Clojure: How to to recur upon exception? - clojure

I am trying to execute a func several times before giving up upon exceptions.
But it is not valid in Clojure to recur from catch block.
How can this be achieved ?
(loop [tries 10]
(try
(might-throw-exception)
(catch Exception e
(when (pos? tries) (recur (dec tries))))))
java.lang.UnsupportedOperationException: Cannot recur from catch/finally
The best I could find is the following clumsy solution (wrapping in func and calling it)
(defn do-it []
(try
(might-throw-exception)
(catch Exception e nil)))
(loop [times 10]
(when (and (nil? (do-it)) (pos? times))
(recur (dec times))))

Macros are calling...
How about this:
(defn try-times*
"Executes thunk. If an exception is thrown, will retry. At most n retries
are done. If still some exception is thrown it is bubbled upwards in
the call chain."
[n thunk]
(loop [n n]
(if-let [result (try
[(thunk)]
(catch Exception e
(when (zero? n)
(throw e))))]
(result 0)
(recur (dec n)))))
(defmacro try-times
"Executes body. If an exception is thrown, will retry. At most n retries
are done. If still some exception is thrown it is bubbled upwards in
the call chain."
[n & body]
`(try-times* ~n (fn [] ~#body)))

kotarak's idea is the way to go, but this question tickled my fancy so I'd like to provide a riff on the same theme that I prefer because it doesn't use loop/recur:
(defn try-times* [thunk times]
(let [res (first (drop-while #{::fail}
(repeatedly times
#(try (thunk)
(catch Throwable _ ::fail)))))]
(when-not (= ::fail res)
res)))
And leave the try-times macro as it is.
If you want to allow the thunk to return nil, you can drop the let/when pair, and let ::fail represent "the function failed n times", while nil means "the function returned nil". This behavior would be more flexible but less convenient (the caller has to check for ::fail to see if it worked rather than just nil), so perhaps it would be best implemented as an optional second parameter:
(defn try-times* [thunk n & fail-value]
(first (drop-while #{fail-value} ...)))

A try-times macro is elegant, but for a one-off, just pull your when out of the try block:
(loop [tries 10]
(when (try
(might-throw-exception)
false ; so 'when' is false, whatever 'might-throw-exception' returned
(catch Exception e
(pos? tries)))
(recur (dec tries))))

My proposal:
(defmacro try-times
"Retries expr for times times,
then throws exception or returns evaluated value of expr"
[times & expr]
`(loop [err# (dec ~times)]
(let [[result# no-retry#] (try [(do ~#expr) true]
(catch Exception e#
(when (zero? err#)
(throw e#))
[nil false]))]
(if no-retry#
result#
(recur (dec err#))))))
Will print "no errors here" once:
(try-times 3 (println "no errors here") 42)
Will print "trying" 3 times, then throw Divide by zero:
(try-times 3 (println "trying") (/ 1 0))

One more solution, without macro
(defn retry [& {:keys [fun waits ex-handler]
:or {ex-handler #(log/error (.getMessage %))}}]
(fn [ctx]
(loop [[time & rem] waits]
(let [{:keys [res ex]} (try
{:res (fun ctx)}
(catch Exception e
(when ex-handler
(ex-handler e))
{:ex e}))]
(if-not ex
res
(do
(Thread/sleep time)
(if (seq rem)
(recur rem)
(throw ex))))))))

This allows catching multiple more then one exception and provides some feedback about the causes for the retries.
(defmacro try-n-times
"Try running the body `n` times, catching listed exceptions."
{:style/indent [2 :form :form [1]]}
[n exceptions & body]
`(loop [n# ~n
causes# []]
(if (> n# 0)
(let [result#
(try
~#body
~#(map (partial apply list 'catch) exceptions (repeat `(e# e#))))]
(if (some #(instance? % result#) ~exceptions)
(recur (dec n#) (conj causes# result#))
result#))
(throw (ex-info "Maximum retries exceeded!"
{:retries ~n
:causes causes#})))))

If you add a result arg to your loop, you can nest the (try) block inside of the (recur). I solved it like this:
(loop [result nil tries 10]
(cond (some? result) result
(neg? tries) nil
:else (recur (try (might-throw-exception)
(catch Exception e nil))
(dec tries))))

