(defn to-percentage [wins total]
(if (= wins 0) 0
(* (/ wins total) 100)))
(defn calc-winrate [matches]
(let [data (r/atom [])]
(loop [wins 0
total 0]
(if (= total (count matches))
#data
(recur (if (= (get (nth matches total) :result) 1)
(inc wins))
(do
(swap! data conj (to-percentage wins total))
(inc total)))))))
(calc-winrate [{:result 0} {:result 1} {:result 0} {:result 1} {:result 1}])
I got the following code, calc-winrate on the last line returns [0 0 50 0 25]. I'm trying to make it return [0 50 33.33333333333333 50 60].
Am I doing the increment for wins wrong? When I print the value of wins for each iteration I get
0
nil
1
nil
1
so I'm guessing I somehow reset or nil wins somehow?
Also, could this whole loop be replaced with map/map-indexed or something? It feels like map would be perfect to use but I need to keep the previous iteration wins/total in mind for each iteration.
Thanks!
Here's a lazy solution using reductions to get a sequence of running win totals, and transducers to 1) join the round numbers with the running totals 2) divide the pairs 3) convert fractions to percentages:
(defn calc-win-rate [results]
(->> results
(map :result)
(reductions +)
(sequence
(comp
(map-indexed (fn [round win-total] [win-total (inc round)]))
(map (partial apply /))
(map #(* 100 %))
(map float)))))
(calc-win-rate [{:result 0} {:result 1} {:result 0} {:result 1} {:result 1}])
=> (0.0 50.0 33.333332 50.0 60.0)
You can calculate the running win rates as follows:
(defn calc-winrate [matches]
(map
(comp float #(* 100 %) /)
(reductions + (map :result matches))
(rest (range))))
For example,
=> (calc-winrate [{:result 0} {:result 1} {:result 0} {:result 1} {:result 1}])
(0.0 50.0 33.333332 50.0 60.0)
The map operates on two sequences:
(reductions + (map :result matches)) - the running total of wins;
(rest (range)))) - (1 2 3 ... ), the corresponding number of matches.
The mapping function, (comp float #(* 100 %) /),
divides the corresponding elements of the sequences,
multiplies it by 100, and
turns it into floating point.
Here's a solution with reduce:
(defn calc-winrate [matches]
(let [total-matches (count matches)]
(->> matches
(map :result)
(reduce (fn [{:keys [wins matches percentage] :as result} match]
(let [wins (+ wins match)
matches (inc matches)]
{:wins wins
:matches matches
:percentage (conj percentage (to-percentage wins matches))}))
{:wins 0
:matches 0
:percentage []}))))
So the thing here is to maintain (and update) the state of the calculation thus far.
We do that in the map that's
{:wins 0
:matches 0
:percentage []}
Wins will contain the wins so far, matches are the number of matches we've analysed, and percentage is the percentage for so far.
(if (= (get (nth matches total) :result) 1)
(inc wins))
your if shall be written as follows:
(if (= (get (nth matches total) :result) 1)
(inc wins)
wins ; missing here , other wise it will return a nil, binding to wins in the loop
)
if you go with a reductions ,
(defn calc-winrate2 [ x y ]
(let [ {total :total r :wins } x
{re :result } y]
(if (pos? re )
{:total (inc total) :wins (inc r)}
{:total (inc total) :wins r}
)
)
)
(reductions calc-winrate2 {:total 0 :wins 0} [ {:result 0} {:result 1} {:result 0} {:result 1} {:result 1}])
Related
I'd like to have a version of sequence that doesn't do the chunking of 32 elements. Currently, this code will output
(def peek #(doto % (print " ")))
(def pause #(do (Thread/sleep 10)
%))
(take 2 (->> (range 100)
(sequence (comp (map peek)
(map pause)
(map inc)))))
;; prints 0 1 2 3 4 <..etc..> 32
;; => (0, 1)
I'd like a version of it so that it only iterates through the elements that it needs
(take 2 (->> (range 100)
(iter-sequence (comp (map peek)
(map pause)
(map inc)))))
;; prints 0 1
;; => (0, 1)
Is there a way to do this?
I had to change a couple of things to get it working. The first is to cut and paste sequence code and replace clojure.lang.RT/chunkIteratorSeq with an alternative version of clojure.lang.IteratorSeq that has exposed public constructor methods.
The reason being is that the clojure.lang.IteratorSeq/create has a check to iter.next() on L27 which will block if the source is blocking.
(defn seqiter
{:added "1.0"
:static true}
([coll] coll)
([xform coll]
(IteratorSeq.
(TransformerIterator/create xform (clojure.lang.RT/iter coll))))
([xform coll & colls]
(IteratorSeq.
