UP my solution #14
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UP my solution #14
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tomMoral
reviewed
Dec 18, 2023
numpy_questions.py
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| for k in range(X.shape[0]) : | ||
| for l in range(X.shape[1]) : | ||
| if X[k,l] > max : | ||
| max = X[k,l] | ||
| i = k | ||
| j = l |
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No for loop, look at numpy.unravel_index
numpy_questions.py
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| for k in range(1,n_terms+1) : | ||
| k_square = k**2 | ||
| pi *= (4*k_square/(4*k_square -1)) |
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I like the first optimization (don't compute the square twice), but you can do even better and without for loop using vectorized numpy.
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