I have a map:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3"))
I want to remove all None
elements and flatten the map. What is the easiest way to accomplish that? I only found this way:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3")).filter(_._2.nonEmpty).map(item => (item._1 -> item._2.getOrElse(Nil)))
The result is:
Map(key1 -> value1, key3 -> value3)
Do you know a better way?
My take using pattern matching is:
Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3")).collect {
case (key, Some(value)) => key -> value
}
// Map(key1 -> value1, key3 -> value3)
Collect acts like combined map + filter
You can use for-comprehension + pattern-matching:
for((k, Some(v)) <- yourMap) yield k -> v
Using partition over the map, like this,
val (flattened,_) = map.partition(_._2.isDefined)
My take using for comprehensions:
val m = Map("key1" -> Some("value1"), "key2" -> None, "key3" -> Some("value3"))
for( (key,value) <- m if(value.isDefined)) yield (key,value.get)
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