平坦化Javascriptのオブジェクト配列

Question

は私がオブジェクトを持っているように：私はこれは、日付が同じであるTOTALキーに参加し、0で埋める/連結され、以下のような何かにそれを縮小することができますどのように

[ 
    { 
    "DATA": "2016-01-22", 
    "TOTAL": "7" 
    }, 
    { 
    "DATA": "2016-01-25", 
    "TOTAL": "3" 
    }, 
    { 
    "DATA": "2016-01-26", 
    "TOTAL": "1" 
    }, 
    { 
    "DATA": "2016-01-27", 
    "TOTAL": "2" 
    }, 
    { 
    "DATA": "2016-01-22", 
    "TOTAL": "1" 
    }, 
    { 
    "DATA": "2016-01-25", 
    "TOTAL": "1" 
    }, 
    { 
    "DATA": "2016-01-27", 
    "TOTAL": "1" 
    }, 
... 
] 

ケースには日付が繰り返されない？：

[ 
    { 
    "DATA": "2016-01-22", 
    "TOTAL": ["7", "1"] 
    }, 
    { 
    "DATA": "2016-01-25", 
    "TOTAL": ["3", "1"] 
    }, 
    { 
    "DATA": "2016-01-26", 
    "TOTAL": ["1", "0"] 
    }, 
    { 
    "DATA": "2016-01-27", 
    "TOTAL": ["2", "1"] 
    } 
] 

を、私はこのコードのブロックをしようとしてきたが、TOTALキーにすべて同じ次元を得ることができない - ゼロで埋め罰金だろう。

var output = []; 
d.forEach(function(value) { 
    var existing = output.filter(function(v, i) { 
     return v.DATA == value.DATA; 
    }); 
    if (existing.length) { 
     var existingIndex = output.indexOf(existing[0]); 
     output[existingIndex].TOTAL = output[existingIndex].TOTAL.concat(value.TOTAL); 
    } else { 
     if (typeof value.TOTAL == 'string') 
      value.TOTAL = [value.TOTAL]; 
     output.push(value); 
    } 
}); 
console.log(JSON.stringify(output, null, 4)); 

Answer 1

var someData = [] // <- your instantiated array in question. 
var transformedData = []; 

var highestCount = 0; 

someData.forEach(x => { 
    var foundIndex = transformedData.findIndex((ele) => ele.DATA === x.DATA); 
    if (foundIndex < 0) { 
     transformedData 
      .push({DATA : x.DATA, TOTAL : [x.TOTAL]}); 
    } else { 
     transformedData[foundIndex] 
      .TOTAL.push(x.TOTAL); 
     var currentCountAtIndex = transformedData[foundIndex].TOTAL.length; 
     if (highestCount < transformedData[foundIndex].TOTAL.length) highestCount = currentCountAtIndex; 
    } 
}); 

// fill any indicies in array that are lower than the highest count with 0 
transformedData 
    .forEach(x => { 
     if (x.TOTAL.length < highestCount) { 
      while(x.TOTAL.length < highestCount) { 
       x.TOTAL.push(0); 
      } 
     } 
    }); 

Answer 2

それは、このような単純なことができます

var result = {}; 
 
var test = [ 
 
    { 
 
    "DATA": "2016-01-22", 
 
    "TOTAL": "7" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-25", 
 
    "TOTAL": "3" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-26", 
 
    "TOTAL": "1" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-27", 
 
    "TOTAL": "2" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-22", 
 
    "TOTAL": "1" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-25", 
 
    "TOTAL": "1" 
 
    }, 
 
    { 
 
    "DATA": "2016-01-27", 
 
    "TOTAL": "1" 
 
    }]; 
 

 
console.log("test array: ", test); 
 

 
var len = 0, 
 
sorted; 
 

 
// Flatten the object. 
 
test.forEach(d => { 
 
    result[d.DATA] == undefined ? result[d.DATA] = [d.TOTAL] : result[d.DATA].push(d.TOTAL); 
 
}); 
 

 
// Sort so we get the max length to know how many zeros to add. 
 
sorted = Object.keys(result).sort((k, b) => { 
 
    return result[k].length - result[b].length; 
 
}); 
 

 
// Max length from the sorted array. 
 
len = result[sorted[sorted.length - 1]].length; 
 

 
// push zeros 
 
Object.keys(result).forEach(k => { 
 
    if(result[k].length < len){ 
 
    for(var i = result[k].length; i < len; i++){ 
 
    result[k].push("0"); 
 
    } 
 
    } 
 
}); 
 

 
console.log("result: ", result);