1 1 0 0 1#i.5
0 1 4
   1 1 0 0 1 #>:i.5
1 2 5

Linda   

-----Original Message-----
From: programming-boun...@forums.jsoftware.com 
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Don Kelly
Sent: Saturday, August 30, 2014 7:31 PM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] locating indices of string in list of strings

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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>>>>>>
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