S is the previous count, if any. Seq[V] are potentially many new
counts. All of them have to be added together to keep an accurate
total. It's as if the count were 3, and I tell you I've just observed
2, 5, and 1 additional occurrences -- the new count is 3 + (2+5+1) not
1 + 1.
I butted in since I'd like to ask a different question about the same
line of code. Why:
val currentCount = values.foldLeft(0)(_ + _)
instead of
val currentCount = values.sum
This happens a few places in the code. sum seems equivalent and likely
quicker. Same with things like "filter(_ == 200).size" instead of
"count(_ == 200)"... pretty trivial but hey.
On Wed, Apr 30, 2014 at 9:23 PM, Adrian Mocanu
<amoc...@verticalscope.com> wrote:
> Hi TD,
>
> Why does the example keep recalculating the count via fold?
>
> Wouldn’t it make more sense to get the last count in values Seq and add 1 to
> it and save that as current count?
>
>
>
> From what Sean explained I understand that all values in Seq have the same
> key. Then when a new value for that key is found it is added to this Seq
> collection and the update function is called.
>
>
>
> Is my understanding correct?
