我有一个问题,因为我找不到解决问题的方法。
gen是一个生成器(difflib.Differ.compare()的结果):
通常通过迭代生成我可以读取每一行。问题是在每次迭代时我需要读取当前行和下两行。
示例(逐行迭代的正常输出):
iteration 1:
line = 'a'
iteration 2:
line = 'b'
iteration 3:
line = 'c'
iteration 4:
line = 'd'
iteration 5:
line = 'e'
iteration 6:
line = 'f'
iteration 7:
line = 'g'
但在我的情况下,我需要得到这个:
iteration 1:
line = 'a'
next1 = 'b'
next2 = 'c'
iteration 2:
line = 'b'
next1 = 'c'
next2 = 'd'
iteration 3:
line = 'c'
next1 = 'd'
next2 = 'e'
iteration 4:
line = 'd'
next1 = 'e'
next2 = 'f'
iteration 5:
line = 'e'
next1 = 'f'
next2 = 'g'
iteration 6:
line = 'f'
next1 = 'g'
next2 = None
iteration 7:
line = 'g'
next1 = None
next2 = None
我试图使用gen.send(),itertools.islice(),但我找不到合适的解决方案。我不想将这个生成器转换成一个列表(然后我可以读取next1作为gen [i + 1],next2作为gen [i + 2],但是当diff输出很大时,这是非常低效的。
答案 0 :(得分:5)
这是我建议作为任何迭代器/生成器的通用解决方案。我认为这种方式效率最高。
def genby3(gen):
it = iter(gen) # Make it a separate iterator, to avoid consuming it totally
L1 = it.next() # Get the first two elements
L2 = it.next()
for L3 in it:
yield [L1, L2, L3] # Get the results grouped in 3
L1, L2 = L2, L3 # Update the last 2 elements
yield [L2, L3, None] # And take care of the last 2 cases
yield [L3, None, None]
print list(genby3(xrange(10)))
如果它是您正在阅读的文件,您可以seek
,readline
然后返回,但它可能会变得混乱,因此您可以将其视为任何其他迭代器。
更新:每次迭代不仅仅有3个项目,它的工作效果非常好,它可以像其他项目一样工作。
def genby(gen, n):
assert n>=1, 'This does not make sense with less than one element'
it = iter(gen)
last = list(it.next() for i in xrange(n-1))
for nth_item in it:
last = last+[nth_item]
yield last
last.pop(0)
for i in xrange(n-1):
last = last+[None]
yield last
last.pop(0)
r = xrange(10)
for i, n in enumerate(genby(r, 3)):
print i, 'iteration'
print '\t', n
编辑2 :在yield语句之前移动列表的串联,以避免必须两次。明智的改善表现。
答案 1 :(得分:2)
尝试保留临时变量。
line = iterator.next()
next1 = iterator.next()
for next2 in iterator:
#do stuff
line = next1
next1 = next2
答案 2 :(得分:2)
有一个recipe in the itertools
docs,pairwise()
。它可以适应:
from itertools import tee, izip_longest
def triplewise(iterable):
xs, ys, zs = tee(iterable, 3)
next(ys, None)
next(zs, None)
next(zs, None)
return izip_longest(xs, ys, zs)
for line, next1, next2 in triplewise(gen):
...
它也可以概括:
from itertools import tee, izip, izip_longest, islice
no_fillvalue = object()
def nwise(iterable, n=2, fillvalue=no_fillvalue):
iters = (islice(each, i, None) for i, each in enumerate(tee(iterable, n)))
if fillvalue is no_fillvalue:
return izip(*iters)
return izip_longest(*iters, fillvalue=fillvalue)
for line, next1, next2 in nwise(gen, 3, None):
...
答案 3 :(得分:1)
如何将三个序列压缩在一起?
izip_longest(gen, islice(gen,1,None), islice(gen,2,None), fillvalue=None)
答案 4 :(得分:1)
您可以使用以下内容:
def genTriplets(a):
first = a.next()
second = a.next()
third = a.next()
while True:
yield (first, second, third)
first = second
second = third
try:
third = a.next()
except StopIteration:
third = None
if (first is None and second is None and third is None):
break