How to look ahead one element in a Python generator?

Question

I can't figure out how to look ahead one element in a Python generator. As soon as I look it's gone.

Here is what I mean:

gen = iter([1,2,3])
next_value = gen.next()  # okay, I looked forward and see that next_value = 1
# but now:
list(gen)  # is [2, 3]  -- the first value is gone!

Here is a more real example:

gen = element_generator()
if gen.next_value() == 'STOP':
  quit_application()
else:
  process(gen.next())

Can anyone help me write a generator that you can look one element forward?

Thanks, Boda Cydo.

Answer 1

Unfortunately you cannot "peek" a value from a generator. For a workaround please see Easy way to 'unget' a value from a generator using send()

Answer 2

The Python generator API is one way: You can't push back elements you've read. But you can create a new iterator using the itertools module and prepend the element:

import itertools

gen = iter([1,2,3])
peek = gen.next()
print list(itertools.chain([peek], gen))

Answer 3

This will work -- it buffers an item and calls a function with each item and the next item in the sequence.

Your requirements are murky on what happens at the end of the sequence. What does "look ahead" mean when you're at the last one?

def process_with_lookahead( iterable, aFunction ):
    prev= iterable.next()
    for item in iterable:
        aFunction( prev, item )
        prev= item
    aFunction( item, None )

def someLookaheadFunction( item, next_item ):
    print item, next_item