group-by replacement when an item has many groups?

[Colin Yates]

Hi,

I have a sequence of items and want to group them into categories, the value of which is a function of the item.  This sounds exactly what group-by is after.

The kicker is that the function could return multiple values.  Imagine each item was a date range and I wanted to group them by the number of days in that date range.  

For example, (group-by #(range 1 (inc %)) [1 2 3] => {(1) [1] (1 2) [2] (1 2 3) [3]}.  I want {1 [1] 2 [1 2] 3 [1 2 3]}.

Any ideas?

Thanks!

Col

[Jim - FooBar();]

you can use group-by as you showed and then reduce-kv over them map
replacing each key with '(count key)'...I don't see a way of doing this
in one-pass using group-by alone...

Jim

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[Ben Wolfson]

You could do something like this, which is just a generalization of group-by to the multiple value case (except that group-by actually uses transients):

user> (defn groups-by [f coll]
        (reduce (fn [acc x]
                    (let [ks (f x)]
                      (reduce (fn [acc' k] (update-in acc' [k] #(conj (or % []) x))) acc ks)))
                {} coll))
#'user/groups-by
user> (groups-by #(range 1 (inc %)) [1 2 3])
{3 [3], 2 [2 3], 1 [1 2 3]}

(This isn't the return value you say you want, but it seems to be the right return value, unless I've misunderstood the problem: 1, 2, and 3 all include 1, but only 3 includes 3.)

You can define regular group-by in terms of this function as (defn group-by [f coll] (groups-by (comp vector f) coll)).

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[Colin Yates]

Hi Jim,

I don't think that would give the result I need.  Let me give a more realistic example:

(defn every-day [[start :start end: end]] ...) ; returns [date1 date2 date3]

(def m [{:id 1 :start 1/1/2010 :end 3/1/2010} {:id 2 :start 3/1/2010 :end 3/1/2010}]

(group-by every-day m) would give something like:

{[1/1/2010 2/1/2010 3/1/2010] [{:id1 ...}] 

 [3/1/2010] [{:id 2...}]}

I now want to transform that result to the following:

{1/1/2010 [{:id 1...}]

 3/1/2010 [{:id 1 ...} {:id 3}]}

I am thinking I probably need to process each key/value in the original map (m) and then use assoc-in to build up the map but that feels pretty imperative and given Clojure's magic and transforming shapes I figure there must be a better way...

On 8 July 2013 20:35, Jim - FooBar(); <jimpi...@gmail.com> wrote:

you can use group-by as you showed and then reduce-kv over them map replacing each key with '(count key)'...I don't see a way of doing this in one-pass using group-by alone...

Jim

On 08/07/13 20:25, Colin Yates wrote:

Hi,

I have a sequence of items and want to group them into categories, the value of which is a function of the item.  This sounds exactly what group-by is after.

The kicker is that the function could return multiple values.  Imagine each item was a date range and I wanted to group them by the number of days in that date range.

For example, (group-by #(range 1 (inc %)) [1 2 3] => {(1) [1] (1 2) [2] (1 2 3) [3]}.  I want {1 [1] 2 [1 2] 3 [1 2 3]}.

Any ideas?

Thanks!

Col
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[Colin Yates]

Perfect, and bonus points for spotting my stupid "not slept in days" muppetry.  The results I want are as you infer: {1 [1 2 3] 2 [1 2] 3 [3]}.

Now all I have to do is understand your code - time for more coffee I think (not implying your code is difficult, rather the difficultly is a function of my ignorance).

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[Michał Marczyk]

Another implementation:

(defn groups-by [f coll]
(apply merge-with into
(map (fn [k] (zipmap (f k) (repeat [k])))
coll)))

Or with ->>:

(defn groups-by [f coll]
(->> coll
(map (fn [k] (zipmap (f k) (repeat [k]))))
(apply merge-with into)))

Could use #(zipmap (f %) (repeat [%])) in map, of course.

In any case,

(groups-by #(range 1 (inc %)) [1 2 3])

returns

{3 [3], 2 [2 3], 1 [1 2 3]}

Cheers,
Michał

[Colin Yates]

Wow - I need to get some sleep:  {1 [1 2 3] 2 [2 3] 3 [3]}

[Yoshinori Kohyama]

Hi Colin,

One more solution, with example data in the process commented

(let [f #(range 1 (inc %))

      coll '(1 2 3)]

  (->>

    (for [x coll    ; x = 1, 2, 3

          y (f x)]  ; y = 1       (where x = 1),

                    ;     1, 2    (where x = 2),

                    ;     1, 2, 3 (where x = 3)

      [y x])        ; [y, x] = [1, 1], [1, 2], [2, 2], [1, 3], [2, 3], [3, 3]

    (reduce         ; for example where

      (fn [a [y x]] ;   a = {1 [1 2 3], 2 [2]}, [y x] = [2 3]

                    ; => (a y []) = [2]

                    ;    (conj ... x) = [2 3]

        (assoc a y (conj (a y []) x)))

      {})))         ;    (assoc a y ...) = {1 [1 2 3], 2 [2 3]}

Hope this helps.

Y.Kohyama

[Colin Yates]

Thanks Y.Kohyama - it does.

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