Here's yet another approach:
(loop [tries 10]
(let [res (try
(might-throw-exception)
(catch Exception e
(if (pos? tries)
::retry
(throw e))))]
(if (#{::retry} res)
(recur (dec tries))
res)))
But may I also recommend a cool little trick, instead of having a number of retries, provide a seq of times to sleep for:
(loop [tries [10 10 100 1000]]
(let [res (try
(might-throw-exception)
(catch Exception e
(if tries
::retry
(throw e))))]
(if (#{::retry} res)
(do
(Thread/sleep (first tries))
(recur (next tries)))
res)))
And finally put it all into a macro if you want it to be less verbose:
(defmacro with-retries
[retries & body]
`(loop [retries# ~retries]
(let [res# (try ~#body
(catch Exception e#
(if retries#
'retry#
(throw e#))))]
(if (= 'retry# res#)
(do (Thread/sleep (first retries#))
(recur (next retries#)))
res#))))
(with-retries [10 10 100 1000]
(might-throw-exception))

Related

Return an else value when using recur

I am new to Clojure, and doing my best to forget all my previous experience with more procedural languages (java, ruby, swift) and embrace Clojure for what it is. I am actually really enjoying the way it makes me think differently -- however, I have come up against a pattern that I just can't seem to figure out. The easiest way to illustrate, is with some code:
(defn char-to-int [c] (Integer/valueOf (str c)))
(defn digits-dont-decrease? [str]
(let [digits (map char-to-int (seq str)) i 0]
(when (< i 5)
(if (> (nth digits i) (nth digits (+ i 1)))
false
(recur (inc i))))))
(def result (digits-dont-decrease? "112233"))
(if (= true result)
(println "fit rules")
(println "doesn't fit rules"))
The input is a 6 digit number as a string, and I am simply attempting to make sure that each digit from left to right is >= the previous digit. I want to return false if it doesn't, and true if it does. The false situation works great -- however, given that recur needs to be the last thing in the function (as far as I can tell), how do I return true. As it is, when the condition is satisfied, I get an illegal argument exception:
Execution error (IllegalArgumentException) at clojure.exercise.two/digits-dont-decrease? (four:20).
Don't know how to create ISeq from: java.lang.Long
How should I be thinking about this? I assume my past training is getting in my mental way.
This is not answering your question, but also shows an alternative. While the (apply < ...) approach over the whole string is very elegant for small strings (it is eager), you can use every? for an short-circuiting approach. E.g.:
user=> (defn nr-seq [s] (map #(Integer/parseInt (str %)) s))
#'user/nr-seq
user=> (every? (partial apply <=) (partition 2 1 (nr-seq "123")))
true
You need nothing but
(apply <= "112233")
Reason: string is a sequence of character and comparison operator works on character.
(->> "0123456789" (mapcat #(repeat 1000 %)) (apply str) (def loooong))
(count loooong)
10000
(time (apply <= loooong))
"Elapsed time: 21.006625 msecs"
true
(->> "9123456789" (mapcat #(repeat 1000 %)) (apply str) (def bad-loooong))
(count bad-loooong)
10000
(time (apply <= bad-loooong))
"Elapsed time: 2.581750 msecs"
false
(above runs on my iPhone)
In this case, you don't really need loop/recur. Just use the built-in nature of <= like so:
(ns tst.demo.core
(:use demo.core tupelo.core tupelo.test))
(def true-samples
["123"
"112233"
"13"])
(def false-samples
["10"
"12324"])
(defn char->int
[char-or-str]
(let [str-val (str char-or-str)] ; coerce any chars to len-1 strings
(assert (= 1 (count str-val)))
(Integer/parseInt str-val)))
(dotest
(is= 5 (char->int "5"))
(is= 5 (char->int \5))
(is= [1 2 3] (mapv char->int "123"))
; this shows what we are going for
(is (<= 1 1 2 2 3 3))
(isnt (<= 1 1 2 1 3 3))
and now test the char sequences:
;-----------------------------------------------------------------------------
; using built-in `<=` function
(doseq [true-samp true-samples]
(let [digit-vals (mapv char->int true-samp)]
(is (apply <= digit-vals))))
(doseq [false-samp false-samples]
(let [digit-vals (mapv char->int false-samp)]
(isnt (apply <= digit-vals))))
if you want to write your own, you can like so:
(defn increasing-equal-seq?
"Returns true iff sequence is non-decreasing"
[coll]
(when (< (count coll) 2)
(throw (ex-info "coll must have at least 2 vals" {:coll coll})))
(loop [prev (first coll)
remaining (rest coll)]
(if (empty? remaining)
true
(let [curr (first remaining)
prev-next curr
remaining-next (rest remaining)]
(if (<= prev curr)
(recur prev-next remaining-next)
false)))))
;-----------------------------------------------------------------------------
; using home-grown loop/recur
(doseq [true-samp true-samples]
(let [digit-vals (mapv char->int true-samp)]
(is (increasing-equal-seq? digit-vals))))
(doseq [false-samp false-samples]
(let [digit-vals (mapv char->int false-samp)]
(isnt (increasing-equal-seq? digit-vals))))
)
with result
-------------------------------
Clojure 1.10.1 Java 13
-------------------------------
Testing tst.demo.core
Ran 2 tests containing 15 assertions.
0 failures, 0 errors.
Passed all tests
Finished at 23:36:17.096 (run time: 0.028s)
You an use loop with recur.
Assuming you require following input v/s output -
"543221" => false
"54321" => false
"12345" => true
"123345" => true
Following function can help
;; Assuming char-to-int is defined by you before as per the question
(defn digits-dont-decrease?
[strng]
(let [digits (map char-to-int (seq strng))]
(loop [;;the bindings in loop act as initial state
decreases true
i (- (count digits) 2)]
(let [decreases (and decreases (>= (nth digits (+ i 1)) (nth digits i)))]
(if (or (< i 1) (not decreases))
decreases
(recur decreases (dec i)))))))
This should work for numeric string of any length.
Hope this helps. Please let me know if you were looking for something else :).
(defn non-decreasing? [str]
(every?
identity
(map
(fn [a b]
(<= (int a) (int b)))
(seq str)
(rest str))))
(defn non-decreasing-loop? [str]
(loop [a (seq str) b (rest str)]
(if-not (seq b)
true
(if (<= (int (first a)) (int (first b)))
(recur (rest a) (rest b))
false))))
(non-decreasing? "112334589")
(non-decreasing? "112324589")
(non-decreasing-loop? "112334589")
(non-decreasing-loop? "112324589")

