Sorry to be so slow getting back. Because they are maps, and a set can
contain many maps which contain identical key value pairs - I assume
because although they do indeed have identical key value pairs they are
nevertheless not the same map.
On Tuesday, 24 March 2015 21:55:05 UTC, Ben wrote:
>
> why not just accumulate the interested customers in a set rather than a
> seq in the first place?
>
> On Tue, Mar 24, 2015 at 2:50 PM, Simon Brooke <stil...@googlemail.com
> <javascript:>> wrote:
>
>> I'm rewriting a very odd e-commerce website I first wrote in 1996, for a
>> friend.
>>
>> The website features categories, which are supposed to be arranged into
>> an acyclic directed graph (actually there are currently cycles in the
>> graph, but there shouldn't be, and that is not the problem I'm trying to
>> solve here; assume there are not). Customers can register interest in a
>> category or in several categories. When an item is added to the site, we
>> want to email every customer who may be interested in that item; which is
>> to say
>>
>> customers who are interested in the category to which the item has been
>> added, and
>> customers who are interested in the parent of that category, and
>> so on recursively to the root of the graph.
>>
>> Now, I can recurse up the graph asking each category in turn for the
>> customers interested in it; and by concatenating the lists, arrive at one
>> list of all the interested customers. The problem is that if one customer
>> has registered interest in both a category and its parent, he will appear
>> on the composite list twice. I need to construct from this a list of
>> records in which each customer appears only once, otherwise they'll receive
>> duplicate emails, which no-one likes.
>>
>> My solution, which I give below, is to recurse across the composite list
>> constructing a map whose keys are email addresses and whose values are the
>> customer records as maps; and then to take the values of that map. It
>> works, but it doesn't feel like idiomatic clojure:
>>
>> (defn map-by-key
>> "From this `list-of-maps`, produce a map keyed on the values for this
>> `key`"
>> [list-of-maps key]
>> (if (empty? list-of-maps) {}
>> (let [record (first list-of-maps)]
>> (merge {(record key) record} (map-by-key (rest list-of-maps)
>> key)))))
>>
>>
>> (defn interested-customers
>> "Return a list of the customers interested in the category with this
>> `category-id`. Recurses up the tree of categories; `path` is passed to
>> protect against cycles in the tree (which shouldn't be there by are
>> not impossible)"
>> ([category-id]
>> (vals (map-by-key (interested-customers category-id #{}) :email)))
>> ([category-id path]
>> (if (contains? path category-id) ()
>> (let
>> [id (if (integer? category-id) category-id
>> (Integer/valueOf (str category-id)))
>> category (categories/fetch id)
>> parent (:parent category)
>> upstream (if (not (nil? parent)) (interested-customers parent
>> (conj path id)) ())
>> interested (select schema/customer
>> (with schema/category
>> (where {:id id})))]
>> (concat interested upstream)))))
>>
>> Can anyone re-express that in a more idiomatic form?
>>
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>
>
>
> --
> Ben Wolfson
> "Human kind has used its intelligence to vary the flavour of drinks, which
> may be sweet, aromatic, fermented or spirit-based. ... Family and social
> life also offer numerous other occasions to consume drinks for pleasure."
> [Larousse, "Drink" entry]
>
>
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