I'm trying to figure out how to remove an item from a list in groovy from within a loop.
static main(args) {
def list1 = [1, 2, 3, 4]
for(num in list1){
if(num == 2)
list1.remove(num)
}
println(list1)
}
views:
64answers:
3
+1
A:
I think you can do:
list - 2;
or...
list.remove(2)
There's no loop required.
If you want to use a loop I guess you could look at using the iterator to actually remove the item.
import java.util.Iterator;
static main(args) { def list1 = [1, 2, 3, 4]
Iterator i = list1.iterator();
while (i.hasNext()) {
n = i.next();
if (n == 2) i.remove();
}
println(list1)
}
but I don't see why you'd want to do it that way.
Jon
2010-10-28 20:17:09
The example was really meant to only be an example of some real world situation. I ended up using the iterator to do the remove and it worked great. Thanks!
ScArcher2
2010-10-29 13:06:45
+1
A:
list = [1, 2, 3, 4]
newList = list.findAll { it != 2 }
Should give you all but the 2
Of course you may have a reason for requiring the loop?
tim_yates
2010-10-28 20:32:54
That doesn't explain Java's ConcurrentModificationException that the OP (original poster) had received (the most likely solution on that issue would be to use a CopyOnWriteArrayList in the Java world), but, in my eyes, this is the most practical way in Groovy. - Please, may, additionally, everyone note that the List.remove(int) method differs from the List.remove((Object) int) method.
robbbert
2010-10-28 23:51:27
+1
A:
If you want to remove the item with index 2, you can do
list = [1,2,3,4]
list.remove(2)
assert list == [1,2,4]
// or with a loop
list = [1,2,3,4]
i = list.iterator()
2.times {
i.next()
}
i.remove()
assert list == [1,2,4]
If you want to remove the (first) item with value 2, you can do
list = [1,2,3,4]
list.remove(list.indexOf(2))
assert list == [1,3,4]
// or with a loop
list = [1,2,3,4]
i = list.iterator()
while (i.hasNext()) {
if (i.next() == 2) {
i.remove()
break
}
}
assert list == [1,3,4]
ataylor
2010-10-28 20:39:39