Insert item before each element of a list

[Terry Reedy]

On 10/11/2012 6:21 PM, Hans Mulder wrote:
> On 9/10/12 04:39:28, rusi wrote:
>> On Oct 9, 7:34 am, rusi <rustompm... at 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.

-- 
Terry Jan Reedy