Question

我有一组对象， baseAsset键和“体积”是每个对象的一部分   但是每个对象的音量都不同。

我想匹配baseAsset键并返回带有   最高音量值。   效率和速度很重要，因为数组有3000多个对象

$ curl http://my.site.net/ -v -I --http1.0 --Header 'Host:'
*   Trying 100.101.102.103...
* TCP_NODELAY set
* Connected to my.site.net (100.101.102.103) port 80 (#0)
> HEAD / HTTP/1.0
> User-Agent: curl/7.63.0
> Accept: */*
>
< HTTP/1.1 301 Moved Permanently
< Server: awselb/2.0
< Date: Fri, 10 May 2019 23:46:54 GMT
< Content-Type: text/html
< Content-Length: 150
< Connection: close
< Location: https://internal-mysite-1234567890.us-west-2.elb.amazonaws.com:443/

函数的预期收益

let tickerA = [{
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.74,
    volume: 1000
}, {
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.76,
    volume: 2000
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.25,
    volume: 1200
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.19,
    volume: 1500
}]

Answer 1

您可以通过循环并将最大的项目保存到对象来在O（n）时间内完成此操作。最后，您的值将在组的Object.values中

let tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}]

let groups = tickerA.reduce((largest, {baseAsset, lastPriceUSD, volume}) => {
    /* 
     * if it's a new baseAsset or bigger than a previous one, save it
     * to the group under the baseAsset key 
    */
    if (!largest[baseAsset] || largest[baseAsset]['volume'] < volume ) {
        largest[baseAsset] = {baseAsset, lastPriceUSD, volume}
    }

    return largest
}, {})

TickerB = Object.values(groups)
console.log(TickerB)

Answer 2

一种方法是迭代tickerA的值，并将其映射到baseAsset键，如果该项的volume值大于该{{1}的当前项的值}键：

baseAsset

Answer 3

此替代组baseAsset并最终提取分组的值。

这是O(n)的时间复杂度。

let tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}];
let result = Object.values(tickerA.reduce((a, {baseAsset, lastPriceUSD, volume}) => {
    let {volume: current} = a[baseAsset] || {volume: Number.MAX_SAFE_INTEGER};
    
    if (current < volume) a[baseAsset].volume = volume;
    else a[baseAsset] = {baseAsset, lastPriceUSD, volume};
    
    return a;
}, Object.create(null)));

console.log(result);

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Answer 4

使用reduce和Object.values：

let tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}];
const res = Object.values(tickerA.reduce((acc, { baseAsset, lastPriceUSD, volume }) => {
  acc[baseAsset] = (!acc[baseAsset] || acc[baseAsset].volume < volume) ? { baseAsset, lastPriceUSD, volume } : acc[baseAsset];
  return acc;
}, {}));
console.log(res);

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匹配对象数组中的对象键并返回键，具有最大音量值的值

4 个答案: