I wrote a simple enough randomize function that loops through the elements in an array and displays them one after the other.
function changeSubTitle() {
var whatAmI = ["Webdesigner", "Drummer", "Techie", "Linguistics student", "Photographer", "Geek", "Coder", "Belgian", "Batman", "Musician", "StackExchanger", "AI student"];
setTimeout(function () {
$(".page-header > h2").animate({
"opacity": 0
}, 700, function () {
$(this).text(whatAmI[Math.floor(Math.random() * whatAmI.length)]);
$(this).animate({
"opacity": 1
}, 700, changeSubTitle);
});
}, 1000);
}
However, obviously it is very well possible that the same element is displayed twice, one immediately after the other. This happens because I randomize the array each time I call the function. How would I prevent an element to be displayed two times right after each other?
I suppose the most straightforward way to do this is to get the randomize function out the loop, run the loop and each time an element is called, remove the index from the array and refill the array when it's empty. A lot of questions on SO consider this problem, but not as specific as mine: I'm not sure how to do this in the loop to display each element.
Stop upvoting please :-D
The following answer is better: https://stackoverflow.com/a/32395535/1636522. The extended discussion between me and Tim enlighten on why you should avoid using my solutions. My answer is only interesting from this point of view, hence, it does not deserve any upvote :-D
You could save the last integer and "while" until the next one is different:
Or even simpler, just add 1... :-D Modulo n to prevent index overflow:
First solution applied to your code:
To give an even chance of all elements except the previous one being used do the following:
The comments in the code should explain how this works. The key thing to remember is that you are actually selecting from 9 possible values, not from 10.
You should initialize i to be a random element in the array before starting.
For a simple walk through on a 3 element array with the second element selected:
i=1, n=3
The random result gives us either 0 or 1.
If it is 0 then
j >= i
returns false and we select element zeroIf it is 1 then
j >= i
returns true and we select the third element.You can do the same walk through with i being 0 and and i being 2 to see that it never overruns the buffer and always has an equal chance to select all other elements.
You can then extend that same logic to an array of any size. It works exactly the same way.
Try this method.