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generate chars order from given array
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// Question from | |
// https://twitter.com/RplusTW/status/1498288300934971401 | |
// inspired from @esp10mm | |
// want to get all order => [a, c, b, d] | |
genOrder([ | |
['c', 'b'], | |
['a', 'b', 'd'], | |
['a', 'c'], | |
]); | |
function genOrder(data) { | |
var relations = data.reduce((all, i) => { | |
let prevChars = []; | |
i.forEach(j => { | |
if (!all[j]) { | |
all[j] = { | |
value: j, | |
prev: [], | |
}; | |
} | |
// collect prev chars & remove dup item | |
all[j].prev = [...new Set(all[j].prev.concat(prevChars))]; | |
prevChars.push(j); | |
}); | |
return all; | |
}, {}); | |
// { | |
// "c": { "value": "c", "prev": ["a"] }, | |
// "b": { "value": "b", "prev": ["c", "a"] }, | |
// "a": { "value": "a", "prev": [] }, | |
// "d": { "value": "d", "prev": ["a", "b"] } | |
// } | |
let calcOrder = (prevArr) => { | |
if (!prevArr.length) { | |
return 0; | |
} | |
return prevArr.reduce((all, i) => { | |
return all + getOrder(i); | |
}, 1 /* base */); | |
} | |
// reduce duplicated calc, but pass order = 0 | |
let getOrder = (char) => { | |
let order = relations[char].order ?? -1; | |
if (order === -1) { | |
order = relations[char].order = calcOrder(relations[char].prev) ; | |
} | |
return order; | |
} | |
for (let i in relations) { | |
relations[i].order = calcOrder(relations[i].prev); | |
} | |
return Object.values(relations) | |
.sort((a, b) => a.order - b.order) | |
.map(i => i.value); | |
} |
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genOrder([ | |
['c', 'b'], | |
['a', 'b', 'd'], | |
['a', 'c'], | |
]); | |
function genOrder(data) { | |
// 將所有關係打散為前後兩兩一組 | |
// 再直接遞迴遞增序號 | |
const spliter = '___'; | |
const relations = data.map(i => { | |
let str = i.join(spliter); | |
let reg = new RegExp(`(?=([^${spliter}]${spliter}[^${spliter}]))`, 'g'); | |
return Array.from(str.matchAll(reg), x => x[1].split(spliter)); | |
}) | |
.flat(); | |
const chars = [...new Set(relations.flat())] | |
.reduce((all, char) => { | |
all[char] = { | |
char: char, | |
prev: [], | |
} | |
return all; | |
}, {}); | |
relations.forEach(relation => { | |
chars[relation[1]].prev = [...new Set([ | |
...chars[relation[1]].prev, relation[0] | |
])]; | |
}); | |
for (let i in chars) { | |
chars[i].order = 0; | |
chars[i].order = calcCharOrder(i); | |
} | |
let output = Object.values(chars).sort(sortCharOrder); | |
// check | |
{ | |
if (output.findIndex(i => i.order === 0) === -1) { | |
console.warn('contradiction for 1st item', output); | |
} | |
let checking = output.reduce((all, i) => { | |
if (!all[i.order]) { | |
all[i.order] = { | |
order: i.order, | |
chars: [], | |
} | |
} | |
all[i.order].chars.push(i.char); | |
return all; | |
}, []) | |
.filter(i => i.chars.length > 1); | |
if (checking.length) { | |
console.warn('dup. order', checking); | |
} | |
} | |
return output.map(i => i.char); | |
function calcCharOrder(char) { | |
let prevArr = chars[char].prev; | |
if (!prevArr.length) { return 0; } | |
return prevArr.reduce((all, i) => { | |
let order = chars[i].order ?? calcCharOrder(i); | |
// save order to reduce calculation | |
if (isNaN(chars[i].order)) { | |
chars[i].order = order; | |
} | |
return all + order; | |
}, 1 /* upup! */ ) | |
} | |
function sortCharOrder(a, b) { | |
return a.order - b.order; | |
} | |
} |
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There is still an issue of incomplete condition.
[aa,bb,dd] [aa,cc] [cc,dd]
=> can't make sure order ofbb
&cc
.