How can I improve my solution to the Hackerrank Maximum Element challenge to avoid timeouts?

I am trying to complete the Hackerrank Maximum Element challenge found here: https://www.hackerrank.com/challenges/maximum-element/problem
My solution produces the correct output, but times out on the final test cases beginning with #17.
Initially, I used a list and loop/recur to get my answer:
(defn get-query []
(map #(Integer/parseInt %) (clojure.string/split (read-line) #" ")))
(defn stack-stepper [query stack]
(condp = (first query)
1 (conj stack (second query))
2 (rest stack)
3 (do (println (apply max stack)) stack)))
(loop [stack '()
queries-left (Integer/parseInt (read-line))]
(if (> queries-left 0)
(recur (stack-stepper (get-query) stack) (dec queries-left))))
After some research and feedback from other channels, I tried a vector instead of a list, and reduce instead of loop/recur, but the results were the same.
(defn get-query []
(map #(Integer/parseInt %) (clojure.string/split (read-line) #" ")))
(defn get-queries []
(loop [queries []
queries-left (Integer/parseInt (read-line))]
(if (= queries-left 0)
queries
(recur (conj queries (get-query)) (dec queries-left)))))
(defn stack-stepper [stack query]
(condp = (first query)
1 (conj stack (second query))
2 (pop stack)
3 (do (println (apply max stack)) stack)))
(reduce stack-stepper [] (get-queries))
I am still new to FP and Clojure and I would really like to understand what I am missing. I greatly appreciate your time and help!
HackerRank problems are often very demanding from a performance point of view.
The obvious thing to try first is using a transient vector so see if that helps. I tried this:
(let [in (clojure.string/split (slurp *in*) #"\s")
tests (first in)
input-data (map #(Integer/parseInt %) (rest in))]
(loop [v (transient [])
d input-data]
(when (seq d)
(condp = (first d)
1 (recur (conj! v (second d)) (drop 2 d))
2 (recur (pop! v) (rest d))
3 (let [pv (persistent! v)] (println (apply max pv)) (recur (transient pv) (rest d)))))))
If failed at the same point as your solution. Clearly they're looking for something cleverer than that.
The obvious bottleneck is the calculation of the max value on the current stack, which gets re-calculated each time. We can instead save the previous max value on the stack, and recover it as the current max value when we pop the stack:
(defn peek! [tvec] (get tvec (dec (count tvec))))
(let [in (clojure.string/split (slurp *in*) #"\s")
tests (first in)
input-data (map #(Integer/parseInt %) (rest in))]
(loop [v (transient [])
m 0
d input-data]
(when (seq d)
(condp = (first d)
1 (let [snd (second d)
max-now (max m snd)]
(recur (conj! v {:val snd :max-prev m}) max-now (drop 2 d)))
2 (let [popped (peek! v)
max (if popped (:max-prev popped) 0)]
(recur (pop! v) max (rest d)))
3 (do
(println m)
(recur v m (rest d)))))))
Which puts me at rank 1 on the leaderboard :)

