I have an array:
var items = [[1,1,2,2,3,3,4,4,5,5],[0,0,0,1,0,1,0,1,0,1]];
Put another way, I want to loop through the elements of items[0] and items[1] and collect all the elements until I've found three unique values for items[0].
The output I want is:
output = [[1,1,2,2,3,3],[0,0,0,1,0,1]];
If I understand you correctly, then the following will do what you want.
function threeDistincts(items) {
var distincts = 0, inArr = false;
for (var i = 0; i < items[0].length; i++) {
inArr = false;
for (var j = 0; j < i; j++) {
if (items[0][j] === items[0][i]) {
inArr = true;
break;
}
}
if (!inArr) distincts++;
if (distincts === 4) break;
}
items[0].length = items[1].length = i;
}
Call it like:
var items = [[1,1,2,2,3,3,3,4,5,5],[0,0,0,1,0,1,0,1,0,1]];
threeDistincts(items);
// items = [[1,1,2,2,3,3],[0,0,0,1,0,1]];
Another version of this function, using indexOf (as DTing suggested), will limit the inner loop (in the original function) to the distinct elements only, but considering the high complexity of Array.indexOf and Array.push implementation, better performance is not guaranteed.
function threeDistincts(items) {
var arr = [];
for (var i = 0; i < items[0].length; i++) {
if (arr.indexOf(items[0][i]) === -1) arr.push(items[0][i]);
if (arr.length === 4) break;
}
items[0].length = items[1].length = i;
}
indexOf but coded with a for loop. O(n^2) time complexity.
for (...; j < i; ...)), if you want to use indexOf you will have to create a sub-array, which will possibly add greater overhead.1,2,..<repeating 2's>...,2,3,...<repeating 3's>...,3,4 The nested for loop will have to look through all the 2's before finding the 3's .Using the newer Set and Array.prototype.findIndex JavaScript features, which can be polyfilled via es6-shim for older browsers.
// limitTo3DistinctFromFirst :: [[a], [b]] -> [[a], [b]]
function limitTo3DistinctFromFirst(items) {
var uniques = new Set(),
indexOfFourth = items[0].findIndex(fourthUnique);
return items.map(sliceToFourth);
// fourthUnique :: a -> Boolean
function fourthUnique(el) {
if (uniques.has(el)) return false;
uniques.add(el);
return uniques.size === 4;
}
// sliceToFourth :: [a] -> [a]
function sliceToFourth(a) {
return a.slice(0, indexOfFourth);
}
}
You can use an object to track how many unique items you have seen if your objects are hashable:
If you want to stop after you add a third unique value:
var items = [[1,1,2,2,3,3,4,4,5,5],[0,0,0,1,0,1,0,1,0,1]];
function solve(items) {
var obj = Object.create(null);
var seen = 0;
var i = -1;
while (++i < items[0].length && seen < 3) {
var element = items[0][i];
if (element in obj) continue;
obj[element] = true;
seen++;
}
return [items[0].slice(0,i), items[1].slice(0,i)];
};
document.getElementById('result').innerHTML = JSON.stringify(solve(items));<pre id="result"></pre>If you want to break out of the loop when the 4th unique value is seen, you can do it like this:
var items = [[1,1,2,2,3,3,4,4,5,5],[0,0,0,1,0,1,0,1,0,1]];
function solve(items) {
var obj = Object.create(null);
var seen = 0;
var i = -1;
while (++i < items[0].length && seen < 4) {
var element = items[0][i];
if (element in obj) continue;
obj[element] = true;
seen++;
}
return [items[0].slice(0,i-1), items[1].slice(0,i-1)];
};
document.getElementById('result').innerHTML = JSON.stringify(solve(items));<pre id="result"></pre>You could use an object as a set to keep track of the number of unique values already selected. This has the benefit of preventing unnecessary iterations.
var func = function (data) {
var selectedValues = {};
var sliceIndex = 0;
while (Object.keys(selectedValues).length < 3 && sliceIndex < data[0].length) {
selectedValues[data[0][sliceIndex]] = true;
sliceIndex++;
}
return data.map(function (axis) {
return axis.slice(0, sliceIndex);
});
};
console.log(func([[1,1,2,2,3,3,4,4,5,5],[0,0,0,1,0,1,0,1,0,1]]));
items[1]? It just have 2 distinct elements .. I think you just truncated it to the same length ofitems[0].itemsever contain more than two arrays? If it does do you want to do the same thing to the rest of them or do you always want to just process the first two items?