I'm trying to learn functional programming with SICP. I want to use Clojure.
Clojure is a dialect of Lisp but I'm very unfamiliar with Lisp. This code snippet unclean and unreadable. How to write more efficient code with Lisp dialects ?
And how to pass multiple parameters function from other function ?
(defn greater [x y z]
(if (and (>= x y) (>= x z))
(if (>= y z)
[x,y]
[x,z])
(if (and (>= y x) (>= y z))
(if (>= x z)
[y,x]
[y,z])
(if (and (>= z x) (>= z y))
(if (>= y x)
[z,y]
[z,x])))))
(defn sum-of-squares [x y]
(+ (* x x) (* y y)))
(defn -main
[& args]
(def greats (greater 2 3 4))
(def sum (sum-of-squares greats)))
You are asking two questions, and I will try to answer them in reverse order.
Applying Collections as Arguments
To use a collection as an function argument, where each item is a positional argument to the function, you would use the apply function.
(apply sum-of-squares greats) ;; => 25
Readability
As for the more general question of readability:
You can gain readability by generalizing the problem. From your code sample, it looks like the problem consists of performing the sum, of the squares, on the two largest numbers in a collection. So, it would be visually cleaner to sort the collection in descending order and take the first two items.
(defn greater [& numbers]
(take 2 (sort > numbers)))
(defn sum-of-squares [x y]
(+ (* x x) (* y y)))
You can then use apply to pass them to your sum-of-squares function.
(apply sum-of-squares (greater 2 3 4)) ;; => 25
Keep in Mind: The sort function is not lazy. So, it will both realize and sort the entire collection you give it. This could have performance implications in some scenarios. But, in this case, it is not an issue.
One Step Further
You can further generalize your sum-of-squares function to handle multiple arguments by switching the two arguments, x and y, to a collection.
(defn sum-of-squares [& xs]
(reduce + (map #(* % %) xs)))
The above function creates an anonymous function, using the #() short hand syntax, to square a number. That function is then lazily mapped, using map, over every item in the xs collection. So, [1 2 3] would become (1 4 9). The reduce function takes each item and applies the + function to it and the current total, thus producing the sum of the collection. (Because + takes multiple parameters, in this case you could also use apply.)
If put it all together using one of the threading macros, ->>, it starts looking very approachable. (Although, an argument could be made that, in this case, I have traded some composability for more readability.)
(defn super-sum-of-squares [n numbers]
(->> (sort > numbers)
(take n)
(map #(* % %))
(reduce +)))
(super-sum-of-squares 2 [2 3 4]) ;;=> 25
(defn greater [& args] (take 2 (sort > args)))
(defn -main
[& args]
(let [greats (greater 2 3 4)
sum (apply sum-of-squares greats)]
sum))
A key to good clojure style is to use the built in sequence operations. An alternate approach would have been a single cond form instead of the deeply nested if statements.
def should not be used inside function bodies.
A function should return a usable result (the value returned by -main will be printed if you run the project).
apply uses a list as the args for the function provided.
To write readable code, use the functions provided by the language as much as possible:
e.g. greater can be defined as
(defn greater [& args]
(butlast (sort > args)))
To make sum-of-squares work on the return value from greater, use argument destructuring
(defn sum-of-squares [[x y]]
(+ (* x x) (* y y)))
which requires the number of elements in the argument sequence to be known,
or define sum-of-squares to take a single sequence as argument
(defn sum-of-squares [args]
(reduce + (map (fn [x] (* x x)) args)))
Related
I'm trying to estimate the mean distance of all pairs of points in a unit square.
This transducer returns a vector of the distances of x randomly selected pairs of points, but the final step would be to take the mean of all values in that vector. Is there a way to use mean as the final reducing function (or to include it in the composition)?
(defn square [x] (* x x))
(defn mean [x] (/ (reduce + x) (count x)))
(defn xform [iterations]
(comp
(partition-all 4)
(map #(Math/sqrt (+ (square (- (first %) (nth % 1)))
(square (- (nth % 2) (nth % 3))))))
(take iterations)))
(transduce (xform 5) conj (repeatedly #(rand)))
[0.5544757422041136
0.4170515673848907
0.7457675423415904
0.5560901974277822
0.6053573945754688]
(transduce (xform 5) mean (repeatedly #(rand)))
Execution error (ArityException) at test.core/eval19667 (form-init9118116578029918666.clj:562).
