| Re: How to extract sub-string from array [message #61805 is a reply to message #61759] |
Thu, 31 July 2008 00:36  |
Ian Dean
Messages: 26
Registered: January 2000
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Junior Member |
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Hi again,
Although I had some good help on this problem, it has now got a bit more
complicated with more than one semicolon in a string:
[ 'Test;1;02;1234', 'Another_test;;2'...]
Has anyone come across an array version of STRSPLIT?
It would be great to be able to produce n string arrays of each part of
the original:
names = ['Test','Another_test'...]
sub_1 = ['1',''...]
sub_2 = ['02','2'...]
sub_3 = ['1234',''...]
Regards,
Ian
"Ian Dean" <ian.d.dean@baesystems.com> wrote in message
news:488ed1bf$1_1@glkas0286.greenlnk.net...
>
> "F�LDY Lajos" <foldy@rmki.kfki.hu> wrote in message
> news:Pine.LNX.4.64.0807281111470.17704@bifur.rmki.kfki.hu...
>>
>> On Mon, 28 Jul 2008, Ian Dean wrote:
>>
>>> Hi,
>>> I've got an array such as ['Test_data;01', 'Test_data;005',
>>> 'New_dat;100','Old_dat'....]
>>>
>>> What I'd like to do but haven;t achieved is create an array of the
>>> extension
>>> past the ';'
>>> i.e. ['01','005','100',''....]
>>> I've tried this:
>>>
>>> semi_colon = STREGEX(Data_array, ';')
>>> ; This gives an array of positions [9, 9, 7,-1...]
>>>
>>> I can replace the -1s with string lengths easily enough giving [9, 9, 7,
>>> 7...]
>>> However, I can't seem to extract the sub-strings
>>> I tried
>>> Extension = STRMID(Data_array, semi_colon + 1)
>>> but this gives me a 2-d array and no clue which elements I require.
>>> I suspect that there is some arcane use of histogram, but I am not brave
>>> enough to try.
>>>
>>> I know why the STRMID is failing, because the help says so, because both
>>> arguments are arrays.
>>> Looping though the list is not an option as in reality I have an array
>>> length in the order of 500,000
>>>
>>> Any help would be appreciated,
>>> Regards,
>>> Ian
>>
>>
>> Try this: STRMID(STREGEX(Data_array, ';.*', /extract), 1)
>>
>> regards,
>> lajos
>>
> Thanks to all. It's so simple when you know how!!
> Regards,
> Ian
>
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