Properly Removing an Integer from a List≪Integer≫

Properly removing an Integer from a List<Integer>

Java always calls the method that best suits your argument. Auto boxing and implicit upcasting is only performed if there's no method which can be called without casting / auto boxing.

The List interface specifies two remove methods (please note the naming of the arguments):

• remove(Object o)
• remove(int index)

That means that list.remove(1) removes the object at position 1 and remove(new Integer(1)) removes the first occurrence of the specified element from this list.

How to remove integer from list?

This is normal, there are two versions of the .remove() method for lists: one which takes an integer as an argument and removes the entry at this index, the other which takes a generic type as an argument (which, at runtime, is an Object) and removes it from the list.

And the lookup mechanism for methods always picks the more specific method first...

You need to:

list.remove(Integer.valueOf(300));

in order to call the correct version of .remove().

Java, how to remove an Integer item in an ArrayList

try this

list.removeAll(Arrays.asList(2));

it will remove all elements with value = 2

you can also use this

list.remove(Integer.valueOf(2));

but it will remove only first occurence of 2

list.remove(2) does not work because it matches List.remove(int i) which removes element with the specified index

Remove From ArrayList - Remove an integer from the list

For scenario 1, use remove(int index). Since the third item has index 2 (it is zero-based), you may remove it like this:

// Scenario 1:
numbers.remove(2);

For scenario 2, use remove(Object o). Just cast it to Integer and it will search for the object that equals 3:

// Scenario 2:
numbers.remove((Integer)3);

Test it:

List<Integer> numbers = new ArrayList<Integer>();
numbers.add(1);
numbers.add(2);
numbers.add(3);
numbers.add(4);

// Scenario 1:
numbers.remove(2);

// Scenario 2:
numbers.remove((Integer)3);

System.out.println(numbers);

Note: In both scenarios, there is no need to iterate the list.

Update: In scenario 2, if the list may contain more than one number 3, then to remove all of them (Java 8) I would do:

numbers.removeIf(((Integer)3)::equals);

Removing integers from a list+python

The problem is:

1. You are iterating over the list as you remove items from it, and
2. Your counter variable x is not taking into account the positions of removed items.

Try this:

list_a = ["Michell",123,"Apple","Food",456,3231,"Another"]
list_e = []
x = 0

for item in list_a[:]:
    print "x is: ", x
    print "item is: ", item, type(item)
    if isinstance(item,int):
        list_e.append(item)
        list_a.pop(list_a.index(item))

    x+=1
print list_a
print list_e

See also:

(1), (2), (3)

Removing an integer from a list

List<Integer> list = new ArrayList<Integer>(Arrays.asList(5, 10, 42));
if (list.contains(10)) {
    list.remove(10); // IOOBE
}

The problem with the above code is that you're actually not calling List#remove(Object) but List#remove(int), which removes the element at given index (and there's no element at index 10).

Use instead:

List<Integer> list = new ArrayList<Integer>(Arrays.asList(5, 10, 42));
if (list.contains(10)) {
    list.remove((Integer) 10);
}

That way, you force the compiler to use the List#remove(Object) method.

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