How to refer to the whole array within reduce()?

Question

What I have:

array = original_array[:]
result = reduce(lambda a,b: some_function(b,array), array)

What I want:

I want to get rid of the array = original_array[:] statement. Ideally I would simply replace the array parameter inside reduce() with original_array[:], but I need it inside lambda as well. Is there a way to refer to the array parameter from within lambda?

The following is not an acceptable solution, because it makes a new array copy for every element:

result = reduce(lambda a,b: some_function(b,original_array[:]), original_array[:])

I need something like this:

result = reduce(lambda a,b: some_function(b,reduce_parameter), original_array[:])

Answer 1

You could wrap the whole thing in another lambda:

result = (lambda array: reduce(lambda a,b: some_function(b,array), array))(original_array[:])

But your original solution is in my opinion preferable because it's more readable.

Answer 2

Here's a way to remove that outer lambda

result = reduce(lambda a,b,array=array[:]: some_function(b,array), array)

edit: Whoops, misread the question

This of course assumes you actually need to copy the array, and that it isn't sufficient to use

result = reduce(lambda a,b: some_function(b, array), array)

Also, this is an incorrect use of reduce - you're not using the a argument, so result holds some_function(array[-1], array)

Answer 3

Try:

result = reduce(lambda a,b, array=array: some_function(b,array), array)