Question

我有来自javascript对象的绘制邻接矩阵的问题。

我的对象：

var result = {
    "D5":  ["D#5", "A#4", "D#5", "A#4"],
    "D#5": ["D5", "D5"],
    "A#4": ["G4", "D5", "F5"],
    "G4":  ["A#4"],
    "F5":  ["A#4"]
}

我想获得输出：

        D#5 A#4 D5 G4 F5

D5      2   2   0   0  0

D#5     0   0   2   0  0

A#4     0   0   1   1  1

G4      0   1   0   0  0

F5      0   1   0   0  0

如何计算密钥中的值？

我做了草稿：

for (var key in result){
         console.log('Key: ' + key + ' values: ' + result[key])


            for(var val in result[key]){
                // console.log(result[key][val]);

                var counter = 0;
                for(var v in result[key]){
                    if (result[key][v] === result[key][val]){
                        counter++;
                    }
                }
                console.log('Value ' + result[key][val] + 'count' + counter)
            }
        }

Answer 1

您可以使用reduce

＆＃13;

let result = {"D5":["D#5","A#4","D#5","A#4"],"D#5":["D5","D5"],"A#4":["G4","D5","F5"],"G4":["A#4"],"F5":["A#4"]};
let order = Object.keys(result).reduce((c,v)=>Object.assign(c,{[v]:0}),{});		//Make Dynamic Order

let count = Object.keys(result).reduce((c, v) => {
  return Object.assign(c, {[v]: result[v].reduce((p, o)=>{
      p[o] += 1;
      return p;
    }, Object.assign({}, order))});
}, {});

console.log(count);

＆＃13;

Answer 2

这可行，forEach使用result[key]

var result = {
	"D5":["D#5","A#4","D#5","A#4"],
  "D#5":["D5","D5"],
  "A#4":["G4","D5","F5"],
  "G4":["A#4"],"F5":["A#4"]
 };
var final = {};
for (var x in result){
	final[x] = {};
  for(var y in result){
    final[x][y]=0;
  }
	result[x].forEach(function(z){
    final[x][z]++;
  });
}
console.log(final);

Answer 3

只是ES5

var result = {"D5":["D#5","A#4","D#5","A#4"],
"D#5":["D5","D5"],"A#4":["G4","D5","F5"],"G4":["A#4"],"F5":["A#4"]}

/***************************************************************************/
/*********************** THIS IS WHAT YOU ASKED ****************************/
/***************************************************************************/

function myCustomFunction(array){
  // VAR USED TO STORE THE UNIQUE KEYS
  var inner_keys = [];

  // LOOP TO GET THE UNIQUE KEYS
  Object.keys(array).map(function(a){inner_keys = inner_keys.concat(array[a])})
  inner_keys = inner_keys.filter(function(a,b,c){return c.indexOf(a) === b});

  //  VAR USED TO STORE THE FINAL RESULT
  var result = {};

  // LOOP TO GET THE AMOUNT OF REPETITION FOR EACH KEY IN THE ORIGINAL ARRAY
  Object.keys(array).map(function(a){
    result[a] = inner_keys.map(function(b){return array[a].filter(function(c){return c==b}).length})
  })
  
  // RETURNS THE UNIQUE KEYS AND THE RESULT
  return {orderOfKeys: inner_keys, result: result};
}

/**************************************************************************/
/**************************************************************************/
/**************************************************************************/



/**************************************************************************/
/******************** THIS IS JUST TO PRINT IT INTO A TABLE ***************/
/**************************************************************************/

function fillTable(array){
  var table_header = array.orderOfKeys;  
  var table_body = array.result;
  var thead = document.getElementById("thead");  
  var tbody = document.getElementById("tbody");
  
  createElement('TR', thead)    

  for(var i = 0; i < table_header.length; i++){
    if(i == 0){
      createElement('TH', thead.children[0], '')
    }
    createElement('TH', thead.children[0], table_header[i])
  } 
  
  for(var k in table_body){
    var row = createElement('TR', tbody);
    for(var i = 0; i < table_body[k].length; i++){
      if(i == 0){
        createElement('TD', row, k)
      }
      createElement('TD', row, table_body[k][i])
    }
  }  
  
}

function createElement(el, parent, value){
  var element = document.createElement(el);
  if(value != undefined){
    var element_text = document.createTextNode(value);
    element.appendChild(element_text);
  }
  parent.appendChild(element);
  return element;
}

/**************************************************************************/
/**************************************************************************/
/**************************************************************************/


fillTable(myCustomFunction(result))

<table>
  <thead id="thead">
  </thead>
  <tbody id="tbody">
  </tbody>
</table>

Answer 4

您可以使用一个Set来获取所有唯一的子数组元素，然后使用reduce方法获取具有这些子数组中每个元素的计数的对象。然后，您可以将该数据添加到一个表中。

const data = {"D5":["D#5","A#4","D#5","A#4"],
"D#5":["D5","D5"],"A#4":["G4","D5","F5"],"G4":["A#4"],"F5":["A#4"]}
const table = document.querySelector('table tbody');

const keys = new Set();
const output = Object.keys(data).reduce((r, key) => {
  data[key].forEach(e => {
    keys.add(e);
    if (!r[key]) r[key] = {}
    r[key][e] = (r[key][e] || 0) + 1;
  })
  return r;
}, {})


table.innerHTML += '<tr><td></td>' + Array.from(keys).map(e => `<td>${e}</td>`).join('') +'<tr>';
Object.keys(output).forEach(key => {
  let data = `<td>${key}</td>`;
  data += Array.from(keys, e => '<td>' + (output[key][e] ? output[key][e] : 0) + '</td>').join('')
  table.innerHTML += `<tr>${data}</tr>`;
})

<table><tbody></tbody></table>

来自对象的javascript邻接矩阵

4 个答案: