the Heavy Hitters using the Lossy Count with Forgetting factor algorithm1.

Keep track of the most frequent item(set)s in a data stream and apply a forgetting factor to discard previous frequent items that do not often appear anymore. This is an approximation algorithm designed to work with a limited amount of memory rather than accounting for every possible solution (thus using an unbounded memory footprint). Any hashable type can be passed as input, hence tuples or frozensets can also be monitored.

Considering a data stream where n elements were observed so far, the Lossy Count algorithm has the following properties:

  • All item(set)s whose true frequency exceeds support * n are output. There are no

false negatives;

  • No item(set) whose true frequency is less than (support - epsilon) * n is outputted;

  • Estimated frequencies are less than the true frequencies by at most epsilon * n.

Parameters

  • support

    Type → float

    Default → 0.001

    The support threshold used to determine if an item is frequent. The value of support must be in \([0, 1]\). Elements whose frequency is higher than support times the number of observations seen so far are outputted.

  • epsilon

    Type → float

    Default → 0.005

    Error parameter to control the accuracy-memory tradeoff. The value of epsilon must be in \((0, 1]\) and typically epsilon \(\ll\) support. The smaller the epsilon, the more accurate the estimates will be, but the count sketch will have an increased memory footprint.

  • fading_factor

    Type → float

    Default → 0.999

    Forgetting factor applied to the frequency estimates to reduce the impact of old items. The value of fading_factor must be in \((0, 1]\).

Examples

import random
import string
from river import sketch

rng = random.Random(42)
hh = sketch.HeavyHitters()

We will feed the counter with printable ASCII characters:

for _ in range(10_000):
    hh = hh.update(rng.choice(string.printable))

We can retrieve estimates of the n top elements and their frequencies. Let's try n=3 

hh.most_common(3)

[(',', 122.099142...), ('[', 116.049510...), ('W', 115.013402...)]

We can also access estimates of individual elements:

hh['A']

99.483575...

Unobserved elements are handled just fine: 

hh[(1, 2, 3)]

0.0

Methods

most_common
update

  1. Veloso, B., Tabassum, S., Martins, C., Espanha, R., Azevedo, R., & Gama, J. (2020). Interconnect bypass fraud detection: a case study. Annals of Telecommunications, 75(9), 583-596. ↩