[python] Skip first entry in for loop in python?

In python, How do I do something like:

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

If cars is a sequence you can just do

for car in cars[1:-1]:
    pass

An alternative method:

for idx, car in enumerate(cars):
    # Skip first line.
    if not idx:
        continue
    # Skip last line.
    if idx + 1 == len(cars):
        continue
    # Real code here.
    print car

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

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)

The other answers only work for a sequence.

For any iterable, to skip the first item:

itercars = iter(cars)
next(itercars)
for car in itercars:
    # do work

If you want to skip the last, you could do:

itercars = iter(cars)
# add 'next(itercars)' here if you also want to skip the first
prev = next(itercars)
for car in itercars:
    # do work on 'prev' not 'car'
    # at end of loop:
    prev = car
# now you can do whatever you want to do to the last one on 'prev'

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.

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

The more_itertools project extends itertools.islice to handle negative indices.

Example

import more_itertools as mit

iterable = 'ABCDEFGH'
list(mit.islice_extended(iterable, 1, -1))
# Out: ['B', 'C', 'D', 'E', 'F', 'G']

Therefore, you can elegantly apply it slice elements between the first and last items of an iterable:

for car in mit.islice_extended(cars, 1, -1):
    # do something

Here is a more general generator function that skips any number of items from the beginning and end of an iterable:

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()

Example usage:

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

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

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

Example:

mylist=['one','two','three','four','five']
for i in mylist[1:]:
   print(i)

In python index start from 0, We can use slicing operator to make manipulations in iteration.

for i in range(1,-1):