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>
47 lines
1.1 KiB
JavaScript
47 lines
1.1 KiB
JavaScript
'use strict';
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// do not edit .js files directly - edit src/index.jst
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module.exports = function equal(a, b) {
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if (a === b) return true;
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if (a && b && typeof a == 'object' && typeof b == 'object') {
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if (a.constructor !== b.constructor) return false;
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var length, i, keys;
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if (Array.isArray(a)) {
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length = a.length;
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if (length != b.length) return false;
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for (i = length; i-- !== 0;)
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if (!equal(a[i], b[i])) return false;
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return true;
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}
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if (a.constructor === RegExp) return a.source === b.source && a.flags === b.flags;
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if (a.valueOf !== Object.prototype.valueOf) return a.valueOf() === b.valueOf();
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if (a.toString !== Object.prototype.toString) return a.toString() === b.toString();
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keys = Object.keys(a);
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length = keys.length;
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if (length !== Object.keys(b).length) return false;
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for (i = length; i-- !== 0;)
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if (!Object.prototype.hasOwnProperty.call(b, keys[i])) return false;
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for (i = length; i-- !== 0;) {
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var key = keys[i];
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if (!equal(a[key], b[key])) return false;
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}
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return true;
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}
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// true if both NaN, false otherwise
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return a!==a && b!==b;
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};
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