96 lines
2.3 KiB
C
96 lines
2.3 KiB
C
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/**
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* @license Apache-2.0
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*
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* Copyright (c) 2020 The Stdlib Authors.
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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#include "stdlib/stats/base/dnanvarianceyc.h"
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#include <stdint.h>
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/**
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* Computes the variance of a double-precision floating-point strided array ignoring `NaN` values and using a one-pass algorithm proposed by Youngs and Cramer.
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*
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* ## Method
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*
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* - This implementation uses a one-pass algorithm, as proposed by Youngs and Cramer (1971).
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*
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* ## References
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*
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* - Youngs, Edward A., and Elliot M. Cramer. 1971. "Some Results Relevant to Choice of Sum and Sum-of-Product Algorithms." _Technometrics_ 13 (3): 657–65. doi:[10.1080/00401706.1971.10488826](https://doi.org/10.1080/00401706.1971.10488826).
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*
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* @param N number of indexed elements
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* @param correction degrees of freedom adjustment
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* @param X input array
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* @param stride stride length
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* @return output value
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*/
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double stdlib_strided_dnanvarianceyc( const int64_t N, const double correction, const double *X, const int64_t stride ) {
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double sum;
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int64_t ix;
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double nc;
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double n;
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double S;
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double v;
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double d;
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double i;
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if ( N <= 0 ) {
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return 0.0 / 0.0; // NaN
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}
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if ( N == 1 || stride == 0 ) {
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v = X[ 0 ];
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if ( v == v && (double)N-correction > 0.0 ) {
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return 0.0;
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}
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return 0.0 / 0.0; // NaN
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}
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if ( stride < 0 ) {
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ix = (1-N) * stride;
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} else {
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ix = 0;
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}
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// Find the first non-NaN element...
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for ( i = 0; i < N; i++ ) {
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v = X[ ix ];
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if ( v == v ) {
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break;
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}
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ix += stride;
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}
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if ( i == N ) {
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return 0.0 / 0.0; // NaN
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}
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ix += stride;
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sum = v;
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S = 0.0;
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n = 1.0;
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i += 1;
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for (; i < N; i++ ) {
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v = X[ ix ];
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if ( v == v ) {
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n += 1.0;
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sum += v;
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d = (n*v) - sum;
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S += (1.0/(n*(n-1.0))) * d * d;
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}
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ix += stride;
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}
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nc = n - correction;
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if ( nc <= 0.0 ) {
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return 0.0 / 0.0; // NaN
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}
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return S / nc;
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}
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