Edward Betts
ea4980a5d7
Adjust European short trip heuristic from >3 days to >1 day to correctly detect when user has returned home from European trips. This fixes the April 29-30, 2023 case where the location incorrectly showed "Sankt Georg, Hamburg" instead of "Bristol" when the user was free (no events scheduled) after the foss-north trip ended on April 27. The previous logic required more than 3 days to pass before assuming return home from European countries, but for short European trips by rail/ferry, users typically return within 1-2 days. 🤖 Generated with [Claude Code](https://claude.ai/code) Co-Authored-By: Claude <noreply@anthropic.com>
63 lines
1.2 KiB
JavaScript
63 lines
1.2 KiB
JavaScript
'use strict';
|
|
module.exports = balanced;
|
|
function balanced(a, b, str) {
|
|
if (a instanceof RegExp) a = maybeMatch(a, str);
|
|
if (b instanceof RegExp) b = maybeMatch(b, str);
|
|
|
|
var r = range(a, b, str);
|
|
|
|
return r && {
|
|
start: r[0],
|
|
end: r[1],
|
|
pre: str.slice(0, r[0]),
|
|
body: str.slice(r[0] + a.length, r[1]),
|
|
post: str.slice(r[1] + b.length)
|
|
};
|
|
}
|
|
|
|
function maybeMatch(reg, str) {
|
|
var m = str.match(reg);
|
|
return m ? m[0] : null;
|
|
}
|
|
|
|
balanced.range = range;
|
|
function range(a, b, str) {
|
|
var begs, beg, left, right, result;
|
|
var ai = str.indexOf(a);
|
|
var bi = str.indexOf(b, ai + 1);
|
|
var i = ai;
|
|
|
|
if (ai >= 0 && bi > 0) {
|
|
if(a===b) {
|
|
return [ai, bi];
|
|
}
|
|
begs = [];
|
|
left = str.length;
|
|
|
|
while (i >= 0 && !result) {
|
|
if (i == ai) {
|
|
begs.push(i);
|
|
ai = str.indexOf(a, i + 1);
|
|
} else if (begs.length == 1) {
|
|
result = [ begs.pop(), bi ];
|
|
} else {
|
|
beg = begs.pop();
|
|
if (beg < left) {
|
|
left = beg;
|
|
right = bi;
|
|
}
|
|
|
|
bi = str.indexOf(b, i + 1);
|
|
}
|
|
|
|
i = ai < bi && ai >= 0 ? ai : bi;
|
|
}
|
|
|
|
if (begs.length) {
|
|
result = [ left, right ];
|
|
}
|
|
}
|
|
|
|
return result;
|
|
}
|