What's the idiomatic way in Clojure to implement take-while-and-n-more below:
=> (take-while-and-n-more #(<= % 3) 1 (range 10))
(0 1 2 3 4)
My try is:
(defn take-while-and-n-more [pred n coll]
(let
[take-while-result (take-while pred coll)
n0 (count take-while-result)]
(concat
take-while-result
(into [] (take n (drop n0 coll))))))
I would use split-with, which is equivalent of getting results of both take-while and drop-while for the same parameters:
(defn take-while-and-n-more [pred n coll]
(let [[head tail] (split-with pred coll)]
(concat head (take n tail))))
Yet another way:
(defn take-while-and-n-more [pred n coll]
(let [[a b] (split-with pred coll)]
(concat a (take n b))))
The following code is a modified version of Clojures take-while. Where Clojures take-while returns nil as a default case (when the predicate does not match), this one invokes take to take the the additional items after the predicate fails.
Note that unlike versions using split-with, this version traverses the sequence only once.
(defn take-while-and-n-more
[pred n coll]
(lazy-seq
(when-let [s (seq coll)]
(if (pred (first s))
(cons (first s) (take-while-and-n-more pred n (rest s)))
(take n s)))))
Related
I have function
(defn goneSeq [inseq uptil]
(loop [counter 0 newSeq [] orginSeq inseq]
(if (== counter uptil)
newSeq
(recur (inc counter) (conj newSeq (first orginSeq)) (rest orginSeq)))))
(defn insert [sorted-seq n]
(loop [currentSeq sorted-seq counter 0]
(cond (empty? currentSeq) (concat sorted-seq (vector n))
(<= n (first currentSeq)) (concat (goneSeq sorted-seq counter) (vector n) currentSeq)
:else (recur (rest currentSeq) (inc counter)))))
that takes in a sorted-sequence and insert the number n at its appropriate position for example: (insert [1 3 4] 2) returns [1 2 3 4].
Now I want to use this function with reduce to sort a given sequence so something like:
(reduce (insert seq n) givenSeq)
What is thr correct way to achieve this?
If the function works for inserting a single value, then this would work:
(reduce insert [] givenSeq)
for example:
user> (reduce insert [] [0 1 2 30.5 0.88 2.2])
(0 0.88 1 2 2.2 30.5)
Also, it should be noted that sort and sort-by are built in and are better than most hand-rolled solutions.
May I suggest some simpler ways to do insert:
A slowish lazy way is
(defn insert [s x]
(let [[fore aft] (split-with #(> x %) s)]
(concat fore (cons x aft))))
A faster eager way is
(defn insert [coll x]
(loop [fore [], coll coll]
(if (and (seq coll) (> x (first coll)))
(recur (conj fore x) (rest coll))
(concat fore (cons x coll)))))
By the way, you had better put your defns in bottom-up order, if possible. Use declare if there is mutual recursion. You had me thinking your solution did not compile.
I have the following functions that check for odd parity in sequence
(defn countOf[a-seq elem]
(loop [number 0 currentSeq a-seq]
(cond (empty? currentSeq) number
(= (first currentSeq) elem) (recur (inc number) (rest currentSeq))
:else (recur number (rest currentSeq))
)
)
)
(defn filteredSeq[a-seq elemToRemove]
(remove (set (vector (first a-seq))) a-seq)
)
(defn parity [a-seq]
(loop [resultset [] currentSeq a-seq]
(cond (empty? currentSeq) (set resultset)
(odd? (countOf currentSeq (first currentSeq))) (recur (concat resultset (vector(first currentSeq))) (filteredSeq currentSeq (first currentSeq)))
:else (recur resultset (filteredSeq currentSeq (first currentSeq)))
)
)
)
for example (parity [1 1 1 2 2 3]) -> (1 3) that is it picks odd number of elements from a sequence.
Is there a better way to achieve this?