(TransformerIterator/createMulti
xform
(map #(clojure.lang.RT/iter %) (cons coll colls))))))
I'm trying to write a function with recur that cut the sequence as soon as it encounters a repetition ([1 2 3 1 4] should return [1 2 3]), this is my function:
(defn cut-at-repetition [a-seq]
(loop[[head & tail] a-seq, coll '()]
(if (empty? head)
coll
(if (contains? coll head)
coll
(recur (rest tail) (conj coll head))))))
The first problem is with the contains? that throws an exception, I tried replacing it with some but with no success. The second problem is in the recur part which will also throw an exception
You've made several mistakes:
You've used contains? on a sequence. It only works on associative
collections. Use some instead.
You've tested the first element of the sequence (head) for empty?.
Test the whole sequence.
Use a vector to accumulate the answer. conj adds elements to the
front of a list, reversing the answer.
Correcting these, we get
(defn cut-at-repetition [a-seq]
(loop [[head & tail :as all] a-seq, coll []]
(if (empty? all)
coll
(if (some #(= head %) coll)
coll
(recur tail (conj coll head))))))
(cut-at-repetition [1 2 3 1 4])
=> [1 2 3]
The above works, but it's slow, since it scans the whole sequence for every absent element. So better use a set.
Let's call the function take-distinct, since it is similar to take-while. If we follow that precedent and make it lazy, we can do it thus:
(defn take-distinct [coll]
(letfn [(td [seen unseen]
(lazy-seq
(when-let [[x & xs] (seq unseen)]
(when-not (contains? seen x)
(cons x (td (conj seen x) xs))))))]
(td #{} coll)))
We get the expected results for finite sequences:
(map (juxt identity take-distinct) [[] (range 5) [2 3 2]]
=> ([[] nil] [(0 1 2 3 4) (0 1 2 3 4)] [[2 3 2] (2 3)])
And we can take as much as we need from an endless result:
(take 10 (take-distinct (range)))
=> (0 1 2 3 4 5 6 7 8 9)
I would call your eager version take-distinctv, on the map -> mapv precedent. And I'd do it this way:
(defn take-distinctv [coll]
(loop [seen-vec [], seen-set #{}, unseen coll]
(if-let [[x & xs] (seq unseen)]
(if (contains? seen-set x)
seen-vec
(recur (conj seen-vec x) (conj seen-set x) xs))
seen-vec)))
Notice that we carry the seen elements twice:
as a vector, to return as the solution; and
as a set, to test for membership of.
Two of the three mistakes were commented on by #cfrick.
There is a tradeoff between saving a line or two and making the logic as simple & explicit as possible. To make it as obvious as possible, I would do it something like this:
(defn cut-at-repetition
[values]
(loop [remaining-values values
result []]
(if (empty? remaining-values)
result
(let [found-values (into #{} result)
new-value (first remaining-values)]
(if (contains? found-values new-value)
result
(recur
(rest remaining-values)
(conj result new-value)))))))
(cut-at-repetition [1 2 3 1 4]) => [1 2 3]
Also, be sure to bookmark The Clojure Cheatsheet and always keep a browser tab open to it.
I'd like to hear feedback on this utility function which I wrote for myself (uses filter with stateful pred instead of a loop):
(defn my-distinct
"Returns distinct values from a seq, as defined by id-getter."
[id-getter coll]
(let [seen-ids (volatile! #{})
seen? (fn [id] (if-not (contains? #seen-ids id)
(vswap! seen-ids conj id)))]
(filter (comp seen? id-getter) coll)))
(my-distinct identity "abracadabra")
; (\a \b \r \c \d)
(->> (for [i (range 50)] {:id (mod (* i i) 21) :value i})
(my-distinct :id)
pprint)
; ({:id 0, :value 0}
; {:id 1, :value 1}
; {:id 4, :value 2}
; {:id 9, :value 3}
; {:id 16, :value 4}
; {:id 15, :value 6}
; {:id 7, :value 7}
; {:id 18, :value 9})
Docs of filter says "pred must be free of side-effects" but I'm not sure if it is ok in this case. Is filter guaranteed to iterate over the sequence in order and not for example take skips forward?
In Clojure I want to find the result of multiple reductions while only consuming the sequence once. In Java I would do something like the following:
double min = Double.MIN_VALUE;
double max = Double.MAX_VALUE;
for (Item item : items) {
double price = item.getPrice();
if (price > min) {
min = price;
}
if (price < max) {
max = price;
}
}
In Clojure I could do much the same thing by using loop and recur, but it's not very composable - I'd like to do something that lets you add in other aggregation functions as needed.
I've written the following function to do this:
(defn reduce-multi
"Given a sequence of fns and a coll, returns a vector of the result of each fn
when reduced over the coll."
