I am stuck at the Pascal's Trapezoid from 4Clojure site, where you need to build a lazy sequence of the trapezoid's numbers.
My first shot was this:
(defn pascal [x]
(cons x
(lazy-seq
(pascal
(map +
(cons 0 x)
(conj x 0)
)))))
Which didn't work:
user=> (take 5 (pascal [1 1]))
([1 1] (1 2 1) (0 2 4 2) (0 0 4 8 4) (0 0 0 8 16 8))
Writing it this way works, however:
(defn pascal2 [x]
(cons x
(lazy-seq
(pascal2
(map +
(concat [0] x)
(concat x [0])
)))))
user=> (take 5 (pascal2 [1 1]))
([1 1] (1 2 1) (1 3 3 1) (1 4 6 4 1) (1 5 10 10 5 1))
So, what exactly am I doing wrong here? What is the difference between cons/conj and concat?
As others stated conj inserts the element(s) it receives in a different position depending on the concrete collection type, see this SO question for more detailed information about the difference between conj and cons.
In the first version of your pascal function you are providing a vector as the initial argument so the expression (conj x 0) will insert the 0 at the end of the vector for the computation of the second element in the series, but since map returns a lazy sequence, when the third element is computed the insertion happens at the beginning ((conj (map inc '(0)) 2) ;= (2 1)), which results in wrong elements in the series from then on.
In order to use the cons and conj approach you have to make sure you return a vector by using mapv instead of map.
(defn pascal [x]
(cons x
(lazy-seq
(pascal
(mapv +
(cons 0 x)
(conj x 0))))))
(take 5 (pascal [1 1]))
;= ([1 1] [1 2 1] [1 3 3 1] [1 4 6 4 1] [1 5 10 10 5 1])
The drawback with mapv is that it is eager so it will compute all members in the pascal element, instead of just holding it back until you actually need them.
On the other hand, when using concat you do ensure you append the element at the end and that everything is lazy, but the append is not done cheaply like with vectors, see here for more information.
Regardless of these factors you can still use cons in both cases, since what it does is what you need in either case (i.e. have an element inserted at the beginning of a collection).
(defn pascal2 [x]
(cons x
(lazy-seq
(pascal2
(map +
(cons 0 x) ; Replaced concat with cons
(concat x [0]))))))
(take 5 (pascal2 [1 1]))
;= ([1 1] (1 2 1) (1 3 3 1) (1 4 6 4 1) (1 5 10 10 5 1))
According to ClojureDocs
conj
conj clojure.core
(conj coll x)
(conj coll x & xs)
conj[oin]. Returns a new collection with the xs 'added'. (conj nil
item) returns (item). The 'addition' may happen at different 'places'
depending on the concrete type.
conj accepts the first argument as a collection, which means coll must be a collection type. conj will return a new collection with x added into coll, and the place of x added is depending on the type of coll.
e.g.
> (conj [1] [0])
[1 [0]] ; See [0] is added into [1] as an element. Instead of returning [1 0], it returns [1 [0]]
> (conj [1] 0)
[1 0]
> (conj '(1) 0)
(0 1) ;See the element `0` position is different in each case.
concat
concat clojure.core
(concat)
(concat x)
(concat x y)
(concat x y & zs)
Returns a lazy seq representing the concatenation of the elements in
the supplied colls.
concat accepts all the argument as collection types, which is different from conj. concat returns the concatenation of arguments.
e.g.
> (concat [0] [1])
(0 1)
> (concat [0] [[1]])
(0 [1])
> (concat [0] 1) ;See the second argument is not a collection type, thus the function throws an exception.
java.lang.IllegalArgumentException: Don't know how to create ISeq from: java.lang.Long
>
cons
cons clojure.core
(cons x seq)
Returns a new seq where x is the first element and seq is the rest.
The doc of cons states clearly how cons would work. The second argument of cons must be a seq.
e.g.
> (cons [1] [0])
([1] 0) ; [1] is the first element and (0) is the rest
> (first (cons [1] [0]))
[1]
> (rest (cons [1] [0]))
(0)
> (cons 1 [0]) ; 1 is the first element and (0) is the rest
(1 0)
> (cons [1] 0) ;the second argument is not a seq, throwing exception
java.lang.IllegalArgumentException: Don't know how to create ISeq from: java.lang.Long
conj on a list adds the element to the front of the list.
If you convert the list to a vector it will work.
user=> (conj '(1 2 3) 4)
(4 1 2 3)
user=> (conj [1 2 3] 4)
[1 2 3 4]
Related
I want to write a function that inserts elements between existing elements in a vector. The inserted elements are a function of the elements that precede and succeed it, with the first and last elements remaining unaffected.
E.g. I want inserted elements to be the mean of the elements that precede and succeed it:
Input:
[1 10 15]
Output:
[1 5.5 10 12.5 15]
What is the best way to do this in Clojure?
Here's another way:
(defn insert-mean-between [xs]
(let [f (fn [x y]
[(* (+ x y) 0.5) y])]
(->> xs
(partition 2 1)
(mapcat (partial apply f))
(cons (first xs))
vec)))
(insert-mean-between [1 10 15])
;;=> [1 5.5 10 12.5 15]
The main trick is that f is returning the answer and the RHS input. This way later on they will all compose together without repeats. The only problem you will have is that the first element is missing. So we just cons it onto the front. From the outset we had to know that cons would be a convenient operation when we chose to be returning the RHS rather than the LHS.
As calculating the mean was just an example, an improved solution would be for the inserting to be independent of the mean/whatever function:
(defn calc-mean [x y] (* (+ x y) 0.5)
(insert-between calc-mean [1 10 15])
Then a more general inserting function might be:
(defn insert-between [g xs]
(->> xs
(partition 2 1)
(mapcat (fn [[x y]] [(g x y) y]))
(cons (first xs))))
and the list of variants won't be complete without the recursive lazy sequence generation:
(defn with-avg [[x1 & [x2 :as tail] :as items]]
(when (seq items)
(if (seq tail)
(lazy-cat [x1 (/ (+ x1 x2) 2)] (with-avg tail))
[x1])))
user> (with-avg [1 2 3 4 5])
;;=> (1 3/2 2 5/2 3 7/2 4 9/2 5)
user> (with-avg [1])
;;=> [1]
user> (with-avg [])
;;=> nil
user> (with-avg [1 2])
;;=> (1 3/2 2)
user> (with-avg [1 2 3])
;;=>(1 3/2 2 5/2 3)
One way I could solve it is pattern matching Vector as f s t, I'm assuming it has 3 elements
Then create variable to assign first median first + second / 2 and second median second + third /2.
At the end return a new Vector with a combination you want.
Example, (I'm using lein REPL)
user=> (defn insert_medians[vect]
#_=> (let [[f s t] vect
#_=> m1 (float (/ (+ f s) 2))
#_=> m2 (float (/ (+ s t) 2))]
#_=> [f m1 s m2 t]))
#'user/insert_medians
user=> (insert_medians [1 10 15])
[1 5.5 10 12.5 15]
If a vector is larger than 3 elems, you need to find all the medians first and then insert into the original vector using interleave fn.
(defn insert-between
"Inserts elements between existing elements in a vector v. The inserted
elements are a result of applying the function f to the elements that precede
and succeed it, with the first and last elements of v remaining unaffected."
[f [x & xs :as v]]
(->> (partition 2 1 v)
(mapcat (fn [[a b]] [(f a b) b]))
(cons x)
(into [])))
(defn mean [& numbers]
(float (/ (apply + numbers) (count numbers))))
(insert-between mean [1 10 15]) ; => [1 5.5 10 10 12.5 15]
(insert-between + [1 10 15 20 25]) ; => [1 11 10 25 15 35 20 45 25]
(insert-between mean []) ; => [nil] :(
I am a newbie to clojure (and functional programming for that matter) and I was trying to do some basic problems. I was trying to find the nth element in a sequence without recursion.
so something like
(my-nth '(1 2 3 4) 2) => 3
I had a hard time looping through the list and returning when i found the nth element. I tried a bunch of different ways and the code that I ended up with is
(defn sdsu-nth
[input-list n]
(loop [cnt n tmp-list input-list]
(if (zero? cnt)
(first tmp-list)
(recur (dec cnt) (pop tmp-list)))))
This gives me an exception which says "cant pop from empty list"
I dont need code, but if someone could point me in the right direction it would really help!
You are using the function pop, which has different behavior for different data structures.
user> (pop '(0 1 2 3 4))
(1 2 3 4)
user> (pop [0 1 2 3 4])
[0 1 2 3]
user> (pop (map identity '(0 1 2 3 4)))
ClassCastException clojure.lang.LazySeq cannot be cast to clojure.lang.IPersistentStack clojure.lang.RT.pop (RT.java:640)
Furthermore, you are mixing calls to pop with calls to first. If iterating, use peek/pop or first/rest as pairs, mixing the two can lead to unexpected results. first / rest are the lowest common denominator, if you want to generalize over various sequential types, use those, and they will coerce the sequence to work if they can.
user> (first "hello")
\h
user> (first #{0 1 2 3 4})
0
user> (first {:a 0 :b 1 :c 2})
[:c 2]
With your function, replacing pop with rest, we get the expected results:
user> (defn sdsu-nth
[input-list n]
(loop [cnt n tmp-list input-list]
(if (zero? cnt)
(first tmp-list)
(recur (dec cnt) (rest tmp-list)))))
#'user/sdsu-nth
user> (sdsu-nth (map identity '(0 1 2 3 4)) 2)
2
user> (sdsu-nth [0 1 2 3 4] 2)
2
user> (sdsu-nth '(0 1 2 3 4) 2)
2
user> (sdsu-nth "01234" 2)
\2
given a list as list_nums, take up to n + 1 then from that return the last element which is nth.
(fn [list_nums n] (last (take (inc n) list_nums)))
and alternatively:
#(last (take (inc %2) %1))
proof:
(= (#(last (take (inc %2) %1)) '(4 5 6 7) 2) 6) ;; => true
What you would really want to do is use the built-in nth function as it does exactly what you're asking:
http://clojuredocs.org/clojure_core/clojure.core/nth
However, since you're learning this is still a good exercise. Your code actually works for me. Make sure you're giving it a list and not a vector -- pop does something different with vectors (it returns the vector without the last item rather than the first -- see here).
Your code works fine for lists if supplied index is not equal or greater then length of sequence (you've implemented zero indexed nth). You get this error when tmp-list gets empty before your cnt gets to the zero.
It does not work so well with vectors:
user> (sdsu-nth [1 2 3 4] 2)
;; => 1
user> (sdsu-nth [10 2 3 4] 2)
;; => 10
it seems to return 0 element for every supplied index. As noisesmith noticed it happens because pop works differently for vectors because of their internal structure. For vectors pop will remove elements form the end, and then first returns first value of any vector.
How to fix: use rest instead of pop, to remove differences in behavior of your function when applied to lists and vectors.
(fn [xs n]
(if (= n 0)
(first xs)
(recur (rest xs) (dec n))))
One more way that I thought of doing this and making it truly non recursive (ie without for/recur) is
(defn sdsu-nth
[input-list n]
(if (zero? (count input-list))
(throw (Exception. "IndexOutOfBoundsException"))
(if (>= n (count input-list))
(throw (Exception. "IndexOutOfBoundsException"))
(if (neg? n)
(throw (Exception. "IndexOutOfBoundsException"))
(last (take (+ n 1) input-list))))))
If I use the reductions function like so:
(reductions + [1 2 3 4 5])
Then I get
(1 3 6 10 15)
Which is great - but I'd like to apply a binary function in the same way without the state being carried forward - something like
(magic-hof + [1 2 3 4 5])
leads to
(1 3 5 7 9)
ie it returns the operation applied to the first pair, then steps 1 to the next pair.
Can someone tell me the higher-order function I'm looking for? (Something like reductions)
This is my (non-working) go at it:
(defn thisfunc [a b] [(+ a b) b])
(reduce thisfunc [1 2 3 4 5])
You can do it with map:
(map f coll (rest coll))
And if you want a function:
(defn map-pairwise [f coll]
(map f coll (rest coll)))
And if you really need the first element to remain untouched (thanx to juan.facorro's comment):
(defn magic-hof [f [x & xs :as s]]
(cons x (map f s xs)))
partition will group your seq:
user> (->> [1 2 3 4 5] (partition 2 1) (map #(apply + %)) (cons 1))
(1 3 5 7 9)
So, you want to apply a function to subsequent pairs of elements?
