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Multithreaded standard deviation is consistent now

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Erica Fischer
2023-02-09 12:14:54 -08:00
parent 07ef9f4a3c
commit b024c515eb
1 changed files with 20 additions and 27 deletions
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mbtiles.cpp
+20 -27
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@@ -856,38 +856,31 @@ std::map<std::string, layermap_entry> merge_layermaps(std::vector<std::map<std::
fk2->second.count += fk->second.count;
if (fk->second.numeric_count + fk2->second.numeric_count > 0) {
// Why is this standard deviation not exactly right?
// Because the subsamples had different means?
// Because the subsamples have different counts?
#if 0
size_t n = fk->second.numeric_count;
size_t m = fk2->second.numeric_count;
double s2x = fk->second.m2;
double s2y = fk2->second.m2;
double xmean = fk->second.mean;
double ymean = fk2->second.mean;
fk2->second.m2 = ((n - 1) * s2x + (m - 1) * s2y) / (n + m - 1) +
((n * m) * pow(xmean - ymean, 2) /
((n + m) * (n + m - 1)));
#endif
// https://math.stackexchange.com/questions/2971315/how-do-i-combine-standard-deviations-of-two-groups
size_t n1 = fk->second.numeric_count;
size_t n2 = fk2->second.numeric_count;
double var_x1 = fk->second.m2 / n1;
double var_x2 = fk2->second.m2 / n2;
double mean_x1 = fk->second.mean;
double mean_x2 = fk2->second.mean;
double mean_x = (n1 * mean_x1 + n2 * mean_x2) / (n1 + n2);
double q1 = (n1 - 1) * var_x1 + n1 * mean_x1 * mean_x1;
double q2 = (n2 - 1) * var_x2 + n2 * mean_x2 * mean_x2;
double qc = q1 + q2;
double sc = sqrt((qc - (n1 + n2) * mean_x * mean_x) / (n1 + n2 - 1));
if (n1 == 0) {
// fk2 is already correct
} else if (n2 == 0) {
// move fk to fk2
fk2->second.m2 = fk->second.m2;
} else {
double var_x1 = fk->second.m2 / n1;
double var_x2 = fk2->second.m2 / n2;
double mean_x1 = fk->second.mean;
double mean_x2 = fk2->second.mean;
double mean_x = (n1 * mean_x1 + n2 * mean_x2) / (n1 + n2);
fk2->second.m2 = sc * sc * (n1 + n2);
// stack overflow uses n1 - 1 and n2 - 1
double q1 = (n1) * var_x1 + n1 * mean_x1 * mean_x1;
double q2 = (n2) * var_x2 + n2 * mean_x2 * mean_x2;
double qc = q1 + q2;
// stack overflow uses n1 + n2 - 1
double sc = sqrt((qc - (n1 + n2) * mean_x * mean_x) / (n1 + n2));
fk2->second.m2 = sc * sc * (n1 + n2);
}
fk2->second.mean = (
fk2->second.mean * fk2->second.numeric_count +