How to look ahead one element (peek) in a Python generator?

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?


Solution 1:

For sake of completeness, the more-itertools package (which should probably be part of any Python programmer's toolbox) includes a peekable wrapper that implements this behavior. As the code example in the documentation shows:

>>> p = peekable(['a', 'b'])
>>> p.peek()
'a'
>>> next(p)
'a'

However, it's often possible to rewrite code that would use this functionality so that it doesn't actually need it. For example, your realistic code sample from the question could be written like this:

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

(reader's note: I've preserved the syntax in the example from the question as of when I'm writing this, even though it refers to an outdated version of Python)

Solution 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))

Solution 3:

Ok - two years too late - but I came across this question, and did not find any of the answers to my satisfaction. Came up with this meta generator:

class Peekorator(object):

    def __init__(self, generator):
        self.empty = False
        self.peek = None
        self.generator = generator
        try:
            self.peek = self.generator.next()
        except StopIteration:
            self.empty = True

    def __iter__(self):
        return self

    def next(self):
        """
        Return the self.peek element, or raise StopIteration
        if empty
        """
        if self.empty:
            raise StopIteration()
        to_return = self.peek
        try:
            self.peek = self.generator.next()
        except StopIteration:
            self.peek = None
            self.empty = True
        return to_return

def simple_iterator():
    for x in range(10):
        yield x*3

pkr = Peekorator(simple_iterator())
for i in pkr:
    print i, pkr.peek, pkr.empty

results in:

0 3 False
3 6 False
6 9 False
9 12 False    
...
24 27 False
27 None False

i.e. you have at any moment during iteration access to the next item in the list.