[Lodash] 明日方舟公开招募计算器的简单实现

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Head Pic：【明日方舟】「series」/「Alcxome」的插画 [pixiv]

差不多一周前还是经不住劝诱被基友拉入了坑，对刚入坑还不熟悉角色池的人来说，公开招募计算器是非常有用且必要的工具

之前我一直用的是 graueneko.github.io，做的非常好，而后面我懒癌发作，因此想能不能将这个功能也整合进我自己写的QQ机器人里，利用现成的 OCR 做到：直接识别截图获取词条->计算可能的词条组合及其包含的干员->输出

但一开始的词条排列组合问题就让我感到头疼，如果要自己用动态规划来降低计算量和复杂度又感觉很麻烦，因为当时脑子瓦特了想不到什么简便的储存词条组合中间结果的方法（后来才想起可以排序后用 hash 或者直接拼接作为 key 啊真蠢）

在神的指引下我想起了还有 Lodash，并且发现其很适合写这类需要对数组进行大量操作的算法，于是决定写一个来练习一下

1. 数据预处理#

还是要感谢 graueneko 等人的工作，已经将干员数据整理成了 json 格式，可以方便的直接拿来使用

那么首先需要明确，要做的是一个公开招募计算器，得到最终结果的算法大致如下

计算指定词条的所有排列组合方案
对每种方案，找到满足这些词条的干员，并对方案进行评分
按评分对所有有效方案进行排序

实现的重点是第2步，而干员数据的结构是干员->词条，每当我们想寻找满足某些词条的干员，我们都必须把列表遍历一遍，这是非常麻烦的

因此我们可以换一种储存方式，使结构变为词条->干员，这样比如我们想寻找同时拥有词条A和词条B的干员，只需要求A和B的交集就行了

以下是示例：

1
import { get } from "axios";
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import _ from "lodash";
3

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const akhr = "https://graueneko.github.io/akhr.json";
5

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async function pullData() {
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  let json = await get(akhr).then((r) => r.data);
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  //将干员按照稀有度从高到低排序，后续输出的时候就不用再排序了
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  json.sort((a, b) => b.level - a.level);
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  let characters = []; //角色列表
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  let data = {}; //词条数据
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  let charTagSum = 0;
15

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  for (let character of json) {
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    if (character.hidden) continue;
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    let { level, name, sex, tags, type } = character;
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    //将性别和干员种类也加入词条中
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    tags.push(`${sex}性干员`);
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    tags.push(`${type}干员`);
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    //将干员放入 characters 数组，后续直接用其在数组中的 index 指代
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    let p =
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      characters.push({
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        n: name, //名字
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        r: level, //稀有度
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      }) - 1;
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    //储存：词条->干员
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    for (let tag of tags) {
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      if (!data[tag]) data[tag] = [];
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      data[tag].push(p);
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    }
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    charTagSum += tags.length;
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  }
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  let tagCount = _.size(data);
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  return {
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    characters,
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    data,
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    //这是后续计算评分用的数据（来自 graueneko）
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    avgCharTag: charTagSum / tagCount,
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  };
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}

