This is probably a common problem or I shouldn't be using flatten but I dont exactly understand whats happening?
The idea is to take a two-dimensions Vector [[x][y][z]] and return a single dimensional collection (dont really care what form for now e.g. (x y z) and [x y z] are both fine for now.
However if I do the following (subbed is the 2d vector:
(let [flat-coll (flatten subbed)
header-less (subvec flat-coll 0 (count flat-coll))]
...
)
The value of flat-col is realized = 0, and I was expecting it to be the value of the flattened collection.
I guess its down to it being Lazy, is there a 'better' way to achieve what I want?
flat-coll is a seq since flatten returns a seq, therfor subvec does not work
how about (take 3 (flatten [ [1] [2] [3] ])) to get header-less
although I do not entierly understand what you are aiming to do, since you in any case get all elements.
maybe you want to do (vec (flatten [ [1] [2] [3] ])) to get a vector ?
If you just want a vector as the result you can use transducers to flatten:
(into [] cat [[2] [3]])
;=> [2 3]
If you also want to get rid of the head:
(into [] (comp cat (drop 1)) [[2] [3] [8]])
;=> [3 8]
If you don't care about vector or list can do also:
user=> (apply concat [[:x][:y][:z]])
(:x :y :z)
which returns a lazy seq.
Related
I have two sequences, which can be vector or list. Now I want to return a sequence whose elements are not in common to the two sequences.
Here is an example:
(removedupl [1 2 3 4] [2 4 5 6]) = [1 3 5 6]
(removeddpl [] [1 2 3 4]) = [1 2 3 4]
I am pretty puzzled now. This is my code:
(defn remove-dupl [seq1 seq2]
(loop [a seq1 b seq2]
(if (not= (first a) (first b))
(recur a (rest b)))))
But I don't know what to do next.
I encourage you to think about this problem in terms of set operations
(defn extrasection [& ss]
(clojure.set/difference
(apply clojure.set/union ss)
(apply clojure.set/intersection ss)))
Such a formulation assumes that the inputs are sets.
(extrasection #{1 2 3 4} #{2 4 5 6})
=> #{1 6 3 5}
Which is easily achieved by calling the (set ...) function on lists, sequences, or vectors.
Even if you prefer to stick with a sequence oriented solution, keep in mind that searching both sequences is an O(n*n) task if you scan both sequences [unless they are sorted]. Sets can be constructed in one pass, and lookup is very fast. Checking for duplicates is an O(nlogn) task using a set.
I'm still new to Clojure but I think the functional mindset is more into composing functions than actually doing it "by hand", so I propose the following solution:
(defn remove-dupl [seq1 seq2]
(concat
(remove #(some #{%} seq1) seq2)
(remove #(some #{%} seq2) seq1)))
EDIT: I think it is better if we define that remove part as a local function and reuse it:
(defn remove-dupl [seq1 seq2]
(let [removing (fn [x y] (remove #(some #{%} x) y))]
(concat (removing seq1 seq2) (removing seq2 seq1))))
EDIT2: As commented by TimothyPratley
(defn remove-dupl [seq1 seq2]
(let [removing (fn [x y] (remove (set x) y))]
(concat (removing seq1 seq2) (removing seq2 seq1))))
There are several problems with your code.
It doesn't test for the end of either sequence argument.
It steps through b but not a.
It implicitly returns nil when any two sequences have the same
first element.
You want to remove the common elements from the concatenated sequences. You have to work out the common elements first, otherwise you don't know what to remove. So ...
We use
clojure.set/intersection to find the common elements,
concat to stitch the collections together.
remove to remove (1) from (2).
vec to convert to a vector.
Thus
(defn removedupl [coll1 coll2]
(let [common (clojure.set/intersection (set coll1) (set coll2))]
(vec (remove common (concat coll1 coll2)))))
... which gives
(removedupl [1 2 3 4] [2 4 5 6]) ; [1 3 5 6]
(removedupl [] [1 2 3 4]) ; [1 2 3 4]
... as required.
