I need to do some special operation for the last element in a list. Is there any better way than this?
array = [1,2,3,4,5]
for i, val in enumerate(array):
if (i+1) == len(array):
// Process for the last element
else:
// Process for the other element
for item in list[:-1]:
print "Not last: ", item
print "Last: ", list[-1]
If you don't want to make a copy of list, you can make a simple generator:
# itr is short for "iterable" and can be any sequence, iterator, or generator
def notlast(itr):
itr = iter(itr) # ensure we have an iterator
prev = itr.next()
for item in itr:
yield prev
prev = item
# lst is short for "list" and does not shadow the built-in list()
# 'L' is also commonly used for a random list name
lst = range(4)
for x in notlast(lst):
print "Not last: ", x
print "Last: ", lst[-1]
Another definition for notlast:
import itertools
notlast = lambda lst:itertools.islice(lst, 0, len(lst)-1)
list and iter as variable names as they shadow the builtins
If your sequence isn't terribly long then you can just slice it:
for val in array[:-1]:
do_something(val)
else:
do_something_else(array[-1])
using itertools
>>> from itertools import repeat, chain,izip
>>> for val,special in izip(array, chain(repeat(False,len(array)-1),[True])):
... print val, special
...
1 False
2 False
3 False
4 False
5 True
Version of liori's answer to work on any iterable (doesn't require len() or slicing)
def last_flagged(seq):
seq = iter(seq)
a = next(seq)
for b in seq:
yield a, False
a = b
yield a, True
mylist = [1,2,3,4,5]
for item,is_last in last_flagged(mylist):
if is_last:
print "Last: ", item
else:
print "Not last: ", item
izip became the built-in zip so use that instead.If your logic is never breaking out of the loop then a for...else construct might work:
In [1]: count = 0
...: for i in [1, 2, 3]:
...: count +=1
...: print("any item:", i)
...: else:
...: if count:
...: print("last item: ", i)
...:
any item: 1
any item: 2
any item: 3
last item: 3
You need the count variable just in case the iterable is empty, otherwise the variable i won't be defined.
Use more_itertools:
import more_itertools
array = [1,2,3,4,5]
peekable = more_itertools.peekable(array)
last = object()
for val in peekable:
if peekable.peek(last) is last:
print('last', val)
else:
print(val)
Gives:
1
2
3
4
last 5
for i in items:
if i == items[-1]:
print 'The last item is: '+i
items = [1,2,1]
enumerateto do pretty much the same thing in the link you gave. He's asking for a better and more pythonic way to detect the last element, rather than using the approach he listed. Should not be considered a duplicate.