然后就能得到这样子的数据结构

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{
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  "characters": [
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    { "n": "推进之王", "r": 6 },
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    { "n": "闪灵", "r": 6 },
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    { "n": "夜莺", "r": 6 },
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    { "n": "伊芙利特", "r": 6 },
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    { "n": "能天使", "r": 6 },
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    { "n": "银灰", "r": 6 },
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    { "n": "星熊", "r": 6 },
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    { "n": "塞雷娅", "r": 6 },
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    { "n": "德克萨斯", "r": 5 },
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    { "n": "凛冬", "r": 5 },
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    { "n": "白面鸮", "r": 5 },
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    { "n": "赫默", "r": 5 },
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    { "n": "华法琳", "r": 5 },
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    { "n": "红", "r": 5 },
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    { "n": "狮蝎", "r": 5 },
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    { "n": "崖心", "r": 5 },
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    { "n": "食铁兽", "r": 5 },
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    { "n": "普罗旺斯", "r": 5 },
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    { "n": "蓝毒", "r": 5 },
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    { "n": "守林人", "r": 5 },
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    { "n": "陨星", "r": 5 },
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    { "n": "白金", "r": 5 },
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    { "n": "初雪", "r": 5 },
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    { "n": "真理", "r": 5 },
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    { "n": "梅尔", "r": 5 },
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    { "n": "幽灵鲨", "r": 5 },
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    { "n": "因陀罗", "r": 5 },
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    { "n": "临光", "r": 5 },
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    { "n": "雷蛇", "r": 5 },
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    { "n": "火神", "r": 5 },
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    { "n": "可颂", "r": 5 },
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    { "n": "清道夫", "r": 4 },
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    { "n": "红豆", "r": 4 },
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    { "n": "末药", "r": 4 },
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    { "n": "调香师", "r": 4 },
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    { "n": "夜烟", "r": 4 },
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    { "n": "远山", "r": 4 },
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    { "n": "砾", "r": 4 },
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    { "n": "暗索", "r": 4 },
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    { "n": "阿消", "r": 4 },
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    { "n": "白雪", "r": 4 },
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    { "n": "流星", "r": 4 },
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    { "n": "杰西卡", "r": 4 },
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    { "n": "地灵", "r": 4 },
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    { "n": "杜宾", "r": 4 },
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    { "n": "艾丝黛尔", "r": 4 },
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    { "n": "慕斯", "r": 4 },
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    { "n": "霜叶", "r": 4 },
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    { "n": "缠丸", "r": 4 },
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    { "n": "蛇屠箱", "r": 4 },
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    { "n": "古米", "r": 4 },
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    { "n": "角峰", "r": 4 },
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    { "n": "芬", "r": 3 },
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    { "n": "香草", "r": 3 },
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    { "n": "翎羽", "r": 3 },
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    { "n": "芙蓉", "r": 3 },
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    { "n": "安赛尔", "r": 3 },
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    { "n": "炎熔", "r": 3 },
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    { "n": "史都华德", "r": 3 },
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    { "n": "克洛丝", "r": 3 },
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    { "n": "安德切尔", "r": 3 },
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    { "n": "梓兰", "r": 3 },
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    { "n": "玫兰莎", "r": 3 },
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    { "n": "米格鲁", "r": 3 },
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    { "n": "夜刀", "r": 2 },
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    { "n": "杜林", "r": 2 },
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    { "n": "12F", "r": 2 },
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    { "n": "巡林者", "r": 2 },
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    { "n": "黑角", "r": 2 },
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    { "n": "Lancet-2", "r": 1 },
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    { "n": "Castle-3", "r": 1 }
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  ],
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  "data": {
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    "近战位": [
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      0, 5, 6, 7, 8, 9, 13, 14, 15, 16, 25, 26, 27, 28, 29, 30, 31, 32, 37, 38,
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      39, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 62, 63, 64, 68, 70
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    ],
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    "费用回复": [0, 8, 9, 31, 32, 52, 53, 54],
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    "输出": [
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      0, 4, 5, 6, 14, 15, 17, 18, 19, 21, 23, 26, 28, 29, 31, 32, 35, 41, 42,
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      44, 46, 47, 48, 54, 58, 59, 60, 62
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    ],
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    "高级资深干员": [0, 1, 2, 3, 4, 5, 6, 7],
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    "女性干员": [
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      0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,
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      22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39,
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      40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 57, 59, 61,
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      62, 63, 64, 65, 69
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    ],
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    "先锋干员": [0, 8, 9, 31, 32, 52, 53, 54, 64],
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    "远程位": [
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      1, 2, 3, 4, 10, 11, 12, 17, 18, 19, 20, 21, 22, 23, 24, 33, 34, 35, 36,
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      40, 41, 42, 43, 55, 56, 57, 58, 59, 60, 61, 65, 66, 67, 69
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    ],
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    "支援": [1, 2, 5, 7, 9, 10, 12, 44, 70],
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    "治疗": [1, 2, 7, 10, 11, 12, 27, 33, 34, 50, 55, 56, 69],
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    "医疗干员": [1, 2, 10, 11, 12, 33, 34, 55, 56, 69],
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    "群攻": [3, 20, 25, 36, 40, 45, 57],
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    "削弱": [3, 20, 22, 35, 41],
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    "术师干员": [3, 35, 36, 57, 58, 65, 66],
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    "狙击干员": [4, 17, 18, 19, 20, 21, 40, 41, 42, 59, 60, 67],
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    "男性干员": [5, 51, 56, 58, 60, 66, 67, 68, 70],
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    "近卫干员": [5, 25, 26, 44, 45, 46, 47, 48, 62, 70],
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    "防护": [6, 7, 27, 28, 29, 30, 37, 49, 50, 51, 63],
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    "重装干员": [6, 7, 27, 28, 29, 30, 49, 50, 51, 63, 68],
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    "控场": [8, 13, 24],
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    "资深干员": [
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      8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26,
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      27, 28, 29, 30
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    ],
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    "快速复活": [13, 37],
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    "特种干员": [13, 14, 15, 16, 37, 38, 39],
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    "生存": [14, 25, 26, 29, 42, 45, 48, 62],
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    "位移": [15, 16, 30, 38, 39],
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    "减速": [16, 23, 40, 43, 47, 61],
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    "爆发": [19],
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    "辅助干员": [22, 23, 24, 43, 61],
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    "召唤": [24],
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    "新手": [64, 65, 66, 67, 68]
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  },
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  "avgCharTag": 12.275862068965518
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}

