Say I have a Numpy array of N = 10 random float numbers: import numpy as np np.random.seed(99) N = 10 arr = np.random.uniform(0., 10., size=(N,)) print(arr) out[1]: [6.72278559 4.88078399 8.25495174 0.31446388 8.08049963 5.6561742 2.97622499 0.46695721 9.90627399 0.06825733] I want to find all unique pairs of numbers that are not different from each other more than a tolerance tol = 1. (i.e. absolute difference <= 1). Specifically, I want to get all unique pairs of indexes. The indexes of each close pair should be sorted, and all close pairs should be sorted by the first index. I managed to write the following working code: def all_close_pairs(arr, tol=1.): res = set() for i, x1 in enumerate(arr): for j, x2 in enumerate(arr): if i == j: continue if np.isclose(x1, x2, rtol=0., atol=tol): res.add(tuple(sorted([i, j]))) res = np.array(list(res)) return res[res[:,0].argsort()] print(all_close_pairs(arr, tol=1.)) out[2]: [[1 5] [2 4] [3 7] [3 9] [7 9]] However, in reality I have an array of N = 1000 numbers, and my code becomes extremely slow due to the nested for loops. I believe there are much more efficient ways to do this with Numpy vectorization. Does anyone know the fastest way to do this in Numpy? Continue reading...
