Insert item before each element of a list

[moorem...@gmail.com]

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

[Ian Kelly]

Using the "roundrobin" recipe from the itertools documentation:

x = [1, 2, 3]

y = list(roundrobin(itertools.repeat('insertme', len(x)), x))

[MRAB]

Slightly better is:

y = list(itertools.chain.from_iterable(('insertme', i) for i in x))

[Ian Kelly]

On Mon, Oct 8, 2012 at 1:52 PM, Joshua Landau
<joshua.l...@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.

[Agon Hajdari]

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.

y = [item for tup in y for item in tup]

[Peter Otten]

moorem...@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

[Prasad, Ramit]

Agon Hajdari wrote:
> Sent: Monday, October 08, 2012 3:12 PM
> To: pytho...@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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[Agon Hajdari]

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)]

[Prasad, Ramit]

Agon Hajdari wrote:
> On 10/08/2012 11:15 PM, Prasad, Ramit wrote:
> > Agon Hajdari wrote:
> >>
> >> 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]
>

> 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)]

You are absolutely correct. I missed that when I tried it.
Instead of the nested list comprehension, I might have used
map instead.

>>> y = [('insertme', i) for i in x]
>>> z = []
>>> _ = map( z.extend, y )
>>> z
['insertme', 0, 'insertme', 1, 'insertme', 2, 'insertme', 3, 'insertme', 4]

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.

[Paul Rubin]

moorem...@gmail.com writes:
>>>> x = [1, 2, 3] ..

>>>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]

def ix(prefix, x):
for a in x:
yield prefix
yield a

y = list(ix('insertme', x))

================

from itertools import *
y = list(chain.from_iterable(izip(repeat('insertme'), x)))

================

etc.

[Nobody]

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)]

[Alex]

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."

[Terry Reedy]

On 10/8/2012 3:28 PM, moorem...@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)

>>> lst
['insert me', 1, 'insert me', 2, 'insert me', 3]

Paul Rubin's list(gfunc(prefix, lst)) is similar in execution.

--
Terry Jan Reedy

[Roy Smith]

In article <mailman.1976.1349747...@python.org>,

Terry Reedy <tjr...@udel.edu> wrote:

> On 10/8/2012 3:28 PM, moorem...@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.

[rusi]

On Oct 9, 7:06 am, Roy Smith <r...@panix.com> wrote:
> In article <mailman.1976.1349747963.27098.python-l...@python.org>,
>  Terry Reedy <tjre...@udel.edu> 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]

[rusi]

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))

[alex23]

On Oct 9, 12:06 pm, Roy Smith <r...@panix.com> wrote:
> I'm going to go with this one.  I think people tend to over-abuse list
> comprehensions.

I weep whenever I find `_ = [...]` in other people's code.

[Duncan Booth]

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
['insertme', 1, 'insertme', 2, 'insertme', 3]

--
Duncan Booth

[Steven D'Aprano]

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


Then call it like this:

py> list(mux(itertools.repeat("insert me"), range(5)))
['insert me', 0, 'insert me', 1, 'insert me', 2, 'insert me', 3, 'insert
me', 4]



--
Steven

[Peter Otten]

Duncan Booth wrote:

> moorem...@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:

My post on gmane

http://thread.gmane.org/gmane.comp.python.general/718940/focus=718947

apparently didn't make it through to the list.

>>>> x = [1,2,3]
>>>> y = ['insertme']*(2*len(x))
>>>> y[1::2] = x
>>>> y
> ['insertme', 1, 'insertme', 2, 'insertme', 3]

An advantage of this approach -- it is usually much faster.

[moorem...@gmail.com]

On Monday, October 8, 2012 10:06:50 PM UTC-4, Roy Smith wrote:
> In article <mailman.1976.1349747...@python.org>,
>

(big snip)

>
>
> > 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.

Fair enough. I admit I was looking for a tricky one-liner, which rarely leads to good code...I should know better.

Thanks for all the feedback from everyone. It's amazing how much Python one can learn just asking about a small section of code!

[Hans Mulder]

Or, if you don't want to import the operator module:

sum((['insert', a] for a in x), [])

-- HansM

[Terry Reedy]

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.

--
Terry Jan Reedy

[Steven D'Aprano]

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.



--
Steven