当使用for循环迭代时，我如何处理输入的最后一个元素?特别是，如果有代码应该只出现在元素之间(而不是在最后一个元素之后)，我该如何构造代码?
目前，我写的代码是这样的:
for i, data in enumerate(data_list):
code_that_is_done_for_every_element
if i != len(data_list) - 1:
code_that_is_done_between_elements
我如何简化或改进它?
当前回答
如果你正在浏览这个列表，这对我来说也是有效的:
for j in range(0, len(Array)):
if len(Array) - j > 1:
notLast()
其他回答
这类似于Ants Aasma的方法，但没有使用itertools模块。它也是一个滞后迭代器，它可以提前查找迭代器流中的单个元素:
def last_iter(it):
# Ensure it's an iterator and get the first field
it = iter(it)
prev = next(it)
for item in it:
# Lag by one item so I know I'm not at the end
yield 0, prev
prev = item
# Last item
yield 1, prev
def test(data):
result = list(last_iter(data))
if not result:
return
if len(result) > 1:
assert set(x[0] for x in result[:-1]) == set([0]), result
assert result[-1][0] == 1
test([])
test([1])
test([1, 2])
test(range(5))
test(xrange(4))
for is_last, item in last_iter("Hi!"):
print is_last, item
我只是遇到了这个问题，我的通用解决方案使用迭代器:
from typing import TypeVar, Iterable
E = TypeVar('E')
def metait(i: Iterable[E]) -> Iterable[tuple[E, bool, bool]]:
first = True
previous = None
for elem in i:
if previous:
yield previous, first, False
first = False
previous = elem
if previous:
yield previous, first, True
您将收到一个元组，其中包含第一项和最后一项的原始元素和标志。它可以用于每个可迭代对象:
d = {'a': (1,2,3), 'b': (4,5,6), 'c': (7,8,9)}
for (k,v), is_first, is_last in metait(d.items()):
print(f'{k}: {v} {is_first} {is_last}')
这将给你:
a: (1, 2, 3) True False
b: (4, 5, 6) False False
c: (7, 8, 9) False True
将最后一项的特殊处理延迟到循环之后。
>>> for i in (1, 2, 3):
... pass
...
>>> i
3
除了向上数，你也可以向下数:
nrToProcess = len(list)
for s in list:
s.doStuff()
nrToProcess -= 1
if nrToProcess==0: # this is the last one
s.doSpecialStuff()
这是一个老问题，已经有很多很好的回答了，但我觉得这很python:
def rev_enumerate(lst):
"""
Similar to enumerate(), but counts DOWN to the last element being the
zeroth, rather than counting UP from the first element being the zeroth.
Since the length has to be determined up-front, this is not suitable for
open-ended iterators.
Parameters
----------
lst : Iterable
An iterable with a length (list, tuple, dict, set).
Yields
------
tuple
A tuple with the reverse cardinal number of the element, followed by
the element of the iterable.
"""
length = len(lst) - 1
for i, element in enumerate(lst):
yield length - i, element
这样用:
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
if not num_remaining:
print(f'This is the last item in the list: {item}')
或者你想做相反的事情:
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
if num_remaining:
print(f'This is NOT the last item in the list: {item}')
或者，只是想知道当你走的时候还剩下多少……
for num_remaining, item in rev_enumerate(['a', 'b', 'c']):
print(f'After {item}, there are {num_remaining} items.')
我认为现有枚举的通用性和熟悉性使其最具python性。
注意，与enumerate()不同，rev_enumerate()要求输入实现__len__，但这包括列表、元组、字典和集。