How can this be done with reduce function of clojure
First, I decided to make more idiomatic versions of your code, so I could really see what it was doing:
;; idiomatic naming
;; no need to rewrite count and filter for this code
;; putting item and collection in idiomatic argument order
(defn count-of [elem a-seq]
(count (filter #(= elem %) a-seq)))
;; idiomatic naming
;; putting item and collection in idiomatic argument order
;; actually used the elem-to-remove argument
(defn filtered-seq [elem-to-remove a-seq]
(remove #(= elem-to-remove %) a-seq))
;; idiomatic naming
;; if you want a set, use a set from the beginning
;; destructuring rather than repeated usage of first
;; use rest to recur when the first item is guaranteed to be dropped
(defn idiomatic-parity [a-seq]
(loop [result-set #{}
[elem & others :as current-seq] a-seq]
(cond (empty? current-seq)
result-set
(odd? (count-of elem current-seq))
(recur (conj result-set elem) (filtered-seq elem others))
:else
(recur result-set (filtered-seq elem others)))))
Next, as requested, a version that uses reduce to accumulate the result:
;; mapcat allows us to return 0 or more results for each input
(defn reducing-parity [a-seq]
(set
(mapcat
(fn [[k v]]
(when (odd? v) [k]))
(reduce (fn [result item]
(update-in result [item] (fnil inc 0)))
{}
a-seq))))
But, reading over this, I notice that the reduce is just frequencies, a built in clojure function. And my mapcat was really just a hand-rolled keep, another built in.
(defn most-idiomatic-parity [a-seq]
(set
(keep
(fn [[k v]]
(when (odd? v) k))
(frequencies a-seq))))
In Clojure we can refine our code, and as we recognize places where our logic replicates the built in functionality, we can simplify the code and make it more clear. Also, there is a good chance the built in is better optimized than our own work-alikes.
Is there a better way to achieve this?
(defn parity [coll]
(->> coll
frequencies
(filter (fn [[_ v]] (odd? v)))
(map first)
set))
For example,
(parity [1 1 1 2 1 2 1 3])
;#{1 3}
How can this be done with reduce function of clojure.
We can use reduce to rewrite frequencies:
(defn frequencies [coll]
(reduce
(fn [acc x] (assoc acc x (inc (get acc x 0))))
{}
coll))
... and again to implement parity in terms of it:
(defn parity [coll]
(let [freqs (frequencies coll)]
(reduce (fn [s [k v]] (if (odd? v) (conj s k) s)) #{} freqs)))
I'm working on a project to learn Clojure in practice. I'm doing well, but sometimes I get stuck. This time I need to transform sequence of the form:
[":keyword0" "word0" "word1" ":keyword1" "word2" "word3"]
into:
[[:keyword0 "word0" "word1"] [:keyword1 "word2" "word3"]]
I'm trying for at least two hours, but I know not so many Clojure functions to compose something useful to solve the problem in functional manner.
I think that this transformation should include some partition, here is my attempt:
(partition-by (fn [x] (.startsWith x ":")) *1)
But the result looks like this:
((":keyword0") ("word1" "word2") (":keyword1") ("word3" "word4"))
Now I should group it again... I doubt that I'm doing right things here... Also, I need to convert strings (only those that begin with :) into keywords. I think this combination should work:
(keyword (subs ":keyword0" 1))
How to write a function which performs the transformation in most idiomatic way?
Here is a high performance version, using reduce
(reduce (fn [acc next]
(if (.startsWith next ":")
(conj acc [(-> next (subs 1) keyword)])
(conj (pop acc) (conj (peek acc)
next))))
[] data)
Alternatively, you could extend your code like this
(->> data
(partition-by #(.startsWith % ":"))
(partition 2)
(map (fn [[[kw-str] strs]]
(cons (-> kw-str
(subs 1)
keyword)
strs))))
what about that:
(defn group-that [ arg ]
(if (not-empty arg)
(loop [list arg, acc [], result []]
(if (not-empty list)
(if (.startsWith (first list) ":")
(if (not-empty acc)
(recur (rest list) (vector (first list)) (conj result acc))
(recur (rest list) (vector (first list)) result))
(recur (rest list) (conj acc (first list)) result))
(conj result acc)
))))
Just 1x iteration over the Seq and without any need of macros.
Since the question is already here... This is my best effort:
(def data [":keyword0" "word0" "word1" ":keyword1" "word2" "word3"])
(->> data
(partition-by (fn [x] (.startsWith x ":")))
(partition 2)
(map (fn [[[k] w]] (apply conj [(keyword (subs k 1))] w))))
I'm still looking for a better solution or criticism of this one.
First, let's construct a function that breaks vector v into sub-vectors, the breaks occurring everywhere property pred holds.
(defn breakv-by [pred v]
(let [break-points (filter identity (map-indexed (fn [n x] (when (pred x) n)) v))
starts (cons 0 break-points)
finishes (concat break-points [(count v)])]
(mapv (partial subvec v) starts finishes)))
For our case, given
(def data [":keyword0" "word0" "word1" ":keyword1" "word2" "word3"])
then
(breakv-by #(= (first %) \:) data)
produces
[[] [":keyword0" "word0" "word1"] [":keyword1" "word2" "word3"]]
Notice that the initial sub-vector is different:
It has no element for which the predicate holds.
