Reputation: 3007
What is the best way to remove null items from a list in Groovy?
ex: [null, 30, null]
want to return: [30]
Upvotes: 71
Views: 60079
Reputation: 2004
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 theIDENTITY
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]]
Upvotes: 21
Reputation: 3179
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 }
Upvotes: 2
Reputation: 7619
This can also be achieved by grep:
assert [null, 30, null].grep() == [30]
or
assert [null, 30, null].grep {it} == [30]
or
assert [null, 30, null].grep { it != null } == [30]
Upvotes: 12
Reputation: 1349
Simply [null].findAll{null != it}
if it
is null then it return false so it will not exist in new collection.
Upvotes: 2
Reputation: 120188
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
Upvotes: 89
Reputation: 26029
The findAll method should do what you need.
[null, 30, null].findAll {it != null}
Upvotes: 54