tags:

views:

184

answers:

5

So I have a java hashmap like below:

hMap.put("1", "One");
hMap.put("2", "Two");
hMap.put("3", "Two");

I would like to remove ALL items where the value is "Two"

If I do something like:

hmap.values().remove("Two");

Only the first one is deleted, I want to remove them all, how can this be done?

+9  A: 
for (Iterator<Map.Entry<String,String>> it = hMap.entrySet().iterator(); it.hasNext();) {
 Map.Entry<String,String> e = it.next();
 if ("Two".equals(e.getValue())) {
  it.remove();
 }
}

Ron
This solution is certainly valid, but Kevin's answer is much more concise.
Jon Quarfoth
+2  A: 

You have to iterate through the list, look at the value object, and conditionally do the remove. Note you'll get an exception if you try to remove an object while iterating over a HashMap. Would have to make a copy of the map or use ConcurrentHashMap.

Marcus
+6  A: 

You can use while( hmap.value().remove("Two") ); since the remove call returns true if the collection was changed as a result of the call.

stickynips
this works very well, and i think is a good part of the reason why remove() returns boolean.
oedo
This has quadratic performance, though.
Kevin Bourrillion
@kevin, what is "quadratic performance?" bad/good/?
Anthony Webb
@Anthony: Kevin is talking about worst-case performance. See http://en.wikipedia.org/wiki/Big_O_notation#Orders_of_common_functionsTo answer without (much) math - this solution will iterate through the elements of map n+1 times, where n is the number of values that are removed. Another solution, say Kevin's or Ron's, will only iterate through the values in the Map once. The runtime of the method would be much slower using this solution, if you were operating on very large maps.
Jon Quarfoth
+12  A: 

hmap.values().removeAll(Collections.singleton("Two"));

EDIT: the (significant) disadvantage with this concise approach is that you are basically forced to comment it, saying something like

// remove("Two") would only remove the first one

otherwise, some well-meaning engineer will try to simplify it for you someday and break it. This happens... sometimes the well-intentioned do-gooder is even Future You!

Kevin Bourrillion
This looks good kevin, is there any way to write a debug line above to tell which keys are about to be whacked?
Anthony Webb
You might want to look into Ron's solution (or mine, if you're willing to add google-collections) if you need to log removed keys.
Jon Quarfoth
Yep, Ron's and Jon's (ha) answers are both viable if you want to know the keys you're whacking. I would tend to prefer Ron's, which traverses the map only once.
Kevin Bourrillion
+4  A: 

(Updated solution for logging of removed values)

This solution uses the google-collections library [LINK]

import static com.google.common.collect.Maps.filterValues;
import static com.google.common.base.Predicates.equalTo;

...

Map<String, String> removedValues = filterValues(hMap, equalTo("Two"));      
System.out.println(removedValues); //Log Removed Values
removedValues.clear(); //Removes from original map, since this is a view.

Note - This solution takes advantage of the fact that the Map returned by the filterValues call is a view of the elements in the original HashMap. This allows us to examine them and log out the keys that were removed, and then remove them from the original map with a simple call to clear().

You may have reasons for not wanting to use the google-collections library in your project, but if you don't, I'd suggest checking it out.

Jon Quarfoth