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Re: itertools.groupby

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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():

readlines() slurps the whole file into memory! Don't do that, iterate over 
the file directly instead:

          for line in reader:

>             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.

If you were to drop the empty groups you could simplify it to

def is_header(line):
    return line.startswith("Starting a new group")

with open("my_data") as lines:
    stripped_lines = (line.strip() for line in lines)
    for header, group in itertools.groupby(stripped_lines, key=is_header):
        if not header:
            print(list(group))

And here's a refactoring for your initial code. The main point is the use of 
nonlocal instead of global state to make the function reentrant.

def split_groups(items, header):
    odd = True
    def group_key(item):
        nonlocal odd
        if header(item):
            odd = not odd
        return odd

    for _key, group in itertools.groupby(items, key=group_key):
        yield itertools.islice(group, 1, None)
    
def is_header(line):
    return line.startswith("Starting a new group")

with open("my_data") as lines:
    stripped_lines = map(str.strip, lines)
    for group in split_groups(stripped_lines, header=is_header):
        print(list(group))

One remaining problem with that code is that it will silently drop the first 
line of the file if it doesn't start with a header:

$ cat my_data 
alpha
beta
gamma
Starting a new group
a
b
c
Starting a new group
Starting a new group
1
2
3
4
Starting a new group
X
Y
Z
Starting a new group
$ python3 group.py 
['beta', 'gamma'] # where's alpha?
['a', 'b', 'c']
[]
['1', '2', '3', '4']
['X', 'Y', 'Z']
[]

How do you want to handle that case?

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Re: itertools.groupby Peter Otten <__peter__@web.de> - 2013-04-21 10:28 +0200

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