一下子就变得很直观了呢

2. 计算#

需要用到 lodash.combinations

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import "lodash.combinations";
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import _ from "lodash";
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let AKDATA; //假设这是上面得到的处理过后的数据
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function getChar(i) {
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  return AKDATA.characters[i];
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}
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//tags 是词条数组，例如 ['治疗','新手','削弱']
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function getCombinations(tags) {
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  //_.combinations 得到所有排列组合方案
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  let combs = _.flatMap([1, 2, 3], (v) => _.combinations(tags, v));
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  let result = [];
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  for (let comb of combs) {
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    //获取需要取交集的词条
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    let need = [];
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    for (let tag of comb) need.push(AKDATA.data[tag]);
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    //_.intersection 取交集
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    let chars = _.intersection(...need);
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    //如果词条中没有“高级资深干员”则无法招募6星干员，用 _.remove 按指定条件移除
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    if (!comb.includes("高级资深干员"))
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      _.remove(chars, (i) => AKDATA.characters[i].r == 6);
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    if (chars.length == 0) continue;
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    //计算方案评分，在 graueneko 的基础上有所改进
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    //因为多数情况下我们只招募3星及以上干员，因此在评分时可以忽略1星2星
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    //_.sumBy 按指定方法求和，_.filter 按指定条件过滤
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    let scoreChars = _.filter(chars, (i) => getChar(i).r >= 3);
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    if (scoreChars.length == 0) scoreChars = chars;
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    let score =
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      _.sumBy(scoreChars, (i) => getChar(i).r) / scoreChars.length -
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      comb.length / 10 -
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      scoreChars.length / AKDATA.avgCharTag;
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    //得到保底稀有度用于排序
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    let minI = _.minBy(scoreChars, (i) => getChar(i).r);
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    result.push({
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      comb,
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      chars,
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      min: AKDATA.characters[minI].r,
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      score,
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    });
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  }
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  //按保底稀有度和评分进行排序
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  result.sort((a, b) => (a.min == b.min ? b.score - a.score : b.min - a.min));
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  return result;
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}

3. OCR 以及结果生成#

可以使用 ocr.space 的免费 API，每日 500 次够用了，但是效果不太好
可以使用百度OCR，高精度识别每日也免费 500 次，效果更好

至于展示结果的方法，我最开始用的是直接纯文字输出，但是发现这样很不直观甚至还会眼花（

后来我决定仿照 graueneko 的排版生成图片，考虑到 javascript 特色……我去花了半小时现学了下 canvas 就直接用上了，效果大概这样

最终成果可以参考 akhr.js 和 akhr.draw.js

[Lodash] 明日方舟公开招募计算器的简单实现

1. 数据预处理#

2. 计算#

3. OCR 以及结果生成#

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