Jason Friedman <jsf80238 <at> gmail.com> writes:
>
> I have a file such as:
>
> $ cat my_data
> Starting a new group
>
> a
> b
> c
> Starting a new group
> 1
> 2
> 3
>
> 4
> Starting a new group
> X
> Y
> Z
> Starting a new group
>
>
> I am wanting a list of lists:
> ['a', 'b', 'c']
>
> ['1', '2', '3', '4']
> ['X', 'Y', 'Z']
> []
>
> I wrote this:
>
> ------------------------------------
> #!/usr/bin/python3
> from itertools import groupby
>
> def get_lines_from_file(file_name):
> with open(file_name) as reader:
> for line in reader.readlines():
> yield(line.strip())
>
> counter = 0
> def key_func(x):
> if x.startswith("Starting a new group"):
> global counter
> counter += 1
> return counter
>
> for key, group in groupby(get_lines_from_file("my_data"), key_func):
> print(list(group)[1:])
> ------------------------------------
>
>
>
>
> I get the output I desire, but I'm wondering if there is a solution
without the global counter.
>
Here's a solution that makes use of groupby (which is a good idea I think),
but avoids the counter (actually this is trivial; you just return the result
of startswith directly). It also provides you with the rest of the separator
line (you're using startswith in your code, so I figured you expect more on
these lines). I replaced the startswith() with slicing though as this is
usually faster.
def separate_on(iterable, separator):
sep_len=len(separator)
grouped_iter = (x[1] for x in groupby(iterable,
lambda line: line[:sep_len] == separator))
for separator_line in grouped_iter:
rest_of_separator_line = next(separator_line)[sep_len:].strip()
yield (rest_of_separator_line,
[s.strip() for s in next(grouped_iter)])
then
for sep_tail, group in separate_on(your_input,your_separator):
do_what_ever()
Hope it's what you want,
Wolfgang
--
http://mail.python.org/mailman/listinfo/python-list
