On Mon, Oct 8, 2012 at 1:28 PM, <mooremath...@gmail.com> wrote:
> What's the best way to accomplish this? Am I over-complicating it? My gut feeling is there is a better way than the following:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in range(len(x))))
>>>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
> I appreciate any and all feedback.
Using the "roundrobin" recipe from the itertools documentation:
x = [1, 2, 3]
y = list(roundrobin(itertools.repeat('insertme', len(x)), x))
On 2012-10-08 20:28, mooremath...@gmail.com wrote:
> What's the best way to accomplish this? Am I over-complicating it? My gut feeling is there is a better way than the following:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in range(len(x))))
>>>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
> I appreciate any and all feedback.
Slightly better is:
y = list(itertools.chain.from_iterable(('insertme', i) for i in x))
<joshua.landau...@gmail.com> wrote:
> But it's not far. I wouldn't use Ian Kelly's method (no offence), because of
> len(x): it's less compatible with iterables. Others have ninja'd me with
> good comments, too.
That's fair, I probably wouldn't use it either. It points to a
possible need for a roundrobin variant that truncates like zip when
one of the iterables runs out. It would have to do some look-ahead,
but it would remove the need for the len(x) restriction on
itertools.repeat.
mooremath...@gmail.com wrote:
> What's the best way to accomplish this? Am I over-complicating it? My
> gut feeling is there is a better way than the following:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in
>>>> range(len(x)))) y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
Less general than chain.from_iterable(izip(repeat("insertme"), x)):
>>> x = [1, 2, 3]
>>> y = 2*len(x)*["insertme"]
>>> y[1::2] = x
>>> y
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> Agon Hajdari wrote:
>> Sent: Monday, October 08, 2012 3:12 PM
>> To: python-l...@python.org
>> Subject: Re: Insert item before each element of a list
>> On 10/08/2012 09:45 PM, Chris Kaynor wrote:
>>> [('insertme', i) for i in x]
>> This is not enough, you have to merge it afterwards.
> Why do you say that? It seems to work just fine for me.
>>>> x
> [0, 1, 2, 3, 4]
>>>> [('insertme', i) for i in x]
> [('insertme', 0), ('insertme', 1), ('insertme', 2), ('insertme', 3), ('insertme', 4)]
>> y = [item for tup in y for item in tup]
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> conditions including on offers for the purchase or sale of
> securities, accuracy and completeness of information, viruses,
> confidentiality, legal privilege, and legal entity disclaimers,
> available at http://www.jpmorgan.com/pages/disclosures/email.
I think he wanted to have a 'plain' list
a = [0, 1, 0, 2, 0, 3]
and not
a = [(0, 1), (0, 2), (0, 3)]
I am not sure which is more Pythonic, but to me map + list.extend tells
me more explicitly that I am dealing with an iterable of iterables.
It might make more sense to only to me though.
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On Mon, 08 Oct 2012 12:28:43 -0700, mooremathewl wrote:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in
>>>> range(len(x)))) y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
>>> [i for j in [1,2,3] for i in ('insertme', j)]
['insertme', 1, 'insertme', 2, 'insertme', 3]
mooremath...@gmail.com wrote:
> What's the best way to accomplish this? Am I over-complicating it?
> My gut feeling is there is a better way than the following:
> >>> import itertools
> >>> x = [1, 2, 3]
> >>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i
> in range(len(x)))) >>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
> I appreciate any and all feedback.
> --Matt
Just like the Zen of Python (http://www.python.org/dev/peps/pep-0020/)
says . . . "There should be at least ten-- and preferably more --clever
and obscure ways to do it."
On 10/8/2012 3:28 PM, mooremath...@gmail.com wrote:
> What's the best way to accomplish this? Am I over-complicating it? My gut feeling is there is a better way than the following:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in range(len(x))))
>>>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
The straightforward, crystal-clear, old-fashioned way
>>> lst = []
>>> for item in [1,2,3]:
lst.append('insert me')
lst.append(item)
> On 10/8/2012 3:28 PM, mooremath...@gmail.com wrote:
> > What's the best way to accomplish this? Am I over-complicating it? My gut > > feeling is there is a better way than the following:
> >>>> import itertools
> >>>> x = [1, 2, 3]
> >>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in > >>>> range(len(x))))
> >>>> y
> > ['insertme', 1, 'insertme', 2, 'insertme', 3]
> The straightforward, crystal-clear, old-fashioned way
> >>> lst = []
> >>> for item in [1,2,3]:
> lst.append('insert me')
> lst.append(item)
I'm going to go with this one. I think people tend to over-abuse list comprehensions. They're a great shorthand for many of the most common use cases, but once you stray from the simple examples, you quickly end up with something totally obscure.
> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in range(len(x))))
A statement ending in four close parens is usually going to be pretty difficult to figure out. This is one where I had to pull out my pencil and start pairing them off manually to figure out how to parse it.
> In article <mailman.1976.1349747963.27098.python-l...@python.org>,
> Terry Reedy <tjre...@udel.edu> wrote:
> > On 10/8/2012 3:28 PM, mooremath...@gmail.com wrote:
> > > What's the best way to accomplish this? Am I over-complicating it? My gut
> > > feeling is there is a better way than the following:
> > >>>> import itertools
> > >>>> x = [1, 2, 3]
> > >>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in
> > >>>> range(len(x))))
> > >>>> y
> > > ['insertme', 1, 'insertme', 2, 'insertme', 3]
> > The straightforward, crystal-clear, old-fashioned way
> > >>> lst = []
> > >>> for item in [1,2,3]:
> > lst.append('insert me')
> > lst.append(item)
> I'm going to go with this one. I think people tend to over-abuse list
> comprehensions. They're a great shorthand for many of the most common
> use cases, but once you stray from the simple examples, you quickly end
> up with something totally obscure.
