[英]Most pythonic way to get the previous element
I would like an enumerate
-like functional on iterators which yields the pair (previous_element, current_element)
. 我希望在迭代器上有类似
enumerate
的函数,它产生对(previous_element, current_element)
。 That is, given that iter
is 也就是说,考虑到
iter
是
i0, i1, i1, ...
I would like offset(iter)
to yield 我想
offset(iter)
屈服
(None, i0), (i0, i1), (i1, i2) ...
What about the simple (obvious) solution? 简单(明显)的解决方案呢?
def offset(iterable):
prev = None
for elem in iterable:
yield prev, elem
prev = elem
To put more itertools on the table: 把更多的itertool放在桌子上:
from itertools import tee, izip, chain
def tee_zip(iterable):
a, b = tee(iterable)
return izip(chain([None], a), b)
def pairwise(iterable):
"""s -> (s0,s1), (s1,s2), (s2, s3), ...
see http://docs.python.org/library/itertools.html
"""
a, b = itertools.tee(iterable)
b.next()
return itertools.izip(a, b)
EDIT moved doc string into the function EDIT将doc字符串移动到函数中
def offset(iter, n=1, pad=None):
i1, i2 = itertools.tee(iter)
i1_padded = itertools.chain(itertools.repeat(pad, n), i1)
return itertools.izip(i1_padded, i2)
@bpgergo + @user792036 = this. @bpgergo + @ user792036 =这个。 Best of two worlds :).
最好的两个世界:)。
The best answer I have (and this requires itertools
) is 我有最好的答案(这需要
itertools
)
def offset(iter, n=1):
# returns tuples (None, iter0), (iter0, iter1), (iter1, iter2) ...
previous = chain([None] * n, iter)
return izip(previous, iter)
but I would be interested in seeing if someone has a one-liner (or a better name than offset for this function)! 但是我有兴趣看看是否有人有一个单行(或者这个功能的偏移比一个更好的名字)!
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