Wrong number of args (0) passed to: test.core/mean
If you implement your mean function differently, you won't have to collect all the values before computing the mean. Here is how you can implement it, based on this Java code:
(defn mean
([] [0 1]) ;; <-- Construct an empty accumulator
([[mu n]] mu) ;; <-- Get the mean (final step)
([[mu n] x] ;; <-- Accumulate a value to the mean
[(+ mu (/ (- x mu) n)) (inc n)]))
And you use it like this:
(transduce identity mean [1 2 3 4])
;; => 5/2
or like this:
(transduce (xform 5) mean (repeatedly #(rand)))
;; => 0.582883812837961
From the docs of transduce:
If init is not supplied, (f) will be called to produce it. f should be
a reducing step function that accepts both 1 and 2 arguments, if it
accepts only 2 you can add the arity-1 with 'completing'.
To disect this:
Your function needs 0-arity to produce an initial value -- so conj
is fine (it produces an empty vector).
You need to provide a 2-arity function to do the actual redudcing
-- again conj is fine here
You need to provide a 1-arity function to finalize - here you want
your mean.
So as the docs suggest, you can use completing to just provide that:
(transduce (xform 5) (completing conj mean) (repeatedly #(rand)))
; → 0.4723186070904141
If you look at the source of completing you will see how it produces
all of this:
(defn completing
"Takes a reducing function f of 2 args and returns a fn suitable for
transduce by adding an arity-1 signature that calls cf (default -
identity) on the result argument."
{:added "1.7"}
([f] (completing f identity))
([f cf]
(fn
([] (f))
([x] (cf x))
([x y] (f x y)))))
A reduce call has its f argument first. Visually speaking, this is often the biggest part of the form.
e.g.
(reduce
(fn [[longest current] x]
(let [tail (last current)
next-seq (if (or (not tail) (> x tail))
(conj current x)
[x])
new-longest (if (> (count next-seq) (count longest))
next-seq
longest)]
[new-longest next-seq]))
[[][]]
col))
The problem is, the val argument (in this case [[][]]) and col argument come afterward, below, and it's a long way for your eyes to travel to match those with the parameters of f.
It would look more readable to me if it were in this order instead:
(reduceb val col
(fn [x y]
...))
Should I implement this macro, or am I approaching this entirely wrong in the first place?
You certainly shouldn't write that macro, since it is easily written as a function instead. I'm not super keen on writing it as a function, either, though; if you really want to pair the reduce with its last two args, you could write:
(-> (fn [x y]
...)
(reduce init coll))
Personally when I need a large function like this, I find that a comma actually serves as a good visual anchor, and makes it easier to tell that two forms are on that last line:
(reduce (fn [x y]
...)
init, coll)
Better still is usually to not write such a large reduce in the first place. Here you're combining at least two steps into one rather large and difficult step, by trying to find all at once the longest decreasing subsequence. Instead, try splitting the collection up into decreasing subsequences, and then take the largest one.
(defn decreasing-subsequences [xs]
(lazy-seq
(cond (empty? xs) []
(not (next xs)) (list xs)
:else (let [[x & [y :as more]] xs
remainder (decreasing-subsequences more)]
(if (> y x)
(cons [x] remainder)
(cons (cons x (first remainder)) (rest remainder)))))))
Then you can replace your reduce with:
(apply max-key count (decreasing-subsequences xs))
Now, the lazy function is not particularly shorter than your reduce, but it is doing one single thing, which means it can be understood more easily; also, it has a name (giving you a hint as to what it's supposed to do), and it can be reused in contexts where you're looking for some other property based on decreasing subsequences, not just the longest. You can even reuse it more often than that, if you replace the > in (> y x) with a function parameter, allowing you to split up into subsequences based on any predicate. Plus, as mentioned it is lazy, so you can use it in situations where a reduce of any sort would be impossible.
Speaking of ease of understanding, as you can see I misunderstood what your function is supposed to do when reading it. I'll leave as an exercise for you the task of converting this to strictly-increasing subsequences, where it looked to me like you were computing decreasing subsequences.