I've tried this for so many nights that I've finally given up on myself. Seems like an extremely simple problem, but I guess I'm just not fully understanding Clojure as well as I should be (I partially attribute that to my almost sole experience with imperative languages). The problem is from hackerrank.com
Here is the problem:
Problem Statement
Given a list repeat each element of the list n times. The input and output
portions will be handled automatically by the grader.
Input Format
First line has integer S where S is the number of times you need to repeat
elements. After this there are X lines, each containing an integer. These are the
X elements of the array.
Output Format
Repeat each element of the original list S times. So you have to return
list/vector/array of S*X integers. The relative positions of the values should be
same as the original list provided as input.
Constraints
0<=X<=10
1<=S<=100
So, given:
2
1
2
3
Output:
1
1
2
2
3
3
I've tried:
(fn list-replicate [num list]
(println (reduce
(fn [element seq] (dotimes [n num] (conj seq element)))
[]
list))
)
But that just gives me an exception. I've tried so many other solutions, and this probably isn't one of my better ones, but it was the quickest one I could come up with to post something here.
(defn list-replicate [num list]
(mapcat (partial repeat num) list))
(doseq [x (list-replicate 2 [1 2 3])]
(println x))
;; output:
1
1
2
2
3
3
The previous answer is short and it works, but it is very "compressed" and is not easy for new people to learn. I would do it in a simpler and more obvious way.
First, look at the repeat function:
user=> (doc repeat)
-------------------------
clojure.core/repeat
([x] [n x])
Returns a lazy (infinite!, or length n if supplied) sequence of xs.
user=> (repeat 3 5)
(5 5 5)
So we see how to easily repeat something N times.
What if we run (repeat n ...) on each element of the list?
(def N 2)
(def xvals [1 2 3] )
(for [curr-x xvals]
(repeat N curr-x))
;=> ((1 1) (2 2) (3 3))
So we are getting close, but we have a list-of-lists for output. How to fix? The simplest way is to just use the flatten function:
(flatten
(for [curr-x xvals]
(repeat N curr-x)))
;=> (1 1 2 2 3 3)
Note that both repeat and for are lazy functions, which I prefer to avoid unless I really need them. Also, I usually prefer to store my linear collections in a concrete vector, instead of a generic "seq" type. For these reasons, I include an extra step of forcing the results into a single (eagar) vector for the final product:
(defn list-replicate [num-rep orig-list]
(into []
(flatten
(for [curr-elem xvals]
(repeat N curr-elem)))))
(list-replicate N xvals)
;=> [1 1 2 2 3 3]
I would suggest building onto Alan's solution and instead of flatten use concat as this will preserve the structure of the data in case you have input sth like this [[1 2] [3 4]].
((fn [coll] (apply concat (for [x coll] (repeat 2 x)))) [[1 2] [3 4]])
output: => ([1 2] [1 2] [3 4] [3 4])
unlike with flatten, which does the following
((fn [coll] (flatten (for [x coll] (repeat 2 x)))) [[1 2] [3 4]])
output: => (1 2 1 2 3 4 3 4)
as for simple lists e.g. '(1 2 3), it works the same:
((fn [coll] (apply concat (for [x coll] (repeat 2 x)))) '(1 2 3))
output => (1 1 2 2 3 3)
(reduce #(count (map println (repeat %1 %2))) num list)
The question doesn't really explain what I want to do but I couldn't think of anything else.
I have an empty map in the outer let function in a piece of code, and an integer array.
I want to iterate through the integer array, perform a simple task, and keep appending the resulting map to the variables in the outer variables.
(let [a {} ;outer variables
b {}]
(doseq [x [1 2 3]]
(let [r (merge a {x (* x x)}) ;I want to append this to a
s (merge b {x (+ x x)})] ;and this to b
(println (str "--a--" r "--b--" s)))))
But as soon as I get out of doseq, my a and b vars are still empty. I get that the scope of a and b doesn't extend outside of doseq for it to persist any changes done from within and that they are immutable.
How do I calculate the values of a and b in such cases, please? I tried to extract the functionality of doseq into another function and calling let with:
(let [a (do-that-function)])
etc but even then I couldn't figure out a way to keep track of all the modifications within doseq loop to then send back as a whole.
