Feature Requests item #1185121, was opened at 2005-04-18 07:11
Message generated for change (Comment added) made by rhettinger
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=355470&aid=1185121&group_id=5470
Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Jurjen N.E. Bos (jneb)
Assigned to: Raymond Hettinger (rhettinger)
Summary: itertools.imerge: merge sequences
Initial Comment:
(For the itertools library, so Python 2.2 and up)
This is a suggested addition to itertools, proposed name imerge.
usage: imerge(seq0, seq1, ..., [key=<key function>])
result: imerge assumes the sequences are all in sorted order, and
produces a iterator that returns pairs of the form (value, index),
where value is a value of one of the sequences, and index is the
index number of the given sequence.
The output the imerge is in sorted order (taking into account the
key function), so that identical values in the sequences will be
produced from left to right.
The code is surprisingly short, making use of the builtin heap
module.
(You may disagree with my style of argument handling; feel free to
optimize it.)
def imerge(*iterlist, **key):
"""Merge a sequence of sorted iterables.
Returns pairs [value, index] where each value comes from
iterlist[index], and the pairs are sorted
if each of the iterators is sorted.
Hint use groupby(imerge(...), operator.itemgetter(0)) to get
the items one by one.
"""
if key.keys() not in ([], ["key"]): raise TypeError, "Excess
keyword arguments for imerge"
key = key.get("key", lambda x:x)
from heapq import heapreplace, heappop
#initialize the heap containing (inited, value, index,
currentItem, iterator)
#this automatically makes sure all iterators are initialized,
then run, and finally emptied
heap = [(False, None, index, None, iter(iterator)) for index,
iterator in enumerate(iterlist)]
while heap:
inited, item, index, value, iterator = heap[0]
if inited: yield value, index
try: item = iterator.next()
except StopIteration: heappop(heap)
else: heapreplace(heap, (True, key(item), index, item,
iterator))
If you find this little routine worth its size, please put it into
itertools.
- Jurjen
----------------------------------------------------------------------
>Comment By: Raymond Hettinger (rhettinger)
Date: 2005-04-19 13:58
Message:
Logged In: YES
user_id=80475
For your specific application, it is better to use sorted().
When the underlying data consists of long runs of
previously ordered data, sorted() will take advantage of
that ordering and run in O(n) time. In contrast, using a
heap will unnecessarily introduce O(n log n) behavior and
not exploit the underlying data order.
Recommend that you close this request. This discussion thus
far confirms the original conclusion that imerge() use cases
are dominated by sorted(chain(*iterlist)) which gives code
that is shorter, faster, and easier to understand.
----------------------------------------------------------------------
Comment By: Jurjen N.E. Bos (jneb)
Date: 2005-04-19 03:19
Message:
Logged In: YES
user_id=446428
Well, I was optimizing a piece of code with reasonbly long sorted lists (in
memory, I agree) that were modified in all kinds of ways. I did not want
the nlogn behaviour of sort, so I started writing a merge routine.
I found out that the boundary cases of a merge implementation are a
mess, until I disccovered the heap trick. Then I decided to clean it up
and and put it up for a library routine.
The fact that it uses iterators is obnly to make it more general, not
specifically for the "lazy" properties.
- Jurjen
----------------------------------------------------------------------
Comment By: Raymond Hettinger (rhettinger)
Date: 2005-04-18 17:43
Message:
Logged In: YES
user_id=80475
I had previously looked at an imerge() utility and found
that it had only a single application (isomorphic to lazy
mergesorting) and that the use cases were dominated by the
in-memory alternative: sorted(chain(*iterlist)).
Short of writing an external mergesort, what applications
did you have in mind? What situations have you encountered
where you have multiple sources of sorted data being
generated on the fly (as opposed to already being
in-memory), have needed one element at a time sequential
access to a combined sort of that data, needed that combined
sort only once, and could not afford to have the dataset
in-memory?
----------------------------------------------------------------------
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=355470&aid=1185121&group_id=5470
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
