我不知道如何在javascript中實現此功能

Question

非常感謝您所做的貢獻，我與所有人共同做出了混合，部分解決了問題，但是我只剩下一件事了……我完成了此功能，添加了num_tweets，當添加了“ words”時是1，但是當它匹配並且應該+ 1時，該變量是未定義的，我不明白原因..問題必須從這些JSON執行此操作

{
        "numtweets" : 1,
        "dic_words" : {
                "word1" : 1,
                "word2" : 1,
                "word3" : 1,
                "word4" : 1,
                "word5" : 1,
                "word6" : 1,
                "word7" : 1
        }
}

{
        "numtweets" : 1,
        "dic_words" : {
                "word1" : 1,
                "word2" : 1,
                "word9" : 1
        }
}

{
        "numtweets" : 1,
        "dic_words" : {
                "word1" : 1,
                "word3" : 1,
                "word10" : 1
        }
}

我希望最后我返回一個僅添加所有字段，添加contweet的json，而且在這種情況下匹配的單詞也會這樣顯示...

{
            "numtweets" : 3, //add numtweets
            "dic_words" : {
                    "word1" : 3, //num_occurs word1 in array
                    "word2" : 2, //num_occurs word2 in array
                    "word3" : 2, //num_occurs word3 in array
                    "word4" : 1,  //num_occurs word3 in array
                    "word5" : 1,
                    "word6" : 1,
                    "word7" : 1,
                    "word9" : 1,
                    "word10" : 1
            }
    }

我的功能是...

var r =  function(key, values) {
    result = { "numtweets" : 0,
          "dic_words" : {}
        };
      //   print("  entrada: " + tojson(values));
   for (var idx = 0; idx < values.length; idx++) {
     result.numtweets += values[idx].numtweets;
   //  print("  json: " + tojson(values[idx].dic_words));
        for (paraula in values[idx].dic_words) {
            if(values[idx].dic_words.hasOwnProperty(paraula)) {
            if(values[idx].dic_words[paraula]) {
                           result.dic_words[paraula] = result.dic_words[paraula] + values[idx].dic_words[paraula]; //IS these **part result.dic_words[paraula]** is ***undefined***
           }
            else 
               result.dic_words[paraula] = 1; /^* this part is correct, and if the word appers one only it appears ocurrs

               }
        }   
        }
return result;
};

輸出是

{
                "numtweets" : 3, //add numtweets
                "dic_words" : {
                        **"word1" : NaN, //num_occurs word1 in array MUST BE 3
                        "word2" : NaN, //num_occurs word2 in array MUST BE 2
                        "word3" : NaN, //num_occurs word3 in array MUST BE 2**
                        "word4" : 1,  //num_occurs word3 in array
                        "word5" : 1,
                        "word6" : 1,
                        "word7" : 1,
                        "word9" : 1,
                        "word10" : 1
                }
        }

請任何建議或幫助將受到歡迎...

Answer 1

 var data = [{ "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word3" : 1, "word4" : 1, "word5" : 1, "word6" : 1, "word7" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word9" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word3" : 1, "word10" : 1 } }]; function processData(input) { var result = {numtweets: 0, dic_words:{}}; input.forEach(function(item){ result['numtweets'] += item['numtweets']; for(var i in item['dic_words']) { result['dic_words'][i] = (result['dic_words'][i]) ? result['dic_words'][i]+item['dic_words'][i] : item['dic_words'][i]; } }) return result; } console.log(processData(data)) 

Answer 2

您可以使用以下功能：

 function getSummary(data) { return data.reduce(function (acc, o) { return { numtweets: acc.numtweets + (+o.numtweets || 0), dic_words: Object.keys(o.dic_words || {}).reduce(function (words, word) { words[word] = (words[word] || 0) + o.dic_words[word]; return words; }, acc.dic_words) }; }, { numtweets: 0, dic_words: {} }); } // Sample data var data = [{ "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word3" : 1, "word4" : 1, "word5" : 1, "word6" : 1, "word7" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word9" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word3" : 1, "word10" : 1 } }, { // empty object added to test algorithm will survive and ignore this... }]; console.log(getSummary(data)); 

 .as-console-wrapper { max-height: 100% !important; top: 0; } 

Answer 3

  var arr = [ { "numtweets": 1, "dic_words": { "word1": 1, "word2": 1, "word3": 1, "word4": 1, "word5": 1, "word6": 1, "word7": 1 } }, { "numtweets": 1, "dic_words": { "word1": 1, "word2": 1, "word9": 1 } }, { "numtweets": 1, "dic_words": { "word1": 1, "word3": 1, "word10": 1 } } ] var newObj = { numtweets: 0, dic_words: {} }; arr.forEach(function (item, index, array) { newObj.numtweets += item.numtweets; for (var prop in item.dic_words) { if (!newObj.dic_words[prop]) { newObj.dic_words[prop] = 0 }; newObj.dic_words[prop] += item.dic_words[prop]; } }) console.log(newObj); 

Answer 4

將Array.map（）方法與ES6 Arrow函數一起使用。

DEMO

 var CountObjVal = [{ "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word3" : 1, "word4" : 1, "word5" : 1, "word6" : 1, "word7" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word2" : 1, "word9" : 1 } }, { "numtweets" : 1, "dic_words" : { "word1" : 1, "word3" : 1, "word10" : 1 } }]; var r = function (countObjVals) { var obj = {numtweets: 0, dic_words: {}}; CountObjVal.map(elem => Object.keys(elem.dic_words).map(item => obj.dic_words[item] = 0)); for (var i in CountObjVal) { obj.numtweets += CountObjVal[i].numtweets; Object.keys(CountObjVal[i].dic_words).map(item => obj.dic_words[item] += CountObjVal[i].dic_words[item]); } return obj; }; console.log(r(CountObjVal));