Am I approaching this in a wrong way?
Thanks
edit
Really, what I'm trying to do is this:
(let [a (doseq [x [1 2 3]] {x (* x x)})]
(println a))
but doseq returns nil so a is going to be nil :-s
All variables in clojure are immutable. If you need a mutable state you should use atoms or refs.
But in your case you can simply switch from doseq to for:
(let [a (for [x [1 2 3]] {x (* x x)})]
(println a))
Here is an example of solving your problem with atoms:
(let [a (atom {})
b (atom {})]
(doseq [x [1 2 3]]
(swap! a assoc x (* x x))
(swap! b assoc x (+ x x)))
(println "a:" #a)
(println "b:" #b))
But you should avoid using mutable state as far as possible:
(let [l [1 2 3]
a (zipmap l (map * l l))
b (zipmap l (map + l l))]
(println "a:" a)
(println "b:" b))
The trick is to think in terms of flows of data adding to existing data making new data, instead of changing past data. For your specific problem, where a data structure is being built, reduce is typically used:
(reduce (fn [result x] (assoc result x (* x x))) {} [1 2 3])
hehe, I just noticed that "reduce" might seem confusing given that it's building something, but the meaning is that a collection of things is "reduced" to one thing. In this case, we give reduce an empty map to begin with, which binds to result in the fn, and each successive mapping over the collection results in a new result, which we add to again with assoc.
You could also say:
(into {} (map (fn [x] [x (* x x)]) [1 2 3]))
In your question you wanted to make multiple things at once from a single collection. Here's one way to do that:
(reduce (fn [[a b] x] [(assoc a x (* x x)) (assoc b x (+ x x))]) [{} {}] [1 2 3])
Here we used destructuring syntax to refer to our two result structures - just make a picture of the data [with [vectors]]. Note that reduce is still only returning one thing - a vector in this case.
And, we could generalize that:
(defn xfn [n fs]
(reduce
(fn [results x] (map (fn [r f] (assoc r x (f x x))) results fs))
(repeat (count fs) {}) (range n)))
=> (xfn 4 [* + -])
({3 9, 2 4, 1 1, 0 0} {3 6, 2 4, 1 2, 0 0} {3 0, 2 0, 1 0, 0 0})
The result is a list of maps. And if you wanted to take intermediate steps in the building of these results, you could change reduce to reductions. Generally, map for transforming collections, reduce for building a single result from a collection.
In Clojure, I have a collection coll of 2-elements vectors. I would like to create the collection obtained by applying f and g on the first and second elements on every vector of the collection, respectively. I think this is related to the list comprehension construct.
(def coll [[1 1000] [2 2000] [3 3000]])
IS there an idiomatic way for creating the following result?
[[f(1) g(1000)] [f(2) g(2000)] [f(3) g(3000)]]
Again, list comprehension FTW:
(vec (for [[x y] [[1 1000] [2 2000] [3 3000]]] [(f x) (g y)]))
Yes,
(vec (map (fn [[p1 p2]] [(f p1) (g p2)])
[[1 1000] [2 2000] [3 3000]]))
To write this from scratch, I would do exactly what skuro did - it's simple, easy, and readable. But I also wrote a higher-order function to abstract this some time ago, named knit. So now I would write this as
(map (knit f g) [[1 1000] [2 2000] [3 3000]])
I have this deeply nested list (list of lists) and I want to replace a single arbitrary element in the list. How can I do this ? (The built-in replace might replace many occurrences while I need to replace only one element.)
As everyone else already said, using lists is really not a good idea if you need to do this kind of thing. Random access is what vectors are made for. assoc-in does this efficiently. With lists you can't get away from recursing down into the sublists and replacing most of them with altered versions of themselves all the way back up to the top.
