On Mon, 19 Sep 2005 22:58:40 GMT, [EMAIL PROTECTED] (Bengt Richter) wrote: >On 19 Sep 2005 00:02:34 -0700, "malv" <[EMAIL PROTECTED]> wrote: > >>Simple case: >>In this list, how to find all occurences of intervals of n adjacent >>indexes having at least one list-member with a value between given >>limits. >>Visualizing the list as a two-dimensional curve, this is like >>horizontally dragging a given rectangle over the curve and finding the >>x coordinates where the curve passes through the rectangle.(Define such >>a x-index coordinate as the left corner of the rectangle.) >> >>More complicated case: >>Given a pair of rectangles spaced relatively to each other in a fixed >>manner. Drag this rectangle pair horizontally over the above >>two-dimensional curve and list the indexes of the occurences where the >>curve passes simultaneously through both rectangles. >>(Define such a x-index coordinate as the leftmost corner of the >>rectangle pair). >> >>These problems can be solved by programming a naive search advancing >>index by index. It seems obvious that due to the localized properties >>searched for, much more efficient searches should be possible. >>After having found the occurence-indexes for one particular rectangle >>set, how to find the pattern occurences after changing one or more >>rectangle parameters? >> > >for the first problem, maybe (untested beyond the below): > > >>> from itertools import groupby > >>> data = [5+i%5 for i in xrange(40)] > >>> data > [5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, > 5, 6, 7, 8, 9, 5, 6, > 7, 8, 9, 5, 6, 7, 8, 9] > >>> [g.next()[0] for k,g in groupby(enumerate(data), lambda t: 6<t[1]<9) if k] > [2, 7, 12, 17, 22, 27, 32, 37] > >By the time you figure a more efficient way, this can have solved it for quite >a long sequence. >But since an efficient solution seems a lot like homework, I'll leave it to >you ;-) >Or is this a bioinfo problem in disguise? > Wrong problem, sorry. I misread too quickly. I'll try to misread more slowly next time ;-/
Regards, Bengt Richter -- http://mail.python.org/mailman/listinfo/python-list
