Question

I’ve spent hours trying to debug this code. I want to get the second-to-last element of a list.

for x, y in itertools.groupby(range(0,10), lambda x: int(x / 3)):
    print("the group's key is %d and values are %s" % (x, ','.join(map(str,y))))

temp = itertools.groupby(range(0,10), lambda x: int(x / 3))
the_last_one = None
second_to_last = None
for x,y in temp:
    second_to_last = the_last_one
    the_last_one = y
print(next(iter(second_to_last)))

The output of the first part, for demonstration, is:

the group's key is 0 and values are 0,1,2
the group's key is 1 and values are 3,4,5
the group's key is 2 and values are 6,7,8
the group's key is 3 and values are 9

The goal in the second part is to output the first element in the second-to-last group. I expect 6 but instead, I get the exception StopIteration. If I change the last line to:

print(next(the_last_one))

then I get the expected result of 9. Using a list of tuples with the same structure as the output of groupby works, too. This code only fails on iterators.

Answer 1

According to the documentation on itertools.groupby:

The returned group is itself an iterator that shares the underlying
iterable with groupby(). Because the source is shared, when the
groupby() object is advanced, the previous group is no longer visible.
So, if that data is needed later, it should be stored as a list:

This means that the iterable is being consumed the first time you iterate through it.

Change

for x,y in temp:
    second_to_last = the_last_one
    the_last_one = y

To

for x,y in temp:
    second_to_last = the_last_one
    the_last_one = list(y)

to store the values as they are iterated over.