set-difference works as a filter function, but only for lists. What's about arrays and strings? Are there analogous functions for these types of data? If there are no such functions, what is the proper way to implement them?
For now I use this macro to process any sequence as a list (sometimes it's useful):
(defmacro treat-as-lists (vars &body body)
(let ((type (gensym)))
`(let ((,type (etypecase ,(car vars)
(string 'string)
(vector 'vector)
(list 'list)))
,#(mapcar (lambda (x) `(,x (coerce ,x 'list)))
vars))
(coerce (progn ,#body) ,type))))
My filter:
(defun filter (what where &key key (test #'eql))
(treat-as-lists (what where)
(set-difference where what :key key :test test)))
Examples:
CL-USER> (filter "cat" "can you take this cat away?")
"n you ke his wy?"
CL-USER> (filter #(0 1) #(1 5 0 1 9 8 3 0))
#(5 9 8 3)
Since writing functions that works on all sequences types often means writing separate versions for lists and vectors, it's worthwhile to use standard functions that operate on sequences where you can. In this case, we can use position and remove-if. I've reversed the order of your arguments, in order to make this sequence-difference more like set-difference where the second argument is subtracted from the first.
(defun sequence-difference (seq1 seq2 &key (start1 0) end1 (start2 0) end2
key (key1 key) (key2 key)
test test-not)
"Returns a new sequence of the same type of seq1 that contains the
elements of the subsequence of seq1 designated by start1 and end1, and
in the same order, except for those that appear in the subsequence of
seq2 designated by start2 and end2. Test and test-not are used in the
usual way to elements produced by applying key1 (which defaults to
key) to elements from seq1 and by applying key2 (which defaults to
key) to elements from seq2."
(flet ((in-seq2 (x)
(not (null (position x seq2
:start start2 :end end2
:key key2
:test test :test-not test-not)))))
(remove-if #'in-seq2
(subseq seq1 start1 end1)
:key key1)))
(sequence-difference "can you take this cat away?" #(#\c #\a #\t))
;=> "n you ke his wy?"
(sequence-difference "can you take this cat away?" #(#\c #\a #\t) :start1 3 :start2 1)
" you ke his c wy?"
Note that the standard also includes find, which works on arbitrary sequences, but find returns "an element of the sequence, or nil." This leads to ambiguity if nil is a member of the sequence. Position, on the other hand, returns either an index (which will be a number, and thus not nil) or null, so we can reliably determine whether an element is a in sequence.
There is one important difference here in that you're always getting a copy back here. The reason for that is subjective: Since sequence functions often take start and end index arguments, it's nice to include that functionality here. But, if we ask for (sequence-difference "foobar" "boa" :start1 2) then we want to remove the characters b, o, and a from the "foobar"'s subsequence "obar". What should we return though? "for" or "r"? That is, do we include the portion of seq1 that's outside the indices? In this solution, I've made the decision not to, and thus I'm doing (remove-if … (subseq seq1 …) …), and subseq always makes a copy. Set-difference, on the other hand, may return its list-1 or list-2 argument, if appropriate. This implementation generally won't return seq1 or seq2, except in some pathological cases (e.g., the empty list).
Related
I need a help to convert this Pascal code to Scheme code:
program reverseorder;
type
arraytype = array [1..5] of integer;
var
arr:arraytype;
i:integer;
begin
for i:=1 to 5 do
arr[i]:=i;
for i:=5 downto 1 do
writeln(arr[i]);
end.
I want to accesses to a specific atom and its seems there aren't iteration method in Scheme.
There are many ways to tackle this problem. With the online interpreter you're using you'll be limited to vanilla Scheme, and the solution will be more verbose than needed, using recursion:
(define lst '(1 2 3 4 5))
(let loop ((rev (reverse lst)))
(when (not (null? rev))
(display (car rev))
(newline)
(loop (cdr rev))))
With Racket, which is aimed at beginners, you can write a much simpler (albeit non-standard) solution:
(define lst (range 1 6))
(for ([e (reverse lst)])
(displayln e))
Either way, notice that the procedure for reversing a list is already built in the language, and you don't need to reimplement it - naturally, it's called reverse. And if it wasn't obvious already, in Scheme we prefer to use lists, not arrays to represent a sequence of elements - it's recommended to stop thinking in terms of indexes, array lengths, etc. because that's not how things are done in Scheme.
