Using lookahead with generators

Question

Hi all,

I have implemented a generator-based scanner in Python that tokenizes a string into tuples of the form (token type, token value):

for token in scan("a(b)"):
    print token

would print

("literal", "a")
("l_paren", "(")
...

The next task implies parsing the token stream and for that, I need be able to look one item ahead from the current one without moving the pointer ahead as well. The fact that iterators and generators do not provide the complete sequence of items at once but each item as needed makes lookaheads a bit trickier compared to lists, since the next item is not known unless __next__() is called.

What could a straightforward implementation of a generator-based lookahead look like? Currently I'm using a workaround which implies making a list out of the generator:

token_list = [token for token in scan(string)]

The lookahead then is easily implemented by something like that:

try:
    next_token = token_list[index + 1]
except: IndexError:
    next_token = None

Of course this just works fine. But thinking that over, my second question arises: Is there really a point of making scan() a generator in the first place?

Thanks in advance!

Regards,

jena

Answer 1

It's not pretty, but this may do what you want:

def paired_iter(it):
    token = it.next()
    for lookahead in it:
        yield (token, lookahead)
        token = lookahead
    yield (token, None)

def scan(s):
    for c in s:
        yield c

for this_token, next_token in paired_iter(scan("ABCDEF")):
    print "this:%s next:%s" % (this_token, next_token)

Prints:

this:A next:B
this:B next:C
this:C next:D
this:D next:E
this:E next:F
this:F next:None

Answer 2

You can write a wrapper that buffers some number of items from the generator, and provides a lookahead() function to peek at those buffered items:

class Lookahead:
    def __init__(self, iter):
        self.iter = iter
        self.buffer = []

    def __iter__(self):
        return self

    def next(self):
        if self.buffer:
            return self.buffer.pop(0)
        else:
            return self.iter.next()

    def lookahead(self, n):
        """Return an item n entries ahead in the iteration."""
        while n >= len(self.buffer):
            try:
                self.buffer.append(self.iter.next())
            except StopIteration:
                return None
        return self.buffer[n]

Answer 3

Paul's is a good answer. A class based approach with arbitrary lookahead might look something like:

class lookahead(object):
    def __init__(self, generator, lookahead_count=1):
        self.gen = iter(generator)
        self.look_count = lookahead_count

    def __iter__(self):
        self.lookahead = []
        self.stopped = False
        try:
            for i in range(self.look_count):
                self.lookahead.append(self.gen.next())
        except StopIteration:
            self.stopped = True
        return self

    def next(self):
        if not self.stopped:
            try:
                self.lookahead.append(self.gen.next())
            except StopIteration:
                self.stopped = True
        if self.lookahead != []:
            return self.lookahead.pop(0)
        else:
            raise StopIteration

x = lookahead("abcdef", 3)
for i in x:
    print i, x.lookahead

Answer 4

Here is an example that allows a single item to be sent back to the generator

def gen():
    for i in range(100):
        v=yield i           # when you call next(), v will be set to None
        if v:
            yield None      # this yields None to send() call
            v=yield v       # so this yield is for the first next() after send()

g=gen()

x=g.next()
print 0,x

x=g.next()
print 1,x

x=g.next()
print 2,x # oops push it back

x=g.send(x)

x=g.next()
print 3,x # x should be 2 again

x=g.next()
print 4,x

Answer 5

Pretty good answers there, but my favorite approach would be to use itertools.tee -- given an iterator, it returns two (or more if requested) that can be advanced independently. It buffers in memory just as much as needed (i.e., not much, if the iterators don't get very "out of step" from each other). E.g.:

import itertools
import collections

class IteratorWithLookahead(collections.Iterator):
  def __init__(self, it):
    self.it, self.nextit = itertools.tee(iter(it))
    self._advance()
  def _advance(self):
    self.lookahead = next(self.nextit, None)
  def __next__(self):
    self._advance()
    return next(self.it)

You can wrap any iterator with this class, and then use the .lookahead attribute of the wrapper to know what the next item to be returned in the future will be. I like to leave all the real logic to itertools.tee and just provide this thin glue!-)

Answer 6

Since you say you are tokenizing a string and not a general iterable, I suggest the simplest solution of just expanding your tokenizer to return a 3-tuple: (token_type, token_value, token_index), where token_index is the index of the token in the string. Then you can look forward, backward, or anywhere else in the string. Just don't go past the end. Simplest and most flexible solution I think.

Also, you needn't use a list comprehension to create a list from a generator. Just call the list() constructor on it:

 token_list = list(scan(string))