Python 2.5’s first alpha has been released. From a quick look at the What’s New in Python 2.5 document, what I thought was interesting…

I’m using Computer Organization and Design, 3rd ed (David Patterson and John Hennessy) for an electrical engineering course (that I have to take to graduate) in computer systems architecture. Assigned as homework, from the For More Practice section, is problem 2.28:

I’ve a lot of interactive websites I’ve been wanting to do lately. After learning Python earlier this year, I’m wanting to write all these projects in Python versus something conventional like PHP. I’ve come to the conclusion that web development with Python sucks; it’s no wonder it’s not being used very much. There are too many frameworks, many doing similar things in different ways, or doing totally dissimilar things in totally different ways. I’ve been spending the last week reviewing several different frameworks and deciding which ones I thought were best.

I’m writing a load balancing algorithm for this cluster I’m working on, and have a bunch of things in a Python list sequence. So came the issue, how to divide the list into sublists for each node in the cluster to compute? I came up with the algorithm:

def split_seq_stride(l, n):
  """Splits a sequence l into a list of subsequences containing at least n
  elements each, not preserving order. The first few subsequences may contain
  n+1 elements, containing the last few elements. (Algorithm is good for load
  balancing)"""
  r=[]
  k=len(l)/n
  [r.append(l[i::k]) for i in range(k)]
  return r

which behaves like:

Assuming each task is an equal amount of work, this algorithm is a good load balancing algorithm. it prevents any node from having to do any less significant work than the others.

Probably useful for someone: what about splitting up a sequence in-order, with the last subsequence containing fewer elements? This code segment does just that:

def split_seq(l, n):
  """Splits a sequence l into a list of subsequences containing at most n
  elements each, preserving order. The last subsequence may contain less
  than n elements."""
  r=[]
  [r.append(l[s:s+n]) for s in range(0, len(l), n)]
  return r

which behaves like:

Thanks to yason and others on EFnet’s #python for pointers.