有没有Pythonic等同于Ruby的#each_cons
?
在Ruby中你可以这样做:
array = [1,2,3,4]
array.each_cons(2).to_a
=> [[1,2],[2,3],[3,4]]
对于这样的事情,itertools
是你应该看的模块:
from itertools import tee, izip
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = tee(iterable)
next(b, None)
return izip(a, b)
然后:
>>> list(pairwise([1, 2, 3, 4]))
[(1, 2), (2, 3), (3, 4)]
有关更一般的解决方案,请考虑以下事项:
def split_subsequences(iterable, length=2, overlap=0):
it = iter(iterable)
results = list(itertools.islice(it, length))
while len(results) == length:
yield results
results = results[length - overlap:]
results.extend(itertools.islice(it, length - overlap))
if results:
yield results
这允许任意长度的子序列和任意重叠。用法:
>> list(split_subsequences([1, 2, 3, 4], length=2))
[[1, 2], [3, 4]]
>> list(split_subsequences([1, 2, 3, 4], length=2, overlap=1))
[[1, 2], [2, 3], [3, 4], [4]]
我不认为有一个,我查看了内置模块itertools
,这是我期望的。你可以简单地创建一个:
def each_cons(x, size):
return [x[i:i+size] for i in range(len(x)-size+1)]
我的列表解决方案(Python2):
import itertools
def each_cons(xs, n):
return itertools.izip(*(xs[i:] for i in xrange(n)))
编辑:使用Python 3 itertools.izip
不再,所以你使用普通的zip
:
def each_cons(xs, n):
return zip(*(xs[i:] for i in range(n)))
快速单行:
a = [1, 2, 3, 4]
out = [a[i:i + 2] for i in range(len(a) - 1)]
Python肯定可以做到这一点。如果您不想这么急切地使用,请使用itertool的islice和izip。此外,重要的是要记住正常切片将创建一个副本,因此如果内存使用很重要,您还应该考虑itertool等价物。
each_cons = lambda l: zip(l[:-1], l[1:])
更新:没关系我的答案,只需使用toolz.itertoolz.sliding_window()
- 它会做正确的事情。
对于一个真正的惰性实现,当序列/生成器长度不足时保留Ruby的each_cons
的行为:
import itertools
def each_cons(sequence, n):
return itertools.izip(*(itertools.islice(g, i, None)
for i, g in
enumerate(itertools.tee(sequence, n))))
例子:
>>> print(list(each_cons(xrange(5), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]
>>> print(list(each_cons(xrange(5), 5)))
[(0, 1, 2, 3, 4)]
>>> print(list(each_cons(xrange(5), 6)))
[]
>>> print(list(each_cons((a for a in xrange(5)), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]
请注意,在izip的参数上使用的元组解包应用于n
的大小为itertools.tee(xs, n)
的元组(即“窗口大小”),而不是我们想要迭代的序列。
与elias的代码相同,但适用于python 2和3:
try:
from itertools import izip # python 2
except ImportError:
from builtins import zip as izip # python 3
from itertools import islice, tee
def each_cons(sequence, n):
return izip(
*(
islice(g, i, None)
for i, g in
enumerate(tee(sequence, n))
)
)
接近@ Blender的解决方案,但有一个修复:
a = [1, 2, 3, 4]
n = 2
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2], [2, 3], [3, 4]]
要么
a = [1, 2, 3, 4]
n = 3
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2, 3], [2, 3, 4]]