Javascript 计数出现次数

Question

我有以下数组数据，我正在计算使用包含条件匹配的记录数。

var final = 
[
  [
    "61131",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61132",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61133",
    "TOYOTA",
    "STOCKPORT",
    "STORED"
  ],
  [
    "61134",
    "TOYOTA",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61135",
    "NISSAN",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61136",
    "NISSAN",
    null,
    null
  ]
]

代码是这样的：

var obj = {};
var num1 = 0;
var num2 = 0;
for (var i=0; i<final.length; i++){
    if(final[i].join(';').includes('BOLTON') && final[i].join(';').includes('MAINT') && final[i].join(';').includes('NISSAN')) {
        num1++;
    }
}

for (var i=0; i<final.length; i++){
    if(final[i].join(';').includes('STOCKPORT') && final[i].join(';').includes('STORED') && final[i].join(';').includes('TOYOTA')) {
        num2++;
    }
}


obj['BOLTON_MAINT_NISSAN'] = num1
obj['STOCKPORT_STORED_TOYOTA'] = num2

console.log(obj)

输出

  {  "BOLTON_MAINT_NISSAN": 2, "STOCKPORT_STORED_TOYOTA": 1}

我得到了想要的结果，有没有更有效的方法来编写上面的最小代码？

Answer 1

const count = (list, keys) => {
  return keys.reduce(
    (obj, key) => ({
      ...obj,
      [key]: list.filter(sublist =>
        key.split('_').every(keyPart => sublist.includes(keyPart))
      ).length
    }),
    {}
  );
};

count(final, ['BOLTON_MAINT_NISSAN', 'STOCKPORT_STORED_TOYOTA'])

 const final = [ [ "61131", "NISSAN", "BOLTON", "MAINT" ], [ "61132", "NISSAN", "BOLTON", "MAINT" ], [ "61133", "TOYOTA", "STOCKPORT", "STORED" ], [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ], [ "61135", "NISSAN", "STOCKPORT", "MAINT" ], [ "61136", "NISSAN", null, null ] ]; const count = (list, keys) => { return keys.reduce( (obj, key) => ({ ...obj, [key]: list.filter(sublist => key.split('_').every(keyPart => sublist.includes(keyPart)) ).length }), {} ); }; var bmtNsst = count(final, ['BOLTON_MAINT_NISSAN', 'STOCKPORT_STORED_TOYOTA']); console.log(bmtNsst);

Answer 2

使用reduce，slice 去除数字，过滤去除空值，然后join 使其成为key。

 var final = [ ["61131", "NISSAN", "BOLTON", "MAINT"], ["61132", "NISSAN", "BOLTON", "MAINT"], ["61133", "TOYOTA", "STOCKPORT", "STORED"], ["61134", "TOYOTA", "STOCKPORT", "MAINT"], ["61135", "NISSAN", "STOCKPORT", "MAINT"], ["61136", "NISSAN", null, null] ]; var grouped = final.reduce(function(obj, data) { var key = data.slice(1).filter(Boolean).join("_"); obj[key] = (obj[key] || 0) + 1; return obj; }, {}); console.log(grouped)

Answer 3

无需使用String.join 。 Array 也有函数Array.prototype.includes来检查项目是否存在于数组中。

使用它，您可以找到满足您条件的 subArray。 并且使用Array.prototype.filter ，您可以提取子数组以满足以下条件。

 const final = [ [ "61131", "NISSAN", "BOLTON", "MAINT" ], [ "61132", "NISSAN", "BOLTON", "MAINT" ], [ "61133", "TOYOTA", "STOCKPORT", "STORED" ], [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ], [ "61135", "NISSAN", "STOCKPORT", "MAINT" ], [ "61136", "NISSAN", null, null ] ]; const num1Arr = final.filter((item) => item.includes('BOLTON') && item.includes('MAINT') && item.includes('NISSAN')); const num2Arr = final.filter((item) => item.includes('STOCKPORT') && item.includes('STORED') && item.includes('TOYOTA')); const output = { BOLTON_MAINT_NISSAN: num1Arr.length, STOCKPORT_STORED_TOYOTA: num2Arr.length }; console.log(output);

或者简单地说，您可以使用Array.prototype.reduce 。

 const final = [ [ "61131", "NISSAN", "BOLTON", "MAINT" ], [ "61132", "NISSAN", "BOLTON", "MAINT" ], [ "61133", "TOYOTA", "STOCKPORT", "STORED" ], [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ], [ "61135", "NISSAN", "STOCKPORT", "MAINT" ], [ "61136", "NISSAN", null, null ] ]; const output = final.reduce((acc, cur) => { if (cur.includes('BOLTON') && cur.includes('MAINT') && cur.includes('NISSAN')) { acc['BOLTON_MAINT_NISSAN'] ++; } if (cur.includes('STOCKPORT') && cur.includes('STORED') && cur.includes('TOYOTA')) { acc['STOCKPORT_STORED_TOYOTA'] ++; } return acc; }, { BOLTON_MAINT_NISSAN: 0, STOCKPORT_STORED_TOYOTA: 0 }); console.log(output);

Answer 4

 const final = [ [ "61131", "NISSAN", "BOLTON", "MAINT" ], [ "61132", "NISSAN", "BOLTON", "MAINT" ], [ "61133", "TOYOTA", "STOCKPORT", "STORED" ], [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ], [ "61135", "NISSAN", "STOCKPORT", "MAINT" ], [ "61136", "NISSAN", null, null ] ]; console.log({ BOLTON_MAINT_NISSAN: final.filter((value) => { return value[1] === 'NISSAN' && value[2] === 'BOLTON' && value[3] === 'MAINT'; }).length, STOCKPORT_STORED_TOYOTA: final.filter((value) => { return value[1] === 'TOYOTA' && value[2] === 'STOCKPORT' && value[3] === 'STORED'; }).length, });

Answer 5

使用Array.prototype.reduce() ：

 const final = [ [ "61131", "NISSAN", "BOLTON", "MAINT" ], [ "61132", "NISSAN", "BOLTON", "MAINT" ], [ "61133", "TOYOTA", "STOCKPORT", "STORED" ], [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ], [ "61135", "NISSAN", "STOCKPORT", "MAINT" ], [ "61136", "NISSAN", null, null ] ]; let occurrence = final.reduce((obj, item)=>{ var key=item[2]+"_"+item[3]+"_"+item[1]; if(!obj[key]) obj[key]=0; obj[key]++; return obj; },{}); console.log(occurrence);

Answer 6

在您的解决方案中，您在数组中迭代两次，这是不必要的，好像一个对象是您正在搜索的两个对象之一，然后它不是另一个，因此您可以在一个循环中完成所有操作。

 var final = [ ["61131", "NISSAN", "BOLTON", "MAINT"], ["61132", "NISSAN", "BOLTON", "MAINT"], ["61133", "TOYOTA", "STOCKPORT", "STORED"], ["61134", "TOYOTA", "STOCKPORT", "MAINT"], ["61135", "NISSAN", "STOCKPORT", "MAINT"], ["61136", "NISSAN", null, null] ]; var obj = { BOLTON_MAINT_NISSAN: 0, STOCKPORT_STORED_TOYOTA: 0 }; final.forEach(y => { let x = y.join(';'); if(x.includes('BOLTON') && x.includes('MAINT') && x.includes('NISSAN')) obj.BOLTON_MAINT_NISSAN++; else if ( x.includes('STOCKPORT') && x.includes('STORED') && x.includes('TOYOTA') ) obj.STOCKPORT_STORED_TOYOTA++; }) console.log(obj)