You don't have to use reduce or recursion to get the descending (or ascending) sequences. Here we are returning all the descending sequences in order from longest to shortest:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(defn descending-sequences [xs]
(->> xs
(partition 2 1)
(map (juxt (fn [[x y]] (> x y)) identity))
(partition-by first)
(filter ffirst)
(map #(let [xs' (mapcat second %)]
(take-nth 2 (cons (first xs') xs'))))
(sort-by (comp - count))))
(descending-sequences in)
;;=> ((7 6 5 4 3 2) (3 2 1 0 -1) (7 6))
(partition 2 1) gives every possible comparison and partition-by allows you to mark out the runs of continuous decreases. At this point you can already see the answer and the rest of the code is removing the baggage that is no longer needed.
If you want the ascending sequences instead then you only need to change the < to a >:
;;=> ((-1 2 7) (6 7))
If, as in the question, you only want the longest sequence then put a first as the last function call in the thread last macro. Alternatively replace the sort-by with:
(apply max-key count)
For maximum readability you can name the operations:
(defn greatest-continuous [op xs]
(let [op-pair? (fn [[x y]] (op x y))
take-every-second #(take-nth 2 (cons (first %) %))
make-canonical #(take-every-second (apply concat %))]
(->> xs
(partition 2 1)
(partition-by op-pair?)
(filter (comp op-pair? first))
(map make-canonical)
(apply max-key count))))
I feel your pain...they can be hard to read.
I see 2 possible improvements. The simplest is to write a wrapper similar to the Plumatic Plumbing defnk style:
(fnk-reduce { :fn (fn [state val] ... <new state value>)
:init []
:coll some-collection } )
so the function call has a single map arg, where each of the 3 pieces is labelled & can come in any order in the map literal.
Another possibility is to just extract the reducing fn and give it a name. This can be either internal or external to the code expression containing the reduce:
(let [glommer (fn [state value] (into state value)) ]
(reduce glommer #{} some-coll))
or possibly
(defn glommer [state value] (into state value))
(reduce glommer #{} some-coll))
As always, anything that increases clarity is preferred. If you haven't noticed already, I'm a big fan of Martin Fowler's idea of Introduce Explaining Variable refactoring. :)
I will apologize in advance for posting a longer solution to something where you wanted more brevity/clarity.
We are in the new age of clojure transducers and it appears a bit that your solution was passing the "longest" and "current" forward for record-keeping. Rather than passing that state forward, a stateful transducer would do the trick.
(def longest-decreasing
(fn [rf]
(let [longest (volatile! [])
current (volatile! [])
tail (volatile! nil)]
(fn
([] (rf))
([result] (transduce identity rf result))
([result x] (do (if (or (nil? #tail) (< x #tail))
(if (> (count (vswap! current conj (vreset! tail x)))
(count #longest))
(vreset! longest #current))
(vreset! current [(vreset! tail x)]))
#longest)))))))
Before you dismiss this approach, realize that it just gives you the right answer and you can do some different things with it:
(def coll [2 1 10 9 8 40])
(transduce longest-decreasing conj coll) ;; => [10 9 8]
(transduce longest-decreasing + coll) ;; => 27
(reductions (longest-decreasing conj) [] coll) ;; => ([] [2] [2 1] [2 1] [2 1] [10 9 8] [10 9 8])
Again, I know that this may appear longer but the potential to compose this with other transducers might be worth the effort (not sure if my airity 1 breaks that??)
I believe that iterate can be a more readable substitute for reduce. For example here is the iteratee function that iterate will use to solve this problem:
(defn step-state-hof [op]
(fn [{:keys [unprocessed current answer]}]
(let [[x y & more] unprocessed]
(let [next-current (if (op x y)
(conj current y)
[y])
next-answer (if (> (count next-current) (count answer))
next-current
answer)]
{:unprocessed (cons y more)
:current next-current
:answer next-answer}))))
current is built up until it becomes longer than answer, in which case a new answer is created. Whenever the condition op is not satisfied we start again building up a new current.
iterate itself returns an infinite sequence, so needs to be stopped when the iteratee has been called the right number of times:
(def in [3 2 1 0 -1 2 7 6 7 6 5 4 3 2])
(->> (iterate (step-state-hof >) {:unprocessed (rest in)
:current (vec (take 1 in))})
(drop (- (count in) 2))
first
:answer)
;;=> [7 6 5 4 3 2]
Often you would use a drop-while or take-while to short circuit just when the answer has been obtained. We could so that here however there is no short circuiting required as we know in advance that the inner function of step-state-hof needs to be called (- (count in) 1) times. That is one less than the count because it is processing two elements at a time. Note that first is forcing the final call.