If you don't care about returned value (it's #<undef>) and just want to produce output, you can use for-each:
(for-each print (reverse (list 1 2 3 4 5)))
Output:
5
4
3
2
1
Not idiomatic Scheme, but a literal translation of the Pascal code would be:
(let ([arr (make-vector 5)])
(do ([i 0 (+ i 1)]) ((= i 5)) (vector-set! arr i i))
(do ([i 4 (- i 1)]) ((negative? i))
(display (vector-ref arr i))
(newline)))
I'm working on a Newton algorithm to count n-th root and I've encountered a problem with non-working loop. Here is the code:
(defn root [nth guess]
(if (<= guess 0) "Root doesn't exists" (count_root nth guess))
)
(defn count_root nth guess [nth guess]
(def result guess)
(def last_result result)
(def temp (power nth result))
(while (> (absolute (- result last_result)) 0.01)
(do
(def last_result result)
(def result ('fn [nth result guess temp] (* (/ 1.0 nth) (+ (* (- nth 1) result) (/ guess temp)))))
(def temp (power nth result))
)
)
(str "Result: " result)
)
(defn power [nth result]
(* result (- nth 1))
)
(defn absolute [x]
(if (>= x 0) x (- x))
)
When I'll comment out (while...) line, it counts one loop pass and the result is correct. But when there is (while...) line included, all code below is ignored.
I've rebuilt above code to something like this:
(defn power [nth result]
(* result (- nth 1))
)
(defn absolute [x]
(if (>= x 0) x (- x))
)
(defn is-good? [prev-result result]
(< (absolute (- prev-result result)) 0.01)
)
(defn improve [nth result temp]
(* (/ 1.0 nth) (+ (* (- nth 1) result) (/ result temp)))
)
(defn count-root [nth number]
(loop [result number
prev-result result
temp (power nth result)]
(let [next-result (* (/ 1.0 nth) (+ (* (- nth 1) result) (/ number temp)))])
(if (is-good? (result next-result)) result (recur next-result)))
)
(defn root [nth number]
(if (<= number 0) "Root doesn't exists" (count-root nth number))
)
But compiler throw error that next-result cannot be resolved...What's going on now?
There are a number of issues with this code. To provide you with a working example let's rewrite the root and count_root functions:
(defn power [nth result]
(* result (- nth 1)))
(defn count-root [nth guess]
(loop [result guess
last_result guess
temp (power nth result)]
(cond
(> (Math/abs (- result last_result)) 0.01)
(let [next-result (* (/ 1.0 nth) (+ (* (- nth 1) result) (/ guess temp)))]
(recur next-result
result
(power nth next-result)))
:else
(str "Result: " result))))
(defn root [nth guess]
(cond
(<= guess 0) "Root doesn't exists"
:else (count-root nth guess)))
Let's look at what was changed here, and try to explain it.
First, in the count-root function:
I suggest that you not use underscores in names. The most commonly followed Clojure convention is to use dashes to separate elements of function and definition names.
To perform a loop-with-definitions, use the loop construct to define where the loop begins, along with any needed definitions, and use recur to perform a recursive invocation of the loop with new values specified for the definitions. Note that if all the necessary definitions were given as parameters to the function count-root then no loop would be necessary and the recursion could go back to the start of the function.
I recommend that you get in the habit of using cond instead of if. The if function looks a whole lot like an if statement in other languages, and that gets us (well, OK, ME :-) into difficulties from time to time. The issue is that when you use if you have to remember to specify BOTH the "true" and "false" actions, because an if form can't be "short-cut". That is, if this was so in "other" languages you'd always have to write:
IF something
THEN true-stuff
ELSE false-stuff
ENDIF
or, in more Clojurist fashion
(if something
true-stuff
false-stuff)
In Clojure you shouldn't write
(if something
true-stuff)
because I've found that this doesn't work correctly. If you invoke the if function and specify only the true argument, as you would often do in "other" programming languages, it will compile and run, but you'll get odd results, things that can't happen will happen, and generally you'll get confused until you realize you've got an if without a "false" half or something like that. Save yourself some trouble - use cond instead of if.
4. I suggest using Math/abs instead of defining your own absolute function.
5. Use a let form to define temporary values where you won't need to loop inside the form. Here I've defined next-result as a temporary value which I'll use in the next form.
