Increment operator returns NaN

Question

I am trying to increment a variable using the ++ operator but I keep getting NaN as a result and I'm not sure why. Here is my code:

var wordCounts = { };
var x = 0
var compare = "groove is in the heart";
        var words = compare.split(/\b/);
        for(var i = 1; i < words.length; i++){
            if(words[i].length > 2){
                wordCounts["_" + words[i]]++;
            }
        }


alert(wordCounts.toSource());

Answer 1

The value of wordCounts["_" + words[i]] is initially undefined so when you ++ it, it gives you NaN. Just change your code to:

if (wordCounts["_" + words[i]]) {
    wordCounts["_" + words[i]]++;
} else {
    wordCounts["_" + words[i]] = 1;
}

Answer 2

Try something like...

var key = "_" + words[i];

if (wordCounts[key]) {
    wordCounts[key]++
} else {
    wordCounts[key] = 1;
}

You are trying to increment undefined which is giving you NaN.

Answer 3

To be able to use the ++ operator (which takes a number and increments it by one) the target needs to have a number first.

Attempt a check to see if the object is defined, and if not initialize it by setting it's value to 1.

if ('undefined' === typeof wordCounts["_" + words[i]]) {
            wordCounts["_" + words[i]] = 0;
}

Something like:

var wordCounts = {};
var x = 0
var compare = "groove is in the heart";
var words = compare.split(/\b/);
for (var i = 1; i < words.length; i++) {
    if ('undefined' === typeof wordCounts["_" + words[i]]) {
        wordCounts["_" + words[i]] = 0;
    }
    if (words[i].length > 2) {
        wordCounts["_" + words[i]]++;
    }
}
alert( JSON.stringify( wordCounts ) );

Answer 4

What you're basically doing is

undefined++

Which will result in...

NaN

Try...

wordCounts["_" + words[i]] = (wordCounts["_" + words[i]]++ || 1);

Since NaN is a "falsey" value the || will fall back to 1.

Answer 5

You're trying to increment an object (wordCounts[]++) it's not a number so it can't be incremented, which is why you're getting that error. What are you actually trying to do (in plain English)?