This code will do it though, albeit inefficiently and clumsily. Borrowing from dermatthias:
(defn replace-in-list [coll n x]
(concat (take n coll) (list x) (nthnext coll (inc n))))
(defn replace-in-sublist [coll ns x]
(if (seq ns)
(let [sublist (nth coll (first ns))]
(replace-in-list coll
(first ns)
(replace-in-sublist sublist (rest ns) x)))
x))
Usage:
user> (def x '(0 1 2 (0 1 (0 1 2) 3 4 (0 1 2))))
#'user/x
user> (replace-in-sublist x [3 2 0] :foo)
(0 1 2 (0 1 (:foo 1 2) 3 4 (0 1 2)))
user> (replace-in-sublist x [3 2] :foo)
(0 1 2 (0 1 :foo 3 4 (0 1 2)))
user> (replace-in-sublist x [3 5 1] '(:foo :bar))
(0 1 2 (0 1 (0 1 2) 3 4 (0 (:foo :bar) 2)))
You'll get IndexOutOfBoundsException if you give any n greater than the length of a sublist. It's also not tail-recursive. It's also not idiomatic because good Clojure code shies away from using lists for everything. It's horrible. I'd probably use mutable Java arrays before I used this. I think you get the idea.
Edit
Reasons why lists are worse than vectors in this case:
user> (time
(let [x '(0 1 2 (0 1 (0 1 2) 3 4 (0 1 2)))] ;'
(dotimes [_ 1e6] (replace-in-sublist x [3 2 0] :foo))))
"Elapsed time: 5201.110134 msecs"
nil
user> (time
(let [x [0 1 2 [0 1 [0 1 2] 3 4 [0 1 2]]]]
(dotimes [_ 1e6] (assoc-in x [3 2 0] :foo))))
"Elapsed time: 2925.318122 msecs"
nil
You also don't have to write assoc-in yourself, it already exists. Look at the implementation for assoc-in sometime; it's simple and straightforward (compared to the list version) thanks to vectors giving efficient and easy random access by index, via get.
You also don't have to quote vectors like you have to quote lists. Lists in Clojure strongly imply "I'm calling a function or macro here".
Vectors (and maps, sets etc.) can be traversed via seqs. You can transparently use vectors in list-like ways, so why not use vectors and have the best of both worlds?
Vectors also stand out visually. Clojure code is less of a huge blob of parens than other Lisps thanks to widespread use of [] and {}. Some people find this annoying, I find it makes things easier to read. (My editor syntax-highlights (), [] and {} differently which helps even more.)
Some instances I'd use a list for data:
If I have an ordered data structure that needs to grow from the front, that I'm never going to need random-access to
Building a seq "by hand", as via lazy-seq
Writing a macro, which needs to return code as data
For the simple cases a recursive substitution function will give you just what you need with out much extra complexity. when things get a little more complex its time to crack open clojure build in zipper functions: "Clojure includes purely functional, generic tree walking and editing, using a technique called a zipper (in namespace zip)."
adapted from the example in: http://clojure.org/other_libraries
(defn randomly-replace [replace-with in-tree]
(loop [loc dz]
(if (zip/end? loc)
(zip/root loc)
(recur
(zip/next
(if (= 0 (get-random-int 10))
(zip/replace loc replace-with)
loc)))))
these will work with nested anything (seq'able) even xmls
It sort of doesn't answer your question, but if you have vectors instead of lists:
user=> (update-in [1 [2 3] 4 5] [1 1] inc)
[1 [2 4] 4 5]
user=> (assoc-in [1 [2 3] 4 5] [1 1] 6)
[1 [2 6] 4 5]
So if possible avoid lists in favour of vectors for the better access behaviour. If you have to work with lazy-seq from various sources, this is of course not much of an advice...
You could use this function and adapt it for your needs (nested lists):
(defn replace-item
"Returns a list with the n-th item of l replaced by v."
[l n v]
(concat (take n l) (list v) (drop (inc n) l)))
A simple-minded suggestion from the peanut gallery:
copy the inner list to a vector;
fiddle that vector's elements randomly and to your heart's content using assoc;
copy the vector back to a list;
replace the nested list in the outer list.
This might waste some performance; but if this was a performance sensitive operation you'd be working with vectors in the first place.