假设有400.000个这样的大数据:
var data = [
 {
  "kDefinition": "ford stream, wade across", 
  "kMandarin": "shè", 
  "char": "涉", 
  "kCantonese": "sip3", 
  "kJapaneseKun": "WATARU KAKAWARU", 
 }, 
 {
  "kDefinition": "to pierce, to stab; to take", 
  "kMandarin": "jué", 
  "char": "鈌", 
  "kCantonese": "kyut3", 
  "kJapaneseKun": "SASU"
 },
 ...
]并给出与char字段匹配的较小的1000个单元子集列表,如下所示:
var list = ['早', '维', '基', '涉', '百', ..., '科', '月']  如何减少较大的数据,使其只保留obj.char在list中存在的1000个对象?
编辑:、list和data没有按照相同的顺序排序。
发布于 2018-01-11 23:56:13
您可以将list数组转换为Set,然后按照@Amadan的建议,在过滤器中使用has()方法。
var data = [{"kDefinition":"ford stream, wade across","kMandarin":"shè","char":"涉","kCantonese":"sip3","kJapaneseKun":"WATARU KAKAWARU"},{"kDefinition":"to pierce, to stab; to take","kMandarin":"jué","char":"鈌","kCantonese":"kyut3","kJapaneseKun":"SASU"}]
var list = ['早', '维', '基', '涉', '百', '科', '月']  
list = new Set(list)
const result = data.filter(({char}) => list.has(char));
console.log(result)
发布于 2018-01-12 00:03:04
对于这样大的列表,我会考虑使用旧的循环。它通常是最快的,而且几乎和任何更现代的解决方案一样容易。
var data = [
 {
  "kDefinition": "ford stream, wade across", 
  "kMandarin": "shè", 
  "char": "涉", 
  "kCantonese": "sip3", 
  "kJapaneseKun": "WATARU KAKAWARU", 
 }, 
 {
  "kDefinition": "to pierce, to stab; to take", 
  "kMandarin": "jué", 
  "char": "鈌", 
  "kCantonese": "kyut3", 
  "kJapaneseKun": "SASU"
 }
];
var list = [ '早', '维', '基', '涉', '百', '科', '月' ];
// First, make an object from the char list
var chars = {};
for( var i = 0, n = list.length;  i < n;  ++i ) {
    chars[ list[i] ] = true;
}
// Now we can match against the data
var output = [];
for( var i = 0, n = data.length;  i < n;  ++i ) {
    if( chars.hasOwnProperty( data[i].char ) ) {
        output.push( data[i] );
    }
}
console.log( JSON.stringify( output, null, 4 ) );
https://stackoverflow.com/questions/48217612
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