Related
//code1
let a= [1, 3 , 4, 6];
[7, 8 , 9].forEach(l => a.push(l));
console.log(a);
// [1, 3, 4, 6, 7, 8, 9 ]
1.it worked for push() function
//code2
let a= [1, 3 , 4, 6];
a.forEach(l => a.pop(l));
console.log(a);
//[ 1, 3 ]
2. didn't work for pop() though
Javascript Array.pop() removes the last element from the array and returns that.
Example:
var arr = [1,2,3]
arr.pop(); // returns 3
Reference
If you want to remove a element with specific value than try something like:
var arr = [1, 2, 3];
var index = arr.indexOf(1);
if (index > -1) {
array.splice(index, 1);
}
var arr = [1, 2, 3, 4];
console.log(arr.pop());
var index = arr.indexOf(2);
if (index > -1) {
arr.splice(index, 1);
}
console.log(arr)
forEach automatically extracts the elements one by one and gives them to you
It starts from the beginning of the array, and does them all.
It doesn't delete elements from the array.
a = [1, 3, 4, 6];
a.forEach(item => console.log(item));
// output is in forwards order
// and 'a' retains original contents
pop() extracts and deletes one element for you
It starts from the end of the array, and does only one.
It deletes the element from the array.
a = [1, 3, 4, 6];
while (a.length > 0) {
console.log(a.pop())
}
// items come out in reverse order
// and 'a' is being emptied so it is [] at the end
Choose your method
Do you want the last element actually removed from the array? This is what you would want if you were implementing a stack, for example. In that case, use ".pop()".
This gets one element from the end of the array and deletes it from the array.
Or do you want to just look at each element in turn from the array (starting at the beginning), without changing the array itself. This is a commoner situation. In this case, use ".forEach"
This is what I'm trying to do, I have an array
var arr = [1, 2, 3, 4, 5];
then I want to create a new array each time by removing an item once i.e when i remove item at index 0 i should have [2, 3, 4, 5]and when i remove an item at index 1, I should have [1, 3, 4, 5] and so on till i get to arr.length-1 and each time i remove an item i still want my arr to be intact unchanged
using javaScript I have tried some array methods like splice, slice but all that changes the value of arr
how do i go about it with either javascript or python.
For Javascript, using ES6 array spread operator and slice method,
var new_array = [...a.slice(0, index), ...a.slice(index + 1)];
const cut = (a, i) => [...a.slice(0, i), ...a.slice(i + 1)];
let arr = [2, 2, 2, 4, 2];
console.log(cut(arr, 3));
console.log(arr);
For Python:
array = [1,2,3,4,5];
newarray = [value for counter, value in enumerate(array) if counter != 0 ]
PS each time you will use this list-comprehension, array will not be modified! so basically you will get the same output for newarray.
If you want to have newarray each time removed one element you need to create a function instead of list-comprehension (of course it's possible but will likely be less readable).
For JavaScript:
Try making a copy with slice() (slice returns a shallow copy of the array that you can manipulate without affecting the original array) and then using splice() to remove the value at your desired index:
newArray = slice(arr).splice(index, 1);
I have an array input and another array indexes. I want to remove item from array input whose index is provided in indexes array.
I have tried it using array.splice in for loop but as item is being removed in each iteration, indexes of other items are being changed.
JavaScript:
var array = [10, 11, 12, 13, 14, 15];
var indexes = [0, 1, 2, 3, 4, 5];
indexes.forEach(function(item) {
array.splice(item, 1);
});
console.log(array);
You can utilize Array.prototype.filter and do the following:
var array = [10, 11, 12, 13, 14, 15];
var indexes = [0, 1, 2, 3, 4, 5];
array = array.filter(function(x, i) {
return indexes.indexOf(i) === -1;
});
console.log(array);
Here you are using forEach loop which give you the item as first argument and the index on second one, so as per my understanding what you want to do can achieve by this, try this hope this solve your problem :)
indexes.forEach(function(item, index) {
array.splice(index, 1);
});
Sort the indexes array from high to low, then spice will only change the index of the numbers you have already removed
I want to replace elements in some array from 0 element, with elements of another array with variable length. Like:
var arr = new Array(10), anotherArr = [1, 2, 3], result;
result = anotherArr.concat(arr);
result.splice(10, anotherArr.length);
Is there some better way?
You can use the splice method to replace part of an array with items from another array, but you have to call it in a special way as it expects the items as parameters, not the array.
