CODEWITHSUNDEEP
pythonmaya
Mathieson
Mathieson

Reputation: 1282

Way to iterate two items at a time in a list?

I am wondering if there is a better way to iterate two items at a time in a list. I work with Maya a lot, and one of its commands (listConnections) returns a list of alternating values. The list will look like [connectionDestination, connectionSource, connectionDestination, connectionSource]. To do anything with this list, I would ideally like to do something similar to:

for destination, source in cmds.listConnections():
    print source, destination

You could, of course just iterate every other item in the list using [::2] and enumerate and source would be the index+1, but then you have to add in extra checks for odd numbered lists and stuff.

The closest thing I have come up with so far is:

from itertools import izip
connections = cmds.listConnections()
for destination, source in izip(connections[::2], connections[1::2]):
    print source, destination

This isn't super important, as I already have ways of doing what I want. This just seems like one of those things that there should be a better way of doing it.

Upvotes: 19

Views: 16848

Answers (4)

Andrew Clark
Andrew Clark

Reputation: 208545

You can use the following method for grouping items from an iterable, taken from the documentation for zip():

connections = cmds.listConnections()
for destination, source in zip(*[iter(connections)]*2):
    print source, destination

Or for a more readable version, use the grouper recipe from the itertools documentation:

def grouper(n, iterable, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Upvotes: 18

chirob
chirob

Reputation: 91

lis = list(range(12))
lg = len(lis)
tup = tuple()
for k in range(0,lg,2):
    tup = (lis[k],lis[k+1])
    print(tup)

(0, 1) (2, 3) (4, 5) (6, 7) (8, 9) (10, 11)

Upvotes: 2

Kevin Kreiser
Kevin Kreiser

Reputation: 714

why not just use range with the step=2?

for i in range(0, len(connections)):
  source, destination = connections[i:i + 2]
  print source, destination

you can make a generic function out of it like:

def tuples(iterable, arity):
  return [iterable[i:i+arity] for i in range(0, len(iterable), arity)]

Upvotes: 3

John C. King
John C. King

Reputation: 141

All, Great question & answer. I'd like to provide another solution that should compliment Andrew Clark's answer (props for using itertools!). His answer returns each value once like this:

iterable = [0, 1, 2, 3, 4, 5, 6,...] n = 2 grouper(n, iterable, fillvalue=None) --> [(0, 1), (2, 3), (3, 4), (5, 6),...]

In the code below each value will appear in n sub-sequences. Like this:

def moving_window(n, iterable):
  start, stop = 0, n
  while stop <= len(iterable):
      yield iterable[start:stop]
      start += 1
      stop += 1

--> [(0, 1), (1, 2), (2, 3), (3, 4),...]

A common application for this type of 'moving window' or 'kernel' is moving averages in the sciences and finance.

Also note that the yield statement allows each sub-sequence to be created as it's needed and not stored in memory.

Upvotes: 4

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