Can I simplify this map/reduce ?

[Bdfy]

I have a map/reduce function on pymongo ( analog sql query: select
count( distinct on (ip,id )) ... ) . Can I simplify this map/reduce ?


map = Code(
"function () {"
"emit({ ip: this.ip, id: this.id }, {});"
"}"
)

reduce = Code("function (key, values) {""}")

results = self.mongo.map_reduce(map, reduce, "test", query =
{ .... }, "field" : "value" } )

count = 0
for result in results.find():
count = count + 1

print count

[Sam Millman]

I dunno I'd say that's pretty darn simple already all your doing is picking a unique record out, saving that and then using your client code to get the count.

I'd say that's as simple as can get.

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[Bdfy]

but list all valies in client ( for example 200 000 rows ) for one
result value - is it very "expensive" for time ?

On 13 апр, 17:34, Sam Millman <sam.mill...@gmail.com> wrote:
> I dunno I'd say that's pretty darn simple already all your doing is picking
> a unique record out, saving that and then using your client code to get the
> count.
>
> I'd say that's as simple as can get.
>

[Sam Millman]

Not really I suppose you could use the distinct() command or pre-aggregate (which might work here) but really distinct would do the same as you do there so meh to that command.

[Sam Millman]

Wait:

"but list all valies in client ( for example 200 000 rows ) for one
result value"

What do you mean? If you explain how amny documents go in and out I can answer that better.

If your iterating over 200k rows in your client side (not sure if the English means that) then it could be time consuming.