在“ for”循环中检测最后一个元素的pythonic方法是什么？

I’d like to know the best way (more compact and “pythonic” way) to do a special treatment for the last element in a for loop. There is a piece of code that should be called only between elements, being suppressed in the last one.

Here is how I currently do it:

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

Is there any better way?

Note: I don’t want to make it with hacks such as using reduce. ;)

Most of the times it is easier (and cheaper) to make the first iteration the special case instead of the last one:

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

    item()

This will work for any iterable, even for those that have no len():

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

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

Apart from that, I don’t think there is a generally superior solution as it depends on what you are trying to do. For example, if you are building a string from a list, it’s naturally better to use str.join() than using a for loop “with special case”.

Using the same principle but more compact:

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

Looks familiar, doesn’t it? :)

For @ofko, and others who really need to find out if the current value of an iterable without len() is the last one, you will need to look ahead:

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

Then you can use it like this:

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

Although that question is pretty old, I came here via google and I found a quite simple way: List slicing. Let’s say you want to put an ‘&’ between all list entries.

s = ""
l = [1, 2, 3]
for i in l[:-1]:
    s = s + str(i) + ' & '
s = s + str(l[-1])

This returns ‘1 & 2 & 3’.

The ‘code between’ is an example of the Head-Tail pattern.

You have an item, which is followed by a sequence of ( between, item ) pairs. You can also view this as a sequence of (item, between) pairs followed by an item. It’s generally simpler to take the first element as special and all the others as the “standard” case.

Further, to avoid repeating code, you have to provide a function or other object to contain the code you don’t want to repeat. Embedding an if statement in a loop which is always false except one time is kind of silly.

def item_processing( item ):
    # *the common processing*

head_tail_iter = iter( someSequence )
head = next(head_tail_iter)
item_processing( head )
for item in head_tail_iter:
    # *the between processing*
    item_processing( item )

This is more reliable because it’s slightly easier to prove, It doesn’t create an extra data structure (i.e., a copy of a list) and doesn’t require a lot of wasted execution of an if condition which is always false except once.

If you’re simply looking to modify the last element in data_list then you can simply use the notation:

L[-1]

However, it looks like you’re doing more than that. There is nothing really wrong with your way. I even took a quick glance at some Django code for their template tags and they do basically what you’re doing.

if the items are unique:

for x in list:
    #code
    if x == list[-1]:
        #code

other options:

pos = -1
for x in list:
    pos += 1
    #code
    if pos == len(list) - 1:
        #code


for x in list:
    #code
#code - e.g. print x


if len(list) > 0:
    for x in list[:-1]
        #code
    for x in list[-1]:
        #code

This is similar to Ants Aasma’s approach but without using the itertools module. It’s also a lagging iterator which looks-ahead a single element in the iterator stream:

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

You can use a sliding window over the input data to get a peek at the next value and use a sentinel to detect the last value. This works on any iterable, so you don’t need to know the length beforehand. The pairwise implementation is from 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

Is there no possibility to iterate over all-but the last element, and treat the last one outside of the loop? After all, a loop is created to do something similar to all elements you loop over; if one element needs something special, it shouldn’t be in the loop.

(see also this question: does-the-last-element-in-a-loop-deserve-a-separate-treatment)

EDIT: since the question is more about the “in between”, either the first element is the special one in that it has no predecessor, or the last element is special in that it has no successor.

I like the approach of @ethan-t, but while True is dangerous from my point of view.

data_list = [1, 2, 3, 2, 1]  # sample data
L = list(data_list)  # destroy L instead of data_list
while L:
    e = L.pop(0)
    if L:
        print(f'process element {e}')
    else:
        print(f'process last element {e}')
del L

Here, data_list is so that last element is equal by value to the first one of the list. L can be exchanged with data_list but in this case it results empty after the loop. while True is also possible to use if you check that list is not empty before the processing or the check is not needed (ouch!).

data_list = [1, 2, 3, 2, 1]
if data_list:
    while True:
        e = data_list.pop(0)
        if data_list:
            print(f'process element {e}')
        else:
            print(f'process last element {e}')
            break
else:
    print('list is empty')

The good part is that it is fast. The bad – it is destructible (data_list becomes empty).

