在 JS 中创建具有多个变量的计算 JSON 数组

Question

首先，如果这通常是一件简单的事情，我深表歉意，但我在 JS 方面并不是特别有经验。我正在尝试在 Plotly.js 中为当前保存原始 JSON 的工作站点制作一些图表。我们有一个 JSON 格式：

stuff = [{"name": "shark", "location": "somewhere", "number":10},
         {"name": "shark", "location": "somewhereelse", "number":50},
         {"name": "shark", "location": "somewhere", "number":25},
         {"name": "turtle", "location": "anotherplace", "number":1},
         {"name": "elephant", "location": "greatplace", "number":50},
         {"name": "elephant", "location": "greatplace", "number":75}

我需要结果看起来像：

computed = [{"name":"shark", "location":"somewhere", "number":35},
            {"name":"shark", "location":"somewhereelse", "number":50},
            {"name":"turtle", "location":"anotherplace", "number":1},
            {"name":"elephant", "location":"greatplace", "number":125}

所有名称都按位置分组，并且该组中的所有数字相加。然后，我将使用这个计算得到的 JSON 与 Plotly.js 一起绘制我已经构建的一些函数。

我想我可以将问题简化为：

forEach((item) => {
x.push(item['location']
y.push(y += stuff.number where stuff.location === x[-1])
}

但这也意味着我将获得与 stuff 中相同的行数，只是计算和绘制。恐怕我想不通了。

任何帮助将不胜感激。

Answer 1

只需循环出原始数组，然后将元素添加到新数组中，除非它们已经添加，或者只是添加 number 将是这里的技巧。这是定义的：

var computed: any = [];

for(var element of stuff){
  let computedElement = computed.find((ele: { location: string; }) => ele.location == element.location);
  if (computedElement){
    computedElement.number += element.number; 
  }
  else{
    computed.push(element);
  }
}

Answer 2

首先，json 以 {} 而不是 [] 开头。您的对象是一个 javascript 对象。我在下面做了一个很长的函数：

const stuffList = [
{"name": "shark", "location": "somewhere", "number":10},
  {"name": "shark", "location": "somewhereelse", "number":50},
  {"name": "shark", "location": "somewhere", "number":25},
  {"name": "turtle", "location": "anotherplace", "number":1},
  {"name": "elephant", "location": "greatplace", "number":50},
  {"name": "elephant", "location": "greatplace", "number":75}
]

//extract all available location
const locations = stuffList.map( stuffItem => {
  return stuffItem.location
})

const filteredStuff = locations.map( location => {
  const number= stuffList.map( stuffItem => {
      if(stuffItem.location === location) {
        return stuffItem.number
      }
      return 0
  }).reduce( (a, b) => a + b)// total all the numbers of mathcing 
  location
  
  //get the first newStuffItem value equals to location 
  const [filteredStuffItem] = stuffList.map( stuffItem => {
    if(stuffItem.location === location) {
      return {
        name: stuffItem.name,
        location,
        number
      }
    }
  }).filter( stuffItem => {
    return stuffItem?.location === location
  })

  return filteredStuffItem
})

const newStuffList = filteredStuff.filter((thing, index) => {
  const _thing = JSON.stringify(thing);
  return index === filteredStuff.findIndex(obj => {
    return JSON.stringify(obj) === _thing;
  })
});

console.log(newStuffList)

或者 Deepak 在 javascript 中的魔法：

const stuff = [
  {"name": "shark", "location": "somewhere", "number":10},
  {"name": "shark", "location": "somewhereelse", "number":50},
  {"name": "shark", "location": "somewhere", "number":25},
  {"name": "turtle", "location": "anotherplace", "number":1},
  {"name": "elephant", "location": "greatplace", "number":50},
  {"name": "elephant", "location": "greatplace", "number":75}
]

var computed = [];

for(var element of stuff){
  let computedElement = computed.find((ele) => ele.location == element.location);
  if (computedElement){
    computedElement.number += element.number; 
  }
  else{
    computed.push(element);
  }
}

console.log(computed)

Answer 3

很抱歉没有尽快回复，感谢评论的人。无论出于何种原因，我无法让其他建议的解决方案发挥作用，但我看到了对所需逻辑的另一种看法，帮助我弄清楚了。

if (!label.includes(item['specimen'] + ':' + item['well_label'] + ':' + item['run'] + ':' + item['group'])) {
    label.push(item['specimen'] + ':' + item['well_label'] + ':' + item['run'] + ':' + item['group'])
    x.push(item['date'])
    y.push(item['sum'])
} else {
    var indx = label.indexOf(item['specimen'] + ':' + item['well_label'] + ':' + item['run'] + ':' + item['group'])
    let summed = y[indx] + item['sum']
    y.splice(indx, 1, summed)

所以本质上这创建了类似于复合键的东西，并推送初始数据以将其添加到标签列表，然后将数据添加到相关列表。如果该复合键再次出现，则该值将使用 splice 方法更新。

唯一的问题是，由于这些列表与输入到它们的数据没有任何关联，因此很少有正确的标签和数据不匹配的机会，但经过大量测试后，我已经整理出来了确实有效。

再次感谢之前提供帮助的评论者。