I wanted this order for the form:
reduce
val, col
f
I was able to figure out that this technically satisfies my requirements:
> (apply reduce
(->>
[0 [1 2 3 4]]
(cons
(fn [acc x]
(+ acc x)))))
10
But it's not the easiest thing to read.
This looks much simpler:
> (defn reduce< [val col f]
(reduce f val col))
nil
> (reduce< 0 [1 2 3 4]
(fn [acc x]
(+ acc x)))
10
(< is shorthand for "parameters are rotated left"). Using reduce<, I can see what's being passed to f by the time my eyes get to the f argument, so I can just focus on reading the f implementation (which may get pretty long). Additionally, if f does get long, I no longer have to visually check the indentation of the val and col arguments to determine that they belong to the reduce symbol way farther up. I personally think this is more readable than binding f to a symbol before calling reduce, especially since fn can still accept a name for clarity.
This is a general solution, but the other answers here provide many good alternative ways to solve the specific problem I gave as an example.
I have a function that produces lazy-sequences called a-function.
If I run the code:
(map a-function a-sequence-of-values)
it returns a lazy sequence as expected.
But when I run the code:
(mapcat a-function a-sequence-of-values)
it breaks the lazyness of my function. In fact it turns that code into
(apply concat (map a-function a-sequence-of-values))
So it needs to realize all the values from the map before concatenating those values.
What I need is a function that concatenates the result of a map function on demand without realizing all the map beforehand.
I can hack a function for this:
(defn my-mapcat
[f coll]
(lazy-seq
(if (not-empty coll)
(concat
(f (first coll))
(my-mapcat f (rest coll))))))
But I can't believe that clojure doesn't have something already done. Do you know if clojure has such feature? Only a few people and I have the same problem?
I also found a blog that deals with the same issue: http://clojurian.blogspot.com.br/2012/11/beware-of-mapcat.html
Lazy-sequence production and consumption is different than lazy evaluation.
Clojure functions do strict/eager evaluation of their arguments. Evaluation of an argument that is or that yields a lazy sequence does not force realization of the yielded lazy sequence in and of itself. However, any side effects caused by evaluation of the argument will occur.
The ordinary use case for mapcat is to concatenate sequences yielded without side effects. Therefore, it hardly matters that some of the arguments are eagerly evaluated because no side effects are expected.
Your function my-mapcat imposes additional laziness on the evaluation of its arguments by wrapping them in thunks (other lazy-seqs). This can be useful when significant side effects - IO, significant memory consumption, state updates - are expected. However, the warning bells should probably be going off in your head if your function is doing side effects and producing a sequence to be concatenated that your code probably needs refactoring.
Here is similar from algo.monads
(defn- flatten*
"Like #(apply concat %), but fully lazy: it evaluates each sublist
only when it is needed."
[ss]
(lazy-seq
(when-let [s (seq ss)]
(concat (first s) (flatten* (rest s))))))
Another way to write my-mapcat:
(defn my-mapcat [f coll] (for [x coll, fx (f x)] fx))
Applying a function to a lazy sequence will force realization of a portion of that lazy sequence necessary to satisfy the arguments of the function. If that function itself produces lazy sequences as a result, those are not realized as a matter of course.
Consider this function to count the realized portion of a sequence
(defn count-realized [s]
(loop [s s, n 0]
(if (instance? clojure.lang.IPending s)
(if (and (realized? s) (seq s))
(recur (rest s) (inc n))
n)
(if (seq s)
(recur (rest s) (inc n))
n))))
Now let's see what's being realized
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
concat-seq (apply concat seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "concat-seq: " (count-realized concat-seq))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 4
; concat-seq: 0
; seqs-in-seq: [0 0 0 0 0 0]
So, 4 elements of the seq-of-seqs got realized, but none of its component sequences were realized nor was there any realization in the concatenated sequence.