(defn pairwise-apply
[f sq]
(when (seq sq)
(->> (map f sq (next sq))
(cons (first sq)))))
Let's try it:
(pairwise-apply + (range 1 6))
;; => (1 3 5 7 9)
This is sufficient:
(#(map + (cons 0 %) %) [1 2 3 4 5])
;; => (1 3 5 7 9)
The comprehension:
(for [i (range 5])] i)
... yields: (0 1 2 3 4)
Is there an idiomatic way to get (0 0 1 1 2 4 3 9 4 16) (i.e. the numbers and their squares) using mostly the for comprehension?
The only way I've found so far is doing a:
(apply concat (for [i (range 5)] (list i (* i i))))
Actually, using only for is pretty simple if you consider applying each function (identity and square) for each value.
(for [i (range 5), ; for every value
f [identity #(* % %)]] ; for every function
(f i)) ; apply the function to the value
; => (0 0 1 1 2 4 3 9 4 16)
Since for loops x times, it will return a collection of x values. Multiple nested loops (unless limited by while or when) will give x * y * z * ... results. That is why external concatenation will always be necessary.
A similar correlation between input and output exists with map. However, if multiple collections are given in map, the number of values in the returned collection is the size of the smallest collection parameter.
=> (map (juxt identity #(* % %)) (range 5))
([0 0] [1 1] [2 4] [3 9] [4 16])
Concatenating the results of map is so common mapcat was created. Because of that, one might argue mapcat is a more idiomatic way over for loops.
=> (mapcat (juxt identity #(* % %)) (range 5))
(0 0 1 1 2 4 3 9 4 16)
Although this is just shorthand for apply concat (map, and a forcat function or macro could be created just as easily.
However, if an accumulation over a collection is needed, reduce is usually considered the most idiomatic.
=> (reduce (fn [acc i] (conj acc i (* i i))) [] (range 5))
[0 0 1 1 2 4 3 9 4 16]
Both the for and map options would mean traversing a collection twice, once for the range, and once for concatenating the resulting collection. The reduce option only traverses the range.
Care to share why "using mostly the for comprehension" is a requirement ?
I think you are doing it right.
A slightly compressed way maybe achieved using flatten
(flatten (for [i (range 5)] [ i (* i i) ] ))
But I would get rid of the for comprehension and just use interleave
(let [x (range 5)
y (map #(* % %) x)]
(interleave x y))
Disclaimer: I am just an amateur clojurist ;)
The clojure document gives the following examples:
(take 10 (iterate (partial + 2) 0))
(def powers-of-two (iterate (partial * 2) 1))
(take 10 powers-of-two)
(def fib (map first (iterate (fn [[a b]] [b (+ a b)]) [1 1])))
(take 10 fib)
Anyone can explain the syntax of clojure's iterate function in more detail? I am very confused with all the usage. Why two brackets are there in (fn [[a b]] [b (+ a b)])?
Another example can be found here:
(defn iter [[x y]]
(vector y (+ x y)))
(nth (iterate iter [0 1]) 10000)
iterate takes a function f and an initial value x and produces a lazy sequence. The first element in the seq is x. Each subsequent element is computed by calling f with the previous element.
Example 1:
(iterate (partial + 2) 0)
This generates a sequence, starting at 0, where each element is the previous element with 2 added to it. I.e.:
0
(+ 2 0) ; => 2
(+ 2 2) ; => 4
(+ 2 4) ; => 6
; etc
Each element in the seq is passed to (partial + 2) when generating the following element.
Example 2:
(iterate (partial * 2) 1)
This generates a sequence, starting at 1, where each element is the previous element multiplied by 2. I.e.:
1
(* 2 1) ; => 2
(* 2 2) ; => 4
(* 2 4) ; => 8
(* 2 8) ; => 16
; etc
Again, you can see how each element feeds into the generation of the next one.
Example 3:
(iterate (fn [[a b]] [b (+ a b)]) [1 1])
Firstly, (fn [[a b]] ...) is a way to destructure a value into parts. In this case, the function accepts a two-element vector and unpacks it into the local variables a and b.
The function returns a two-element vector containing b and the sum of a and b (i.e. the second value in the previous pair and the sum of both values in the previous pair).
With this in mind, this iterate call generates:
[1 1]
[1 (+ 1 1)] ; => [1 2]
[2 (+ 1 2)] ; => [2 3]
[3 (+ 2 3)] ; => [3 5]
[5 (+ 3 5)] ; => [5 8]
; etc
Then (map first ...) grabs the first value in each pair, which gives you your Fibonacci sequence.