/** * @license Apache-2.0 * * Copyright (c) 2020 The Stdlib Authors. * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ #include "stdlib/stats/base/snanmeanpn.h" #include /** * Computes the arithmetic mean of a single-precision floating-point strided array, ignoring `NaN` values and using a two-pass error correction algorithm. * * ## Method * * - This implementation uses a two-pass approach, as suggested by Neely (1966). * * ## References * * - Neely, Peter M. 1966. "Comparison of Several Algorithms for Computation of Means, Standard Deviations and Correlation Coefficients." _Communications of the ACM_ 9 (7). Association for Computing Machinery: 496–99. doi:[10.1145/365719.365958](https://doi.org/10.1145/365719.365958). * - Schubert, Erich, and Michael Gertz. 2018. "Numerically Stable Parallel Computation of (Co-)Variance." In _Proceedings of the 30th International Conference on Scientific and Statistical Database Management_. New York, NY, USA: Association for Computing Machinery. doi:[10.1145/3221269.3223036](https://doi.org/10.1145/3221269.3223036). * * @param N number of indexed elements * @param X input array * @param stride stride length * @return output value */ float stdlib_strided_snanmeanpn( const int64_t N, const float *X, const int64_t stride ) { int64_t ix; int64_t i; int64_t n; int64_t o; double dn; float s; float t; float v; if ( N <= 0 ) { return 0.0f / 0.0f; // NaN } if ( N == 1 || stride == 0 ) { return X[ 0 ]; } if ( stride < 0 ) { ix = (1-N) * stride; } else { ix = 0; } o = ix; // Compute an estimate for the mean... s = 0.0f; n = 0; for ( i = 0; i < N; i++ ) { v = X[ ix ]; if ( v == v ) { s += v; n += 1; } ix += stride; } if ( n == 0 ) { return 0.0f / 0.0f; // NaN } dn = (double)n; s = (double)s / dn; // Compute an error term... t = 0.0f; ix = o; for ( i = 0; i < N; i++ ) { v = X[ ix ]; if ( v == v ) { t += v - s; } ix += stride; } return s + (float)((double)t/dn); }