在Javascript中计算数据集中每个状态的值答案

【问题标题】：Counting the value for each status in a data set in Javascript在Javascript中计算数据集中每个状态的值
【发布时间】：2017-03-26 03:13:27
【问题描述】：

来自变量：

var locations = [
    ['FULL',3.720185, 103.124075,1],
    ['EMPTY',3.719693, 103.123896,0],
    ['FULL',3.720916, 103.12492,1],
    ['EMPTY',3.721032, 103.124532,0],
    ['FULL',3.722299, 103.124587,1],
    ['FULL',3.723189, 103.124706,1],
    ['FULL',3.725067, 103.124593,1]
];

我如何得到这个结果：

已满 = 5 空 = 2

我需要它来根据纬度和经度确定我大学的废物管理

【问题讨论】：

标签： javascript arrays count geolocation maps

【解决方案1】：

考虑到locations 数组中的所有子数组都按原样布局，您可以简单地使用for loop 来检查每个子数组的第一个元素的值是"FULL" 还是@ 987654324@。

例子：

var locations = [
    ['FULL',3.720185, 103.124075,1],
    ['EMPTY',3.719693, 103.123896,0],
    ['FULL',3.720916, 103.12492,1],
    ['EMPTY',3.721032, 103.124532,0],
    ['FULL',3.722299, 103.124587,1],
    ['FULL',3.723189, 103.124706,1],
    ['FULL',3.725067, 103.124593,1]
];

var full = 0;
var empty = 0;

for(var i = 0; i < locations.length; i++){
   if(locations[i][0] == 'FULL') full++;
   if(locations[i][0] == 'EMPTY') empty++;
}

console.log("full " + full + "   empty " + empty );

如果"EMPTY" 或"FULL" 可能位于子数组中的任何位置，即不能保证它们始终位于索引0 处，那么您可以使用以下解决方案：

var locations = [
    ['FULL',3.720185, 103.124075,1],
    ['EMPTY',3.719693, 103.123896,0],
    ['FULL',3.720916, 103.12492,1],
    ['EMPTY',3.721032, 103.124532,0],
    ['FULL',3.722299, 103.124587,1],
    ['FULL',3.723189, 103.124706,1],
    ['FULL',3.725067, 103.124593,1]
];

var full = 0;
var empty = 0;

for(var i = 0; i < locations.length; i++){
   for(var j = 0; j < locations[i].length; j++){
      if(locations[i][j] == 'FULL') full++;
      if(locations[i][j] == 'EMPTY') empty++;
  }
}

console.log("full " + full + "   empty " + empty );

【讨论】：

【解决方案2】：

有很多方法，这是其中之一：

var locations = [
    ['FULL',3.720185, 103.124075,1],
    ['EMPTY',3.719693, 103.123896,0],
    ['FULL',3.720916, 103.12492,1],
    ['EMPTY',3.721032, 103.124532,0],
    ['FULL',3.722299, 103.124587,1],
    ['FULL',3.723189, 103.124706,1],
    ['FULL',3.725067, 103.124593,1]
];

res = {};
locations.forEach(item=>res.hasOwnProperty(item[0]) ? res[item[0]]++ : res[item[0]] = 1);
console.log(res);

JSFiddle：https://jsfiddle.net/k94qfa9r/2/

如果你需要它作为一个函数：

var locations = [
['FULL',3.720185, 103.124075,1],
['EMPTY',3.719693, 103.123896,0],
['FULL',3.720916, 103.12492,1],
['EMPTY',3.721032, 103.124532,0],
['FULL',3.722299, 103.124587,1],
['FULL',3.723189, 103.124706,1],
['FULL',3.725067, 103.124593,1]
];

function countItems(locs) {
  res = {};
  locs.forEach(item=>res.hasOwnProperty(item[0]) ? res[item[0]]++ : res[item[0]] = 1);
  return res;
} 
console.log(countItems(locations));

https://jsfiddle.net/k94qfa9r/4/

如果你需要它作为数组：

var locations = [
['FULL',3.720185, 103.124075,1],
['EMPTY',3.719693, 103.123896,0],
['FULL',3.720916, 103.12492,1],
['EMPTY',3.721032, 103.124532,0],
['FULL',3.722299, 103.124587,1],
['FULL',3.723189, 103.124706,1],
['FULL',3.725067, 103.124593,1]
];

function countItems(locs) {
  res = {};
  resArr = [];
  locs.forEach(item=>res.hasOwnProperty(item[0]) ? res[item[0]]++ : res[item[0]] = 1);
  for(var p in res)
  	resArr.push([p, res[p]])
  return resArr;
} 
console.log(countItems(locations));

JSFiddle：https://jsfiddle.net/k94qfa9r/6/

【讨论】：