比较 2 个 JSON 对象并找到最佳时间 JavaScript答案

【问题标题】：Compare 2 JSON objects and find best time JavaScript比较 2 个 JSON 对象并找到最佳时间 JavaScript
【发布时间】：2020-10-17 21:21:38
【问题描述】：

我试图弄清楚如何根据 JSON 对象中的用户键从 2 个 JSON 对象中找到最佳时间，并构建正确的数组以显示具有最佳结果的块。

第一个 JSON 对象：

[
{
        "user1": "00:31.889",
        "user2": "00:32.739",
        "user3": "01:00.515",
        "user4": "00:28.336",
        "user5": "00:35.745"
    },
    {
        "user1": "00:26.470",
        "user2": "00:30.063",
        "user3": "00:28.696",
        "user4": "00:30.248",
        "user5": "00:35.123"
    },
    {
        "user1": "00:26.956",
        "user2": "00:33.588",
        "user3": "00:30.021",
        "user4": "00:29.154",
        "user5": "00:38.492"
    },
    {
        "user1": "00:27.190",
        "user2": "00:32.307",
        "user3": "00:30.467",
        "user4": "00:30.189",
        "user5": "00:39.669"
    },
    {
        "user1": "00:27.368",
        "user2": "00:28.124",
        "user3": "00:29.960",
        "user4": "00:29.649",
        "user5": "00:42.450"
    }
]

第二个对象：

[
    {
        "key": "user1",
        "label": "Alex"
    },
    {
        "key": "user2",
        "label": "Jane"
    },
    {
        "key": "user3",
        "label": "Frank"
    },
    {
        "key": "user4",
        "label": "Merlyn"
    },
    {
        "key": "user5",
        "label": "Josh"
    }
]

返回应该是这样的：[{name: 'Alex', value: 233433}]

尝试了许多选项，包括 Object.key，但仍然找不到解决方案。

【问题讨论】：

标签： javascript arrays json object

【解决方案1】：

了解我的善良

const results = [
    {
        "user1": "00:31.889",
        "user2": "00:32.739",
        "user3": "01:00.515",
        "user4": "00:28.336",
        "user5": "00:35.745"
    },
    {
        "user1": "00:26.470",
        "user2": "00:30.063",
        "user3": "00:28.696",
        "user4": "00:30.248",
        "user5": "00:35.123"
    },
    {
        "user1": "00:26.956",
        "user2": "00:33.588",
        "user3": "00:30.021",
        "user4": "00:29.154",
        "user5": "00:38.492"
    },
    {
        "user1": "00:27.190",
        "user2": "00:32.307",
        "user3": "00:30.467",
        "user4": "00:30.189",
        "user5": "00:39.669"
    },
    {
        "user1": "00:27.368",
        "user2": "00:28.124",
        "user3": "00:29.960",
        "user4": "00:29.649",
        "user5": "00:42.450"
    }
]

const users = [
    {
        "key": "user1",
        "label": "Alex"
    },
    {
        "key": "user2",
        "label": "Jane"
    },
    {
        "key": "user3",
        "label": "Frank"
    },
    {
        "key": "user4",
        "label": "Merlyn"
    },
    {
        "key": "user5",
        "label": "Josh"
    }
]

const stringToObject = (str) => {
  const m = +str.split(":")[0];
  const s = +str.split(":")[1].split(".")[0];
  const ms = +str.split(":")[1].split(".")[1];
  return {m, s, ms}
}

const stringToMs = (str) => {
  const {m, s, ms} = stringToObject(str)
  return( m * 60 * 1000) + (s * 1000) + ms;
}

const getBestResults = () =>{
  const output = []
  for(let user of users){
     let min = Infinity;
     for(let result of results){
        const time = stringToMs(result[user["key"]])
        min = Math.min(time, min)
     }
     output.push({name: user["label"], value: min})
  }
  return output;
}

console.log(getBestResults())

solution JsBin

这个 O(n^) 但可能你不在乎。

【讨论】：

【解决方案2】：

一步一步的解决方案

我认为你可以用两行 Javascript 来做到这一点。

观察你只需要知道整个时间列表中最低的时间，然后给出那个人的身份。

第1行.查找最短时间的userId和时间

将每个 5 元素的对象转换成一个数组，这样前 5 次就存储为一个数组：[ ["user 1", "00:31.889"], ["user 2", ...] ] ，接下来的 5 个在另一个数组中，等等。一个好的方法是 .map 使用函数 Object.entries 的原始大数组。
将您将拥有的 5 个不同的数组展平为一个长数组。这将包含用户 1 的 5 次、用户 2 的 5 次等。Javascript .flat() 函数将为您执行此操作。
获取时间最短的 userId-and-time 对。由于时间都是包含相同格式的数值的所有字符串，带有前导零，因此您可以按字母顺序对它们进行排序并获取结果列表的元素 #0。一种方便的方法是使用 Javascript .sort() 函数。您想告诉排序程序查看每个微型数组的第 [1] 个元素，即时间，而不是作为用户 ID 的第 [0] 个元素（例如“用户 1”）。

