RE: r/fold combinef and reducef init values

2019-01-26 Thread Sean Corfield 
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  on behalf of Brian 
Craft 
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
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?
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F

Re: r/fold combinef and reducef init values

2019-01-26 Thread Brian Craft 
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 
> *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?
>
> -- 
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>  
>

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RE: r/fold combinef and reducef init values

2019-01-25 Thread Sean Corfield 
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
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?
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Re: r/fold combinef and reducef init values

2019-01-25 Thread Brian Craft 
Looks like it's something that's changed over different clojure releases.

On Friday, January 25, 2019 at 3:35:58 PM UTC-8, Brian Craft wrote:
>
> 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?
>

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