Remove null items from a list in Groovy

Question

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

ex: [null, 30, null]

want to return: [30]

Funny that nobody asked how the nulls ended up in the list in the first place. Seems like you might be addressing the symptom rather than the problem. — Snekse, Jan 28 '15 at 23:13

score 148 · Answer 1 · answered Mar 11 '11 at 08:35

148

Just use minus:

[null, 30, null] - null

answered Mar 11 '11 at 08:35

Sergei

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One can also use [null, 30, null].minus(null).That seems more readable to me. Version with '-' reminds about SBT operator abuse. – Mikhail Boyarsky Dec 08 '15 at 19:50

score 87 · Accepted Answer · edited Jul 20 '10 at 07:55

87

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

edited Jul 20 '10 at 07:55

tim_yates

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answered Jul 19 '10 at 21:35

hvgotcodes

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+1 for the most readable solution (reads like: "remove all nulls") – Pablo Fernandez Jul 19 '10 at 21:40

score 54 · Answer 3 · answered Jul 19 '10 at 21:29

54

The findAll method should do what you need.

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

answered Jul 19 '10 at 21:29

Chris Dail

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`[null, 30, null].findAll {it}` will do. – Thomas Jung Dec 21 '12 at 09:56
6

Thomas, findAll {it} will remove 0 as well as null. And if that is what you want then you might as well just do a findAll() – Dino Fancellu Feb 02 '13 at 16:00

score 18 · Answer 4 · edited Jun 15 '21 at 14:23

18

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]]

edited Jun 15 '21 at 14:23

Maicon Mauricio

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answered Feb 02 '13 at 15:59

Dino Fancellu

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score 12 · Answer 5 · answered Jan 17 '15 at 09:27

12

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]

answered Jan 17 '15 at 09:27

MKB

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New location for docs: [grep](https://docs.groovy-lang.org/latest/html/groovy-jdk/java/util/Collection.html#grep(java.lang.Object)) – RomanMitasov May 19 '22 at 09:24

score 2 · Answer 6 · edited Nov 26 '14 at 15:12

2

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

edited Nov 26 '14 at 15:12

Alexander Suraphel

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answered Mar 19 '14 at 14:49

Kasper Ziemianek

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`assertEquals [], [0, false , null].findAll{it}` – Alexander Suraphel Nov 26 '14 at 11:55

score 2 · Answer 7 · answered Sep 27 '16 at 22:03

2

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

answered Sep 27 '16 at 22:03

Anonymous1

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score 2 · Answer 8 · answered Jun 13 '17 at 20:58

2

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 }

answered Jun 13 '17 at 20:58

B. Anderson

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Remove null items from a list in Groovy

8 Answers8

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