[fns coll]
(let [n (count fns)
r (rest coll)
initial-v (transient (into [] (repeat n (first coll))))
fns (into [] fns)
reduction-fn
(fn [v x]
(loop [v-current v, i 0]
(let [y (nth v-current i)
f (nth fns i)
v-new (assoc! v-current i (f y x))]
(if (= i (- n 1))
v-new
(recur v-new (inc i))))))]
(persistent! (reduce reduction-fn initial-v r))))
This can be used in the following way:
(reduce-multi [max min] [4 3 6 7 0 1 8 2 5 9])
=> [9 0]
I appreciate that it's not implemented in the most idiomatic way, but the main problem is that it's about 10x as slow as doing the reductions one at at time. This might be useful for lots performing lots of reductions where the seq is doing heavy IO, but surely this could be better.
Is there something in an existing Clojure library that would do what I want? If not, where am I going wrong in my function?
that's what i would do: simply delegate this task to a core reduce function, like this:
(defn multi-reduce
([fs accs xs] (reduce (fn [accs x] (doall (map #(%1 %2 x) fs accs)))
accs xs))
([fs xs] (when (seq xs)
(multi-reduce fs (repeat (count fs) (first xs))
(rest xs)))))
in repl:
user> (multi-reduce [+ * min max] (range 1 10))
(45 362880 1 9)
user> (multi-reduce [+ * min max] [10])
(10 10 10 10)
user> (multi-reduce [+ * min max] [])
nil
user> (multi-reduce [+ * min max] [1 1 1000 0] [])
[1 1 1000 0]
user> (multi-reduce [+ * min max] [1 1 1000 0] [1])
(2 1 1 1)
user> (multi-reduce [+ * min max] [1 1 1000 0] (range 1 10))
(46 362880 1 9)
user> (multi-reduce [max min] (range 1000000))
(999999 0)
The code for reduce is fast for reducible collections. So it's worth trying to base multi-reduce on core reduce. To do so, we have to be able to construct reducing functions of the right shape. An ancillary function to do so is ...
(defn juxt-reducer [f g]
(fn [[fa ga] x] [(f fa x) (g ga x)]))
Now we can define the function you want, which combines juxt with reduce as ...
(defn juxt-reduce
([[f g] coll]
(if-let [[x & xs] (seq coll)]
(juxt-reduce (list f g) [x x] xs)
[(f) (g)]))
([[f g] init coll]
(reduce (juxt-reducer f g) init coll)))
For example,
(juxt-reduce [max min] [4 3 6 7 0 1 8 2 5 9]) ;=> [9 0]
The above follows the shape of core reduce. It can clearly be extended to cope with more than two functions. And I'd expect it to be faster than yours for reducible collections.
Here is how I would do it:
(ns clj.core
(:require [clojure.string :as str] )
(:use tupelo.core))
(def data (flatten [ (range 5 10) (range 5) ] ))
(spyx data)
(def result (reduce (fn [cum-result curr-val] ; reducing (accumulator) fn
(it-> cum-result
(update it :min-val min curr-val)
(update it :max-val max curr-val)))
{ :min-val (first data) :max-val (first data) } ; inital value
data)) ; seq to reduce
(spyx result)
(defn -main [] )
;=> data => (5 6 7 8 9 0 1 2 3 4)
;=> result => {:min-val 0, :max-val 9}
So the reducing function (fn ...) carries along a map like {:min-val xxx :max-val yyy} through each element of the sequence, updating the min & max values as required at each step.
While this does make only one pass through the data, it is doing a lot of extra work calling update twice per element. Unless your sequence is very unusual, it is probably more efficient to make two (very efficient) passes through the data like:
(def min-val (apply min data))
(def max-val (apply max data))
(spyx min-val)
(spyx max-val)
;=> min-val => 0
;=> max-val => 9
I'm trying to make a conditional transducer in Clojure as follows:
(defn if-xf
"Takes a predicate and two transducers.
Returns a new transducer that routes the input to one of the transducers
depending on the result of the predicate."
[pred a b]
(fn [rf]
(let [arf (a rf)
brf (b rf)]
(fn
([] (rf))
([result]
(rf result))
([result input]
(if (pred input)
(arf result input)
(brf result input)))))))
It is pretty useful in that it lets you do stuff like this:
;; multiply odd numbers by 100, square the evens.
(= [0 100 4 300 16 500 36 700 64 900]
(sequence
(if-xf odd? (map #(* % 100)) (map (fn [x] (* x x))))
(range 10)))
However, this conditional transducer does not work very well with transducers that perform cleanup in their 1-arity branch:
;; negs are multiplied by 100, non-negs are partitioned by 2
;; BUT! where did 6 go?
;; expected: [-600 -500 -400 -300 -200 -100 [0 1] [2 3] [4 5] [6]]
;;
(= [-600 -500 -400 -300 -200 -100 [0 1] [2 3] [4 5]]
(sequence
(if-xf neg? (map #(* % 100)) (partition-all 2))
(range -6 7)))
Is it possible to tweak the definition of if-xf to handle the case of transducers with cleanup?