6. Use recur to tail-recursively "jump" to the next-earliest recursion point. Here the next-earliest recursion point is the loop form. If we didn't have this loop form, or some other recursion point, the function definition for count-root would serve as a recursion point. When
Next, in the root function:
the definition of the function was incorrect in the original version. I've cleaned it up here such that the function is defined with two arguments, nth and guess.
EDIT
In the original version of count_root you used the while function. Using while requires that its argument has to change, which means that it will have to deal in some manner or other with an atom; that is, it has to deal with a modifiable variable. Most of the time in Clojure, what are usually thought of as "variables" can't actually be changed. This is intentional. Clojure was intended from the outset to be used in writing multi-threaded applications, where coordinating changes to "variables" becomes a significant concern. If "variables" can't actually be changed, the need for that coordination (by means of such constructs as semaphores, critical sections, and etc) disappears. Thus, definitions of values (they're not really "variables" if they can't vary or change) are constant by default. Now, any "real" program has a need for values which change over time, and Clojure provides for such needs - but it's not the default. You must explicitly think about what needs to be modified, and you need to access those modifiable values differently.
Now, an observation for you. You may find that programming in Clojure, or in any language which makes you do things in a different manner than what you're accustomed to, will at first feel very strange, and perhaps even uncomfortably difficult. I know that's what I've gone through when learning new-to-me systems. Quite often, our thought processes get into "ruts", and anything that forces us out of those "ruts" makes us uncomfortable. But sometimes working your way through that "uncomfortable" period ends up with you having learned something valuable. If you stick with Clojure you'll learn a different way to think about software - and having multiple viewpoints from which to consider a problem is in itself a good thing.
When you're learning something new a good guide is helpful. I suggest Clojure For The Brave And True, which has the advantage that it's online edition is free-like-beer.
Best of luck.
How do I translate the loop part of this working Common Lisp (SBCL v.1.2.3) code into Clojure (v.1.6)? I am a bit frustrated after working on it for some hours/days without results. Somewhere I don't get this functional orientation I suppose ...
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Unconditional Entropy
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Probabilities
(setq list_occur_prob '()) ;; init
;; set probabilities for which we want to calculate the entropy
(setq list_occur_prob '(1/2 1/3 1/6)) ;;
;; Function to calculate the unconditional
;; entropy H = -sigma i=0,n (pi*log2(pi)
;; bits persymbol.
(setq entropy 0) ;; init
(setq entropy (loop for i in list_occur_prob
for y = (* (log_base2 i) i)
collect y
))
(setq entropy (* -1 (apply '+ entropy))) ;; change the sign
;; Print the unconditional entropy in bits per symbol.
(print entropy) ;; BTW, here the entropy is 1.4591479 bits per symbol.
Before we dive into the Clojure equivalent of the code, you should take some time to clean up the Common Lisp code. Using setqthe way you're doing it is considered bad style at best and can lead to undefined consequences at worst: setq is intended to assign values to variables, but your variables list_occur_proband entropy aren't defined (via defvar). In addition, this piece of code looks like you're assigning global variables (cf. defvar again), which are dynamic variables, which by convention should be marked with earmuffs, e.g. *entropy*.
However, for this small piece of code, you could just as well use local, non-dynamic variables, introduced via let like this (warning, I don't have any CL or Clojure environment handy):
(let ((list_occur_prob '(1/2 1/3 1/6)))
(loop for i in list_occur_prob
for y = (* (log_base 2 i) i)
collect y into acc
finally (return (* -1 (apply '+ acc)))))
There are ways to optimize the apply clause away into the loop:
(let ((list-occur-prob '(1/2 1/3 1/6)))
(- (loop for i in list-occur-prob
sum (* (log i 2) i))))
Now, Daniel Neal has shown you already a map/reduce based solution, here is one which is more closer to the original looping construct, using a recursive approach:
(defn ent-helper [probs acc]
(if (seq probs)
(recur (rest probs)
(conj acc (* (log_base 2 (first probs)) (first probs))))
acc))
(let [probs 1/2 1/3 1/6
acc (ent-helper probs [])]
(* -1 (apply + acc))
We're using conj instead of collect to gather the results into the accumulator. The call to ent-helper, which is essentially triggered for all values of probs via the recur recursion call, takes an (initially empty) second parameter in which the values build up so far are collected. If we've exhausted all probabilities, we simply return the collected values.