Throttle Functions with core.async

The number of possible executions of a function should be throttled. So after calling a function, any repeated call should be ignored within a time period. If there where calls in the meantime, the last one should be executed after the time period.
Here's my approach with core.async. The problem here is, that additional calls are summing up in the channel c. I'd need a channel with only one position inside, which will be overridden by put! everytime.
(defn throttle [f time]
(let [c (chan 1)]
(go-loop []
(apply f (<! c))
(<! (timeout time))
(recur))
(fn [& args]
(put! c (if args args [])))))
usage:
(def throttled (throttle #(print %) 4000))
(doseq [x (range 10)]
(throttled x))
; 0
;... after 4 seconds
; 9
Does anyone have an idea how to fix this?
Solution
(defn throttle [f time]
(let [c (chan (sliding-buffer 1))]
(go-loop []
(apply f (<! c))
(<! (timeout time))
(recur))
(fn [& args]
(put! c (or args [])))))
To solve your channel question you can use a chan with a sliding buffer:
user> (require '[clojure.core.async :as async])
nil
user> (def c (async/chan (async/sliding-buffer 1)))
#'user/c
user> (async/>!! c 1)
true
user> (async/>!! c 2)
true
user> (async/>!! c 3)
true
user> (async/<!! c)
3
that way only the last value put into the channel will be computed at the next interval.
You can use a debounce function.
I'll copy it out here:
(defn debounce [in ms]
(let [out (chan)]
(go-loop [last-val nil]
(let [val (if (nil? last-val) (<! in) last-val)
timer (timeout ms)
[new-val ch] (alts! [in timer])]
(condp = ch
timer (do (>! out val) (recur nil))
in (recur new-val))))
out))
Here only when in has not emitted a message for ms is the last value it emitted forwarded onto the out channel. While in continues to emit without a long enough pause between emits then all-but-the-last-message are continuously discarded.
I've tested this function. It waits 4 seconds and then prints out 9, which is nearly what you asked for - some tweaking required!
(defn my-sender [to-chan values]
(go-loop [[x & xs] values]
(>! to-chan x)
(when (seq xs) (recur xs))))
(defn my-receiver [from-chan f]
(go-loop []
(let [res (<! from-chan)]
(f res)
(recur))))
(defn setup-and-go []
(let [in (chan)
ch (debounce in 4000)
sender (my-sender in (range 10))
receiver (my-receiver ch #(log %))]))
And this is the version of debounce that will output as required by the question, which is 0 immediately, then wait four seconds, then 9:
(defn debounce [in ms]
(let [out (chan)]
(go-loop [last-val nil
first-time true]
(let [val (if (nil? last-val) (<! in) last-val)
timer (timeout (if first-time 0 ms))
[new-val ch] (alts! [in timer])]
(condp = ch
timer (do (>! out val) (recur nil false))
in (recur new-val false))))
out))
I've used log rather than print as you did. You can't rely on ordinary println/print functions with core.async. See here for an explanation.
This is taken from David Nolens blog's source code:
(defn throttle*
([in msecs]
(throttle* in msecs (chan)))
([in msecs out]
(throttle* in msecs out (chan)))
([in msecs out control]
(go
(loop [state ::init last nil cs [in control]]
(let [[_ _ sync] cs]
(let [[v sc] (alts! cs)]
(condp = sc
in (condp = state
::init (do (>! out v)
(>! out [::throttle v])
(recur ::throttling last
(conj cs (timeout msecs))))
::throttling (do (>! out v)
(recur state v cs)))
sync (if last
(do (>! out [::throttle last])
(recur state nil
(conj (pop cs) (timeout msecs))))
(recur ::init last (pop cs)))
control (recur ::init nil
(if (= (count cs) 3)
(pop cs)
cs)))))))
out))
(defn throttle-msg? [x]
(and (vector? x)
(= (first x) ::throttle)))
(defn throttle
([in msecs] (throttle in msecs (chan)))
([in msecs out]
(->> (throttle* in msecs out)
(filter #(and (vector? %) (= (first %) ::throttle)))
(map second))))
You probably also want to add a dedupe transducer to the channel.
I needed to pass a function to capture the args because I was using it for an input event and it was passing a mutable object. 🤷
(defn throttle-for-mutable-args [time f arg-capture-fn]
(let [c (async/chan (async/sliding-buffer 1))]
(async-m/go-loop []
(f (async/<! c))
(async/<! (async/timeout time))
(recur))
(fn [& args]
(async/put! c (apply arg-capture-fn (or args []))))))
And I use like
[:input
{:onChange (util/throttle-for-mutable-args
500
#(really-use-arg %)
#(-> % .-target .-value))}]

