David Hirschfield wrote: > I have this function: > > def sequentialChunks(l, stride=1): > chunks = [] > chunk = [] > for i,v in enumerate(l[:-1]): > v2 = l[i+1] > if v2-v == stride: > if not chunk: > chunk.append(v) > chunk.append(v2) > else: > if not chunk: > chunk.append(v) > chunks.append(chunk) > chunk = [] > if chunk: > chunks.append(chunk) > return chunks > > Which takes a list of numerical values "l" and splits it into chunks > where each chunk is sequential, where sequential means each value in a > chunk is > separated from the next by "stride". > > So sequentialChunks([1,2,3,5,6,8,12]) returns: > > [[1,2,3],[5,6],[8],[12]] > > I don't think the code above is the most efficient way to do this, but > it is relatively clear. I tried fiddling with list-comprehension ways of > accomplishing it, but kept losing track of things...so if anyone has a > suggestion, I'd appreciate it. >
see the groupby example here: http://docs.python.org/lib/itertools-example.html Gerard -- http://mail.python.org/mailman/listinfo/python-list
