You need to copy the array before you sort it. One way with es6: const sorted = […arr].sort(); The spread-syntax as array literal (copied from mdn): var arr = [1, 2, 3]; var arr2 = […arr]; // like arr.slice() https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_operator
If you are using java-8 there’s also another way to do this. int[] arr = list.stream().mapToInt(i -> i).toArray(); What it does is: getting a Stream<Integer> from the list obtaining an IntStream by mapping each element to itself (identity function), unboxing the int value hold by each Integer object (done automatically since Java 5) getting the … Read more
I think I’ve figured it out: a = [3, 2, 4, 6, 3, 8] a.delete(3) #=> 3 a #=> [2, 4, 6, 8]
One possibility: myArray = myArray.filter(function( obj ) { return obj.field !== ‘money’; }); Please note that filter creates a new array. Any other variables referring to the original array would not get the filtered data although you update your original variable myArray with the new reference. Use with caution.
As of Swift 3 / 4 / 5, this is done as follows. To add a new element to the end of an Array. anArray.append(“This String”) To append a different Array to the end of your Array. anArray += [“Moar”, “Strings”] anArray.append(contentsOf: [“Moar”, “Strings”]) To insert a new element into your Array. anArray.insert(“This String”, at: … Read more
(A==B).all() test if all values of array (A==B) are True. Note: maybe you also want to test A and B shape, such as A.shape == B.shape Special cases and alternatives (from dbaupp’s answer and yoavram’s comment) It should be noted that: this solution can have a strange behavior in a particular case: if either A … Read more
Changing Values: array_map cannot change the values inside input array(s) while array_walk can; in particular, array_map never changes its arguments. Array Keys Access: array_map cannot operate with the array keys, array_walk can. Return Value: array_map returns a new array, array_walk only returns true. Hence, if you don’t want to create an array as a result … Read more
$arr[$newkey] = $arr[$oldkey]; unset($arr[$oldkey]);
Objects There is no benefit to using new Object(); – whereas {}; can make your code more compact, and more readable. For defining empty objects they’re technically the same. The {} syntax is shorter, neater (less Java-ish), and allows you to instantly populate the object inline – like so: var myObject = { title: ‘Frog’, … Read more
If you are going to use a collection that you don’t know the size of in advance, there are better options than arrays. Use a List<string> instead – it will allow you to add as many items as you need and if you need to return an array, call ToArray() on the variable. var listOfStrings … Read more