It can be of length zero.
All the others
start with their only element for which the predicate holds and
are at least of length 1.
So breakv-by behaves properly with data that
doesn't start with a breaking element or
has a succession of breaking elements.
For the purposes of the question, we need to muck about with what breakv-by produces somewhat:
(let [pieces (breakv-by #(= (first %) \:) data)]
(mapv
#(update-in % [0] (fn [s] (keyword (subs s 1))))
(rest pieces)))
;[[:keyword0 "word0" "word1"] [:keyword1 "word2" "word3"]]
I have a seq, (def coll '([:a 20] [:b 30] [:c 50] [:d 90]))
I want to iterate through the seq, and modify only the first element that matches a predicate.
The predicate (def pred (fn [[a b]] (> b 30)))
(f pred (fn [[a b]] [a (+ b 2)]) coll) => ([:a 20] [:b 30] [:c 52] [:d 90])
f is the fn I want, which takes a pred, and a fn to apply to the first elem which matches the pred. All the rest of the elems are not modified and returned in the seq.
What is the idiomatic way to do the above?
One possible way is to split the collection with split-with, apply the function f to the first element of the second collection returned by split-with, and concat the elements together again.
(defn apply-to-first [pred f coll]
(let [[h t] (split-with (complement pred) coll)]
(concat h (list (f (first t))) (rest t))))
Note that the pred function in your example should probably look like this:
(def pred #(> (second %) 30))
As with most problems, there is a number of ways to solve it. This is but one of them.
If you're running Clojure 1.5, give this a try:
(reduce
(fn [acc [a b]]
(if (pred b)
(reduced (concat (:res acc) [[a (+ b 2)]] (rest (:coll acc))))
(assoc acc
:res (conj (:res acc) [a b])
:coll (rest (:coll acc)))))
{:coll coll :res []}
coll)
;; ([:a 20] [:b 30] [:c 52] [:d 90])
The key in this algorithm is the use of the reduced (note the 'd') function - it essentially tells reduce to halt the iteration and return the result. From its doc string:
-------------------------
clojure.core/reduced
([x])
Wraps x in a way such that a reduce will terminate with the value x
The code is a bit terse, but it should give you the basic idea.
Hope this helps.
This function is not hard to write recursively "from scratch". Not only is this a good learning exercise, it also produces the best solution: it is as lazy as possible, and does the absolute minimum amount of computation. So far, only one answer to this question is lazy, and that one calls pred twice on all the items before the update occurs: once in the take-while, and once in the drop-while, parts of split-with.
(defn update-first [pred f coll]
(lazy-seq
(when-let [coll (seq coll)]
(if (pred (first coll))
(cons (f (first coll))
(rest coll))
(cons (first coll)
(update-first pred f (rest coll)))))))
To keep it straightforward: find first element, find its index and use assoc to "update" the element at index:
(let [e (first (filter pred coll))
ind (.indexOf coll e)]
(assoc (vec coll) ind ((fn [[a b]] [a (+ b 2)]) e) ))
Dominic's note about pred applies:
(def pred #(> (second %) 30))
What would be a more idiomatic way to partition a seq based on a seq of integers instead of just one integer?
Here's my implementation:
(defn partition-by-seq
"Return a lazy sequence of lists with a variable number of items each
determined by the n in ncoll. Extra values in coll are dropped."
[ncoll coll]
(let [partition-coll (mapcat #(repeat % %) ncoll)]
(->> coll
(map vector partition-coll)
(partition-by first)
(map (partial map last)))))
Then (partition-by-seq [2 3 6] (range)) yields ((0 1) (2 3 4) (5 6 7 8 9 10)).
Your implementation looks fine, but there could be a more simple solution which uses simple recursion wrapped in lazy-seq(and turns out to be more efficient) than using map and existing partition-by as in your case.
(defn partition-by-seq [ncoll coll]
(if (empty? ncoll)
'()
(let [n (first ncoll)]
(cons (take n coll)
(lazy-seq (partition-by-seq (rest ncoll) (drop n coll)))))))
A variation on Ankur's answer, with a minor addition of laziness and when-let instead of an explicit test for empty?.
(defn partition-by-seq [parts coll]
(lazy-seq
(when-let [s (seq parts)]
(cons
(take (first s) coll)
(partition-by-seq (rest s) (nthrest coll (first s)))))))
(first (reduce (fn [[r l] n]
[(conj r (take n l)) (drop n l)])
[[] (range)]
[2 3 6]))
=> [(0 1) (2 3 4) (5 6 7 8 9 10)]