This package finds close numerical matches fast across two 2D arrays of shape (n, d) and (m, d) (if it can be assumed there will be relatively few matches and d is relatively low). Returns the indices of the matches.
You can install close-numerical-matches
from PyPI:
$ pip install close-numerical-matches
The package is supported on Python 3.8 and above and requires Numpy.
Import find_matches
from close_numerical_matches
and supply two arrays of shape (n, d) and (m, d) and a given tolerance level. Optionally provide your desired distance metric and a bucket tolerance multiplier. The arguments in more detail:
arr0
: np.ndarray
First array to find matches against. Should be of size (n, d).arr1
: np.ndarray
Second array to find matches against. Should be of size (m, d).dist
: {'norm', 'max', 'cos'}
or Callable[[np.ndarray, np.ndarray], np.ndarray]
, default='norm'
Distance metric to calculate distance. 'norm'
, 'max'
and 'cos'
are currently supported. If you want some other distance function, you can supply your own function. It should take two arrays: One array of size (n, d) of entries and another array of size (d,) of a particular entry. Should return an (n,) array of distances.tol
: float
, default=0.1
The tolerance where values are considered the similar enough to count as a match. Should be > 0.bucket_tol_mult
: int
, default=2
The tolerance multiplier to use for assigning buckets. By setting this lower, there are fewer potential close points that need to be checked (faster runtime), but you risk missing some match pairs. By setting it higher, you need to compare more potential matches, but the risk of missing any close pair is lower. For cosine distance, you might need to set this high.
If you supply your own distance function, do take this parameter into consideration.
Should never be less than 1.>>> import numpy as np
>>> from close_numerical_matches import find_matches
>>> arr0 = np.array([[25, 24], [50, 50], [25, 26]])
>>> arr1 = np.array([[25, 23], [25, 25], [50.6, 50.6], [60, 60]])
>>> find_matches(arr0, arr1, tol=1.0001)
array([[0, 0], [0, 1], [1, 2], [2, 1]])
>>> find_matches(arr0, arr1, tol=0.9999)
array([[1, 2]])
>>> find_matches(arr0, arr1, tol=0.60001)
array([], dtype=int64)
>>> find_matches(arr0, arr1, tol=0.60001, dist='max')
array([[1, 2]])
>>> find_matches([[0, 0.05]], [[0, 5], [0, -0.01]], tol=0.1, dist='cos')
array([[0, 0]])
>>> manhatten_dist = lambda arr, row: np.sum(np.abs(arr - row), axis=1)
>>> matches = find_matches(arr0, arr1, tol=1.0001, dist=manhatten_dist)
>>> matches
array([[0, 0], [0, 1], [2, 1]])
>>> indices0, indices1 = matches.T
>>> arr0[indices0]
array([[25, 24], [25, 24], [25, 26]])
More examples can be found in the test cases.
Here is an unscientific example:
from timeit import default_timer as timer
import numpy as np
from close_numerical_matches import naive_find_matches, find_matches
arr0 = np.random.rand(320_000, 2)
arr1 = np.random.rand(44_000, 2)
start = timer()
naive_find_matches(arr0, arr1, tol=0.001)
end = timer()
print(end - start) # 255.335 s
start = timer()
find_matches(arr0, arr1, tol=0.001)
end = timer()
print(end - start) # 5.821 s
Instead of comparing every element in the first array against every element in the second array, resulting in an O(nmd) runtime, all elements are at first assigned to buckets so only elements that are relatively close are compared. In the case of relatively few matches and a low dimensionality d, this cuts the runtime down to almost linear O((n + m)d).
In general, the algorithm runtime of the bucket approach is O((n + m)d + Bd³ + ∑_{b ∈ B} n_b m_b) where B is the number of buckets and n_b and m_b are the number of items assigned to bucket b. As can be seen, it scales bad with dimensionality and also does not improve from the naive approach if all elements are assigned to the same bucket. In case the bucket approach is likely to be slower than the naive approach, this library will fall back to the naive approach.