有內存的迭代器？

Question

我正在使用Markov鏈的應用程序上工作。

此代碼的示例如下：

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)

我正在尋找的是迭代train_seq，但要保留N個最后一個元素。

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)

希望我的問題的描述足夠清楚

Answer 1

window將一次為您提供n個iterable項。

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)]

如果您想要相同數量的商品，即使是前幾次，

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])]

會這樣做的。

Answer 2

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

您還可以執行以下操作來創建任意大小的對：

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

編輯：但是使用迭代zip可能更好：

from itertools import izip

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

我在seq為10,000,000整數的系統上嘗試了此操作，結果相當即時。

Answer 3

改進yan的答案以避免復制：

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