Clojure - map function

Question

As we know, (map f [a b c]) is equivalent to [(f a) (f b) (f c)].

My question is: The evaluation result of (map #(- (int %) (int 0円)) "1234") is (1 2 3 4), why does it return the results of applying #(- (int %) (int 0円)) to every digits of "1234", rather than the string "1234" as a whole? How should I understand this code example?

• M Smith: This has bitten me quite often
• Mars: One qualification: The result of (map f [a b c]) isn't exactly the same as the result of [(f a) (f b) (f c)], since map returns a lazy sequence rather than a vector. I think it's correct to say that the result of (map f [a b c]) is the same as the result of (lazy-seq [(f a) (f b) (f c)]), and that result of (vec (map f [a b c])) is the same as the result of [(f a) (f b) (f c)]. (If I'm wrong about some details, more knowledgeable people will hopefully correct me.)
• DaoWen: You don't need map if you just want to call a function using a single argument. If that's what you want, then don't use map: (#(- (int %) (int 0円)) "1234")
• Thumbnail: @DaoWen Your code produces ClassCastException java.lang.String cannot be cast to java.lang.Character ... - surely not what's intended.
• DaoWen: @Thumbnail - Ha, you're right, thanks. You would need to use Integer/parseInt to convert the string to an int instead of the int "cast" that works for chars. The point is that I don't really understand the OP's question—*if you just want to apply the function to the string (not the individual characters), why don't you just call it directly rather than using map?* Apparently I'm missing something...

Answer 1

map calls seq on all arguments after the first. seq turns a string into a sequence of characters.

Answer 2

Clojure can treat a string as a sequence - of characters. This is useful because you can:

• map things over the string
• partition the string
• get locations by index
• do everything else sequences do.

It's perhaps a bit annoying having to remember to put the resulting sequence back into a string by wrapping the sequence manipulating expression in a call to str.

• noisesmith: it may make sense to merge our answers into yours, since yours is more complete anyway - perhaps making it clear that it is specifically map which calls seq on args, and showing (apply str (map f string)) as the canonical way to get a string back out the other side.
• Mars: I value the simplicity, elegance, and implicit generality of your answer, @noisesmith, as well as Arthur Ulfeldt's discussion of the ramifications and purpose of map's behavior.