Fix WMA weighted_mean NaN handling#2263
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The weighted_mean function in WMA._load_internal uses np.nanmean(w * x), which produces incorrect results when NaN values are present in the input. The problem is that np.nanmean renormalizes the product (w * x) by counting how many non-NaN products there are, but it doesn't renormalize the weights themselves — so the weight denominator still sums over all elements including the NaN positions, leading to a biased result. For example, with x = [1, NaN, 3] and w = [1/6, 2/6, 3/6]: Old: np.nanmean(w * x) = (1/6 + 9/6) / 2 = 0.833 Correct weighted avg: (1*1/4 + 3*3/4) = 2.5 The fix filters out NaN positions from both x and w before computing the weighted sum, then renormalizes w so it sums to 1 over the valid elements.
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Fix WMA weighted_mean NaN handling