Ah, yes… So this line… The reducef function will be called with no arguments to produce an identity value in each partition.
…needs updating/removing. Sean Corfield -- (970) FOR-SEAN -- (904) 302-SEAN An Architect's View -- http://corfield.org/ "If you're not annoying somebody, you're not really alive." -- Margaret Atwood ________________________________ From: clojure@googlegroups.com <clojure@googlegroups.com> on behalf of Brian Craft <craft.br...@gmail.com> Sent: Saturday, January 26, 2019 2:02:42 PM To: Clojure Subject: Re: r/fold combinef and reducef init values hey Sean -- The ones on the reducers reference page: https://clojure.org/reference/reducers On Friday, January 25, 2019 at 4:05:12 PM UTC-8, Sean Corfield wrote: Which docs are you reading? The docstring for r/fold says this – with no indication of calling (reducef) with no arguments (well, unless you do not supply combinef – in which case reducef will be used for that, so (reducef) would be called to seed the reductions): "Reduces a collection using a (potentially parallel) reduce-combine strategy. The collection is partitioned into groups of approximately n (default 512), each of which is reduced with reducef (with a seed value obtained by calling (combinef) with no arguments). The results of these reductions are then reduced with combinef (default reducef). combinef must be associative, and, when called with no arguments, (combinef) must produce its identity element. These operations may be performed in parallel, but the results will preserve order." Sean Corfield -- (970) FOR-SEAN -- (904) 302-SEAN An Architect's View -- http://corfield.org/ "If you're not annoying somebody, you're not really alive." -- Margaret Atwood From: Brian Craft<javascript:> Sent: Friday, January 25, 2019 3:36 PM Subject: r/fold combinef and reducef init values >From the docs: r/fold takes a reducible collection and partitions it into groups of approximately n (default 512) elements. Each group is reduced using the reducef function. The reducef function will be called with no arguments to produce an identity value in each partition. The results of those reductions are then reduced with the combinef (defaults to reducef) function. When called with no arguments, (combinef) must produce its identity element - this will be called multiple times. Operations may be performed in parallel. Results will preserve order. So, this seems to say r/fold will partition the collection and reduce each partition using the (reducef) as the init value. Then, all these intermediate results will be reduced with combinef, using (combinef) as the init value. However, in test it seems (reducef) is never called, and (combinef) is used as the init value for calls to reducef. (defn combinef ([] {:combine :f}) ([acc v] acc)) (defn reducef ([] {:reduce :f}) ([acc v] (println "acc" acc "v" v) v)) (clojure.core.reducers/fold combinef reducef ["foo" "bar"]) ; outputs: acc {:combine :f} v foo acc foo v bar "bar" The accumulator in reducef is the init value from combinef, not the init value from reducef. What's going on? -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clo...@googlegroups.com<javascript:> Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+u...@googlegroups.com<javascript:> For more options, visit this group at http://groups.google.com/group/clojure?hl=en --- You received this message because you are subscribed to the Google Groups "Clojure" group. To unsubscribe from this group and stop receiving emails from it, send an email to clojure+u...@googlegroups.com<javascript:>. 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