Counting a sequence in Clojure

Question

I'm solving Problem 22 from the 4Clojure site asking me to write a function that counts the elements in a sequence. Because I messed around with Haskell at one point, I know that using fold is probably the way to go about doing that. After reading about it, I've come to the understanding that I should be using reduce for the same purpose. Here's the answer I submitted:

#(reduce inc 0 %)

The reasoning behind this is to iterate the list, and call inc each time on the value which is initially 0. However, this does not work. The site is complaining that the "Wrong number of args (2) passed to: core$inc". So I tried adding parens around inc:

#(reduce (inc) 0 %)

Now it thinks zero arguments are being passed to inc. What am I doing wrong here?

score 10 · Accepted Answer · edited Oct 24 '16 at 22:47

10

So I tried adding parens around inc...

Please, don't use parens in clojure ever for the purpose of guiding the compiler/interpretator. In clojure every parenthesis counts and this way you're just calling a function with zero arguments.

Now, take a look at the clojure doc on reduce

If val is supplied, returns the result of applying f to val and the first item in coll, then applying f to that result and the 2nd item, etc.

So when you write

(reduce inc 0 [1 2 3])

what actually happens is

(inc 
   (inc 
      (inc 0 1) 2) 3)

Proper function might look like

#(reduce 
    (fn [c _] (inc c))
     0 %)

edited Oct 24 '16 at 22:47

Dennis Hackethal

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answered May 23 '13 at 21:30

om-nom-nom

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Works when the coll contains numbers, but not characters. – Ross Presser May 23 '13 at 21:46
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@RossPresser oh, I forgot about initial value, fixed now: `(reduce (fn [c _] (inc c)) 0 [\a \b \c])` will yield 3. Thanks – om-nom-nom May 23 '13 at 21:48

score 3 · Answer 2 · answered May 23 '13 at 21:28

Assume the list is

[1 2 3 4]

In your first example, you are telling reduce to execute these:

(inc 0 1)
(inc 0 2)
(inc 0 3)
(inc 0 4)

Clearly inc is being passed two arguments here, when it requires one.

In your second attempt, you are telling reduce that (inc) will return a function which is then to be used with the value and the list. (inc) is a call to the inc function with zero arguments.

What you need for the first argument to reduce is a function that takes two values and returns the first value incremented by one.

score 2 · Answer 3 · edited May 23 '17 at 12:01

Clojure's reduce can be thought as similar to haskell's left-fold function foldl. The haskell way of counting items in a list xs would be something like this

foldl (const . (+1)) 0 xs

which is better understood as foldl (\acc _ -> acc + 1) 0 xs.

The idea is to increment the first operand in the fold function and so i'd be tempted to write it as

(reduce #(inc (%1)) 0 xs)

except that is wrong as the arity of the anonymous function is determined by the highest argument referenced in the expression and our fold function has to have an arity of 2.

So, a clever workaround: (reduce #(inc (first %&)) 0 xs)

score 1 · Answer 4 · answered Jul 22 '17 at 08:48

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I have been going through the problem list learning clojure too. Came up with this answer to that problem:

reduce + (#(for [o %] 1))

answered Jul 22 '17 at 08:48

Travis S

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score 0 · Answer 5 · answered Mar 29 '14 at 14:24

0

There is another approach by converting sequence to Java array and calling alength function:

#(alength (to-array %))

answered Mar 29 '14 at 14:24

Artem

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Counting a sequence in Clojure

5 Answers5