The splice method expects parameters like (0, anotherArr.Length, 1, 2, 3), so you need to create an array with the parameters and use the apply method to call the splice method with the parameters:
Array.prototype.splice.apply(arr, [0, anotherArr.length].concat(anotherArr));
Example:
var arr = [ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'];
var anotherArr = [ 1, 2, 3 ];
Array.prototype.splice.apply(arr, [0, anotherArr.length].concat(anotherArr));
console.log(arr);
Output:
[ 1, 2, 3, 'd', 'e', 'f', 'g', 'h', 'i', 'j']
Demo: http://jsfiddle.net/Guffa/bB7Ey/
In ES6 with a single operation, you can do this to replace the first b.length elements of a with elements of b:
let a = [1, 2, 3, 4, 5]
let b = [10, 20, 30]
a.splice(0, b.length, ...b)
console.log(a) // -> [10, 20, 30, 4, 5]
It could be also useful to replace the entire content of an array, using a.length (or Infinity) in the splice length:
let a = [1, 2, 3, 4, 5]
let b = [10, 20, 30]
a.splice(0, a.length, ...b)
// or
// a.splice(0, Infinity, ...b)
console.log(a) // -> [10, 20, 30], which is the content of b
The a array's content will be entirely replaced by b content.
Note 1: in my opinion the array mutation should only be used in performance-critical applications, such as high FPS animations, to avoid creating new arrays. Normally I would create a new array maintaining immutability.
Note 2: if b is a very large array, this method is discouraged, because ...b is being spread in the arguments of splice, and there's a limit on the number of parameters a JS function can accept. In that case I encourage to use another method (or create a new array, if possible!).
In ES6, TypeScript, Babel or similar you can just do:
arr1.length = 0; // Clear your array
arr1.push(...arr2); // Push the second array using the spread opperator
Simple.
For anyone looking for a way to replace the entire contents of one array with entire contents of another array while preserving the original array:
Array.prototype.replaceContents = function (array2) {
//make a clone of the 2nd array to avoid any referential weirdness
var newContent = array2.slice(0);
//empty the array
this.length = 0;
//push in the 2nd array
this.push.apply(this, newContent);
};
The prototype function takes an array as a parameter which will serve as the new array content, clones it to avoid any weird referential stuff, empties the original array, and then pushes in the passed in array as the content. This preserves the original array and any references.
Now you can simply do this:
var arr1 = [1, 2, 3];
var arr2 = [3, 4, 5];
arr1.replaceContents(arr2);
I know this is not strictly what the initial question was asking, but this question comes up first when you search in google, and I figured someone else may find this helpful as it was the answer I needed.
You can just use splice, can add new elements while removing old ones:
var arr = new Array(10), anotherArr = [1, 2, 3];
arr.splice.apply(arr, [0, anotherArr.length].concat(anotherArr))
If you don't want to modify the arr array, you can use slice that returns a shallow copy of the array:
var arr = new Array(10), anotherArr = [1, 2, 3], result = arr.slice(0);
result.splice.apply(result, [0, anotherArr.length].concat(anotherArr));
Alternatively, you can use slice to cut off the first elements and adding the anotherArr on top:
result = anotherArr.concat(arr.slice(anotherArr.length));
I'm not sure if it's a "better" way, but at least it allows you to choose the starting index (whereas your solution only works starting at index 0). Here's a fiddle.
// Clone the original array
var result = arr.slice(0);
// If original array is no longer needed, you can do with:
// var result = arr;
// Remove (anotherArr.length) elements starting from index 0
// and insert the elements from anotherArr into it
Array.prototype.splice.apply(result, [0, anotherArr.length].concat(anotherArr));
(Damnit, so many ninjas. :-P)
You can just set the length of the array in this case. For more complex cases see #Guffa's answer.
var a = [1,2,3];
a.length = 10;
a; // [1, 2, 3, undefined x 7]
I've got a collection of 20 results (objects), and what I'd like to do when a button is clicked is to:
a) Pick a random object from this collection/array
b) When the button is pressed again - I don't want that object re-picked until the collection is exhausted (i.e. until the 20 items are shown)
I thought of just splicing out the index of that collection, but I'm hoping for a cleaner way using Underscore.js
EXAMPLE:
var data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11...]
var getRand = _.random(0, data.length);
==> 3
Next time I press the button, I don't want the result "3" to re-appear as it's been used
I hope this makes sense
var data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
// cache indexes
var cache = _.map(new Array(data.length + 1).join(), function (item, index) {
return index;
});
// get random from cached array
var rand = _.random(0, cache.length);
// remove random index from cache
cache.splice(rand, 1);
console.log(rand, cache)
var data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
var picked = [];
$("#link").click(function() {
if(data.length == 0) return;
var pick = data.splice(_.random(0,data.length),1);
picked.push(pick);
$("#pick").html(pick);
$("#data").html(data.join(","));
$("#picked").html(picked.join(","));
});
http://jsfiddle.net/Z3vjk/
You could make an array to store the values you've used and check all new random numbers to see if they appear. This would get messy near the end of the array though as the random number generator tries to guess a single number.
If it were me I would just what you alluded to and take the elements out as you use them and place them into a temporary array. Once all elements are used, reassign the temp array to the original variable name.