You're right, thanks !
2009/2/17 Chouser
>
> On Mon, Feb 16, 2009 at 7:54 PM, samppi wrote:
> >
> > Thanks a lot, everyone. Isn't
> > (reduce conj [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
> > equivalent to
> > (into [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
> > though?
>
> Yes.
>
>
On Mon, Feb 16, 2009 at 7:54 PM, samppi wrote:
>
> Thanks a lot, everyone. Isn't
> (reduce conj [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
> equivalent to
> (into [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
> though?
Yes.
--Chouser
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Thanks a lot, everyone. Isn't
(reduce conj [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
equivalent to
(into [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
though?
On Feb 16, 3:55 pm, Laurent PETIT wrote:
> And to get the enclosing vector:
>
> (reduce conj [] (apply map vector [[:a :b] [1 2
And to get the enclosing vector:
(reduce conj [] (apply map vector [[:a :b] [1 2] ['a 'b]]))
2009/2/16 David Nolen
>
> I'm sure it can be done, but it's not clear to me if you have a vector of
>> vectors
>> how Stuart's solution would work:
>>
>> 1:15 user=> (map vector vecs)
>
> ([[:a0 :a1 :a
> I'm sure it can be done, but it's not clear to me if you have a vector of
> vectors
> how Stuart's solution would work:
>
> 1:15 user=> (map vector vecs)
([[:a0 :a1 :a2]] [[:b0 :b1 :b2]])
(apply (partial map vector) [[1 2 3] ['a 'b 'c] ["cat" "dog" "bird"]])
works on a vector of vectors. The
>From the original question it looked like there was a vector of unknown
length of vectors
[[a0 a1 a2] [b0 b1 b2] ...]
So my solution is something like:
1:12 user=> (def vecs [[:a0 :a1 :a2] [:b0 :b1 :b2]])
#'user/vecs
1:13 user=> (partition (count vecs) (interleave (flatten vecs)))
((:a0 :a1) (:
Of course ;) Keep forgetting the obvious things.
On Sun, Feb 15, 2009 at 9:36 PM, Stuart Halloway
wrote:
>
> (map vector [1 2 3] ['a 'b 'c] ["cat" "dog" "bird"])
> -> ([1 a "cat"] [2 b "dog"] [3 c "bird"])
>
> > Actually something closer to your exact expression is this:
> >
> > (apply (partial m
(map vector [1 2 3] ['a 'b 'c] ["cat" "dog" "bird"])
-> ([1 a "cat"] [2 b "dog"] [3 c "bird"])
> Actually something closer to your exact expression is this:
>
> (apply (partial map (fn [& rest] (apply vector rest))) [[1 2 3] ['a
> 'b 'c] ["cat" "dog" "bird"]])
>
> On Sun, Feb 15, 2009 at 7:42 P
Thanks a lot. I didn't know that map could take multiple collections.
On Feb 15, 5:47 pm, David Nolen wrote:
> Actually something closer to your exact expression is this:
>
> (apply (partial map (fn [& rest] (apply vector rest))) [[1 2 3] ['a 'b 'c]
> ["cat" "dog" "bird"]])
>
> On Sun, Feb 15, 2
Actually something closer to your exact expression is this:
(apply (partial map (fn [& rest] (apply vector rest))) [[1 2 3] ['a 'b 'c]
["cat" "dog" "bird"]])
On Sun, Feb 15, 2009 at 7:42 PM, David Nolen wrote:
> (map (fn [& rest] (apply vector rest)) [1 2 3] ['a 'b 'c] ["cat" "dog"
> "bird"])
>
(map (fn [& rest] (apply vector rest)) [1 2 3] ['a 'b 'c] ["cat" "dog"
"bird"])
On Sun, Feb 15, 2009 at 7:16 PM, samppi wrote:
>
> What would I do if I wanted this:
>
> [[a0 a1 a2] [b0 b1 b2] ...] -> [[a0 b0 ...] [a1 b1 ...] [a2 b2 ...]]
>
> I could write a loop, I guess, but is there a nice, id
What would I do if I wanted this:
[[a0 a1 a2] [b0 b1 b2] ...] -> [[a0 b0 ...] [a1 b1 ...] [a2 b2 ...]]
I could write a loop, I guess, but is there a nice, idiomatic,
functional way of doing this? I didn't spot a way in
clojure.contrib.seq-utils either.
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