Question

我在java脚本数组中有以下lat / long集合。

locData=[[32.0770488,34.7958072],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077589,34.774565],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077795,34.774825],[32.077589,34.774565],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.083678,34.775796],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.115119,34.818128],[31.749309,35.236047]];

现在我想从这个数组中找到重复的lat / long并生成包含重复数组列表的新数组。

我的代码：

for (var i = 0; i < locData.length; i++) {
        var title = JSON.stringify(titleData[i]);
        var id = JSON.stringify(ids[i]);
        title = title.replace(/[[\]/"]/g, "");
        id = id.replace(/[[\]/"]/g, "");
        var count = 0;
        var multiArray = [];    

        for(var j=0; j<locData.length; j++){
            if(locData[i][0] === locData[j][0] && locData[i][1] === locData[j][1]){
                count++;

                if(count>1){
                    //alert(locData[i]);
                    //console.log(locData[i]);
                    multiArray.push(locData[i]);

                }

            }
        }
        console.log(multiArray); 
}

但是没有产生结果，因为我希望它每次都是唯一的重复项吗？

Answer 1

请尝试以下代码：

locData=[[32.0770488,34.7958072],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.07849,34.773982],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077589,34.774565],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077795,34.774825],[32.077589,34.774565],[32.077795,34.774825],[32.077795,34.774825],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.077589,34.774565],[32.083678,34.775796],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.079743,34.770179],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.0776696,34.7683079],[32.115119,34.818128],[31.749309,35.236047]];
var dupArr = []
var templocData = locData.join("|");
$.each(locData, function(index, value) { 
    //console.log(index + ': ' + value);     
    //console.log($.inArray( value , locData  ));
    var searchVal = ""+value+"";
    //console.log(countInstances(templocData,searchVal));
    //console.log(templocData);     
    //console.log(searchVal);     
    if ( countInstances(templocData,searchVal) >= 1 && $.inArray( searchVal, dupArr) == -1 ) {
        dupArr.push(searchVal);
    }
});
function countInstances(string, word) {
   var substrings = string.split(word);
   return substrings.length - 1;
}
console.log(dupArr);

因此，结果将是：

["32.0770488,34.7958072", "32.0776696,34.7683079", "32.07849,34.773982", "32.077795,34.774825", "32.077589,34.774565", "32.083678,34.775796", "32.079743,34.770179", "32.115119,34.818128", "31.749309,35.236047"]

Answer 2

试试此代码

var multiArray = [];
for (var i = 0; i < locData.length; i++) {

        var count = 0;
        for(var j=0; j<locData.length; j++){
            if(locData[i][0] === locData[j][0] && locData[i][1] === locData[j][1]){
                count++;
                if(count>1){ 
                      var mult_indx = 0;
                      for(x in multiArray)
                      {
                          if(locData[j][0] == multiArray[x][0] && locData[j][1] == multiArray[x][1])
                          {
                              mult_indx++;
                          }
                      }
                      if(mult_indx==0)
                      {
                          multiArray.push(locData[i]);
                      }                    
                     locData.splice(j,2);
                }              
            }
        }
}

 console.log(multiArray);

您的结果将是

[[32.0776696,34.7683079]
 [32.07849,34.773982]
 [32.077795,34.774825]
 [32.077589,34.774565]
 [32.079743,34.770179]]

在jQuery中从数组中找到重复的lat long

2 个答案: