How do I efficiently filter computed values within

Question

The Python list comprehension syntax makes it easy to filter values within a comprehension. For example:

result = [x**2 for x in mylist if type(x) is int]

Will return a list of the squares of integers in mylist. However, what if the test involves some (costly) computation and you want to filter on the result? One option is:

result = [expensive(x) for x in mylist if expensive(x)]

This will result in a list of non-"false" expensive(x) values, however expensive() is called twice for each x. Is there a comprehension syntax that allows you to do this test while only calling expensive once per x?

Answer 1

The most obvious (and I would argue most readable) answer is to not use a list comprehension or generator expression, but rather a real generator:

def gen_expensive(mylist):
    for item in mylist:
        result = expensive(item)
        if result:
            yield result

It takes more horizontal space, but it's much easier to see what it does at a glance, and you end up not repeating yourself.

Answer 2

You could memoize expensive(x) (and if you are calling expensive(x) frequently, you probably should memoize it any way. This page gives an implementation of memoize for python:

http://code.activestate.com/recipes/52201/

This has the added benefit that expensive(x) may be run less than N times, since any duplicate entries will make use of the memo from the previous execution.

Note that this assumes expensive(x) is a true function, and does not depend on external state that may change. If expensive(x) does depend on external state, and you can detect when that state changes, or you know it wont change during your list comprehension, then you can reset the memos before the comprehension.

Answer 3

If the calculations are already nicely bundled into functions, how about using filter and map?

result = filter (None, map (expensive, mylist))

You can use itertools.imap if the list is very large.

Answer 4

This is exactly what generators are suited to handle:

result = (expensive(x) for x in mylist)
result = (do_something(x) for x in result if some_condition(x))
...
result = [x for x in result if x]  # finally, a list

1. This makes it totally clear what is happening during each stage of the pipeline.
2. Explicit over implicit
3. Uses generators everywhere until the final step, so no large intermediate lists

cf: 'Generator Tricks for System Programmers' by David Beazley

Answer 5

Came up with my own answer after a minute of thought. It can be done with nested comprehensions:

result = [y for y in (expensive(x) for x in mylist) if y]

I guess that works, though I find nested comprehensions are only marginally readable

Answer 6

I will have a preference for:

itertools.ifilter(bool, (expensive(x) for x in mylist))

This has the advantage to:

• avoid None as the function (will be eliminated in Python 3): http://bugs.python.org/issue2186
• use only iterators.