Iterating Through List by 3's

Question

a = [3,5,8,3,9,5,0,3,2,7,5,4]
for o in a[::3]:
    print o

This gets me the first and every 3 item. 3,3,0,7

is there a way i can retrieve the next two items as well?

a = [3,5,8,3,9,5,0,3,2,7,5,4]
for o in a[::3]:
  if o == 0:
    print o
    print o + 1
    print o + 2

output 0 3 2

I know that's not correct but maybe you can see what I'm trying to do. Basically, I have a long list of properties and there are three parts to each property, parent_id, property_type and property_value and I need to retrieve all three parts of the property from the list.

Answer 1

You can do this using the "grouper" recipe from the itertools documentation:

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Example:

>>> a = [3, 5, 8, 3, 9, 5, 0, 3, 2, 7, 5, 4]
>>>  list(grouper(3, a))
[(3, 5, 8), (3, 9, 5), (0, 3, 2), (7, 5, 4)]

Answer 2

Try this:

array = [3,5,8,3,9,5,0,3,2,7,5,4]
for i in xrange(0, len(array), 3):
    a, b, c = array[i:i+3] # partition the array in groups of 3 elements
    print a, b, c

It works because (as stated in the question) there are exactly three parts to each property.