time-to-botec

Benchmark sampling in different programming languages
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sdsnanmeanors.c (1532B)


      1 /**
      2 * @license Apache-2.0
      3 *
      4 * Copyright (c) 2020 The Stdlib Authors.
      5 *
      6 * Licensed under the Apache License, Version 2.0 (the "License");
      7 * you may not use this file except in compliance with the License.
      8 * You may obtain a copy of the License at
      9 *
     10 *    http://www.apache.org/licenses/LICENSE-2.0
     11 *
     12 * Unless required by applicable law or agreed to in writing, software
     13 * distributed under the License is distributed on an "AS IS" BASIS,
     14 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
     15 * See the License for the specific language governing permissions and
     16 * limitations under the License.
     17 */
     18 
     19 #include "stdlib/stats/base/sdsnanmeanors.h"
     20 #include <stdint.h>
     21 
     22 /**
     23 * Computes the arithmetic mean of a single-precision floating-point strided array, ignoring `NaN` values and using ordinary recursive summation with extended accumulation.
     24 *
     25 * @param N       number of indexed elements
     26 * @param X       input array
     27 * @param stride  stride length
     28 * @return        output value
     29 */
     30 float stdlib_strided_sdsnanmeanors( const int64_t N, const float *X, const int64_t stride ) {
     31 	double sum;
     32 	int64_t ix;
     33 	int64_t i;
     34 	int64_t n;
     35 	double v;
     36 
     37 	if ( N <= 0 ) {
     38 		return 0.0 / 0.0; // NaN
     39 	}
     40 	if ( N == 1 || stride == 0 ) {
     41 		return X[ 0 ];
     42 	}
     43 	if ( stride < 0 ) {
     44 		ix = (1-N) * stride;
     45 	} else {
     46 		ix = 0;
     47 	}
     48 	sum = 0.0;
     49 	n = 0;
     50 	for ( i = 0; i < N; i++ ) {
     51 		v = (double)X[ ix ];
     52 		if ( v == v ) {
     53 			sum += v;
     54 			n += 1;
     55 		}
     56 		ix += stride;
     57 	}
     58 	if ( n == 0 ) {
     59 		return 0.0 / 0.0; // NaN
     60 	}
     61 	return sum / (double)n;
     62 }