clojure

问题 Is there a less verbose way of making a spec for a map or a record than the way it is presented in the official spec guide? (defrecord Person [first-name last-name email phone]) (s/def ::first-name string?) (s/def ::last-name string?) (s/def ::email ::email-type) (s/def ::person (s/keys :req-un [::first-name ::last-name ::email] :opt-un [::phone])) ideally it would be nice if I could just write something like (defrecord Person [first-name last-name email phone]) (s/def ::person (s/keys :req

问题 Given: (def seq1 ({:id 1 :val 10} {:id 2 :val 20})) (def seq2 ({:id 1 :val 12} {:id 3 :val 30})) Within each sequence, the :id value is guaranteed to be unique in that sequence, though not necessarily ordered. How can these two sequences of maps be joined by the :id key, so that the result is as shown? {1 [{:id 1 :val 10} {:id 1 :val 12}], 2 [{:id 2 :val 20} nil ], 3 [nil {:id 3 :val 30}]} The ultimate result is a map of pairs. This is similar to a full outer join, where not only the

问题 This may actually be a bit of an XY-problem, so I'll try to explain what the goal is first. I'm building a ClojureScript application which is composed of a set of Reagent components. It provides a user interface where you can dynamically add or remove UI elements. These UI elements (components) have a certain type. For example a Markdown component is-a Text component. Whenever the user is presented with the option to add Text we list all the components that match the type+descendants (in this

问题 I want to append a value to a list in a Clojure atom: (def thing (atom {:queue '()})) I know when it's not an atom, I can do this: (concat '(1 2) '(3)) How can I translate that into a swap! command? Note: I asked a similar question involving maps: Using swap to MERGE (append to) a nested map in a Clojure atom? 回答1: user=> (def thing (atom {:queue '()})) #'user/thing user=> (swap! thing update-in [:queue] concat (list 1)) {:queue (1)} user=> (swap! thing update-in [:queue] concat (list 2)) {

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . There is a Python library for this in the Calibre project, but I would like to find a library that I can use from Clojure or Ruby. Any ideas? 回答1: Check out the KindleGen tool available from Amazon here. It's a command line tool used to build eBooks that can be sold through Amazon's Kindle platform. You can call

问题 Some VMs, most notably the JVM, are said to not support TCO. As a result, language like Clojure require the user to use loop recur instead. However, I can rewrite self-tail calls to use a loop. For example, here's a tail-call factorial: def factorial(x, accum): if x == 1: return accum else: return factorial(x - 1, accum * x) Here's a loop equivalent: def factorial(x, accum): while True: if x == 1: return accum else: x = x - 1 accum = accum * x This could be done by a compiler (and I've

问题 I've recently been playing with clojure and reached a problem that I'm not sure how to handle. I have a doseq with 7 parameters and it expands to a huge block, almost passing the maximum class size. Why does doseq expand to such a huge block of clojure code? Example: (def q '(doseq [p0 (nth (:params operator) 0 (quote (nil))) p1 (nth (:params operator) 1 (quote (nil))) p2 (nth (:params operator) 2 (quote (nil))) p3 (nth (:params operator) 3 (quote (nil))) p4 (nth (:params operator) 4 (quote

问题 I've recently been playing with clojure and reached a problem that I'm not sure how to handle. I have a doseq with 7 parameters and it expands to a huge block, almost passing the maximum class size. Why does doseq expand to such a huge block of clojure code? Example: (def q '(doseq [p0 (nth (:params operator) 0 (quote (nil))) p1 (nth (:params operator) 1 (quote (nil))) p2 (nth (:params operator) 2 (quote (nil))) p3 (nth (:params operator) 3 (quote (nil))) p4 (nth (:params operator) 4 (quote

问题 Is there a simple way in specter to collect all the structure satisfying a predicate ? (./pull '[com.rpl/specter "1.0.0"]) (use 'com.rpl.specter) (def data {:items [{:name "Washing machine" :subparts [{:name "Ballast" :weight 1} {:name "Hull" :weight 2}]}]}) (reduce + (select [(walker :weight) :weight] data)) ;=> 3 (select [(walker :name) :name] data) ;=> ["Washing machine"] How can we get all the value for :name, including ["Ballast" "Hull"] ? 回答1: Here's one way, using recursive-path and

问题 Is there a simple way in specter to collect all the structure satisfying a predicate ? (./pull '[com.rpl/specter "1.0.0"]) (use 'com.rpl.specter) (def data {:items [{:name "Washing machine" :subparts [{:name "Ballast" :weight 1} {:name "Hull" :weight 2}]}]}) (reduce + (select [(walker :weight) :weight] data)) ;=> 3 (select [(walker :name) :name] data) ;=> ["Washing machine"] How can we get all the value for :name, including ["Ballast" "Hull"] ? 回答1: Here's one way, using recursive-path and