I have an array like this
var array = [1,2,3,4,5,6,7,8,9,10];
Looping code is like this using _.each function in underscore.js
_.each(array,function(item,index){
console.log(item);
});
But I want remove some items in array when looping. For example I need to remove number 5 from array and loop does not print number 5. The question is, is it possible to remove items in array when looping on this array?
It's usually a very bad idea to modify an array while iterating over it. The best solution is to store the indexes in a separate array and remove them afterwards (remember to iterate over that array from last to first so you don't have to deal with changing indexes).
2 ways, I'd recommend the first one.
var array = [1,2,3,4,5,6,7,8],
items_to_remove = [], i;
_.each(array, function(item,index){
if(item === 'something'){
items_to_remove.push(index);
}
});
while((i = items_to_remove.pop()) != null){
array.splice(i, 1);
}
// OR
for(i = array.length - 1; i > -1; --i) {
if(array[i] === 'something') {
array.splice(i, 1);
}
}
With underscore you could do this:
var array = [1,2,3,4,5,6,7,8,9,10];
var filteredList = _.filter(array, function(item){
// do something with each item - here just log the item
console.log(item);
// only want even numbers
return item % 2 == 0;
});
// show filtered list
_.each(filteredList, function(item){
console.log(item);
});
var filteredList = _.filter(...) with array = _.filter(...) and the original array is mutated.You can use Undersocre to reduce the line of code. Underscore is an handy-tool.
var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
array = _.without(array, _.findWhere(array, {
id: 3
}));
console.log(arr);Here's a simple inline solution that worked for me using underscore's reject function:
_.each(_.reject(array, function(arrayItem) { return arrayItem === 5}), function(item){
console.log(item);
});
.each()because (again I'm not sure about underscore, but for some other equivalents) the length of the array may be cached at the beginning so then if you remove items the iterator will run off the end. It's no problem if iterating with a traditionalforloop.i--whenever he removes the current element ;)