Removing all items of a given value from a hashmap

Question

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?

Solution 1

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!

Solution 2

In Java 8

hmap.values().removeIf(val -> "Two".equals(val));

Solution 3

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();
 }
}

---

Solution 4

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

Solution 5

(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.

View more solutions

Comments

[chris]

this works very well, and i think is a good part of the reason why remove() returns boolean.

[Jon Quarfoth]

This solution is certainly valid, but Kevin's answer is much more concise.

[Anthony Webb]

This looks good kevin, is there any way to write a debug line above to tell which keys are about to be whacked?

[Jon Quarfoth]

@Anthony: Kevin is talking about worst-case performance. See en.wikipedia.org/wiki/Big_O_notation#Orders_of_common_functi‌​ons To 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]

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.

[Kevin Bourrillion]

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.

[Display Name]

It can be put into a separate method, and then it doesn't matter if it's too verbose. :P

[Alex]

Kevin, your last sentence make me think you already fell into the trap left by the Past-you :p

[Ashraf Alshahawy]

It requires API level 24

[Graeme Moss]

Rather than being forced to add a comment, adding a unit test would also help to "document" the reason for requiring removeAll