On 04/04/17 00:37, D.V.N.Sarma డి.వి.ఎన్.శర్మ wrote: > I will go for this modification of the original code.
> count = 0 > b= '3'+b[2:] > n = len(b) > for i in range(n-4): > if b[i:i+4] == get_year: > count += 1 While I think this works OK, I would probably suggest that this is one of the rare valid cases for using a regex(*) with a simple string. The findall() method with a suitable pattern and flags should catch overlaps. And is probably faster being written in C. (*)Assuming you want to include overlapping cases, otherwise just use b.count()... -- 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 _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
