从Python列表中获得2乘2的项

Question

What is the most elegant way to obtain items 2 by 2 from a list ?

from:
my_list = ['I', 'swear', 'I', 'googled', 'first']

to:
res_list = ['I swear', 'swear I', 'I googled', 'googled first']

Answer 1

def window(seq, n=2):
    """
    Returns a sliding window (of width n) over data from the sequence
    s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...
    """
    for i in xrange(len(seq)-n+1):
        yield tuple(seq[i:i+n])

my_list = ['I', 'swear', 'I', 'googled', 'first']
res_list = [' '.join(group) for group in window(my_list, 2)]
print(res_list)
# ['I swear', 'swear I', 'I googled', 'googled first']

Answer 2

I would say a standard case for zip :

def pairs(i):
    return list(zip(i, i[1:]))

my_list = ['I', 'swear', 'I', 'googled', 'first']
res_list = pairs(my_list)
print(res_list)
# [('I', 'swear'), ('swear', 'I'), ('I', 'googled'), ('googled', 'first')]
print([' '.join(a) for a in res_list])
# ['I swear', 'swear I', 'I googled', 'googled first']

Just for completeness's sake, the same with arbitrary window width:

def window(i, n = 2):
    return list(zip(*(i[p:] for p in range(n))))

Answer 3

def pairwise( iterable, n=2 ):
    from itertools import tee, izip, islice
    return izip(*(islice(it,pos,None) for pos,it in enumerate(tee(iterable, n))))

my_list = ['I', 'swear', 'I', 'googled', 'first']

print list(pairwise(my_list))
#[('I', 'swear'), ('swear', 'I'), ('I', 'googled'), ('googled', 'first')]

Answer 4

A neat little one-liner in the form of a generator expression:

>>> my_list = ['I', 'swear', 'I', 'googled', 'first']
>>> res_list = (' '.join((x, y)) for x, y in zip(my_list, my_list[1:]))
>>> list(res_list)
['I swear', 'swear I', 'I googled', 'googled first']