Why call map in macro will fail in my Clojure code?

I want to transfer code from
("AAA" ("BB" 11 #"XXX"))
to
("AAA" ("BB" 11 "YYY"))
I just want to change #"XXX" to "YYY".
I write a function it works.
(defn tt [clause]
(cond (not (sequential? clause)) clause
(and (sequential? clause)
(= 2 (count clause))
(= `deref (first clause))
(string? (second clause)))
"YYY"
:else (map tt clause)))
there is my result:
(tt '("AAA" ("BB" 11 #"XXX"))) --> ("AAA" ("BB" 11 "YYY"))
But when I change the function to macro , it raise exception.
(defmacro test [& clause]
(let [f (fn tt [clause]
(cond (not (sequential? clause)) clause
(and (sequential? clause)
(= 2 (count clause))
(= `deref (first clause))
(string? (second clause)))
"YYY"
:else (map tt clause)))]
(f clause)))
and it raise exception like this
(test "AAA" ("BB" 11 #"XXX")) --> ClassCastException java.lang.String cannot be cast to clojure.lang.IFn
I have test map for prewalk function. Both of them raise exception.
I have no idea what is wrong with it and how to fix this error in macro??

macro always throw "UnmatchedDelimiter" if given anon function

I wrote a macro to handle http response
(defmacro defhandler
[name & args]
(let [[docstring args] (if (string? (first args))
[(first args) (next args)]
[nil args])
args (apply hash-map :execute-if true (vec args))]
`(do
(def ~name
(with-meta (fn [scope# promise#]
(let [e# (:execute-if ~args)
ei# (if (fn? e#)
(e# scope#)
(boolean e#))]
(when ei#
(.then promise# (fn [result#]
(let [{:strs [http-status# value#]} result#
the-func# ((keyword http-status#) ~args)]
(the-func# scope# value#))))))) {:structure ~args}))
(alter-meta! (var ~name) assoc :doc ~docstring))))
So I can do
(defhandler my-handler
:200 (fn [$scope value] (set! (.-content $scope) value)))
But that throws "UnmatchedDelimiter" at line 1, but if I try with a named function:
(defn my-func [$scope value] (set! (.-content $scope) value))
(defhandler my-handler
:200 my-func)
It works ok. I'm just curious, is that a normal behaviour?
That is not the behavior I see when I try your example, nor does it seem very likely. I suggest checking that the forms you pasted here are exactly the ones that produce an error; I suspect your actual anonymous function included one too many )s.