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Split a list into two parts based on a filter in python

0 votes
178 views

I have a list, songs, which I want to divide into two groups.
Essentially, I want:

new_songs = [s for s in songs if s.is_new()]
old_songs = [s for s in songs if not s.is_new()]

but I don't want to make two passes over the list. I could do:

new_songs = []
old_songs = []
for s in songs:
 if s.is_new():
 new_songs.append(s)
 else:
 old_songs.append(s)

Which works, but is klunky compared to the two-liner above. This seems like a common enough thing that I was expecting to find something in itertools which did this. I'm thinking something along the lines of:

matches, non_matches = isplit(lambda s: s.is_new, songs)

Does such a thing exist?

posted Jun 10, 2013 by anonymous

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2 Answers

+1 vote

You could do something like:

new_songs, old_songs = [], []
[(new_songs if s.is_new() else old_songs).append(s) for s in songs]

But I'm not sure that that's any better than the long version.

answer Jun 10, 2013 by anonymous
+1 vote

itertools.groupby() is kinda similar, but unfortunately doesn't fit the bill due to its sorting requirement.
There is regrettably no itertools.partition(). And given how dead-set Raymond seems to be against adding things to the itertools module, there will likely never be.
Maybe more-itertools ( https://pypi.python.org/pypi/more-itertools )
would accept a patch?

answer Jun 10, 2013 by anonymous
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+3 votes

Input:
[1 7 15 29 11 9]

Output:
[9 15] [1 7 11 29]

Average of first part: (15+9)/2 = 12,
Average of second part: (1 + 7 + 11 + 29) / 4 = 12

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