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如何在隨機(jī)存儲(chǔ)到另一個(gè)數(shù)組后找到與哪個(gè)數(shù)組相關(guān)的值（例如：數(shù)組a，數(shù)組b）

如何在隨機(jī)存儲(chǔ)到另一個(gè)數(shù)組后找到與哪個(gè)數(shù)組相關(guān)的值（例如：數(shù)組a，數(shù)組b）

qq_遁去的一_1 2019-04-18 15:15:57
我有兩種不同的數(shù)組，我從兩個(gè)不同的數(shù)組中選擇了一些隨機(jī)值并存儲(chǔ)在另一個(gè)數(shù)組中，如何找到與哪個(gè)數(shù)組相關(guān)的值例....var firstDbArray=[1,2,7,9,3,4,0,6,5,10]var secondDbArray=[1,2,9,7,0,4,8,11,15,30,10,12]var uniquefromtwoarrays= [];   for (var i = 0; i < this.firstDbArray.length; i++) {     if (this.secondDbArray.indexOf(this.firstDbArray[i]) === -1) {       uniquefromtwoarrays.push(this.firstDbArray[i]);     }   }   for (i = 0; i < this.secondDbArray.length; i++) {     if (this.firstDbArray.indexOf(this.secondDbArray[i]) === -1) {       uniquefromtwoarrays.push(this.secondDbArray[i]);     }   }uniquefromtwoarrays的輸出是----> [3,6,5,8,11,15,30,12]我的問(wèn)題是---->如何在此輸出中找到哪個(gè)數(shù)組的值。
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3 回答

?
Helenr

TA貢獻(xiàn)1780條經(jīng)驗(yàn) 獲得超4個(gè)贊

您可以遍歷uniquefromtwoarrays并檢查數(shù)組中是否存在值


var firstDbArray=[1,2,7,9,3,4,0,6,5,10]

var secondDbArray=[1,2,9,7,0,4,8,11,15,30,10,12]


var uniquefromtwoarrays= [];


for (var i = 0; i < this.firstDbArray.length; i++) {

? if (this.secondDbArray.indexOf(this.firstDbArray[i]) === -1) {

? ? uniquefromtwoarrays.push(this.firstDbArray[i]);

? }

}

for (i = 0; i < this.secondDbArray.length; i++) {

? if (this.firstDbArray.indexOf(this.secondDbArray[i]) === -1) {

? ? uniquefromtwoarrays.push(this.secondDbArray[i]);

? }

}


let object = {firstDbArray,secondDbArray}

uniquefromtwoarrays.forEach(value=>{

? let includeIn = Object.entries(object).filter(([key,array])=>array.includes(value)).map(([key])=> key)

? console.log(`${value} is from in ${includeIn.join(', ')}`)

})


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?
動(dòng)漫人物

TA貢獻(xiàn)1815條經(jīng)驗(yàn) 獲得超10個(gè)贊

像這樣的東西：


var firstDbArray=[1,2,7,9,3,4,0,6,5,10]

var secondDbArray=[1,2,9,7,0,4,8,11,15,30,10,12]


var uniquefromtwoarrays= [];


//create a new array were will be store the origin of the value

var uniqueArrayOrigin = [];


  for (var i = 0; i < this.firstDbArray.length; i++) {

    if (this.secondDbArray.indexOf(this.firstDbArray[i]) === -1) {

      //each time you push a new value to the array of unique values,

      //push to the new array of origins the value origins

      uniquefromtwoarrays.push(this.firstDbArray[i]);

      uniqueArrayOrigin.push(0);

    }

  }

  for (i = 0; i < this.secondDbArray.length; i++) {

    if (this.firstDbArray.indexOf(this.secondDbArray[i]) === -1) {

      uniquefromtwoarrays.push(this.secondDbArray[i]);

      uniqueArrayOrigin.push(1);

    }

  }

  

for(i = 0; i<uniquefromtwoarrays.length; i++) {

 //Then, you can print the origin value by access to the same position as the unique values array (for this case 0 is for the first array and 1 is for the second array

 console.log(`origin: ${uniqueArrayOrigin[i]}; value: 

 ${uniquefromtwoarrays[i]}`);

}


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?
犯罪嫌疑人X

TA貢獻(xiàn)2080條經(jīng)驗(yàn) 獲得超4個(gè)贊

if ( secondDbArray.indexOf(uniquefromtwoarrays[i] === -1 ) 

 // uniquefromtwoarrays[i] is related to the first array

else 

// uniquefromtwoarrays[i] is related to the second array


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