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Re: itertools.groupby

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On 4/20/2013 1:09 PM, Jason Friedman wrote:
> 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.
>
>
>

def separate_on(lines, separator):
     group = None
     for line in lines:
         if line.strip() == separator:
             if group is not None:
                 yield group
             group = []
         else:
             assert group is not None # Should have gotten a separator first
             group.append(line)
     yield group

with open("my_data") as my_data:
     for group in separate_on(my_data, "Starting a new group"):
         print group


The handling of the first separator line feels delicate to me, but this 
provides the output you want.

--Ned.

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Re: itertools.groupby Ned Batchelder <ned@nedbatchelder.com> - 2013-04-20 13:27 -0400

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