Why 4? Because the applicable arity overloaded version of concat takes 4 arguments [x y & xs] (count the &).
Compare to
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
foo-seq (apply (fn foo [& more] more) seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 2
; seqs-in-seq: [0 0 0 0 0 0]
(let [seq-of-seqs (map range (list 1 2 3 4 5 6))
foo-seq (apply (fn foo [a b c & more] more) seq-of-seqs)]
(println "seq-of-seqs: " (count-realized seq-of-seqs))
(println "seqs-in-seq: " (mapv count-realized seq-of-seqs)))
;=> seq-of-seqs: 5
; seqs-in-seq: [0 0 0 0 0 0]
Clojure has two solutions to making the evaluation of arguments lazy.
One is macros. Unlike functions, macros do not evaluate their arguments.
Here's a function with a side effect
(defn f [n] (println "foo!") (repeat n n))
Side effects are produced even though the sequence is not realized
user=> (def x (concat (f 1) (f 2)))
foo!
foo!
#'user/x
user=> (count-realized x)
0
Clojure has a lazy-cat macro to prevent this
user=> (def y (lazy-cat (f 1) (f 2)))
#'user/y
user=> (count-realized y)
0
user=> (dorun y)
foo!
foo!
nil
user=> (count-realized y)
3
user=> y
(1 2 2)
Unfortunately, you cannot apply a macro.
The other solution to delay evaluation is wrap in thunks, which is exactly what you've done.
Your premise is wrong. Concat is lazy, apply is lazy if its first argument is, and mapcat is lazy.
user> (class (mapcat (fn [x y] (println x y) (list x y)) (range) (range)))
0 0
1 1
2 2
3 3
clojure.lang.LazySeq
note that some of the initial values are evaluated (more on this below), but clearly the whole thing is still lazy (or the call would never have returned, (range) returns an endless sequence, and will not return when used eagerly).
The blog you link to is about the danger of recursively using mapcat on a lazy tree, because it is eager on the first few elements (which can add up in a recursive application).
I have a function that I basically yanked from a discussion in the Clojure google group, that takes a collection and a list of functions of arbitrary length, and filters it to return a new collection containing all elements of the original list for which at least one of the functions evaluates to true:
(defn multi-any-filter [coll & funcs]
(filter #(some true? ((apply juxt funcs) %)) coll))
I'm playing around with making a generalizable solution to Project Euler Problem 1, so I'm using it like this:
(def f3 (fn [x] (= 0 (mod x 3))))
(def f5 (fn [x] (= 0 (mod x 5))))
(reduce + (multi-any-filter (range 1 1000) f3 f5))
Which gives the correct answer.
However, I want to modify it so I can pass ints to it instead of functions, like
(reduce + (multi-any-filter (range 1 1000) 3 5))
where I can replace 3 and 5 with an arbitrary number of ints and do the function wrapping of (=0 (mod x y)) as an anonymous function inside the multi-any-filter function.
Unfortunately this is past the limit of my Clojure ability. I'm thinking that I would need to do something with map to the list of args, but I'm not sure how to get map to return a list of functions, each of which is waiting for another argument. Clojure doesn't seem to support currying the way I learned how to do it in other functional languages. Perhaps I need to use partial in the right spot, but I'm not quite sure how.
In other words, I want to be able to pass an arbitrary number of arguments (that are not functions) and then have each of those arguments get wrapped in the same function, and then that list of functions gets passed to juxt in place of funcs in my multi-any-filter function above.
Thanks for any tips!
(defn evenly-divisible? [x y]
(zero? (mod x y)))
(defn multi-any-filter [col & nums]
(let [partials (map #(fn [x] (evenly-divisible? x %)) nums)
f (apply juxt partials)]
(filter #(some true? (f %)) col)))
I coudn't use partial because it applies the arg in the first position of the fn. We want it in the second position of evenly-divisible? We could re-arrange in evenly-divisible? but then it would not really look correct when using it standalone.
user=> (reduce + (multi-any-filter (range 1 1000) 3 5))
233168
Coming from imperative programming languages, I am trying to wrap my head around Clojure in hopes of using it for its multi-threading capability.
