输入：
"The boy is running on the train"
预期输出：
["The boy", "boy is", "is running", "running on", "on the", "the train"]
在 python 中实现这一目标的最简单的解决方案是什么。
【问题讨论】：
line="The boy is running on the train"
words=line.split()
k=[words[index]+' '+words[index+1] for index in xrange(len(words)-1)]
print k
输出
['The boy', 'boy is', 'is running', 'running on', 'on the', 'the train']
【讨论】：
c 来指定一个整数是一个小缺陷：这会造成暂时的混乱，因为人们首先认为它指定了一个字符。而且，正如 Paul McGuire 指出的那样，xrange 会更好
你在所有空格上拆分，然后重新加入对：
words = inputstr.split()
secondwords = iter(words)
next(secondwords)
output = [' '.join((first, second))
for first, second in zip(words, secondwords)]
演示：
>>> inputstr = "The boy is running on the train"
>>> words = inputstr.split()
>>> secondwords = iter(words)
>>> next(secondwords) # output is ignored
'The'
>>> [' '.join((first, second)) for first, second in zip(words, secondwords)]
['The boy', 'boy is', 'is running', 'running on', 'on the', 'the train']
【讨论】：
import re
s = "The boy is running on the train"
print map(' '.join,re.findall('([^ \t]+)[ \t]+(?=([^ \t]+))',s))
Koustav Ghosal 的解决方案是最快的：
import re
from time import clock
from itertools import izip
from collections import defaultdict
s = "The boy is running on the train"
z = 200
p = '%-9.6f %6.1f%% %s'
rgx = re.compile('([^ \t]+)[ \t]+(?=([^ \t]+))')
R = defaultdict(list)
for rep in xrange(3000):
t0 = clock()
for i in xrange(z):
map(' '.join,re.findall('([^ \t]+)[ \t]+(?=([^ \t]+))',s))
te1 = clock()-t0
R['e1'].append(te1)
t0 = clock()
for i in xrange(z):
map(' '.join,rgx.findall(s))
te2 = clock()-t0
R['e2'].append(te2)
t0 = clock()
for i in xrange(z):
words = s.split()
secondwords = iter(words)
next(secondwords)
[' '.join((first, second))
for first, second in zip(words, secondwords)]
tM1 = clock()-t0
R['M1'].append(tM1)
t0 = clock()
for i in xrange(z):
words = s.split()
secondwords = iter(words)
next(secondwords)
[' '.join((first, second))
for first, second in izip(words, secondwords)]
tM2 = clock()-t0
R['M2'].append(tM2)
t0 = clock()
for i in xrange(z):
words = s.split()
secondwords = iter(words)
next(secondwords)
[' '.join(x)
for x in izip(words, secondwords)]
tM3 = clock()-t0
R['M3'].append(tM3)
t0 = clock()
for i in xrange(z):
words=s.split()
[words[c]+' '+words[c+1] for c in range(len(words)-1)]
tK1 = clock() - t0
R['K1'].append(tK1)
t0 = clock()
for i in xrange(z):
words=s.split()
[words[c]+' '+words[c+1] for c in xrange(len(words)-1)]
tK2 = clock() - t0
R['K2'].append(tK2)
tmax = min(R['e1'])
for k,s in (('e1','eyquem with re.findall(pat,string)'),
('e2','eyquem with compiled_regex.findall(string)'),
('M1','Martijn Pieters'),
('M2','Martijn Pieters with izip'),
('M3','Martijn Pieters with izip and direct join'),
('K1','Koustav Ghosal'),
('K2','Koustav Ghosal with xrange')):
t = min(R[k])
print p % (t,t/tmax*100,s)
Python 2.7 的结果
0.007127 100.0% eyquem with re.findall(pat,string)
0.004045 56.8% eyquem with compiled_regex.findall(string)
0.003887 54.5% Martijn Pieters
0.002522 35.4% Martijn Pieters with izip
0.002152 30.2% Martijn Pieters with izip and direct join
0.002030 28.5% Koustav Ghosal
0.001856 26.0% Koustav Ghosal with xrange
【讨论】：
Timer.timeit 的确切含义 3/ Timer.repeat 中有注释请注意，在文档中，代码 sn-p 执行时间的可能最佳度量是 sn-p 多次重复的结果时间中的最小值，并且 “您应该查看整个向量和应用常识而不是统计数据” 因此，我更喜欢使用完全合理的时钟，而不是 Timer.repeat 完成的统计数据，IMO，通过以下引用
time.clock() “这是用于对 Python 或计时算法进行基准测试的函数。” 对于 Linux 系统，“分辨率通常优于一微秒。” 适用于 Windows 系统 (docs.python.org/2/library/time.html#time.clock)
clock() 测量时间时，我不会仅根据运行代码得出结论。即使我给出的 SO 答案不会重复几个措施，我也会运行它几次，然后我选择在我看来为所有执行时间提供最低值的运行；甚至，我在多次重复时为每个代码 sn-p 选择最低的执行时间，然后将其分组以比较多次运行获得的最小值。
或者，itertools.combinations 的解决方案：
>>> s = "The boy is running on the train"
>>> seen = set()
>>> new = []
>>> for tup in itertools.combinations(s.split(), 2):
... if tup[0] not in seen:
... new.append(' '.join(tup))
... seen.add(tup[0])
...
>>> print new
['The boy', 'boy is', 'is running', 'running on', 'on the', 'the train']
虽然这确实不是 itertools.combinations 应该用于的：p。
【讨论】：