Again, summing up the values so far could be optimized into the loop, instead of mapping over the values.
They key operation you need is map which transforms a sequence using a function.
In the entropy example you gave, the following should work:
(def probabilities [1/2 1/3 1/6])
(defn log [base x]
(/ (Math/log x) (Math/log base)))
(defn entropy [probabilities]
(->> probabilities
(map #(* (log 2 %) %)) ; note - #(f %) is shorthand for (fn [x] (f x))
(reduce +)
(-)))
(entropy probabilities) ; => 1.459
When working with collections, the pipeline operator (->>) is often used
to clearly show a sequence of operations. I personally find it much easier to read than the nested bracket syntax, especially if there are lots of operations.
Here, we're first mapping the pi * log2(pi) function over the sequence,
and then summing it using (reduce +)
I would start with more functional Common Lisp code:
(- (reduce #'+
'(1/2 1/3 1/6)
:key (lambda (i)
(* (log i 2) i))))
You can write imperative code in Lisp, with lots of operations setting variable values, but it is not the best style.
Even a tight LOOP can look okay:
(- (loop for i in '(1/2 1/3 1/6)
sum (* (log i 2) i)))
I endorse the general flavor of schaueho's answer, but if you prefer you can get something closer to the "feel" of the looping approach with Clojure's for macro:
(apply - 0
(for [prob [1/2 1/3 1/6]]
(* (log prob 2) prob)))
I find this much easier to read than schaueho's version with manual recursion, and it also performs much better, in that it doesn't traverse the list twice, doesn't accumulate results into a temporary vector, and so on.
Note that (- (apply + xs)) is the same as (apply - 0 xs), although which one you find clearer is probably a matter of taste. Also, I'm assuming you already have a suitable log function defined elsewhere.
Preface
Firstly, I'm new to Clojure and programming, so I thought I'd try to create a function that solves a non-trivial equation using my natural instincts. What resulted is the desire to find a square root.
The question
What's the most efficient way to stop my square-n-map-maker function from iterating past a certain point? I'd like to fix square-n-map-maker so that I can comment out the square-maker function which provides me with the results and format I currently want to see but not the ability to recall the square-root answer (insofar as I know).
I.e. I want it to stop when it is greater than or equal to my input value
My initial thought was that instead of a keyword list, I would want it to be a map. But I'm having a very difficult time getting my function to give me a map. The whole reason I wanted a map where one member of a pair is n and another is n^2 so that I could extract the actual square root from it and it give it back to the user as the answer.
Any ideas on the best way to accomplish this? (below is the function I want to fix)
;; attempting to make a map so that I can comb over the
;; map later and recall a value that meets
;; my criteria to terminate and return result if (<= temp-var input)
(defn square-n-map-maker [input] (for [temp-var {remainder-culler input}]
(map list(temp-var) (* temp-var temp-var))
)
)
(square-n-map-maker 100) => clojure.lang.ArityException: Wrong number of args (0) passed to: MapEntry
AFn.java:437 clojure.lang.AFn.throwArity
AFn.java:35 clojure.lang.AFn.invoke
/Users/dbennett/Dropbox/Clojure Files/SquareRoot.clj:40 sqrt-range-high-end/square-n-map-maker[fn]
The following is the rest of my code
;; My idea on the best way to find a square root is simple.
;; If I want to find the square root of n, divide n in half
;; Then find all numbers in 0...n that return only a remainder of 0.
;; Then find the number that can divide by itself with a result of 1.
;; First I'll develop a function that works with evens and then odds
(defn sqrt-range-high-end [input] (/ input 2))
(sqrt-range-high-end 100) => 50
(defn make-sqrt-range [input] (range (sqrt-range-high-end (+ 1 input))))
(make-sqrt-range 100) =>(0 1 2 3 4 5 6 ... 50)
(defn zero-culler [input] (remove zero? (make-sqrt-range input)))
(zero-culler 100) =>(1 2 3 4 5 6 ... 50)
(defn odd-culler [input] (remove odd? (zero-culler input)))
(odd-culler 100) => (2 4 6 8 10...50)
(defn even-culler [input] (remove even? (zero-culler input)))
(even-culler 100) => (1 3 5 7...49)
(defn remainder-culler [input] (filter #(zero? (rem input %)) (odd-culler input)))
(remainder-culler 100) => (2 4 6 12 18)
(defn square-maker [input] (for [temp-var (remainder-culler input)]
(list (keyword (str
temp-var" "
(* temp-var temp-var)
)
)
)
)
(square-maker 100) => ((:2 4) (:4 16) (:10 100) (:20 400) (:50 2500))
Read the Error Messages!
