CODEWITHSUNDEEP
iteratorpython
Dog eat cat world
Dog eat cat world

Reputation: 760

Iterator with memory?

I'm working on an application which use a markov chain.

An example on this code follows:

chain = MarkovChain(order=1)
train_seq = ["","hello","this","is","a","beautiful","world"]

for i, word in enum(train_seq):
 chain.train(previous_state=train_seq[i-1],next_state=word)

What I am looking for is to iterate over train_seq, but to keep the N last elements.

for states in unknown(train_seq,order=1):
 # states should be a list of states, with states[-1] the newest word,
 # and states[:-1] should be the previous occurrences of the iteration.
 chain.train(*states)

Hope the description of my problem is clear enough for

Upvotes: 2

Views: 575

Answers (3)

Karl Knechtel
Karl Knechtel

Reputation: 61607

Improving upon yan's answer to avoid copies:

from itertools import *

def staggered_iterators(sequence, count):
  iterator = iter(sequence)
  for i in xrange(count):
    result, iterator = tee(iterator)
    yield result
    next(iterator)

tuple_size = 4
for w in izip(*(i for i in takewhile(staggered_iterators(seq, order)))):
  print w

Upvotes: 0

yan
yan

Reputation: 20992

seq = [1,2,3,4,5,6,7]
for w in zip(seq, seq[1:]):
  print w

You can also do the following to create an arbitrarily-sized pairs:

tuple_size = 2
for w in zip(*(seq[i:] for i in range(tuple_size)))
  print w

edit: But it's probably better using the iterative zip:

from itertools import izip

tuple_size = 4
for w in izip(*(seq[i:] for i in range(tuple_size)))
  print w

I tried this on my system with seq being 10,000,000 integers and the results were fairly instant.

Upvotes: 1

agf
agf

Reputation: 176900

window will give you n items from iterable at a time.

from collections import deque

def window(iterable, n=3):
    it = iter(iterable)
    d = deque(maxlen = n)
    for elem in it:
        d.append(elem)
        yield tuple(d)


print [x for x in window([1, 2, 3, 4, 5])]
# [(1,), (1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5)]

If you want the same number of items even the first few times,

from collections import deque
from itertools import islice

def window(iterable, n=3):
    it = iter(iterable)
    d = deque((next(it) for Null in range(n-1)), n)
    for elem in it:
        d.append(elem)
        yield tuple(d)


print [x for x in window([1, 2, 3, 4, 5])]

will do that.

Upvotes: 6

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