使用 lodash 查找对象数组上的所有重复项

Question

我有以下对象数组：

[
 { ip: 1, name: 'examplel' },
 { ip: 1, name: 'examplel' },
 { ip: 202.164.171.184, name: 'example2' },
 { ip: 202.164.171.184, name: 'example2' },
 { ip: 202.164.171.184, name: 'example3' },
 { ip: 127.0.0.1, name: 'example4' },
 { ip: 127.0.0.1, name: 'example5' }
]

如果它们具有相同的ip但不相同的name ，我想为它们添加颜色，所以像这样。

[
 { ip: 1, name: 'examplel', color: '' },
 { ip: 1, name: 'examplel', color: '' },
 { ip: 202.164.171.184, name: 'example2', color: 'red' },
 { ip: 202.164.171.184, name: 'example2', color: 'red' },
 { ip: 202.164.171.184, name: 'example3', color: 'red' },
 { ip: 127.0.0.1, name: 'example4', color: 'black' },
 { ip: 127.0.0.1, name: 'example5', color: 'black' }
]

您如何使用 lodash 实现这一目标？ 还是香草？

---

编辑：我试图编码它，但它产生不同的输出。

[
 { ip: 1, name: 'examplel', color: 'red' },
 { ip: 1, name: 'examplel', color: 'red' },
 { ip: 202.164.171.184, name: 'example2', color: 'red' },
 { ip: 202.164.171.184, name: 'example2', color: 'red' },
 { ip: 202.164.171.184, name: 'example3', color: 'red' },
 { ip: 127.0.0.1, name: 'example4', color: 'black' },
 { ip: 127.0.0.1, name: 'example5', color: 'black' }
]

这是我的代码。

      let list = _.groupBy(data, 'ip')
      const colorList = ['pink', 'blue', 'pink', 'red']
      let logsList = []

      _.keys(list).forEach(key => {
        const color = colorList[Math.floor(Math.random() * colorList.length)]

        if (Array.isArray(list[key])) {
          list[key].forEach(data => {
            data.color = color
            logsList.push(data)
          })
        }
      })

Answer 1

在设置颜色之前，您可以尝试使用every来检查 list[key] 的每个元素中的 name 和 ip 是否相同：

https://lodash.com/docs/4.17.15#every

 let data = [ { ip: '1', name: 'examplel' }, { ip: '1', name: 'examplel' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example3' }, { ip: '127.0.0.1', name: 'example4' }, { ip: '127.0.0.1', name: 'example5' } ]; let list = _.groupBy(data, 'ip') const colorList = ['pink', 'blue', 'pink', 'red'] let logsList = [] console.log('list: ' + JSON.stringify(list)); _.keys(list).forEach(key => { if (Array.isArray(list[key])) { let color = ''; if(!_.every(list[key], list[key][0])) { color = colorList[Math.floor(Math.random() * colorList.length)] } list[key].forEach(data => { data.color = color logsList.push(data) }) } }) console.log('logsList: ' + JSON.stringify(logsList));

 <script src="https://cdn.jsdelivr.net/npm/lodash@4.17.15/lodash.min.js"></script>

Answer 2

您的代码中的问题

一个问题可能是生成随机颜色：您可以两次使用颜色

固定代码：

 let data = [ // Example data { ip: 1, name: 'examplel' }, { ip: 1, name: 'examplel' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example3' }, { ip: '127.0.0.1', name: 'example4' }, { ip: '127.0.0.1', name: 'example5' } ]; let list = _.groupBy(data, 'ip') const colorList = ['pink', 'blue', 'red'] // Color should not be in this list twice! let logsList = [] _.keys(list).forEach(key => { if(!colorList.length) throw new Error('Not enough colors'); let color = colorList[Math.floor(Math.random() * colorList.length)] // Chose color colorList.splice(colorList.indexOf(color), 1); // Remove chosen color from the list // So, we cannot chos one color twice // Removed not-needed IF list[key].forEach(data => { data.color = color logsList.push(data) }) }) console.log(logsList)

 <!-- Get lodash --> <script src="https://cdn.jsdelivr.net/npm/lodash@4.17.15/lodash.min.js"></script>

另一个工作解决方案

不使用 Lodash，在您的代码未添加到问题时编写。 悲伤，就删除。

 let datas = [ { ip: 1, name: 'examplel' }, { ip: 1, name: 'examplel' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example2' }, { ip: '202.164.171.184', name: 'example3' }, { ip: '127.0.0.1', name: 'example4' }, { ip: '127.0.0.1', name: 'example5' } ]; let colors = ['green', 'red', 'blue', 'purple', /* And some more... */]; // Colors for diffrent IP's let currentColor = 0, // Which color we are currently using - index in colors[] pastIp = null; // Past Ip let colored = datas.sort((a, b) => ('' + a.ip).localeCompare(b.ip)) // Sort IP's, // Same adresses will be grouped .map(i => { if(pastIp && i.ip != pastIp) currentColor++; // We have started colorig another IP (as example, past was 1.1.1.1, this is 2.2.2.2) // Else, we are looping same IP's (as example, past was 1.1.1.1, this is also 1.1.1.1) pastIp = i.ip; // Current will be past return {color: colors[currentColor], ...i}; // Add color property }) console.log(colored); // Do somethig with result