有多种方法。切片是最快的。再添加一个使用.index()方法的:

>>> l1 = [1,5,2,3,5,1,7,43]                                                 
>>> [i for i in l1 if l1.index(i)+1==len(l1)]                               
[43]

我想到的最简单的解决办法是:

for item in data_list:
    try:
        print(new)
    except NameError: pass
    new = item
print('The last item: ' + str(new))

所以我们总是通过延迟处理一个迭代来提前一个项目。为了在第一次迭代中跳过某些操作，我只需捕获错误。

当然，您需要考虑一下，以便在需要时引发NameError。

还要保留“counstruct”

try:
    new
except NameError: pass
else:
    # continue here if no error was raised

这依赖于之前没有定义名称new。如果你是偏执狂，你可以确保new不存在，使用:

try:
    del new
except NameError:
    pass

当然，你也可以使用if语句(if notfirst: print(new) else: notfirst = True)。但据我所知，开销更大。

Using `timeit` yields:

    ...: try: new = 'test' 
    ...: except NameError: pass
    ...: 
100000000 loops, best of 3: 16.2 ns per loop

所以我认为开销是不可能当选的。

如果要循环List， 使用枚举函数是最好的尝试之一。

for index, element in enumerate(ListObj):
    # print(index, ListObj[index], len(ListObj) )

    if (index != len(ListObj)-1 ):
        # Do things to the element which is not the last one
    else:
        # Do things to the element which is the last one

大多数情况下，让第一次迭代成为特殊情况比最后一次迭代更容易(也更便宜):

first = True
for data in data_list:
    if first:
        first = False
    else:
        between_items()

    item()

这将适用于任何迭代对象，即使是那些没有len()的迭代对象:

file = open('/path/to/file')
for line in file:
    process_line(line)

    # No way of telling if this is the last line!

除此之外，我不认为有更好的解决方案，因为这取决于你想要做什么。例如，如果您正在从列表中构建字符串，那么使用str.join()自然比使用“带有特殊情况”的For循环更好。

使用相同的原理，但更紧凑:

for i, line in enumerate(data_list):
    if i > 0:
        between_items()
    item()

看起来很眼熟，不是吗?：）

对于@ofko，以及其他真正需要找出不带len()的可迭代对象的当前值是否为最后一个值的人，你需要向前看:

def lookahead(iterable):
    """Pass through all values from the given iterable, augmented by the
    information if there are more values to come after the current one
    (True), or if it is the last value (False).
    """
    # Get an iterator and pull the first value.
    it = iter(iterable)
    last = next(it)
    # Run the iterator to exhaustion (starting from the second value).
    for val in it:
        # Report the *previous* value (more to come).
        yield last, True
        last = val
    # Report the last value.
    yield last, False

然后你可以这样使用它:

>>> for i, has_more in lookahead(range(3)):
...     print(i, has_more)
0 True
1 True
2 False

只需检查data是否与data_list (data_list[-1])中的最后一个数据不相同。

for data in data_list:
    code_that_is_done_for_every_element
    if data != data_list[- 1]:
        code_that_is_done_between_elements

假设input是一个迭代器，下面是使用itertools中的tee和izip的方法:

from itertools import tee, izip
items, between = tee(input_iterator, 2)  # Input must be an iterator.
first = items.next()
do_to_every_item(first)  # All "do to every" operations done to first item go here.
for i, b in izip(items, between):
    do_between_items(b)  # All "between" operations go here.
    do_to_every_item(i)  # All "do to every" operations go here.

演示:

>>> def do_every(x): print "E", x
...
>>> def do_between(x): print "B", x
...
>>> test_input = iter(range(5))
>>>
>>> from itertools import tee, izip
>>>
>>> items, between = tee(test_input, 2)
>>> first = items.next()
>>> do_every(first)
E 0
>>> for i,b in izip(items, between):
...     do_between(b)
...     do_every(i)
...
B 0
E 1
B 1
E 2
B 2
E 3
B 3
E 4
>>>