How can I generate 8, say, unique random numbers between 1 and 100 using JavaScript?
相关问题
- Is there a limit to how many levels you can nest i
- How to toggle on Order in ReactJS
- void before promise syntax
- Do the Java Integer and Double objects have unnece
- Keeping track of variable instances
for arrays with holes like this
[,2,,4,,6,7,,]
because my problem was to fill these holes. So I modified it as per my need :)the following modified solution worked for me :)
Same permutation algorithm as The Machine Charmer, but with a prototyped implementation. Better suited to large number of picks. Uses js 1.7 destructuring assignment if available.
Edit: An other proposition, better suited to small number of picks, based on belugabob's answer. To guarantee uniqueness, we remove the picked numbers from the array.
if you need more unique you must generate a array(1..100).
above code is faster:
extractUniqueRandomArray(50)=> [2, 79, 38, 59, 63, 42, 52, 22, 78, 50, 39, 77, 1, 88, 40, 23, 48, 84, 91, 49, 4, 54, 93, 36, 100, 82, 62, 41, 89, 12, 24, 31, 86, 92, 64, 75, 70, 61, 67, 98, 76, 80, 56, 90, 83, 44, 43, 47, 7, 53]
You can also do it with a one liner like this:
[...((add, set) => add(set, add))((set, add) => set.size < 8 ? add(set.add(Math.floor(Math.random()*100) + 1), add) : set, new Set())]
Shuffling the numbers from 1 to 100 is the right basic strategy, but if you need only 8 shuffled numbers, there's no need to shuffle all 100 numbers.
I don't know Javascript very well, but I believe it's easy to create an array of 100 nulls quickly. Then, for 8 rounds, you swap the n'th element of the array (n starting at 0) with a randomly selected element from n+1 through 99. Of course, any elements not populated yet mean that the element would really have been the original index plus 1, so that's trivial to factor in. When you're done with the 8 rounds, the first 8 elements of your array will have your 8 shuffled numbers.
I would do this: