(set abc) will do it...
On Tue, Jul 19, 2011 at 9:20 AM, Tuba Lambanog tuba.lamba...@gmail.comwrote:
Hello,
My apologies for this newbie question. I couldn't find a way to
convert a string to a set, thus:
abc = #{a b c}
Thanks.
tuba
--
You received this message because you are
Hi,
(set abc)
gives me #{\a \b \c}.
I'm expecting instead: #{a b c}
But thanks,
Tuba
On Mon, Jul 18, 2011 at 9:50 PM, Tuba Lambanog tuba.lamba...@gmail.comwrote:
Hello,
My apologies for this newbie question. I couldn't find a way to
convert a string to a set, thus:
abc = #{a b c}
Thanks.
On Mon, Jul 18, 2011 at 9:17 PM, Tuba Lambanog tuba.lamba...@gmail.com wrote:
(set abc)
gives me #{\a \b \c}.
I'm expecting instead: #{a b c}
(set (map abc))
(set (map str Tuba Lambanog))
--
Sean A Corfield -- (904) 302-SEAN
An Architect's View -- http://corfield.org/
World Singles, LLC. --
Tuba Lambanog wrote:
Tuba Lambanog wrote:
Hello, My apologies for this newbie question. I couldn't find a way to
convert a string to a set, thus:
abc = #{a b c}
(set abc) gives me #{\a \b \c}. I'm expecting instead: #{a b c}
Hi Tuba,
Are you quite sure that #{\a \b \c} is not
On 19/07/2011, at 2:29 PM, Sean Corfield wrote:
On Mon, Jul 18, 2011 at 9:17 PM, Tuba Lambanog tuba.lamba...@gmail.com
wrote:
(set abc)
gives me #{\a \b \c}.
I'm expecting instead: #{a b c}
(set (map abc))
(set (map str Tuba Lambanog))
This will produce #{a b c}
I think
(set (map
I'm with Benjamin despite my last post...
On 19/07/2011, at 2:31 PM, Benjamin Esham wrote:
Tuba Lambanog wrote:
Tuba Lambanog wrote:
Hello, My apologies for this newbie question. I couldn't find a way to
convert a string to a set, thus:
abc = #{a b c}
(set abc) gives me #{\a \b \c}.
(thank-you Sean A Corfield)
On Mon, Jul 18, 2011 at 10:29 PM, Sean Corfield seancorfi...@gmail.comwrote:
On Mon, Jul 18, 2011 at 9:17 PM, Tuba Lambanog tuba.lamba...@gmail.com
wrote:
(set abc)
gives me #{\a \b \c}.
I'm expecting instead: #{a b c}
(set (map abc))
(set (map str Tuba
Hi,
I'm clear on what I want ;) (something new to me), but I'm not clear on how
to get there. I'd like to compare str1 and str2, if at least one of the
letters in str1 is in str2. I'm thinking that if I can convert str1 and str2
to sets, then I can use the set intersection operation. It probably
In that case you don't need to convert to a symbol...
(set abc) should be fine...
Using set intersection, something like this is probably what you're looking
for...
(use 'clojure.set)
(if (empty? (intersection (set abc) (set cde))) false true))
I'm sure there's other (better) ways though
On
(some (set str1) str2)
will give you what you want..
Sunil
On Tue, Jul 19, 2011 at 10:06 AM, Tuba Lambanog tuba.lamba...@gmail.comwrote:
(thank-you Sean A Corfield)
On Mon, Jul 18, 2011 at 10:29 PM, Sean Corfield seancorfi...@gmail.comwrote:
On Mon, Jul 18, 2011 at 9:17 PM, Tuba Lambanog
On Tue, Jul 19, 2011 at 12:48 AM, Tuba Lambanog tuba.lamba...@gmail.comwrote:
Hi,
I'm clear on what I want ;) (something new to me), but I'm not clear on how
to get there. I'd like to compare str1 and str2, if at least one of the
letters in str1 is in str2. I'm thinking that if I can convert
Wow, that some function is just what I'd expect from Clojure, simple,
straightforward, elegant. How did I miss it?
Thanks all.
Tuba
On Jul 18, 11:00 pm, David Nolen dnolen.li...@gmail.com wrote:
On Tue, Jul 19, 2011 at 12:48 AM, Tuba Lambanog
tuba.lamba...@gmail.comwrote:
Hi,
I'm clear on
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