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pythonnumpynumpy-ndarraysub-array
Andrea Belle
Andrea Belle

Reputation: 95

How to find N maximum product subarrays of M elements of a Numpy array?

I have a Numpy array, and I need to find the N maximum product subarrays of M elements. For example, I have the array p = [0.1, 0.2, 0.8, 0.5, 0.7, 0.9, 0.3, 0.5] and I want to find the 5 highest product subarrays of 3 elements. Is there a "fast" way to do that?

Upvotes: 3

Views: 141

Answers (2)

Divakar
Divakar

Reputation: 221524

Approach #1

We can create sliding windows and then perform prod reduction and finally np.argpartition to get top N ones among them -

from skimage.util.shape import view_as_windows

def topN_windowed_prod(a, W, N):
    w = view_as_windows(a,W)
    return w[w.prod(1).argpartition(-N)[-N:]]

Sample run -

In [2]: p = np.array([0.1, 0.2, 0.8, 0.5, 0.7, 0.9, 0.3, 0.5])

In [3]: topN_windowed_prod(p, W=3, N=2)
Out[3]: 
array([[0.8, 0.5, 0.7],
       [0.5, 0.7, 0.9]])

Note that the order is not maintained with np.argpartition. So, if we need the top N in descending order of prod values, use range(N) with it. More info.

Approach #2

For smaller window lengths, we can simply slice and get our desired result, like so -

def topN_windowed_prod_with_slicing(a, W, N):
    w = view_as_windows(a,W)
    L = len(a)-W+1
    acc = a[:L].copy()
    for i in range(1,W):
        acc *= a[i:i+L]
    idx = acc.argpartition(-N)[-N:]
    return w[idx]

Upvotes: 1

javidcf
javidcf

Reputation: 59691

Here is another quick way to do it:

import numpy as np

p = [0.1, 0.2, 0.8, 0.5, 0.7, 0.9, 0.3, 0.5]
n = 5
m = 3

# Cumulative product (starting with 1)
pc = np.cumprod(np.r_[1, p])
# Cumulative product of each window
w = pc[m:] / pc[:-m]
# Indices of the first element of top N windows
idx = np.argpartition(w, n)[-n:]
print(idx)
# [1 2 5 4 3]

Upvotes: 1

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