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>
45 lines
1.4 KiB
JavaScript
45 lines
1.4 KiB
JavaScript
/*
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MIT License http://www.opensource.org/licenses/mit-license.php
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Author Tobias Koppers @sokra
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*/
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"use strict";
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/**
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* Compare two arrays or strings by performing strict equality check for each value.
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* @template T [T=any]
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* @param {ArrayLike<T>} a Array of values to be compared
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* @param {ArrayLike<T>} b Array of values to be compared
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* @returns {boolean} returns true if all the elements of passed arrays are strictly equal.
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*/
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exports.equals = (a, b) => {
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if (a.length !== b.length) return false;
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for (let i = 0; i < a.length; i++) {
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if (a[i] !== b[i]) return false;
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}
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return true;
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};
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/**
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* Partition an array by calling a predicate function on each value.
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* @template T [T=any]
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* @param {Array<T>} arr Array of values to be partitioned
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* @param {(value: T) => boolean} fn Partition function which partitions based on truthiness of result.
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* @returns {[Array<T>, Array<T>]} returns the values of `arr` partitioned into two new arrays based on fn predicate.
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*/
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exports.groupBy = (arr = [], fn) => {
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return arr.reduce(
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/**
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* @param {[Array<T>, Array<T>]} groups An accumulator storing already partitioned values returned from previous call.
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* @param {T} value The value of the current element
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* @returns {[Array<T>, Array<T>]} returns an array of partitioned groups accumulator resulting from calling a predicate on the current value.
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*/
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(groups, value) => {
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groups[fn(value) ? 0 : 1].push(value);
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return groups;
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},
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[[], []]
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);
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};
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