将JS数组中的相同元素分组，但仅在连续时答案

【问题标题】：Group same elements in JS array, but only when consecutive将JS数组中的相同元素分组，但仅在连续时
【发布时间】：2018-11-18 17:35:51
【问题描述】：

我有一组对象，代表聊天中的一系列“消息”。我想对存储在每个数组项中的相同“用户”连续创建的消息进行分组。这是我正在尝试做的简化版本：

[
    { message: "One", user: "Bob" },
    { message: "Two", user: "Bob" },
    { message: "Three", user: "Bob" },
    { message: "Hello", user: "Sam" },
    { message: "Hello", user: "Bob" },
    { message: "Hello", user: "Sam" },
    { message: "Hello", user: "Sam" }
]

应该变成：

[
    [
        { message: "One", user: "Bob"},
        { message: "Two", user: "Bob" },
        { message: "Three", user: "Bob" }
    ],
    [
        { message: "Hello", user: "Sam" }
    ],
    [
        { message: "Hello", user: "Bob" }
    ],
    [
        { message: "Hello", user: "Sam" },
        { message: "Hello", user: "Sam" }
    ]
]

有什么简单的方法可以做到这一点吗？我在任何地方都找不到答案/算法。

【问题讨论】：

标签： javascript arrays grouping

【解决方案1】：

您可以使用reduce 方法来做到这一点。

const data = [
  { message: "One", user: "Bob" },
  { message: "Two", user: "Bob" },
  { message: "Three", user: "Bob" },
  { message: "Hello", user: "Sam" },
  { message: "Hello", user: "Bob" },
  { message: "Hello", user: "Sam" },
  { message: "Hello", user: "Sam" }
];

const result = data.reduce((acc, value) => {
  // compare the current value with the last item in the collected array
  if (acc.length && acc[acc.length - 1][0].user == value.user) {
    // append the current value to it if it is matching
    acc[acc.length - 1].push(value);
  } else {
    // append the new value at the end of the collected array
    acc.push([value]);
  }

  return acc;
}, []);
  
console.log(result);

【讨论】：

【解决方案2】：

你可以这样做：

const messages = [
    { message: "One", user: "Bob" },
    { message: "Two", user: "Bob" },
    { message: "Three", user: "Bob" },
    { message: "Hello", user: "Sam" },
    { message: "Hello", user: "Bob" },
    { message: "Hello", user: "Sam" },
    { message: "Hello", user: "Sam" }
]

let currentUser;
let groupedMessages = [];

for (message of messages) {

  if (message.user !== currentUser) {
    groupedMessages.push([]);
    currentUser = message.user;
  }
  groupedMessages[groupedMessages.length - 1].push(message)
}

console.log(groupedMessages);

【讨论】：

【解决方案3】：

您可以检查最后一个元素，如果它没有相同的用户，则创建一个新数组。

var data = [{ message: "One", user: "Bob" }, { message: "Two", user: "Bob" }, { message: "Three", user: "Bob" }, { message: "Hello", user: "Sam" }, { message: "Hello", user: "Bob" }, { message: "Hello", user: "Sam" }, { message: "Hello", user: "Sam" }],
    grouped = data.reduce((r, o, i, a) => {
        if (!i || a[i - 1].user !== o.user) {
            r.push([]);
        }
        r[r.length - 1].push(o);
        return r;
    }, []);
    
console.log(grouped);

【讨论】：

【解决方案4】：

您可以只使用Array.reduce 并使用数组的index 和actual array 的参数来解决这个问题：

const data = [{ message: "One", user: "Bob" }, { message: "Two", user: "Bob" }, { message: "Three", user: "Bob" }, { message: "Hello", user: "Sam" }, { message: "Hello", user: "Bob" }, { message: "Hello", user: "Sam" }, { message: "Hello", user: "Sam" } ]

const group = d => d.reduce((r,c,i,a) => (a[i].user == (a[i-1] && a[i-1].user) 
 ? r[r.length-1].push(c) : r.push([c]), r), [])

console.log(Object.values(group(data)))

【讨论】：