On Wed, Aug 19, 2009 at 2:33 AM, Sean Devlin wrote:
>
> The set is being passed in as a parameter, so that will be a problem.
> Still, I'll be able to re-write my routine with false. Thanks!
When the set is passed in as a parameter, one bulletproof[1] way to do that
is:
user=> (filter (partial
Hi Sean,
I think your solution with (comp not nil? ... ) is very clean already.
But in case you don't like that, here's another way of writing it:
(filter #(contains? #{false} %) [false true false true])
returns (false false)
It probably gets compiled to almost the same thing under the hood. But
Hi,
The problem here is that filter is expecting a predicate and you are
passing a set.
When used as a function like so
user=> (#{1 2 3} 2)
2
user=> (#{1 2 3} 5)
nil
the set returns the argument if it is in the set and nil otherwise.
The problem you are observing is because the i
The set is being passed in as a parameter, so that will be a problem.
Still, I'll be able to re-write my routine with false. Thanks!
On Aug 18, 8:18 pm, "J. McConnell" wrote:
> On Tue, Aug 18, 2009 at 7:37 PM, Sean Devlin wrote:
>
>
>
> > user=> (filter #{false} [true false false true])
> > ()
On Tue, Aug 18, 2009 at 7:37 PM, Sean Devlin wrote:
>
> user=> (filter #{false} [true false false true])
> ()
>
> Obviously, this is not what I intended. The best I could do was the
> following
>
> user=> (filter (comp not nil? #{false}) [true false false true])
> (false false)
>
> Does anyone el
Hello Clojurians,
I commonly use a set in my filtering operations, like the example
below.
user=> (filter #{\a \e \i \o \u} "The quick brown fox jumped over the
lazy dog")
(\e \u \i \o \o \u \e \o \e \e \a \o)
Today I found myself using the following code
user=> (filter #{false} [true false fal
