Is there a idiomatic / built-in way to flip the argument order that is passed to a function in Clojure?
As I do here with the definition of a helper function:
(defn flip [f & xs]
(apply f (reverse xs)))
(vector 1 2) ; [1 2]
(flip vector 1 2) ; [2 1]
I think it makes sense most times to create a flipped function that you use later, so the variation on your function would be this:
(defn flip [f]
(fn [& xs]
(apply f (reverse xs))))
A flip function does exist in other functional languages, but it's easy enough to do in Clojure as well, and you've already done it!
There are other ways to do it as well.
And here is a discussion about it from the Clojure mailing list.
You can also use anonymous inline function to re-define the order of arguments.
(vector "a" "b") ; ["a" "b"]
(#(vector %2 %1) "a" "b") ; ["b" "a"]
Related
So being new to Clojure and functional programming in general, I sometimes (to quote a book) "feel like your favourite tool has been taken from you". Trying to get a better grasp on this stuff I'm doing string manipulation problems.
So knowing the functional paradigm is all about recursion (and other things) I've been using tail recursive functions to do things I'd normally do with loops, then trying to implement using map or reduce. For those more experienced, does this sound like a sane thing to do?
I'm starting to get frustrated because I'm running into problems where I need to keep track of the index of each character when iterating over strings but that's proving difficult because reduce and map feel "isolated". I can't increment a value while a string is being reduced...
Is there something I'm missing; a function for exactly this.. Or can this specific case just not be implemented using these core functions? Or is the way I'm going about it just wrong and un-functional-like which is why I'm stuck?
Here's an example I'm having:
This function takes five separate strings then using reduce, builds a vector containing all the characters at position char-at in each string. How could you change this code so that char-at (in the anonymous function) gets incremented after each string gets passed? This is what I mean by it feels "isolated" and I don't know how to get around this.
(defn new-string-from-five
"This function takes a character at position char-at from each of the strings to make a new vector"
[five-strings char-at]
(reduce (fn [result string]
(conj result (get-char-at string char-at)))
[]
five-strings))
Old :
"abc" "def" "ghi" "jkl" "mno" -> [a d g j m] (always taken from index 0)
Modified :
"abc" "def" "ghi" "jkl" "mno" ->[a e i j n] (index gets incremented and loops back around)
I don't think there's anything insane about writing string manip functions to get your head around things, though it's certainly not the only way. I personally found clojure for the brave and true, 4clojure, and the clojurians slack channel most helpful when learning clojure.
On your question, probably the most common thing to do would be to add an index to your initial collection (in this case a string) using map-indexed
(user=> (map-indexed vector [9 9 9])
([0 9] [1 9] [2 9])
So for your example
(defn new-string-from-five
"This function takes a character at position char-at from each of the strings to make a new vector"
[five-strings char-at]
(reduce (fn [result [string-idx string]]
(conj result (get-char-at string (+ string-idx char-at))))
[]
(map-indexed vector five-strings)))
But how would I build map-indexed? Well
Non-lazily:
(defn map-indexed' [f coll]
(loop [idx 0
res []
rest-coll coll]
(if (empty? rest-coll)
res
(recur (inc idx) (conj res (f idx (first rest-coll))) (rest rest-coll)))))
Lazily (recommend not trying to understand this yet):
(defn map-indexed' [f coll]
(letfn [(map-indexed'' [idx f coll]
(if (empty? coll)
'()
(lazy-seq (conj (map-indexed'' (inc idx) f (rest coll)) (f idx (first coll))))))]
(map-indexed'' 0 f coll)))
You can use reductions:
(defn new-string-from-five
[five-strings]
(->> five-strings
(reductions
(fn [[res i] string]
[(get-char-at string i) (inc i)])
[nil 0])
rest
(mapv first)))
But in this case, I think map, mapv or map-indexed is cleaner. E.g.
(map-indexed
(fn [i s] (get-char-at s i))
["abc" "def" "ghi" "jkl" "mno"])
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.
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.
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
I am going through this article on Tree Visitors in Clojure and came across the below example:
(def data [[1 :foo] [2 [3 [4 "abc"]] 5]])
(walk/postwalk #(do (println "visiting:" %) %) data)
What is the outer form of postwalk doing? I can't understand its utility. How and why is postwalk used? Any explanations will be appreciated.
I'm not sure if you're asking what #() means or what the purpose of do(form1 form2) means, so I'll answer both.
#() is a shorthand for declaring an anonymous function. Anonymous functions are useful when you're passing some function into another function.
To illustrate, look at this in the repl
; define an anonymous function
user=> #(+ %1 %2)
#<user$eval68$fn__69 user$eval68$fn__69#9fe84e>
; is equivalent to
user => (fn [a b] (+ a b))
#<user$eval1951$fn__1952 user$eval1951$fn__1952#118bd3c>
; furthermore, you could then assign your anonymous function to a var
(def f #(+ %1 %2))
; is equivalent to
(defn f [a b] (+ a b))
user=> (#(+ %1 %2) 1 2)
3
user=> (f 1 2)
3
The %n refers to the arguments to positional arguments to a function where n means nth argument, starting at 1 As a further shorthand you can use % to refer to the first argument which works well for single arg anonymous functions. This is what you have in your example.
So you example is equivalent to
(def data [[1 :foo] [2 [3 [4 "abc"]] 5]])
(defn f [x] (do (println "visiting:" x) x))
(walk/postwalk f data)
The do here is a special form, which, from the docs:
(do exprs*)
Evaluates the expressions in order and returns the value of the last. If no expressions are supplied, returns nil.
In fact defn already has an implicit do, so my example above doesn't actually need the do...
; your example is equivalent to:
(def data [[1 :foo] [2 [3 [4 "abc"]] 5]])
(defn f [x] (println "visiting:" x) x)
(walk/postwalk f data)
I had the same question today and find this necrotopik. Maybe later is better the newer. Find this on clojure api reference:
postwalk
function
Usage: (postwalk f form)
Performs a depth-first, post-order traversal of form. Calls f on
each sub-form, uses f's return value in place of the original.
Recognizes all Clojure data structures except sorted-map-by.
Consumes seqs as with doall.
Also this example from clojuredocs makes things clearer:
(use 'clojure.walk)
(let [counter (atom -1)]
(postwalk (fn [x]
[(swap! counter inc) x])
{:a 1 :b 2}))
=> [6 {2 [[0 :a] [1 1]], 5 [[3 :b] [4 2]]}]
So, postwalk replaces every subform with result of the function. It is used for updating nested structures with new values. That is why resulting function contains x or % at the end. Adding input to result leads to even more nested stucture.
In the example above it is seen as it walks through the depths of nested map stucture. It is going first on deepest elements of the map, then go up on higher level, then lurk down to the next form, then again goes up and finishes its last move on the whole form.