> > y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in range(len(x))))
> A statement ending in four close parens is usually going to be pretty
> difficult to figure out. This is one where I had to pull out my pencil
> and start pairing them off manually to figure out how to parse it.
How about a 2-paren version?
>>> x = [1,2,3]
>>> reduce(operator.add, [['insert', a] for a in x])
mooremath...@gmail.com wrote:
> What's the best way to accomplish this? Am I over-complicating it? > My gut feeling is there is a better way than the following:
>>>> import itertools
>>>> x = [1, 2, 3]
>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in
>>>> range(len(x)))) y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]
> I appreciate any and all feedback.
Given the myriad of proposed solutions, I'm surprised nobody has suggested good old list slicing:
>>> x = [1,2,3]
>>> y = ['insertme']*(2*len(x))
>>> y[1::2] = x
>>> y
On Mon, 08 Oct 2012 19:34:26 -0700, rusi wrote:
> How about a 2-paren version?
>>>> x = [1,2,3]
>>>> reduce(operator.add, [['insert', a] for a in x])
> ['insert', 1, 'insert', 2, 'insert', 3]
That works, but all those list additions are going to be slow. It will be an O(N**2) algorithm.
If you're going to be frequently interleaving sequences, a helper function is a good solution. Here's my muxer:
def mux(*iterables):
"""Muxer which yields items interleaved from each iterator or
sequence argument, stopping when the first one is exhausted.
>>> list( mux([1,2,3], range(10, 15), "ABCD") )
[1, 10, 'A', 2, 11, 'B', 3, 12, 'C']
"""
for i in itertools.izip(*iterables): # in Python 3 use builtin zip
for item in i:
yield item
>> What's the best way to accomplish this? Am I over-complicating it?
>> My gut feeling is there is a better way than the following:
>>>>> import itertools
>>>>> x = [1, 2, 3]
>>>>> y = list(itertools.chain.from_iterable(('insertme', x[i]) for i in
>>>>> range(len(x)))) y
>> ['insertme', 1, 'insertme', 2, 'insertme', 3]
>> I appreciate any and all feedback.
> Given the myriad of proposed solutions, I'm surprised nobody has suggested
> good old list slicing:
> On 9/10/12 04:39:28, rusi wrote:
>> On Oct 9, 7:34 am, rusi <rustompm...@gmail.com> wrote:
>>> How about a 2-paren version?
>>>>>> x = [1,2,3]
>>>>>> reduce(operator.add, [['insert', a] for a in x])
>>> ['insert', 1, 'insert', 2, 'insert', 3]
>> Or if one prefers the different parens on the other side:
>>>>> reduce(operator.add, (['insert', a] for a in x))
>> ['insert', 1, 'insert', 2, 'insert', 3]
> Or, if you don't want to import the operator module:
> sum((['insert', a] for a in x), [])
All of the solutions based on adding (concatenating) lists create an unneeded temporary list for each addition except the last and run in O(n**2) time. Starting with one list and appending or extending (which does two appends here) is the 'proper' approach to get an O(N) algorithm.
This does not matter for n=3, but for n = 10000 it would.
expanded = []
expand = expand.append
for item in source:
expand('insert')
expand(item)
is hard to beat for clarity and time.
expanded = []
expand = expand.extend
for item in source:
expand(['insert', item])
might be faster if creating the list is faster than the second expand call. Note that a typical lisp-like version would recursively traverse source to nil and build expanded from tail to head by using the equivalent of
return ['insert' item].extend(expanded)
Extend would be O(1) here also since it would at worst scan the new list of length 2 for each of the items in the source.
def interleave(source):
for item in source:
yield 'insert'
yield item
list(interleave(source))
might also be faster since it avoids the repeated python level call. I prefer it anyway as modern, idiomatic python in that it separates interleaving from creating a list. In many situations, creating a list from the interleaved stream will not be needed.
On Fri, 12 Oct 2012 00:21:57 +0200, Hans Mulder wrote:
> On 9/10/12 04:39:28, rusi wrote:
>> On Oct 9, 7:34 am, rusi <rustompm...@gmail.com> wrote:
>>> How about a 2-paren version?
>>>>>> x = [1,2,3]
>>>>>> reduce(operator.add, [['insert', a] for a in x])
>>> ['insert', 1, 'insert', 2, 'insert', 3]
>> Or if one prefers the different parens on the other side:
>>>>> reduce(operator.add, (['insert', a] for a in x))
>> ['insert', 1, 'insert', 2, 'insert', 3]
> Or, if you don't want to import the operator module:
> sum((['insert', a] for a in x), [])
Which is also O(N**2) like the reduce solution above.
That means that it will seem perfectly fine when you test it using a a hundred or so items, then some day you'll pass it a list with ten million items and it will take 36 hours to complete.
I'm serious by the way. By my tests, increasing the number of items in the list by a factor of ten increases the time taken by between 30 and 300 times.
