Really depends on that you want. First thing coming to my mind is:
(for [l (:langs langs)
[k v] l]
(str k v))
This will give you a flat lazy list of strings.
On Mon, Apr 21, 2014 at 5:32 PM, Hussein B. hubaghd...@gmail.com wrote:
Hi,
For a data structure such as:
(def langs
You can simply call *seq* on your map, and get a sequence of all map
entries.
user (seq {:a 1 :b 2 :c 3})
([:a 1] [:c 3] [:b 2])
Hth
Tim Washington
Interruptsoftware.com http://interruptsoftware.com
On Mon, Apr 21, 2014 at 11:32 AM, Hussein B. hubaghd...@gmail.com wrote:
Hi,
For a data
That doseq is a no-op, because str has no side effects and doseq always
returns nil. Also there is no need to nest doseq calls. for is much like
doseq except it actually returns the result.
user (for [lang (:langs langs)
k (keys lang)]
(str k(k lang)))
(:lang
small correction: sometimes one does need to nest a for or doseq call, but
not for the usual nested iteration case
On Monday, April 21, 2014 8:57:49 AM UTC-7, Justin Smith wrote:
That doseq is a no-op, because str has no side effects and doseq always
returns nil. Also there is no need to