Insert item before each element of a list

[Hans Mulder]

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), [])

-- HansM

 

[Terry Reedy]

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.

 

[Steven D'Aprano]

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.