I have an array of 50 objects as elements.
Each object contains an array of 4 elements:
var all = [{
question: "question 1 goes here",
options: ["A", "B", "C", "D"]
}, ... {
question: "question 50",
options: ["A", "B", "C", "D"]
}]
I want to select 10 elements randomly and save to two other arrays, one of the array I want to shuffle options. but when shuffling both arrays are affected.
var selected = [];
var shuffled = [];
for(let i = 0; i < 10; i++) {
let rand = Math.floor(Math.random() * all.length);
selected.push(all[rand]);
shuffled.push(all[rand]);
all.splice(rand, 1);
for(let j = 3; j > 0; j--) {
let rand2 = Math.floor(Math.random() * j);
[
shuffled[i].options[j],
shuffled[i].options[rand2]
] = [
shuffled[i].options[rand2],
shuffled[i].options[j]
];
}
}
console.log(selected); // is shuffled too
console.log(shuffled);
How do I prevent that?
I feel like I'm missing something pretty simple, but I can't spot it.
You need to create new instances for the chosen objects and their options
arrays:
// shuffle array a in place (Fisher-Yates)
// optional argument len (=a.length):
// reduced shuffling: shuffling is done for the first len returned elements only,
// array a will be shortened to length len.
function shuffle(a, len=a.length){
for(let m=a.length,n=Math.min(len,m-1),i=0,j;i<n;i++){
j=Math.floor(Math.random()*(m-i)+i);
if (j-i) [ a[i],a[j] ] = [ a[j],a[i] ]; // swap 2 array elements
}
a.length=len;
return a;
}
const all=[...new Array(50)].map((_,i)=>({question:"Question "+(i+1), options:["A","B","C","D"]}));
const selected = shuffle([...all],10), // return first 10 shuffled elements only!
shuffled = selected.map(o=>({...o,options:shuffle([...o.options])}));
console.log(selected) // is no longer shuffled!
console.log(shuffled);
I consigned the shuffle algorithm to a separate function (shuffle()
) and applied it twice: first to the all
array to make sure we don't get any duplicates in our "random" selection and then to the options
arrays contained within their sliced-off objects. The function shuffle(a,len)
sorts array a
in place. I made it return the array reference again purely out of convenience, as it helps me keep my code more compact. The optional argument len
will cause the array a
to be shortened to len
shuffled elements (still "in place": the input array will be affected too!).
So, in order to preserve my "input" arrays I created new array instances each time I called the function by applying the ...
operator:
shuffled = shuffle([...all],10);
...
shuffle([...o.options])