Iterator excluding the last element
What is a pythonic way to implement an iterator that excludes the last element without knowing its length?
Example:
>>> list(one_behind(iter(range(10)))
... [0, 1, 2, 3, 4, 5, 6, 7, 8]
>>> iter_ = one_behind(iter((3, 2, 1)))
>>> next(iter_)
... 3
>>> next(iter_)
... 2
>>> next(iter_)
... StopIteration
A simple approach would be to use a loop and keep the previous value, but I would like something a little shorter.
Implementing links using a loop:
def one_behind(iter_):
prev = None
for i, x in enumerate(iter_):
if i > 0:
yield prev
prev = x
+3
simonzack
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3 answers
Using itertools.tee
:
import itertools
def behind(it):
# it = iter(it) # to handle non-iterator iterable.
i1, i2 = itertools.tee(it)
next(i1)
return (next(i2) for x in i1)
using:
>>> list(behind(iter(range(3))))
[0, 1]
+4
falsetru
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Negatively simpler than the reference function:
def lag(iter):
previous_item = next(iter)
for item in iter:
yield previous_item
previous_item = item
+2
Stuart
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Code:
def one_behind(iter):
q = collections.deque()
q.append(next(iter))
while True:
q.append(next(iter))
yield q.popleft()
Runtime:
>>> iterator = iter(range(4))
>>> list(one_behind(iterator))
[0, 1, 2]
0
Bill lynch
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