将数组与对象属性匹配

Question

我有一个对象，其属性如下所示：

{
  maj6: { chromatic: [ 0, 4, 7, 9 ] },
  min6: { chromatic: [ 0, 3, 7, 9 ] },
  maj7: { chromatic: [ 0, 4, 7, 11 ] },
  min7: { chromatic: [ 0, 3, 7, 10 ] },
  minmaj7: { chromatic: [ 0, 3, 7, 11 ] },
  dom7: { chromatic: [ 0, 4, 7, 10 ] },
  maj9: { chromatic: [ 0, 4, 7, 11, 14 ] },
  dom9: { chromatic: [ 0, 4, 7, 10, 14 ] },
  maj: { chromatic: [ 0, 4, 7 ] },
  min: { chromatic: [ 0, 3, 7 ] },
  aug: { chromatic: [ 0, 4, 8 ] },
  dim: { chromatic: [ 0, 3, 6 ] }
}

我正在计算一个数组，例如0,4,7，我试图将其与上述对象属性之一相匹配。我尝试使用find函数来实现这一点，如下面的代码片段所示：

  matchParsedToChord() {
    const result = this.blocks //returns larger payload
    .map(block => block.chromatic) // returns array [0,4,7,4,7,7,7,4]
    .filter((v,i,arr) => arr.indexOf(v) === i) //removes redundancies  [0,4,7]
    .find(v => { //attempts to 
      v === Object.keys(chordDefinitions).map(key => chordDefinitions[key].chromatic)
    })
    console.log(result)
  }

我的问题是Object.keys逻辑返回一个完整的数组，而我喜欢逐个元素进行比较。

我想返回与数组关联的对象属性，例如：

maj: { chromatic: [ 0, 4, 7 ] }

非常感谢

Answer 1

这将返回具有您正在查找的模式的所有和弦。它只是将数组展平，并查找indexOf来组合所有匹配项。可以将其作为严格(精确匹配)或包含(所有部分匹配)运行

​

const a = {
  maj6: { chromatic: [ 0, 4, 7, 9 ] },
  min6: { chromatic: [ 0, 3, 7, 9 ] },
  maj7: { chromatic: [ 0, 4, 7, 11 ] },
  min7: { chromatic: [ 0, 3, 7, 10 ] },
  minmaj7: { chromatic: [ 0, 3, 7, 11 ] },
  dom7: { chromatic: [ 0, 4, 7, 10 ] },
  maj9: { chromatic: [ 0, 4, 7, 11, 14 ] },
  dom9: { chromatic: [ 0, 4, 7, 10, 14 ] },
  maj: { chromatic: [ 0, 4, 7 ] },
  min: { chromatic: [ 0, 3, 7 ] },
  aug: { chromatic: [ 0, 4, 8 ] },
  dim: { chromatic: [ 0, 3, 6 ] }
}

function getPattern(pattern, strict) {
  let b = []
  for (const [key, value] of Object.entries(a)) {
   let set = Object.values(value).flat().join(",") ;
    if ((!strict && set.indexOf(pattern) !== -1) || (strict && set == pattern)) b.push({
      [key]: value
    })
  }
  if (strict && b.length>0) b = b[0];
  return b;
}

// Contains (partial matches)
let matches = getPattern("0,4,7", false);
console.log('Contains', matches);

// strict
matches = getPattern("0,4,7", true);
console.log('Strict', matches);

​

Answer 2

你是说像这样的东西吗？编辑为使用reduce以获得所需的回报

​

const a = {
  maj6: { chromatic: [ 0, 4, 7, 9 ] },
  min6: { chromatic: [ 0, 3, 7, 9 ] },
  maj7: { chromatic: [ 0, 4, 7, 11 ] },
  min7: { chromatic: [ 0, 3, 7, 10 ] },
  minmaj7: { chromatic: [ 0, 3, 7, 11 ] },
  dom7: { chromatic: [ 0, 4, 7, 10 ] },
  maj9: { chromatic: [ 0, 4, 7, 11, 14 ] },
  dom9: { chromatic: [ 0, 4, 7, 10, 14 ] },
  maj: { chromatic: [ 0, 4, 7 ] },
  min: { chromatic: [ 0, 3, 7 ] },
  aug: { chromatic: [ 0, 4, 8 ] },
  dim: { chromatic: [ 0, 3, 6 ] }
}

const find = (obj, match = [0,4,7]) => {
  return Object.keys(a)
  .reduce((acc, k) => {
    // If you want more loose match, just switch search direction
    // like: match.every(m => obj[k].chromatic.includes(m)
    if(obj[k].chromatic.every(m =>  match.includes(m))) {
      acc[k] = obj[k];
    }
    return acc;
  },
  {})
}
console.log(find(a))

​

Answer 3

使用Array#every()并匹配长度。不过，你想要的结果并不是很清楚，所以可能需要根据你的期望进行修改

​

const data ={
  maj6: { chromatic: [ 0, 4, 7, 9 ] },
  min6: { chromatic: [ 0, 3, 7, 9 ] },
  maj7: { chromatic: [ 0, 4, 7, 11 ] },
  min7: { chromatic: [ 0, 3, 7, 10 ] },
  minmaj7: { chromatic: [ 0, 3, 7, 11 ] },
  dom7: { chromatic: [ 0, 4, 7, 10 ] },
  maj9: { chromatic: [ 0, 4, 7, 11, 14 ] },
  dom9: { chromatic: [ 0, 4, 7, 10, 14 ] },
  maj: { chromatic: [ 0, 4, 7 ] },
  min: { chromatic: [ 0, 3, 7 ] },
  aug: { chromatic: [ 0, 4, 8 ] },
  dim: { chromatic: [ 0, 3, 6 ] }
}

const arr =  [0,4,7];

const entries = Object.entries(data).map(([k,v])=> [k, [...new Set(v.chromatic)]]);

const match = entries.find(([k,v])=> v.length === arr.length && arr.every(n => v.includes(n)))

console.log(match)

​