JS：从多个数组元素中提取唯一字符串答案

【问题标题】：JS: Extract unique strings from multiple array elementsJS：从多个数组元素中提取唯一字符串
【发布时间】：2017-12-12 08:37:27
【问题描述】：

我的对象数组的数据结构如下所示：

const data = [ 
  { 
    _id: 'Dn59y87PGhkJXpaiZ',
    title: 'Sample Article',
    slug: 'sample-article',
    created: 1503160075
  },
  { 
    _id: 'ujJCBC2avK8QkR86t',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ'
    reference: [ '9Z7k2wAbXNXY2JWuE' ],
    timestamp: 1513054017
  },
  { 
    _id: 'KRhcfZSWFAawfxAsj',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '8vtFExXqEF4Hghx2b' ],
    timestamp: 1512864671
  }
]

现在我需要获取所有唯一的参考字符串，我试图这样做：

const result = data.filter(doc => doc.reference).map(doc => doc.reference)

但这给了我结果......

[ [ '9Z7k2wAbXNXY2JWuE' ], [ '8vtFExXqEF4Hghx2b' ] ]

...我期待类似的东西

[ '9Z7k2wAbXNXY2JWuE', '8vtFExXqEF4Hghx2b' ]

我还需要消除重复项（此示例数据中未显示）。

【问题讨论】：

标签： javascript ecmascript-6 lodash

【解决方案1】：

使用array.prototype.reduce连接引用数组，Set避免重复，spread operator转换Set em> 返回数组：

var data = [ 
  { 
    _id: 'Dn59y87PGhkJXpaiZ',
    title: 'Sample Article',
    slug: 'sample-article',
    created: 1503160075
  },
  { 
    _id: 'ujJCBC2avK8QkR86t',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '9Z7k2wAbXNXY2JWuE', '9Z7k2wAbXNXY2JWuE' ],
    timestamp: 1513054017
  },
  { 
    _id: 'KRhcfZSWFAawfxAsj',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '8vtFExXqEF4Hghx2b', '8vtFExXqEF4HgDx2b' ],
    timestamp: 1512864671
  }
];

var result = [ ...new Set(data.reduce((m, doc) => m.concat(doc.reference || []), []))];

console.log(result);

【讨论】：

【解决方案2】：

您需要使用map 提取引用然后展平数组：

const data = [ 
  { 
    _id: 'Dn59y87PGhkJXpaiZ',
    title: 'Sample Article',
    slug: 'sample-article',
    created: 1503160075
  },
  { 
    _id: 'ujJCBC2avK8QkR86t',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '9Z7k2wAbXNXY2JWuE' ],
    timestamp: 1513054017
  },
  { 
    _id: 'KRhcfZSWFAawfxAsj',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '8vtFExXqEF4Hghx2b' ],
    timestamp: 1512864671
  }
]

console.log(...[].concat(...data.map(d => d.reference || [])));

【讨论】：

【解决方案3】：

就这样做

let result = data.filter(doc => doc.reference).map(doc => doc.reference)[0]
result = [ ...result ]

展开运算符会将其展开为一个数组。

【讨论】：

result = [["9Z7k2wAbXNXY2JWuE"], ["8vtFExXqEF4Hghx2b"]]
我已经更新了答案，请看一下结果过滤。当您映射对象时，只需在每次数组执行map 时获取第一个索引

【解决方案4】：

使用reduce 代替map

data.filter(doc => doc.reference).reduce( (a,b ) => a.concat( b.reference ),  [] );

演示

var data = [ 
  { 
    _id: 'Dn59y87PGhkJXpaiZ',
    title: 'Sample Article',
    slug: 'sample-article',
    created: 1503160075
  },
  { 
    _id: 'ujJCBC2avK8QkR86t',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '9Z7k2wAbXNXY2JWuE' ],
    timestamp: 1513054017
  },
  { 
    _id: 'KRhcfZSWFAawfxAsj',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '8vtFExXqEF4Hghx2b' ],
    timestamp: 1512864671
  }
];
var output = data.filter(doc => doc.reference).reduce( (a,b ) => a.concat(b.reference),  [] );

console.log( output );

或者如果引用数组中只有一项

data.filter( doc => doc.reference ).map( a => a[0] );

编辑

现在我需要获取我试图获取的所有唯一引用字符串这样做：

如果字符串必须是唯一的，则使用此版本

data.reduce( function( a, b ) {
   if ( b.reference ) 
   {
       a = a.concat(b.filter( i => a.indexOf(i) == -1 ));
   }
   return a;
} , [] );

带箭头功能

data.reduce( ( a, b ) => ( b.reference ? a.concat( b.reference.filter( i => a.indexOf( i ) == -1 ) ) : a) , [] );

演示

var data = [ 
  { 
    _id: 'Dn59y87PGhkJXpaiZ',
    title: 'Sample Article',
    slug: 'sample-article',
    created: 1503160075
  },
  { 
    _id: 'ujJCBC2avK8QkR86t',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '9Z7k2wAbXNXY2JWuE' ],
    timestamp: 1513054017
  },
  { 
    _id: 'KRhcfZSWFAawfxAsj',
    content: 'Lorem ipsum',
    parent: 'Dn59y87PGhkJXpaiZ',
    reference: [ '8vtFExXqEF4Hghx2b' ],
    timestamp: 1512864671
  }
];
var output = data.reduce( ( a, b ) => ( b.reference ? a.concat( b.reference.filter( i => a.indexOf( i ) == -1 ) ) : a) , [] );
console.log( output );

【讨论】：

如果 OP 需要正确映射，Reduce 绝对是矫枉过正。而且更难阅读和理解。
@dfsq 更难阅读和理解同意，如果你以前没有用过的话。