Re: partition-distinct
On 29 March 2012 21:41, Cedric Greevey cgree...@gmail.com wrote:
On Thu, Mar 29, 2012 at 4:18 PM, David Jagoe davidja...@gmail.com wrote:
Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2]
partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)]
!
(defn partition-distinct [s]
(lazy-seq
(loop [seen #{} s (seq s) part []]
(if s
(let [[f r] s]
(if (seen f)
(cons (seq part) (partition-distinct s))
(recur (conj seen f) r (conj part f
(if-let [part (seq part)]
(list part))
Thanks! I had a nasty feeling that it could be done in a one-liner
using partition-by or something similar. But of course you need to be
able to look at previous elements...
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Re: partition-distinct
Here's a version with reduce. It returns your elements as sets, but you could
easily (map seq ,,,) if you really need lists.
user= (reduce (fn [r e] (if ((last r) e) (conj r #{e}) (update-in r [(dec
(count r))] conj e))) [#{}] [1 2 2 3 4 4 1 6])
[#{1 2} #{2 3 4} #{1 4 6}]
Cheers, Jay
On Mar 30, 2012, at 4:29 AM, David Jagoe wrote:
On 29 March 2012 21:41, Cedric Greevey cgree...@gmail.com wrote:
On Thu, Mar 29, 2012 at 4:18 PM, David Jagoe davidja...@gmail.com wrote:
Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2]
partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)]
!
(defn partition-distinct [s]
(lazy-seq
(loop [seen #{} s (seq s) part []]
(if s
(let [[f r] s]
(if (seen f)
(cons (seq part) (partition-distinct s))
(recur (conj seen f) r (conj part f
(if-let [part (seq part)]
(list part))
Thanks! I had a nasty feeling that it could be done in a one-liner
using partition-by or something similar. But of course you need to be
able to look at previous elements...
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Re: partition-distinct
David Jagoe davidja...@gmail.com writes:
Hi David,
Thanks! I had a nasty feeling that it could be done in a one-liner
using partition-by or something similar. But of course you need to be
able to look at previous elements...
Here's a slightly shorter variant.
--8---cut here---start-8---
(defn partition-distinct [c]
(reduce #(if (.contains (last %1) %2)
(conj %1 [%2])
(conj (subvec %1 0 (dec (count %1)))
(conj (last %1) %2)))
[[]] (vec c)))
--8---cut here---end---8---
The difference is that it'll always return a vector of vectors, and of
course the .contains-check is linear for vectors, so it's quadratic to
the size of `c` in total.
If you don't care about the order of elements in each partition, then
you can just go with a vector of sets and have the same performance as
Cedric's recursive solution.
--8---cut here---start-8---
(defn partition-distinct [c]
(reduce #(if ((last %1) %2)
(conj %1 #{%2})
(conj (subvec %1 0 (dec (count %1)))
(conj (last %1) %2)))
[#{}] (vec c)))
--8---cut here---end---8---
Bye,
Tassilo
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partition-distinct
Hi all, I'm sure I'm missing a really simple way of doing this! Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2] partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)] I've been trying to write something generic like partition-by because the values are maps. Also I want to be able to deal with more than 2 distinct values. Thanks! -- David Jagoe davidja...@gmail.com +447535268218 -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
Re: partition-distinct
On Thu, Mar 29, 2012 at 4:18 PM, David Jagoe davidja...@gmail.com wrote:
Hi all,
I'm sure I'm missing a really simple way of doing this!
Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2]
partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)]
I've been trying to write something generic like partition-by because the
values are maps. Also I want to be able to deal with more than 2 distinct
values.
Thanks!
(defn partition-distinct [s]
(lazy-seq
(loop [seen #{} s (seq s) part []]
(if s
(let [[f r] s]
(if (seen f)
(cons (seq part) (partition-distinct s))
(recur (conj seen f) r (conj part f
(if-let [part (seq part)]
(list part))
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Re: partition-distinct
Yeah I don't like that either. Consider (comp vals (partial group-by identity)). On 2012-03-29 16:18 , David Jagoe davidja...@gmail.com wrote: Hi all, I'm sure I'm missing a really simple way of doing this! Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2] partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)] I've been trying to write something generic like partition-by because the values are maps. Also I want to be able to deal with more than 2 distinct values. Thanks! -- David Jagoe davidja...@gmail.com +447535268218 -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
Re: partition-distinct
Meh. Half-assed (mis)reading. Sorry. -Martin On 2012-03-29 16:23 , Weber, Martin S martin.we...@nist.gov wrote: Yeah I don't like that either. Consider (comp vals (partial group-by identity)). On 2012-03-29 16:18 , David Jagoe davidja...@gmail.com wrote: Hi all, I'm sure I'm missing a really simple way of doing this! Given a sequence like this: [1 2 1 2 1 1 2 1 2 2 2] partition it to get this: [(1 2) (1 2) (1) (1 2) (1 2) (2) (2)] I've been trying to write something generic like partition-by because the values are maps. Also I want to be able to deal with more than 2 distinct values. Thanks! -- David Jagoe davidja...@gmail.com +447535268218 -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en -- You received this message because you are subscribed to the Google Groups Clojure group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