You're getting a little ahead of yourself! Your bug has nothing to do with getting for to stop "looping."
(defn square-n-map-maker [input] (for [temp-var {remainder-culler input}]
(map list(temp-var) (* temp-var temp-var))))
(square-n-map-maker 100) => clojure.lang.ArityException: Wrong number of args (0) passed to: MapEntry
AFn.java:437 clojure.lang.AFn.throwArity
AFn.java:35 clojure.lang.AFn.invoke
Pay attention to error messages. They are your friend. In this case, it's telling you that you are passing the wrong number of arguments to MapEntry (search for IPersistentMap). What is that?
{} creates a map literal. {:key :value :key2 :value2} is a map. Maps can be used as if they were functions:
> ({:key :value} :key)
:value
That accesses the entry in the map associated with key. Now, you created a map in your first line: {remainder-culler input}. You just mapped the function remainder-culler to the input. If you grab an item out of the map, it's a MapEntry. Every MapEntry can be used as a function, accepting an index as an argument, just like a Vector:
> ([:a :b :c :d] 2)
:c
Your for is iterating over all MapEntries in {remainder-culler input}, but there's only one: [remainder-culler input]. This MapEntry gets assigned to temp-var.
Then in the next line, you wrapped this map in parentheses: (temp-var). This forms an S-expression, and expressions are evaluated assuming that the first item in the expression is a function/procedure. So it expects an index (valid indices here would be 0 and 1). But you pass no arguments to temp-var. Therefore: clojure.lang.ArityException: Wrong number of args.
Also, note that map is not a constructor for a Map.
Constructing a map
Now, on to your problem. Your square-maker is returning a list nicely formatted for a map, but it's made up of nested lists.
Try this:
(apply hash-map (flatten (square-maker 100)))
Read this page and this page to see how it works.
If you don't mind switching the order of the keys and values, you can use the group-by that I mentioned before:
(defn square-maker [input]
(group-by #(* % %) (remainder-culler input)))
(square-maker 100) => {4 [2], 16 [4], 100 [10], 400 [20], 2500 [50]}
Then you can snag the value you need like so: (first ((square-maker 100) 100)). This uses the map-as-function feature I mentioned above.
Loops
If you really want to stick with the intuitive looping concept, I would use loop, not for. for is lazy, which means that there is neither means nor reason (if you use it correctly) to "stop" it -- it doesn't actually do any work unless you ask for a value from it, and it only does the work it must to give you the value you asked for.
(defn square-root [input]
(let [candidates (remainder-culler input)]
(loop [i 0]
(if (= input (#(* % %) (nth candidates i)))
(nth candidates i)
(recur (inc i))))))
The embedded if determines when the looping will cease.
But notice that loop only returns its final value (acquaint yourself with loop's documentation if that sentence doesn't make sense to you). If you want to build up a hash-map for later analysis, you'd have to do something like (loop [i 0, mymap {}] .... But why analyze later if it can be done right away? :-)
Now, that's a pretty fragile square-root function, and it wouldn't be too hard to get it caught in an infinite loop (feed it 101). I leave it as an exercise to you to fix it (this is all an academic exercise anyway, right?).
I hope that helps you along your way, once again. I think this is a great problem for learning a new language. I should say, for the record, though, that once you are feeling comfortable with your solution, you should search for other Clojure solutions to the problem and see if you can understand how they work -- this one may be "intuitive," but it is not well-suited to Clojure's tools and capabilities. Looking at other solutions will help you grasp Clojure's world a bit better.
For more reading:
Imperative looping with side-effects.
How to position recur with loop
The handy into
Finally, this "not constructive" list of common Clojure mistakes
for is not a loop, and it's not iterating. It lazily creates a list comprehension, and it only realizes values when required (in this case, when the repl tries to print the result of the evaluation). There are two usual ways to do what you want: one is to wrap square-maker in
(first (filter some-predicate (square-maker number))) to obtain the first element in the sequence that complies with some-predicate. E.g.
