How to loop through all but the last item of a list?
PythonListFor LoopPython Problem Overview
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.
Note
freespace answered my actual question, which is why I accepted the answer, but SilentGhost answered the question I should have asked.
Apologies for the confusion.
Python Solutions
Solution 1 - Python
for x in y[:-1]
If y
is a generator, then the above will not work.
Solution 2 - Python
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)
Solution 3 - Python
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])):
Solution 4 - Python
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.
Solution 5 - Python
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
Solution 6 - Python
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 indexstart
*list
or*generator
: passes all elements to the enclosing functionzip
as if it was writtenzip(elem1, elem2, ...)
Note:
AFAIK, this code is as lazy as it can be. Not tested.
Solution 7 - Python
I got you. Suppose this is your list categoricals
, containing the names of the columns you want to iterate over, except the last column ReversedPayment
.
categoricals = ['SeniorCitizen','CollegeDegree','Married','FulltimeJob','ReversedPayment']
Write your for
loop like below and this should give you what you are looking for.
for i in categoricals[:-1]:
print(i)