for loop, how can I loop through all except the last item in a list? I would like to loop through a list checking each item against the one following it. Can I do this without using indices?
for x in y[:-1]
y is a generator, then the above will not work.
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.
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)
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, …
zip(*(l[start:] for start in range(0, n)))
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)))) # == 
l[start:]generates a a list/generator starting from index
*generator: passes all elements to the enclosing function
zipas if it was written
zip(elem1, elem2, ...)
AFAIK, this code is as lazy as it can be. Not tested.