Python can be used to write and test code quickly because it is an interpreted language that types dynamically. However, these are also the reasons it is slow when performing simple tasks repeatedly, for example in loops.
When developing code, there can often be trade-offs between different implementations. However, at the beginning of the development of an algorithm, it is usually counterproductive to worry about the efficiency of the code.
«We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil. Yet we should not pass up our opportunities in that critical 3%.»
In the following, I show examples of the k-means algorithm to form a previously known number of groups from a set of objects. This can be achieved in the following three steps:
Choose the first
kelements as cluster centres
Assign each new element to the cluster with the least increase in variance.
Update the cluster centre
Steps 2 and 3 are repeated until the assignments no longer change.
A possible implementation with pure Python could look like this:
# SPDX-FileCopyrightText: 2021 Veit Schiele # # SPDX-License-Identifier: BSD-3-Clause def dist(x, y): """Calculate the distance""" return sum((xi - yi) ** 2 for xi, yi in zip(x, y)) def find_labels(points, centers): """Assign points to a cluster.""" labels =  for point in points: distances = [dist(point, center) for center in centers] labels.append(distances.index(min(distances))) return labels def compute_centers(points, labels): """Calculate the cluster centres.""" n_centers = len(set(labels)) n_dims = len(points) centers = [[0 for i in range(n_dims)] for j in range(n_centers)] counts = [0 for j in range(n_centers)] for label, point in zip(labels, points): counts[label] += 1 centers[label] = [a + b for a, b in zip(centers[label], point)] return [[x / count for x in center] for center, count in zip(centers, counts)] def kmeans(points, n_clusters): """Calculates the cluster centres repeatedly until nothing changes.""" centers = points[-n_clusters:].tolist() while True: old_centers = centers labels = find_labels(points, centers) centers = compute_centers(points, labels) if centers == old_centers: break return labels
We can create sample data with:
from sklearn.datasets import make_blobs points, labels_true = make_blobs( n_samples=1000, centers=3, random_state=0, cluster_std=0.60 )
And finally, we can perform the calculation with:
Search for existing implementations#
You should not try to reinvent the wheel: If there are existing implementations, you should use them. There are even two implementations for the k-means algorithm:
from sklearn.cluster import KMeans KMeans(10).fit_predict(points)
from dask_ml.cluster import KMeans KMeans(10).fit(points).predict(points)
The best that could be said against these existing solutions is that they could create a considerable overhead in your project if you are not already using scikit-learn or Dask-ML elsewhere. In the following, I will therefore show you further possibilities to optimise your own code.
Then you can use perflint to search your code for the most common performance anti-patterns in Python.
Vectorisations with NumPy#
NumPy moves repetitive operations into a statically typed compiled layer, combining the fast development time of Python with the fast execution time of C. You may be able to use Universal functions (ufunc), vectorisation and Indexing and slicing in all combinations to move repetitive operations into compiled code to avoid slow loops.
With NumPy we can do without some loops:
# SPDX-FileCopyrightText: 2021 Veit Schiele # # SPDX-License-Identifier: BSD-3-Clause import numpy as np def find_labels(points, centers):
The advantages of NumPy are that the Python overhead only occurs per array and not per array element. However, because NumPy uses a specific language for array operations, it also requires a different mindset when writing code. Finally, the batch operations can also lead to excessive memory consumption.
Special data structures#
This way you can also bypass the loop in the
# # SPDX-License-Identifier: BSD-3-Clause diff = points[:, None, :] - centers distances = (diff**2).sum(-1) return distances.argmin(1)
for spatial queries like distances, nearest neighbours, k-Means etc.
find_labelsmethod can then be written more specifically:
import pandas as pd from scipy.spatial import cKDTree
sparse matrices for 2-dimensional structured data.
for N-diemensional structured data.
for graph-like problems, for example searching for shortest paths.
for grouping across multiple dimensions.
At PyCon US in May 2021, Guido van Rossum presented Faster CPython, a project that aims to double the speed of Python 3.11. The cooperation with the other Python core developers is regulated in PEP 659 – Specializing Adaptive Interpreter. There is also an open issue tracker and various tools for collecting bytecode statistics. CPU-intensive Python code in particular is likely to benefit from the changes; code already written in C, I/O-heavy processes and multithreaded code, on the other hand, are unlikely to benefit.
If you don’t want to wait with your project until the release of Python 3.11 in the final version probably on 24 October 2022, you can also have a look at the following Python interpreters:
For intensive numerical operations, Python can be very slow, even if you have avoided all anti-patterns and used vectorisations with NumPy. In this case, translating code into Cython can be helpful. Unfortunately, the code often has to be restructured and thus increases in complexity. Explicit type annotations and the provision of code also become more cumbersome.
Our example could then look like this:
# SPDX-FileCopyrightText: 2021 Veit Schiele # # SPDX-License-Identifier: BSD-3-Clause cimport numpy as np import numpy as np cdef double dist(double[:] x, double[:] y): """Calculate the distance""" cdef double dist = 0 for i in range(len(x)): dist += (x[i] - y[i]) ** 2 return dist def find_labels(double[:, :] points, double[:, :] centers): """Assign points to a cluster.""" cdef int n_points = points.shape cdef int n_centers = centers.shape cdef double[:] labels = np.zeros(n_points) cdef double distance, nearest_distance cdef int nearest_index for i in range(n_points): nearest_distance = np.inf for j in range(n_centers): distance = dist(points[i], centers[j]) if distance < nearest_distance:
Numba is a JIT compiler that translates mainly scientific Python and NumPy code into fast machine code, for example:
# SPDX-FileCopyrightText: 2021 Veit Schiele # # SPDX-License-Identifier: BSD-3-Clause import numba @numba.jit(nopython=True) def dist(x, y): """Calculate the distance""" dist = 0 for i in range(len(x)): dist += (x[i] - y[i]) ** 2 return dist @numba.jit(nopython=True) def find_labels(points, centers): """Assign points to a cluster.""" labels =  min_dist = np.inf min_label = 0 for i in range(len(points)): for j in range(len(centers)): distance = dist(points[i], centers[j])
However, Numba requires LLVM and some Python constructs are not supported.
ipyparallelsimply integrates into a JupyterHub.
Dask imitates pandas, NumPy iterators, Toolz und PySpark when it distributes their tasks.
Ray provides a simple, universal API for building distributed applications.
RLlib will scale reinforcement learning in particular.
Our example could look like this with Dask:
# SPDX-FileCopyrightText: 2021 Veit Schiele # # SPDX-License-Identifier: BSD-3-Clause import numpy as np from dask import array as da from dask import dataframe as dd def find_labels(points, centers): """Assign points to a cluster.""" diff = points[:, None, :] - centers distances = (diff**2).sum(-1) return distances.argmin(1) def compute_centers(points, labels): """Calculate the cluster centres.""" points_df = dd.from_dask_array(points) labels_df = dd.from_dask_array(labels) return points_df.groupby(labels_df).mean() def kmeans(points, n_clusters): """Calculates the cluster centres repeatedly until nothing changes.""" centers = points[-n_clusters:] points = da.from_array(points, 1000) while True: old_centers = centers labels = find_labels(points, da.from_array(centers, 5)) centers = compute_centers(points, labels) centers = centers.compute().values if np.all(centers == old_centers): break return labels.compute()