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Asked: 2024-09-27T07:16:19+05:30

How can I implement a compact K-means algorithm in minimal code characters for a coding challenge?

anonymous user

I recently came across this interesting challenge that revolves around the K-means algorithm, and I thought it might be fun to dive deeper into it. So, here’s the scoop: the K-means algorithm is a popular method for clustering data. It involves dividing a set of data points into K groups based on their features, where each group is represented by the mean of the data points within it.

Now, the challenge takes it up a notch by asking for a compact version of this algorithm—essentially a “golfed” version where you minimize the number of characters in your code. It sounds easy at first, but when you think about it more, it’s actually quite tricky!

Here’s the problem: you are given a set of points in an n-dimensional space and a specified number of clusters (K). You need to implement the K-means algorithm in as few characters as possible while making sure it still fulfills its purpose. The code should include the initialization of cluster centroids, assignment of points to the nearest centroid, and updating centroids based on the assigned points.

But wait, it gets more interesting! You can assume some constraints like a fixed dimension for the points (e.g., 2D or 3D) and a limited range for the data points. Oh, and let’s not forget—your solution has to work within the typical limitations of a code golf challenge, which usually means the output has to be in a specific format, so make sure to consider that too!

So, here’s my question to you: How would you tackle this challenge? What clever tricks or techniques would you use to compress the K-means algorithm without sacrificing its functionality? I’m super curious to see different approaches and see who can come up with the shortest, most elegant solution. Let’s get our coding caps on and see what we can come up with!

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    2 Answers

    1. anonymous user

      import numpy as np
def kmeans(X, k, iters=10):
    # Init centroids
    centroids = X[np.random.choice(X.shape[0], k, replace=False)]
    for _ in range(iters):
        # Assign clusters
        clusters = np.argmin(((X[:, None] - centroids) ** 2).sum(axis=2), axis=1)
        # Update centroids
        centroids = np.array([X[clusters == i].mean(axis=0) for i in range(k)])
    return centroids, clusters

    2. anonymous user

      To tackle the challenge of implementing a compact version of the K-means algorithm, we can leverage Python’s features such as list comprehensions and NumPy for efficient computations. The goal is to minimize character count while retaining the essential components of K-means: initialization, assignment, and update steps. Here’s a golfed version of the K-means algorithm for 2D points:

      import numpy as np
def kmeans(X,K):
 C=X[np.random.choice(range(len(X)),K)]
 for _ in"":C=np.array([C[np.argmin(np.linalg.norm(X-C[:,None],axis=0,ord=2),axis=0)].mean(axis=1) for _ in range(K)]) 
 return C

      This implementation initiates cluster centroids randomly from the input data points using `np.random.choice`. It uses a loop to iteratively assign each point to the nearest centroid and then updates the centroid positions based on the mean of assigned points. The result is a compact function maintaining the core functionality of K-means clustering.

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