(first (filter #(and (odd? %) (< 50 %)) (range)))
=> 51
The above won't realize the infinite range, obviously.
The other one is not to use a list comprehension and do it in a more imperative way: run an actual loop with a termination condition (see loop and recur).
Example:
(loop [x 0]
(if (and (odd? x) (> x 50))
x
(recur (inc x))))
I am stuck in a Clojure loop and need help to get out.
I first want to define a vector
(def lawl [1 2 3 4 5])
I do
(get lawl 0)
And get "1" in return. Now, I want a loop that get each number in the vector, so I do:
(loop [i 0]
(if (< i (count lawl))
(get lawl i)
(recur (inc i))))
In my mind this is supposed to set the value of i to nil, then if i is lower then the count of the lawl vector, it should get each lawl value and then increase the i variable with 1 and try again, getting the next value in the vector.
However, this does not work and I have spent some time trying to get it working and are totally stuck, would appreciate some help. I have also tried changing "if" to "when" with the same result, it doesn't provide any data the REPL just enters a new line and blink.
EDIT: Fixed the recur.
You need to consider what is "to get each lawl value" supposed to mean. Your get call does indeed "get" the appropriate value, but since you never do anything with it, it is simply discarded; Bozhidar's suggestion to add a println is a good one and will allow you to see that the loop does indeed access all the elements of lawl (just replace (get ...) with (println (get ...)), after fixing the (inc) => (inc i) thing Bozhidar mentioned also).
That said, if you simply want to do something with each number in turn, loop / recur is not a good way to go about it at all. Here are some others:
;;; do some side-effecty thing to each number in turn:
(dotimes [i (count lawl)]
(println (str i ": " (lawl i)))) ; you don't really need the get either
;; doseq is more general than dotimes, but doesn't give you equally immediate
;; acess to the index
(doseq [n lawl]
(println n))
;;; transform the lawl vector somehow and return the result:
; produce a seq of the elements of lawl transformed by some function
(map inc lawl)
; or if you want the result to be a vector too...
(vec (map inc lawl))
; produce a seq of the even members of lawl multiplied by 3
(for [n lawl
:when (even? n)]
(* n 3))
This is just the beginning. For a good tour around Clojure's standard library, see the Clojure -- Functional Programming for the JVM article by Mark Volkmann.
(recur (inc)) should be (recur (inc i))
Even so this code will just return 1 in the end, if you want a listing of the number you might add a print expression :-) Btw index based loops are not needed at all in scenarios such as this.
(loop [list [1 2 3 4 5] ]
(if (empty? list)
(println "done")
(do
(println (first list))
(recur (rest list)))))
OK, I'm about 10-1/2 years too late on this, but here goes:
The problem here is a pretty common misunderstanding of how the arguments to the if function are used. if takes three arguments - the condition/predicate, the code to be executed if the predicate is true, and the code to be executed if the predicate is false. In this case both of the true and false cases are supplied. Perhaps if we fix the indentation and add some appropriate comments we'll be able to see what's happening more easily:
(loop [i 0]
(if (< i (count lawl))
(get lawl i) ; then
(recur (inc i)))) ; else
So the problem is not that the code gets "stuck" in the loop - the problem is that the recur form is never executed. Here's how the execution flows:
The loop form is entered; i is set to 0.
The if form is entered.
The predicate form is executed and found to be true.
The code for the then branch of the if is executed, returning 1.
Execution then falls out the bottom of the loop form.
Right now I hear people screaming "Wait! WHAT?!?". Yep - in an if form you can only have a single form in the "then" and "else" branches. "But...THAT'S STUPID!" I hear you say. Well...not really. You just need to know how to work with it. There's a way to group multiple forms together in Clojure into a single form, and that's done by using do. If we want to group (get lawl i) and (recur... together we could write it as
(loop [i 0]
(if (< i (count lawl))
(do
(get lawl i) ; then
(recur (inc i))
)
)
)
As you can see, we have no "else" branch on this if form - instead, the (get... and (recur... forms are grouped together by the (do, so they execute one after the other. So after recurring its way through the lawl vector the above snippet returns nil, which is kind of ugly. So let's have it return something more informative:
(loop [i 0]
(if (< i (count lawl))
(do
(get lawl i) ; then
(recur (inc i)))
(str "All done i=" i) ; else
)
)
Now our else branch returns "All done i=5".