Python: Looping through all but the last item of a list

Question

I would like to loop through a list checking each item against the one following it.

Is there a way I can loop through all but the last item using for x in y? I would prefer to do it without using indexes if I can.

Answer 1

for x in y[:-1]

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

Edit For those people downvoting this answer because you think it doesn't answer the question, read the question carefully. Note the sentence which ends in a question mark, that is the curly thing with a dot below it, like this: ?.

If that does not satisfy you, then perhaps the fact David has accepted this as the answer will.

If that doesn't either, then go right ahead.

David: You are welcome to unaccept this answer so I can delete it. This answer is causing my grief.

Answer 2

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])):

Answer 3

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.

Answer 4

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)

Answer 5

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