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Re: itertools.groupby

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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

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Re: itertools.groupby Wolfgang Maier <wolfgang.maier@biologie.uni-freiburg.de> - 2013-04-20 21:14 +0000

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