I'm trying this, but with weird behavior:
(defn if-xf
"Takes a predicate and two transducers.
Returns a new transducer that routes the input to one of the transducers
depending on the result of the predicate."
[pred a b]
(fn [rf]
(let [arf (a rf)
brf (b rf)]
(fn
([] (rf))
([result]
(arf result) ;; new!
(brf result) ;; new!
(rf result))
([result input]
(if (pred input)
(arf result input)
(brf result input)))))))
Specifically, the flushing happens at the end:
;; the [0] at the end should appear just before the 100.
(= [[-6 -5] [-4 -3] [-2 -1] 100 200 300 400 500 600 [0]]
(sequence
(if-xf pos? (map #(* % 100)) (partition-all 2))
(range -6 7)))
Is there a way to make this branching/conditional transducer without storing the entire input sequence in local state within this transducer (i.e. doing all the processing in the 1-arity branch upon cleanup)?
The idea is to complete every time the transducer switches over. IMO this is the only way to do it without buffering:
(defn if-xf
"Takes a predicate and two transducers.
Returns a new transducer that routes the input to one of the transducers
depending on the result of the predicate."
[pred a b]
(fn [rf]
(let [arf (volatile! (a rf))
brf (volatile! (b rf))
a? (volatile! nil)]
(fn
([] (rf))
([result]
(let [crf (if #a? #arf #brf)]
(-> result crf rf)))
([result input]
(let [p? (pred input)
[xrf crf] (if p? [#arf #brf] [#brf #arf])
switched? (some-> #a? (not= p?))]
(if switched?
(-> result crf (xrf input))
(xrf result input))
(vreset! a? p?)))))))
(sequence (if-xf pos? (map #(* % 100)) (partition-all 2)) [0 1 0 1 0 0 0 1])
; => ([0] 100 [0] 100 [0 0] [0] 100)
I think your question is ill-defined. What exactly do you want to happen when the transducers have state? For example, what do you expect this do:
(sequence
(if-xf even? (partition-all 3) (partition-all 2))
(range 14))
Furthermore, sometimes reducing functions have work to do at the beginning and the end and can't be restarted arbitrarily. For example, here is a reducer that computes the mean:
(defn mean
([] {:count 0, :sum 0})
([result] (double (/ (:sum result) (:count result))))
([result x]
(update-in
(update-in result [:count] inc)
[:sum] (partial + x))))
(transduce identity mean [10 20 40 40]) ;27.5
Now let's take the average, where anything below 20 counts for 20, but everything else is decreased by 1:
(transduce
(if-xf
(fn [x] (< x 20))
(map (constantly 20))
(map dec))
mean [10 20 40 40]) ;29.25
My answer is the following: I think your original solution is best. It works well using map, which is how you stated the usefulness of the conditional transducer in the first place.
Given:
(def my-vec [{:a "foo" :b 10} {:a "bar" :b 13} {:a "baz" :b 7}])
How could iterate over each element to print that element's :a and the sum of all :b's to that point? That is:
"foo" 10
"bar" 23
"baz" 30
I'm trying things like this to no avail:
; Does not work!
(map #(prn (:a %2) %1) (iterate #(+ (:b %2) %1) 0)) my-vec)
This doesn't work because the "iterate" lazy-seq can't refer to the current element in my-vec (as far as I can tell).
TIA! Sean
user> (reduce (fn [total {:keys [a b]}]
(let [total (+ total b)]
(prn a total)
total))
0 my-vec)
"foo" 10
"bar" 23
"baz" 30
30
You could look at this as starting with a sequence of maps, filtering out a sequence of the :a values and a separate sequence of the rolling sum of the :b values and then mapping a function of two arguments onto the two derived sequences.
create sequence of just the :a and :b values with
(map :a my-vec)
(map :b my-vec)
then a function to get the rolling sum:
(defn sums [sum seq]
"produce a seq of the rolling sum"
(if (empty? seq)
sum
(lazy-seq
(cons sum
(recur (+ sum (first seq)) (rest seq))))))
then put them together:
(map #(prn %1 %s) (map :a my-vec) (sums 0 (map :b my-vec)))
This separates the problem of generating the data from processing it. Hopefully this makes life easier.
PS: whats a better way of getting the rolling sum?
Transform it into the summed sequence:
(defn f [start mapvec]
(if (empty? mapvec) '()
(let [[ m & tail ] mapvec]
(cons [(m :a)(+ start (m :b))] (f (+ start (m :b)) tail)))))
Called as:
(f 0 my-vec)
returns:
(["foo" 10] ["bar" 23] ["baz" 30])