One of the problems from 4Clojure is to write a function that generates a list of Fibonacci numbers of length N, for N > 1. I wrote a function, but given my limited background, I would like some input on whether or not this is the best Clojure way of doing things. The code is as follows:
(fn fib [x] (cond
(= x 2) '(1 1)
:else (reverse (conj (reverse (fib (dec x))) (+ (last (fib (dec x))) (-> (fib (dec x)) reverse rest first))))
))
The most idiomatic "functional" way would probably be to create an infinite lazy sequence of fibonacci numbers and then extract the first n values, i.e.:
(take n some-infinite-fibonacci-sequence)
The following link has some very interesting ways of generating fibonnaci sequences along those lines:
http://en.wikibooks.org/wiki/Clojure_Programming/Examples/Lazy_Fibonacci
Finally here is another fun implementation to consider:
(defn fib [n]
(let [next-fib-pair (fn [[a b]] [b (+ a b)])
fib-pairs (iterate next-fib-pair [1 1])
all-fibs (map first fib-pairs)]
(take n all-fibs)))
(fib 6)
=> (1 1 2 3 5 8)
It's not as concise as it could be, but demonstrates quite nicely the use of Clojure's destructuring, lazy sequences and higher order functions to solve the problem.
Here is a version of Fibonacci that I like very much (I took the implementation from the clojure wikibook: http://en.wikibooks.org/wiki/Clojure_Programming)
(def fib-seq (lazy-cat [0 1] (map + (rest fib-seq) fib-seq)))
It works like this: Imagine you already have the infinite sequence of Fibonacci numbers. If you take the tail of the sequence and add it element-wise to the original sequence you get the (tail of the tail of the) Fibonacci sequence
0 1 1 2 3 5 8 ...
1 1 2 3 5 8 ...
-----------------
1 2 3 5 8 13 ...
thus you can use this to calculate the sequence. You need two initial elements [0 1] (or [1 1] depending on where you start the sequence) and then you just map over the two sequences adding the elements. Note that you need lazy sequences here.
I think this is the most elegant and (at least for me) mind stretching implementation.
Edit: The fib function is
(defn fib [n] (nth fib-seq n))
Here's one way of doing it that gives you a bit of exposure to lazy sequences, although it's certainly not really an optimal way of computing the Fibonacci sequence.
Given the definition of the Fibonacci sequence, we can see that it's built up by repeatedly applying the same rule to the base case of '(1 1). The Clojure function iterate sounds like it would be good for this:
user> (doc iterate)
-------------------------
clojure.core/iterate
([f x])
Returns a lazy sequence of x, (f x), (f (f x)) etc. f must be free of side-effects
So for our function we'd want something that takes the values we've computed so far, sums the two most recent, and returns a list of the new value and all the old values.
(fn [[x y & _ :as all]] (cons (+ x y) all))
The argument list here just means that x and y will be bound to the first two values from the list passed as the function's argument, a list containing all arguments after the first two will be bound to _, and the original list passed as an argument to the function can be referred to via all.
Now, iterate will return an infinite sequence of intermediate values, so for our case we'll want to wrap it in something that'll just return the value we're interested in; lazy evaluation will stop the entire infinite sequence being evaluated.
(defn fib [n]
(nth (iterate (fn [[x y & _ :as all]] (cons (+ x y) all)) '(1 1)) (- n 2)))
Note also that this returns the result in the opposite order to your implementation; it's a simple matter to fix this with reverse of course.
Edit: or indeed, as amalloy says, by using vectors:
(defn fib [n]
(nth (iterate (fn [all]
(conj all (->> all (take-last 2) (apply +)))) [1 1])
(- n 2)))
See Christophe Grand's Fibonacci solution in Programming Clojure by Stu Halloway. It is the most elegant solution I have seen.
(defn fibo [] (map first (iterate (fn [[a b]] [b (+ a b)]) [0 1])))
(take 10 (fibo))
Also see
How can I generate the Fibonacci sequence using Clojure?