I'm trying to edit an array and remove elements that do not meet a certain condition. If I use a reverse for loop
combined with .splice(index,n)
, the code works just fine. I'm stuck at implementing the same using the ES6 for...of
loop
let array=[1,2,3];
//reverse for loop
for(var i=array.length - 1; i>=0; i--) {
if(array[i]>=2) {
/*Collect element before deleting it.*/
array.splice(i,1);
}
}
console.log(array);
// returns [1]
Using for..of
let array=[1,2,3];
for(let entry of array) {
if(entry>=2) {
let index = array.indexOf(entry);
/*The array is being re-indexed when I apply
.splice() - the loop will skip over an index
when one element of array is removed*/
array.splice(index, 1);
}
}
console.log(array);
//returns [1,3]
Is there a way to achieve this functionality using a for...of
loop or do I have to stick to the reverse for loop
Update
I need to collect the entries that don't meet the elements removed by either the filter()
or reverse for loop
functions to a secondary array.
Use unset() function to remove array elements in a foreach loop. The unset() function is an inbuilt function in PHP which is used to unset a specified variable.
pop() function: This method is use to remove elements from the end of an array. shift() function: This method is use to remove elements from the start of an array. splice() function: This method is use to remove elements from the specific index of an array.
Find the index of the array element you want to remove using indexOf , and then remove that index with splice . The splice() method changes the contents of an array by removing existing elements and/or adding new elements. The second parameter of splice is the number of elements to remove.
Use the for Loop to Remove Element From Array and Shift in Java. In this approach, we use multiple for loops to loop over all elements of the arr and check for the element that we want to remove. When the element to be deleted is found, we create a new Array newArr with the size arr. length-1 .
You can't reasonably use for-of
for this. If you remove the "current" entry during the iteration, you'll skip the next entry, because of the way array iterators are specified (they use the index of the entry). You can see that with this contrived example:
const array = [1, 2, 3];
for (const entry of array) {
console.log(entry);
if (entry === 2) {
array.splice(1, 1);
}
}
console.log(array);
Notice how there was no loop iteration for entry 3
.
I'd suggest either sticking with your reverse for
loop or using filter
to produce a new array containing only the entries you want to keep.
Using filter
:
let array = [1, 2, 3];
array = array.filter(entry => entry < 2);
console.log(array);
I said "reasonably" above because, of course, there's always a way. You could loop through a copy of the array and maintain the index
outside it:
const array = [1, 2, 3];
let index = 0;
for (const entry of [...array]) {
if (entry >= 2) {
array.splice(index, 1);
} else {
++index;
}
}
console.log(array);
That doesn't seem reasonable compared to the alternatives, unless of course there are constraints pushing you that way. :-)
According to @T.J's answer, when using a for...of loop:
If you remove the "current" entry during the iteration, you'll skip the next entry, because of the way array iterators are specified (they use the index of the entry).
This leaves two other options, using a reverse for loop
and a filter
function. I mentioned earlier that I need to do an operation with the current array element before deleting it.
1. Using .filter()
function
and referring to @T.J's Comment
let array = [1,2,3];
let collection = [];
array = array.filter(function(entry) {
if(entry>=2) {
collection.push(entry);
}
return entry <2;
});
console.log(collection); //returns [2,3]
console.log(array); //returns [1]
2. Using a reverse for loop
let array = [1,2,3];
let collection = [];
for(var i=array.length - 1; i>=0; i--) {
if(array[i]>=2) {
collection.push(array[i]);
array.splice(i,1);
}
}
console.log(collection); //returns [2,3]
console.log(array); //returns [1]
The filter()
function in this case requires an additional step to temporarily hold the elements that do not meet the condition. The reverse for loop
offers a more cleaner way to achieve the same.
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