Remove null items from a list in Groovy

问题

What is the best way to remove null items from a list in Groovy?

ex: [null, 30, null]

want to return: [30]

回答1:

here is an answer if you dont want to keep the original list

void testRemove() {
    def list = [null, 30, null]

    list.removeAll([null])

    assertEquals 1, list.size()
    assertEquals 30, list.get(0)
}

in a handy dandy unit test

回答2:

Just use minus:

[null, 30, null] - null

回答3:

The findAll method should do what you need.

[null, 30, null].findAll {it != null}

回答4:

I think you'll find that this is the shortest, assuming that you don't mind other "false" values also dissappearing:

println([null, 30, null].findAll())

public Collection findAll() Finds the items matching the IDENTITY Closure (i.e. matching Groovy truth). Example:

def items = [1, 2, 0, false, true, '', 'foo', [], [4, 5], null] assert items.findAll() == [1, 2, true, 'foo', [4, 5]]

回答5:

This can also be achieved by grep:

assert [null, 30, null].grep() == [30]

assert [null, 30, null].grep {it} == [30]

assert [null, 30, null].grep { it != null } == [30]

回答6:

Simply [null].findAll{null != it} if it is null then it return false so it will not exist in new collection.

回答7:

This does an in place removal of all null items.

myList.removeAll { !it }

If the number 0 is in your domain you can check against null

myList.removeAll { it == null }

回答8:

Another way to do it is [null, 20, null].findResults{it}.

来源：https://stackoverflow.com/questions/3285241/remove-null-items-from-a-list-in-groovy

标签

groovy