Skip first entry in for loop in python?

Question

In python, How do I do something like:

for car in cars:
   # Skip first and last, do work for rest

Answer 1

The best way to skip the first item(s) is:

from itertools import islice
for car in islice(cars, 1, None):
    # do something

islice in this case is invoked with a start-point of 1, and an end point of None, signifying the end of the iterator.

To be able to skip items from the end of an iterable, you need to know its length (always possible for a list, but not necessarily for everything you can iterate on). for example, islice(cars, 1, len(cars)-1) will skip the first and last items in the cars list.

Answer 2

I do it like this, even though it looks like a hack it works every time:

ls_of_things = ['apple', 'car', 'truck', 'bike', 'banana']
first = 0
last = len(ls_of_things)
for items in ls_of_things:
    if first == 0
        first = first + 1
        pass
    elif first == last - 1:
        break
    else:
        do_stuff
        first = first + 1
        pass

Answer 3

for item in do_not_use_list_as_a_name[1:-1]:
    #...do whatever

Answer 4

Well, your syntax isn't really Python to begin with.

Iterations in Python are over he contents of containers (well, technically it's over iterators), with a syntax for item in container. In this case, the container is the cars list, but you want to skip the first and last elements, so that means cars[1:-1] (python lists are zero-based, negative numbers count from the end, and : is slicing syntax.

So you want

for c in cars[1:-1]:
    do something with c

Answer 5

Here's my preferred choice. It doesn't require adding on much to the loop, and uses nothing but built in tools.

Go from:

for item in my_items:
  do_something(item)

to:

for i, item in enumerate(my_items):
  if i == 0:
    continue
  do_something(item)

Answer 6

Based on @SvenMarnach 's Answer, but bit simpler and without using deque

>>> def skip(iterable, at_start=0, at_end=0):
    it = iter(iterable)
    it = itertools.islice(it, at_start, None)
    it, it1 = itertools.tee(it)
    it1 = itertools.islice(it1, at_end, None)
    return (next(it) for _ in it1)

>>> list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
>>> list(skip(range(10), at_start=2, at_end=5))
[2, 3, 4]

Also Note, based on my timeit result, this is marginally faster than the deque solution

>>> iterable=xrange(1000)
>>> stmt1="""
def skip(iterable, at_start=0, at_end=0):
    it = iter(iterable)
    it = itertools.islice(it, at_start, None)
    it, it1 = itertools.tee(it)
    it1 = itertools.islice(it1, at_end, None)
    return (next(it) for _ in it1)
list(skip(iterable,2,2))
    """
>>> stmt2="""
def skip(iterable, at_start=0, at_end=0):
    it = iter(iterable)
    for x in itertools.islice(it, at_start):
        pass
    queue = collections.deque(itertools.islice(it, at_end))
    for x in it:
        queue.append(x)
        yield queue.popleft()
list(skip(iterable,2,2))
        """
>>> timeit.timeit(stmt = stmt1, setup='from __main__ import iterable, skip, itertools', number = 10000)
2.0313770640908047
>>> timeit.timeit(stmt = stmt2, setup='from __main__ import iterable, skip, itertools, collections', number = 10000)
2.9903135454296716