從多個陣列中發現差異和缺失元素

Question

我有一個包含n數組的數組，每個數組都包含不同數量的字符串元素。

每個字符串包含一個關鍵字如evar#,event#,prop#（其中#是一個數字）。

我需要做到以下幾點：從一個字符串，它是與其他陣列中的至少一個

回報不存在的關鍵詞不同

• 回報所有的關鍵詞至少其他陣列之一。

這裏是一個包含3個數組的數組的例子：

[ 
    [ 
     "overwrite value of evar1 with page_url_query 'int_cmp'", 
     "set event1 to custom value '1'", 
     "set event2 to custom value '1'", 
     "overwrite value of evar2 with page_url", 
     "overwrite value of evar3 with contextdata.user_id", 
     "set event4 to eventid", 
     "set event3 to eventid" 

    ], 
    [ 
     "overwrite value of prop3 with contextdata.phase", 
     "overwrite value of prop2 with contextdata.room", 
     "set event1 to custom value '1'", 
     "set event2 to eventid", 
     "overwrite value of evar5 with contextdata.queue", 
     "set event4 to eventid", 
     "overwrite value of evar6 with contextdata.audience", 
     "set event3 to eventid", 
     "set event5 to custom value '1'" 
    ], 
    [ 
     "overwrite value of evar4 with contextdata.no_challenges", 
     "overwrite value of prop3 with contextdata.user_type", 
     "overwrite value of evar7 with contextdata.interaction", 
     "set event2 to custom value '1'", 
     "set event3 to eventid", 
     "set event4 to eventid", 
     "set event1 to custom value '1'", 
     "set event5 to custom value '1'" 
    ] 
] 

和響應應包含以下陣列：

[evar1, event2, evar2, evar3, prop3, prop2, evar6, event5, evar4, evar7] 

請讓我知道如果我需要提供更多信息。我願意接受任何建議...

以後編輯：

我實現了一個解決方案，但我認爲它需要一些優化。我會很感激，如果有人幫助我...

Answer 1

我實施了一個解決方案，醜陋，但顯然它的作品！有關它的任何想法？

基本上，我找到了最大的數組，然後將其元素與其他數組進行比較。不知道是否有相同長度的多個陣列

var allActions = [ 
 
    [ 
 
     "overwrite value of evar1 with page_url_query 'int_cmp'", 
 
     "set event1 to custom value '1'", 
 
     "set event2 to custom value '1'", 
 
     "overwrite value of evar2 with page_url", 
 
     "overwrite value of evar3 with contextdata.user_id", 
 
     "set event4 to eventid", 
 
     "set event3 to eventid" 
 

 
    ], 
 
    [ 
 
     "overwrite value of prop3 with contextdata.phase", 
 
     "overwrite value of prop2 with contextdata.room", 
 
     "set event1 to custom value '1'", 
 
     "set event2 to eventid", 
 
     "overwrite value of evar5 with contextdata.queue", 
 
     "set event4 to eventid", 
 
     "overwrite value of evar6 with contextdata.audience", 
 
     "set event3 to eventid", 
 
     "set event5 to custom value '1'" 
 
    ], 
 
    [ 
 
     "overwrite value of evar4 with contextdata.no_challenges", 
 
     "overwrite value of prop3 with contextdata.user_type", 
 
     "overwrite value of evar7 with contextdata.interaction", 
 
     "set event2 to custom value '1'", 
 
     "set event3 to eventid", 
 
     "set event4 to eventid", 
 
     "set event1 to custom value '1'", 
 
     "set event5 to custom value '1'" 
 
    ] 
 
] 
 

 
    
 
    var keyWords = ["evar","event","prop"]; 
 
    var resultArray = allActions[0]; 
 
    var arrLen = []; 
 
    var different = []; 
 
    for(var i = 0; i < allActions.length; i++) { 
 
     arrLen.push(allActions[i].length); 
 
    } 
 
    var max = Math.max.apply(null, arrLen) 
 
    var maxArray = arrLen.indexOf(max); 
 

 
    for(elem in allActions[maxArray]) { 
 
     for(var i = 0; i < allActions.length; i++) { 
 
     if(i !== maxArray) { 
 
      for(var j in allActions[i]) { 
 
      var mainElem = allActions[maxArray][elem]; 
 
      var checkElem = allActions[i][j]; 
 
      if(mainElem !== checkElem) { 
 
       for(var k = 0; k < keyWords.length; ++k) { 
 
       if((index = mainElem.indexOf(keyWords[k])) !== -1) { 
 
        splittedStr = mainElem.substring(index, mainElem.length).split(' ', 1); 
 
        if(splittedStr[0].indexOf("evar") !== -1) { 
 
        if(different.indexOf(splittedStr[0]) == -1) { 
 
         different.push(splittedStr[0]); 
 
        } 
 
        } 
 
        if(splittedStr[0].indexOf("prop") !== -1) { 
 
        if(different.indexOf(splittedStr[0]) == -1) { 
 
         different.push(splittedStr[0]); 
 
        } 
 
        } 
 
        if(splittedStr[0].indexOf("event") !== -1) { 
 
        if(different.indexOf(splittedStr[0]) == -1) { 
 
         different.push(splittedStr[0]); 
 
        } 
 
        } 
 
       } 
 
       } 
 
      } 
 
      } 
 
     } 
 
     } 
 
    } 
 
    console.log(different);

Answer 2

如果確定與ES6你可能在功能上做如下

var data = [ 
 
    [ 
 
     "overwrite value of evar1 with page_url_query 'int_cmp'", 
 
     "set event1 to custom value '1'", 
 
     "set event2 to custom value '1'", 
 
     "overwrite value of evar2 with page_url", 
 
     "overwrite value of evar3 with contextdata.user_id", 
 
     "set event4 to eventid", 
 
     "set event3 to eventid" 
 

 
    ], 
 
    [ 
 
     "overwrite value of prop3 with contextdata.phase", 
 
     "overwrite value of prop2 with contextdata.room", 
 
     "set event1 to custom value '1'", 
 
     "set event2 to eventid", 
 
     "overwrite value of evar5 with contextdata.queue", 
 
     "set event4 to eventid", 
 
     "overwrite value of evar6 with contextdata.audience", 
 
     "set event3 to eventid", 
 
     "set event5 to custom value '1'" 
 
    ], 
 
    [ 
 
     "overwrite value of evar4 with contextdata.no_challenges", 
 
     "overwrite value of prop3 with contextdata.user_type", 
 
     "overwrite value of evar7 with contextdata.interaction", 
 
     "set event2 to custom value '1'", 
 
     "set event3 to eventid", 
 
     "set event4 to eventid", 
 
     "set event1 to custom value '1'", 
 
     "set event5 to custom value '1'" 
 
    ] 
 
], 
 

 
interim = data.map(ds => ds.reduce((p,c) => p.concat(c.match(/[a-z]+\d/g)),[])); 
 
result = [...interim.reduce((p,c) => p.filter(f => c.includes(f))) 
 
         .reduce((s,c) => (s.delete(c),s), interim.reduce((s,a) => a.reduce((ss,c) => ss.add(c),s),new Set()))]; 
 
    \t 
 
console.log(result);

它會工作