How to loop through all but the last item of a list?

Question

for x in y[:-1]

If y is a generator, then the above will not work.

the easiest way to compare the sequence item with the following:

for i, j in zip(a, a[1:]):
     # compare i (the current) to j (the following)

If you want to get all the elements in the sequence pair wise, use this approach (the pairwise function is from the examples in the itertools module).

from itertools import tee, izip, chain

def pairwise(seq):
    a,b = tee(seq)
    b.next()
    return izip(a,b)

for current_item, next_item in pairwise(y):
    if compare(current_item, next_item):
        # do what you have to do

If you need to compare the last value to some special value, chain that value to the end

for current, next_item in pairwise(chain(y, [None])):

if you meant comparing nth item with n+1 th item in the list you could also do with

>>> for i in range(len(list[:-1])):
...     print list[i]>list[i+1]

note there is no hard coding going on there. This should be ok unless you feel otherwise.

To compare each item with the next one in an iterator without instantiating a list:

import itertools
it = (x for x in range(10))
data1, data2 = itertools.tee(it)
data2.next()
for a, b in itertools.izip(data1, data2):
  print a, b

This answers what the OP should have asked, i.e. traverse a list comparing consecutive elements (excellent SilentGhost answer), yet generalized for any group (n-gram): 2, 3, ... n:

zip(*(l[start:] for start in range(0, n)))

Examples:

l = range(0, 4)  # [0, 1, 2, 3]

list(zip(*(l[start:] for start in range(0, 2)))) # == [(0, 1), (1, 2), (2, 3)]
list(zip(*(l[start:] for start in range(0, 3)))) # == [(0, 1, 2), (1, 2, 3)]
list(zip(*(l[start:] for start in range(0, 4)))) # == [(0, 1, 2, 3)]
list(zip(*(l[start:] for start in range(0, 5)))) # == []

Explanations:

• l[start:] generates a a list/generator starting from index start
• *list or *generator: passes all elements to the enclosing function zip as if it was written zip(elem1, elem2, ...)

Note:

AFAIK, this code is as lazy as it can be. Not tested.