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()

我想到的一个简单的解决办法是:

for i in MyList:
    # Check if 'i' is the last element in the list
    if i == MyList[-1]:
        # Do something different for the last
    else:
        # Do something for all other elements

第二个同样简单的解决方法是使用计数器:

# Count the no. of elements in the list
ListLength = len(MyList)
# Initialize a counter
count = 0

for i in MyList:
    # increment counter
    count += 1
    # Check if 'i' is the last element in the list
    # by using the counter
    if count == ListLength:
        # Do something different for the last
    else:
        # Do something for all other 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
>>>

您可以在输入数据上使用滑动窗口来查看下一个值，并使用哨兵来检测最后一个值。这适用于任何可迭代对象，所以你不需要事先知道它的长度。成对实现来自itertools recipes。

from itertools import tee, izip, chain

def pairwise(seq):
    a,b = tee(seq)
    next(b, None)
    return izip(a,b)

def annotated_last(seq):
    """Returns an iterable of pairs of input item and a boolean that show if
    the current item is the last item in the sequence."""
    MISSING = object()
    for current_item, next_item in pairwise(chain(seq, [MISSING])):
        yield current_item, next_item is MISSING:

for item, is_last_item in annotated_last(data_list):
    if is_last_item:
        # current item is the last item

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