该步骤序列的代码是：

times.map(Object.entries).flat().sort((a,b)=>a[1].localeCompare(b[1]))[0]

第 2 行。获取获胜用户的用户名

从用户列表中，您要选择与获胜用户具有相同 ID 的用户。您可以为此使用 Javascript .filter 函数。获胜用户的 ID 是 win[0]，即 id-time 对中两个项目中的第一个。获取过滤函数来测试每个用户的ID。
Javascript 会给你一个列表，里面有一个项目。你想要第一个项目，即 [0]。
显示该人的姓名 (.label) 和获胜时间 (winning[1])。

代码如下：

users.filter(user=>user.key===winning[0]) [0] .label,winning[1]

我没有尝试将分钟和秒转换为秒，因为我相信您知道如何做到这一点，并且您的示例正确答案似乎与上面的数据不匹配。大概您只是显示所需信息的类型，而不是指定转换。

下面的代码应该可以满足您的需求。如果有什么需要解释的，请告诉我。

const times = [{
    "user1": "00:31.889",
    "user2": "00:32.739",
    "user3": "01:00.515",
    "user4": "00:28.336",
    "user5": "00:35.745"
  },
  {
    "user1": "00:26.470",
    "user2": "00:30.063",
    "user3": "00:28.696",
    "user4": "00:30.248",
    "user5": "00:35.123"
  },
  {
    "user1": "00:26.956",
    "user2": "00:33.588",
    "user3": "00:30.021",
    "user4": "00:29.154",
    "user5": "00:38.492"
  },
  {
    "user1": "00:27.190",
    "user2": "00:32.307",
    "user3": "00:30.467",
    "user4": "00:30.189",
    "user5": "00:39.669"
  },
  {
    "user1": "00:27.368",
    "user2": "00:28.124",
    "user3": "00:29.960",
    "user4": "00:29.649",
    "user5": "00:42.450"
  }
];

const users = [{
    "key": "user1",
    "label": "Alex"
  },
  {
    "key": "user2",
    "label": "Jane"
  },
  {
    "key": "user3",
    "label": "Frank"
  },
  {
    "key": "user4",
    "label": "Merlyn"
  },
  {
    "key": "user5",
    "label": "Josh"
  }
];

// Line 1. Get the lowest time across all users and all attempts
const winning = times.map(Object.entries).flat().sort((a,b)=>a[1].localeCompare(b[1]))[0]

// Line 2. Display their identity and time
console.log(users.filter(user=>user.key===winning[0])[0].label,winning[1])

【讨论】：

感谢您的精彩解释，现在更清楚了。如何为每个用户输出最佳结果？我试图删除获胜者末尾的索引以输出完整的用户列表，但一直不被拒绝。

【解决方案3】：

如果您想为每个用户获得最好的分数，您需要不同的策略

我之前的回答是基于您想要整个比赛的获胜者的时间，而不是每个跑步者的最佳时间。

我的解决方案是按时间顺序对所有 userIds-and-times 进行排序。那么在该数组中，“user1”的最佳时间是“user1”在该有序列表中第一次出现。

const times = [{
    "user1": "00:31.889",
    "user2": "00:32.739",
    "user3": "01:00.515",
    "user4": "00:28.336",
    "user5": "00:35.745"
  },
  {
    "user1": "00:26.470",
    "user2": "00:30.063",
    "user3": "00:28.696",
    "user4": "00:30.248",
    "user5": "00:35.123"
  },
  {
    "user1": "00:26.956",
    "user2": "00:33.588",
    "user3": "00:30.021",
    "user4": "00:29.154",
    "user5": "00:38.492"
  },
  {
    "user1": "00:27.190",
    "user2": "00:32.307",
    "user3": "00:30.467",
    "user4": "00:30.189",
    "user5": "00:39.669"
  },
  {
    "user1": "00:27.368",
    "user2": "00:28.124",
    "user3": "00:29.960",
    "user4": "00:29.649",
    "user5": "00:42.450"
  }
];

const users = [{
    "key": "user1",
    "label": "Alex"
  },
  {
    "key": "user2",
    "label": "Jane"
  },
  {
    "key": "user3",
    "label": "Frank"
  },
  {
    "key": "user4",
    "label": "Merlyn"
  },
  {
    "key": "user5",
    "label": "Josh"
  }
];

// Line 1. Sort the user-and-time data into order of time, shortest first.

const orderedTimes = times.map(Object.entries).flat().sort((a,b)=>a[1].localeCompare(b[1]))

// Line 2. Loop over each user, using "users.map", and for each user, filter ONLY the user-and-time elements that feature THAT user. Of that list of user-and-time elements, only take the first one, i.e. [0]. That will be that user's best time.

const usersKeyAndBestTime = users.map(user=>orderedTimes.filter(userKeyAndTime=>userKeyAndTime[0]===user.key)[0]
);

// Line 3. Output the result, replacing the user key ("user1") with their actual name.

console.log(
    usersKeyAndBestTime.map(
        keyAndBestTime=>[
            users.filter(
                user=>user.key===keyAndBestTime[0]
             )[0].label,
             keyAndBestTime[1]
         ]
    )
)

【讨论】：