Pythonic iteration over sliding window pairs in li

问题:

What's the most Pythonic efficient way to iterate over a list in sliding pairs? Here's a related example:

>>> l
['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> for x, y in itertools.izip(l, l[1::2]): print x, y
... 
a b
b d
c f

this is iteration in pairs, but how can we get iteration over a sliding pair? Meaning iteration over the pairs:

a b
b c
c d
d e
etc.

which is iteration over the pairs, except sliding the pair by 1 element each time rather than by 2 elements. thanks.

回答1:

How about:

for x, y in itertools.izip(l, l[1:]): print x, y

回答2:

You can go even simpler. Just zip the list and the list offset by one.

In [4]: zip(l, l[1:])
Out[4]: [('a', 'b'), ('b', 'c'), ('c', 'd'), ('d', 'e'), ('e', 'f'), ('f', 'g')]

回答3:

Here is a little generator that I wrote a while back for a similar scenario:

def pairs(items):
    items_iter = iter(items)
    prev = next(items_iter)

    for item in items_iter:
        yield prev, item
        prev = item

回答4:

Here's a function for arbitrarily sized sliding windows that works for iterators/generators as well as lists

def sliding(seq, n):
  return izip(*starmap(islice, izip(tee(seq, n), count(0), repeat(None))))

Nathan's solution is probably more efficient though.

回答5:

The timing, as defined by the addition of two subsequent entries in the list, is displayed below and ordered from fastest to slowest.

Gil

In [69]: timeit.repeat("for x,y in itertools.izip(l, l[1::1]): x + y", setup=setup, number=1000)
Out[69]: [1.029047966003418, 0.996290922164917, 0.998831033706665]

Geoff Reedy

In [70]: timeit.repeat("for x,y in sliding(l,2): x+y", setup=setup, number=1000)
Out[70]: [1.2408790588378906, 1.2099130153656006, 1.207326889038086]

Alestanis

In [66]: timeit.repeat("for i in range(0, len(l)-1): l[i] + l[i+1]", setup=setup, number=1000)
Out[66]: [1.3387370109558105, 1.3243639469146729, 1.3245630264282227]

chmullig

In [68]: timeit.repeat("for x,y in zip(l, l[1:]): x+y", setup=setup, number=1000)
Out[68]: [1.4756009578704834, 1.4369518756866455, 1.5067830085754395]

Nathan Villaescusa

In [63]: timeit.repeat("for x,y in pairs(l): x+y", setup=setup, number=1000)
Out[63]: [2.254757881164551, 2.3750967979431152, 2.302199125289917]

sr2222

Notice the reduced repetition number...

In [60]: timeit.repeat("for x,y in SubsequenceIter(l,2): x+y", setup=setup, number=100)
Out[60]: [1.599524974822998, 1.5634570121765137, 1.608154058456421]

The setup code:

setup="""
from itertools import izip, starmap, islice, tee, count, repeat
l = range(10000)

def sliding(seq, n):
  return izip(*starmap(islice, izip(tee(seq, n), count(0), repeat(None))))

class SubsequenceIter(object):

    def __init__(self, iterable, subsequence_length):

        self.iterator = iter(iterable)
        self.subsequence_length = subsequence_length
        self.subsequence = [0]

    def __iter__(self):

        return self

    def next(self):

        self.subsequence.pop(0)
        while len(self.subsequence) < self.subsequence_length:
            self.subsequence.append(self.iterator.next())
        return self.subsequence

def pairs(items):
    items_iter = iter(items)
    prev = items_iter.next()

    for item in items_iter:
        yield (prev, item)
        prev = item
"""

回答6:

Not exactly the most efficient, but quite flexible:

class SubsequenceIter(object):

    def __init__(self, iterable, subsequence_length):

        self.iterator = iter(iterable)
        self.subsequence_length = subsequence_length
        self.subsequence = [0]

    def __iter__(self):

        return self

    def next(self):

        self.subsequence.pop(0)
        while len(self.subsequence) < self.subsequence_length:
            self.subsequence.append(self.iterator.next())
        return self.subsequence

Usage:

for x, y in SubsequenceIter(l, 2):
    print x, y