Thanks both for your answers.
@Ben, that seems to do it. I was trying to make it a bit "nicer" but failed
@john, that was my original approach but it does not produce what I want.
The result of that is
(("foo" 1) ("bar" 10 20) ("clown" 5))
but I need
(("foo" 1) ("bar" 10) ("bar" 20) ("clown" 5))
Ryan.
On Sunday, December 1, 2013 9:21:14 PM UTC+2, john walker wrote:
>
> Sorry, I spoke without seeing that you were aware of partition-by. Here's
> one that isn't vectorized.
>
> (def v ["foo" 1 "bar" 10 20 "clown" 5])
> (->> v
> (partition-by string?)
> (partition 2)
> (map #(apply concat %)))
>
> On Sunday, December 1, 2013 2:10:04 PM UTC-5, Ben wrote:
>>
>> user=> (def v ["foo" 1 "bar" 2 3 "baz" 4])
>> #'user/v
>> user=> (first (reduce (fn [[res s] e] (if (string? e) [res e] [(conj res
>> [s e]) s])) [[] nil] v))
>> [["foo" 1] ["bar" 2] ["bar" 3] ["baz" 4]]
>>
>> On Sun, Dec 1, 2013 at 10:57 AM, Ryan <areka...@gmail.com> wrote:
>>
>>> Hi all,
>>>
>>> I have a vector which contains an unknown number of repetitions of the
>>> following pattern:
>>>
>>> String, followed by 1 or more integers
>>>
>>> For example:
>>>
>>> String
>>> Integer
>>> String
>>> Integer
>>> Integer
>>> String
>>> Integer
>>> Integer
>>> Integer
>>> String
>>> Integer
>>>
>>> What I am trying to do is to create a vector of pairs which each pair
>>> will be the string and the integers followed.
>>>
>>> Real example:
>>>
>>> foo
>>> 1
>>> bar
>>> 10
>>> 20
>>> clown
>>> 5
>>>
>>> should convert to:
>>>
>>> [[foo 1] [bar 10] [bar 20] [clown 5]]
>>>
>>> I did try to play around with partition-by, partition and instance? but
>>> I couldn't get it quite right.
>>>
>>> Thank you for your time
>>>
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>>
>>
>>
>> --
>> Ben Wolfson
>> "Human kind has used its intelligence to vary the flavour of drinks,
>> which may be sweet, aromatic, fermented or spirit-based. ... Family and
>> social life also offer numerous other occasions to consume drinks for
>> pleasure." [Larousse, "Drink" entry]
>>
>>
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