On 21/01/15 18:14, dw wrote:
- line_array[1] may contain "01/04/2013 10:43 AM 17,410,217
DEV-ALL-01-04-13.rlc\n"
- line_array[2] may contain "01/25/2013 03:21 PM 17,431,230
DEV-ALL-01-25-2013.rlc\n"
- line_array[3] may contain "\n"
I want to retain all elements which are valid (i.e. contains a date
value xx/xx/xxxx)
You need a bit more detail.
Can a row contain anything other than just \n if there is no date?
Is the date always at the start and always the same length?
So I'm using a regex search for the date value located at the start of
each element...this way
misc_int1 =
re.search(r'[0-9]{2}/[0-9]{2}/[0-9]{4}',line_array[x]).start()
You might be better with re.match() if its always at the start of the line.
This method works really well when the regex date characters are found.
It returns the value 0 to misc_int1 because the date regex starts at 0
position.
I was hoping the .start() method would return a -1 value for the
instance in which the regex search string is not found.
You can simply split the code over two lines and check the result:
result = re.search(r'[0-9]{2}/[0-9]{2}/[0-9]{4}',line_array[x])
if result is not None:
misc_intl = result.start()
There are no prizes for writing code on one line, so don't be afraid
to separate it out.
HTH
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
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