查找单词何时在数组中彼此相邻出现

Question

I have a huge array of strings (words) that I'm analyzing for patterns.

I want to create a function to:

• Identify when words are appearing next to each other in a specific order more than once
• For each instance that the words appear in order, combine them into a single array element.

Example

Given the following array

let array = ["john", "smith", "says", "that", "a", "lock", "smith", "can", "open", "the", "lock", "unlike", "john", "smith"]

Desired Result:

["john smith", "says", "that", "a", "lock", "smith", "can", "open", "the", "lock", "unlike", "john smith"]

Ideally the function identifies more than just 2-word combinations (ie identifies when the combination of "white", "house", "press", "secretary" are appearing more than once.

I'm really struggling with the logic to have much to show. I've also been looking for a solution in a library like underscore.js without luck.

Answer 1

Build a "dictionary" of a all words and their immediate successor. Then loop through the original array and for each element, check if all dictionary returns match, and if so, combine the words and skip the immediate successor.

 var arr = ["john", "smith", "says", "that", "a", "lock", "smith", "can", "open", "the", "lock", "unlike", "john", "smith"]; function combineCommon(arr) { var dictionary = {}; for (var a = 0; a < arr.length - 1; a++) { var A = arr[a]; if (dictionary[A] == void 0) { dictionary[A] = []; } dictionary[A].push(arr[a + 1]); } var res = []; for (var index = 0; index < arr.length; index++) { var element = arr[index]; var pass = false; if (dictionary[element].length > 1) { if (dictionary[element] .some(function(a) { return a != dictionary[element][0]; }) == false) { pass = true; } } if (pass) { res.push(arr[index] + " " + dictionary[element][0]); index++; } else { res.push(arr[index]); } } return res; } console.log(combineCommon(arr)); 

Answer 2

You could count the pairs and check for pairs when reassembling the result.

 var array = ["john", "smith", "says", "that", "a", "lock", "foo", "bar", "baz", "smith", "can", "open", "foo", "bar", "baz", "the", "lock", "unlike", "john", "smith"], count = Object.create(null), result; array.forEach(function (a, i, aa) { var key = aa.slice(i, i + 2).join(' '); count[key] = (count[key] || 0) + 1; }); result = array.reduce(function (r, a, i, aa) { var key = aa.slice(i, i + 2).join(' '); if (count[key] > 1) { a = key; } else if (count[aa.slice(i - 1, i + 1).join(' ')] > 1) { a = []; } return r.concat(a); }, []); console.log(result); 

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Answer 3

Please check this.

 var data = ["john", "smith", "says", "that", "a", "lock", "smith", "can", "open", "the", "lock", "unlike", "john", "smith"] var result= []; var flag=0; var n=data.length; var k=0; // Outer main for loop. for(var i=0;i<n;i++){ // Get next word. next_word = data[i+1]; flag=0; // Inner for loop. for(var j=0;j<n;j++){ // john == john && smith == smith // smith == john && smith == smith // .. // .. if(data[j]==data[i] && data[j+1]==next_word){ flag++; temp_word = data[i]+' '+next_word; } } // If flag more than 1 that means same word sequence found more than one time. if(flag>1){ result[k++]=temp_word; // Assign temp_word to result array. i++; // increase outer loop by one so double entry we can restrict. }else{ // If no sequence found then pass outer value to result value as it is. result[k++]=data[i]; } } console.log(result);