My intent was to strip the non essential zeros . the original boolean
vector based on the inside expression
+./"1 'cd' E."1 arr and multiplied by i.#arr gave 0 1 0 3 0 5 0 7 0 9
for which the indices of interest are 0(first one)1 3 5 7
To be able to account for the first index being 0 , I added 1 and
decremented the result.
KEY (/. ) does strip unwanted zeros but gives 2 rows- but the desired
row depends on the first item of the boolean list.
What I would like to see is the equivalent of the APL boolean slash.
e.g.
1 1 0 0 1/⍳5 ( ⍳ being iota)
1 2 5
Don Kelly
On 28/08/2014 4:23 PM, Henry Rich wrote:
Remember that x +./@:E. y, x i.&1@:E. y, etc use Boyer-Moore, so
potentially much faster.
Henry Rich
On 8/28/2014 5:52 PM, Don Kelly wrote:
to complete it
<: ((+./"1 'cd' E."1 arr)*>:i.# arr)-.0
1 3 5 7 9
<: ((+./"1 'ab' E."1 arr)*>:i.# arr)-.0
0 4 8
1 5 9 0 4 8{arr
cd
cd
cd
ab
ab
ab
now make it tacit.
Don Kelly
On 28/08/2014 5:44 AM, Rob Hodgkinson wrote:
Joe, hope this helps to clarify… Bill was suggesting this…/Rob
arr=:10 $ > ;: 'ab cd yyy jcd'
arr
ab
cd
yyy
jcd
ab
cd
yyy
jcd
ab
cd
'cd' E."1 arr
0 0 0
1 0 0
0 0 0
0 1 0
0 0 0
1 0 0
0 0 0
0 1 0
0 0 0
1 0 0
+./"1 'cd' E."1 arr
0 1 0 1 0 1 0 1 0 1
On 28 Aug 2014, at 10:12 pm, Joe Bogner <joebog...@gmail.com> wrote:
Tracy - symbols are a good way to go. Thanks for reminding me of them.
Raul - that's what I was missing... I needed to use e. with the same
shape: eg. arr e. (1 3 $ 'ab ')... Thank you. Your examples are nice
and generic and faster than -:"1
...
bill - Not sure what to do with E.
Bjorn - 'ab' I. arr doesn't seem to help either... I was hoping for a
mask of 0 1 0 1 for whether it was found or the indices
On Thu, Aug 28, 2014 at 7:56 AM, Björn Helgason <gos...@gmail.com>
wrote:
'ab' I. arr
__-------------------_
https://groups.google.com/forum/m/#!forum/havaogskulamal
On 28 Aug 2014 10:58, "Joe Bogner" <joebog...@gmail.com> wrote:
Apologies for the extremely basic question, but I am struggling with
this after searching NuVoc and the dictionary.
How do I locate all the indices of 'ab' in arr?
arr=:10 $ > ;: 'ab cd yyy'
surely this isn't the best way:
] (3 = +/"1 'ab' i. arr) # arr
ab
ab
ab
ab
I have been primarily dealing with boxed strings up to this point,
which seemed easier since it was locating an atom in a list, not a
list in a table (practicing my vocabulary here... I might be wrong
though)
Thanks,
Joe
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