Most intuitive solution:

data_list = [1, 2, 3, 2, 1]  # sample data
for i, e in enumerate(data_list):
    if i != len(data_list) - 1:
        print(f'process element {e}')
    else:
        print(f'process last element {e}')

Oh yes, you have already proposed it!

There is nothing wrong with your way, unless you will have 100 000 loops and wants save 100 000 “if” statements. In that case, you can go that way :

iterable = [1,2,3] # Your date
iterator = iter(iterable) # get the data iterator

try :   # wrap all in a try / except
    while 1 : 
        item = iterator.next() 
        print item # put the "for loop" code here
except StopIteration, e : # make the process on the last element here
    print item

Outputs :

1
2
3
3

But really, in your case I feel like it’s overkill.

In any case, you will probably be luckier with slicing :

for item in iterable[:-1] :
    print item
print "last :", iterable[-1]

#outputs
1
2
last : 3

or just :

for item in iterable :
    print item
print iterable[-1]

#outputs
1
2
3
last : 3

Eventually, a KISS way to do you stuff, and that would work with any iterable, including the ones without __len__ :

item = ''
for item in iterable :
    print item
print item

Ouputs:

1
2
3
3

If feel like I would do it that way, seems simple to me.

Use slicing and is to check for the last element:

for data in data_list:
    <code_that_is_done_for_every_element>
    if not data is data_list[-1]:
        <code_that_is_done_between_elements>

Caveat emptor: This only works if all elements in the list are actually different (have different locations in memory). Under the hood, Python may detect equal elements and reuse the same objects for them. For instance, for strings of the same value and common integers.

if you are going through the list, for me this worked too:

for j in range(0, len(Array)):
    if len(Array) - j > 1:
        notLast()

Google brought me to this old question and I think I could add a different approach to this problem.

Most of the answers here would deal with a proper treatment of a for loop control as it was asked, but if the data_list is destructible, I would suggest that you pop the items from the list until you end up with an empty list:

while True:
    element = element_list.pop(0)
    do_this_for_all_elements()
    if not element:
        do_this_only_for_last_element()
        break
    do_this_for_all_elements_but_last()

you could even use while len(element_list) if you don’t need to do anything with the last element. I find this solution more elegant then dealing with next().

For me the most simple and pythonic way to handle a special case at the end of a list is:

for data in data_list[:-1]:
    handle_element(data)
handle_special_element(data_list[-1])

Of course this can also be used to treat the first element in a special way .

Instead of counting up, you can also count down:

  nrToProcess = len(list)
  for s in list:
    s.doStuff()
    nrToProcess -= 1
    if nrToProcess==0:  # this is the last one
      s.doSpecialStuff()

Delay the special handling of the last item until after the loop.

>>> for i in (1, 2, 3):
...     pass
...
>>> i
3

There can be multiple ways. slicing will be fastest. Adding one more which uses .index() method:

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

Assuming input as an iterator, here’s a way using tee and izip from itertools:

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.

Demo:

>>> 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
>>>

The most simple solution coming to my mind is:

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

So we always look ahead one item by delaying the the processing one iteration. To skip doing something during the first iteration I simply catch the error.

Of course you need to think a bit, in order for the NameError to be raised when you want it.

Also keep the `counstruct

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

This relies that the name new wasn’t previously defined. If you are paranoid you can ensure that new doesn’t exist using:

try:
    del new
except NameError:
    pass

Alternatively you can of course also use an if statement (if notfirst: print(new) else: notfirst = True). But as far as I know the overhead is bigger.

Using `timeit` yields:

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

so I expect the overhead to be unelectable.

Count the items once and keep up with the number of items remaining:

remaining = len(data_list)
for data in data_list:
    code_that_is_done_for_every_element

    remaining -= 1
    if remaining:
        code_that_is_done_between_elements

This way you only evaluate the length of the list once. Many of the solutions on this page seem to assume the length is unavailable in advance, but that is not part of your question. If you have the length, use it.

One simple solution that comes to mind would be:

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

A second equally simple solution could be achieved by using a counter:

# 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