Iterator with memory?

I am working on an application using the brand chain.

Below is an example of this code:

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'm looking for is to iterate over train_seq but keep the N last items.

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

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3 answers


window

will give you n

items from iterable

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 it.

+6


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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 a pair of arbitrary size:

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

      



edit: But it's probably better to use 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 followed by 10,000,000 integers and the results were pretty quick.

+1


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Improving yang's answer to avoid copying:

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

      

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