如何在数组javascript中获取连续的空格

Question

I have this function:

 function insertValue2(cantidad, value, arr){

  var  spaceCount = 0;

    arr.forEach(function(item, index) { 
      if(item == "") {
        console.log(index)
        spacecount++;
      }
    });
  console.log(spaceCount)

My function counts blank spaces in this array : ["", "", "", "B", "B", "B", ""]

So the result for this: 0 1 2 6 are the positions in the array with blank spaces and spacecount = 4

I dont know if this is possible but any idea how to count blank space before i get the first B??

I mean count for consecutive blank spaces like 0 1 2 spacecount = 3 and then count space for 6 spacecount = 1

And if i want to insert in my array quantity = 1 with a value = C that will choose the lowest value for spacecount.

["", "", "", "B", "B", "B", "C"]

EDIT:

Quantity is how many space i will use in the array, i dont want to use more space than the necessary for the quantity

In this array i have blank space in positions 0, 1, 2, 7 , 8, 9, 10

["", "", "", "B", "B", "B", "C", "" , "" ,"" ,""]

if i want to insert quantity = 2 and value = D , the expected result is:

["D", "D", "", "B", "B", "B", "C", "" , "" ,"" ,""]

and if i want to insert quantity = 1 and value = "E" it will choose the position 2 with the spacecount = 1 to save the higher spacecount for a bigger quantity like 3 or 4

["D", "D", "E", "B", "B", "B", "C", "" , "" ,"" ,""]

Thanks for the help :)

Answer 1

You can just iterate over the array once, checking if item and keeping array of arrays, where each item is group of indexes having the desired value, in your case empty string.

One way would be by extending the Array object itself: (see the console for results)

 Array.prototype.groupBy = function(value) { var array = this; var groups = []; var buffer = []; for (var i = 0; i < array.length; i++) { var curItem = array[i]; if (curItem == value) { buffer.push(i); } else if (buffer.length > 0) { groups.push(buffer); buffer = []; } } if (buffer.length > 0) groups.push(buffer); return groups; }; var a = ["", "", "", "B", "B", "B", ""]; var consecutiveBlankSpaces = a.groupBy(""); console.log('total of ' + consecutiveBlankSpaces.length + ' groups of blank spaces'); for (var i = 0; i < consecutiveBlankSpaces.length; i++) { console.log('Found a blank space group consisting of ' + consecutiveBlankSpaces[i].length + ' items, indexes: ' + consecutiveBlankSpaces[i]); } 

Answer 2

 var arr = ["", "", "", "B", "B", "B", "C", "", "", "", ""]; var groups = {}; var groupsIndex = 0; var previouslyUpgraded = false; arr.forEach(function(item, index) { if (item === "") { groups["group" + groupsIndex] = groups["group" + groupsIndex] || []; groups["group" + groupsIndex].push(index); previouslyUpgraded = false; } else if (!previouslyUpgraded) { groupsIndex++; previouslyUpgraded = true; } }); console.log(groups); console.log(Object.keys(groups).length + " groups found !"); for (var key in groups) { console.log(key + " has empty string at indexes : " + groups[key]); } 

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

Answer 3

Try using a counter.

I used the counter for checking if im in a whitespace or an another character for the first time. If counter=0 -> quantity++ and counter=1; Also, if the loop finds a notwhitespace the counter change again to 0.

var arr = ["", "", "", "B", "B", "B", "C", "" , "" ,"" ,""];

var spacecount=0;
var quantity=0;
var contador=0;

arr.forEach(function(item, index) { 
  if(item == "") {
    if(contador==0){
      quantity++;
      contador=1;
      spacecount=0;
    }
    spacecount++;
  }else{
    contador=0;
  }
});

console.log('spacecount: '+spacecount);
console.log('quantity: '+quantity);

Answer 4

This proposal looks first for any spaces, count them and try to minimise the left over empty spaces in a slot. Then apply the value to the slot.

 function insert(array, value, count) { var i, spaces = data.reduce(function (r, a, i, aa) { if (a === '') { if (aa[i - 1] === '') { r[r.length - 1].count++; } else { r.push({ index: i, count: 1 }); } } return r; }, []), index = spaces.reduce(function (r, a) { return a.count < count || r.count < a.count ? r : a; }, {}).index; if (index !== undefined) { for (i = 0; i < count; i++) { array[i + index] = value; } } else { // throw no space error } } var data = ["", "", "", "", "B", "B", "B", "C", "", "", ""]; insert(data, 'X', 5); // throw no space error, if implemented console.log(data); // ["", "", "", "", "B", "B", "B", "C", "", "", ""] insert(data, 'D', 2); console.log(data); // ["", "", "", "", "B", "B", "B", "C", "D", "D", ""] insert(data, 'E', 1); console.log(data); // ["", "", "", "", "B", "B", "B", "C", "D", "D", "E"] 

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

Answer 5

 function CountSpaces(a) { var result = []; var positions = []; for(var i = 0, len = a.length; i<len; i++) { if(a[i] != "" && positions.length > 0) { result.push({ positions: positions, count: positions.length }); positions = []; lastCharacter = a[i]; continue; } if(a[i] == "") positions.push(i); } if(positions.length > 0) { result.push({ positions: positions, count: positions.length }); } return result; } var a1 = ["", "", "B", "", "", "C", "D", "", "", ""]; console.log(a1, CountSpaces(a1));