Possible to capture the returned value from a Python list comprehension for use a condition?

Question

I want to construct a value in a list comprehension, but also filter on that value. For example:

[expensive_function(x) for x in generator where expensive_function(x) < 5]

I want to avoid calling expensive_function twice per iteration.

The generator may return an infinite series, and list comprehensions aren't lazily evaluated. So this wouldn't work:

[y in [expensive_function(x) for x in generator where expensive_function(x)] where y < 5]

I could write this another way, but it feels right for a list comprehension and I'm sure this is a common usage pattern (possible or not!).

Answer 1

If generator may be infinite, you do not want to use a list comprehension. And not everything has to be a one-liner.

def filtered_gen(gen):
    for item in gen:
        result = expensive_function(item)
        if result < 5:
            yield result

Answer 2

I'm going to answer the part of the question about how to capture intermediate results in a list comprehension for use in a condition, and ignore the question of a list comprehension built from an infinite generator (which obviously isn't going to work), just in case anyone looking for an answer to the question in the title comes here.

So, you have a list comprehension like this:

[expensive_function(x) for x in xrange(5) if expensive_function(x) % 2 == 0]

And you want to avoid calculating expensive_function twice when it passes your filter. Languages with more expressive comprehension syntax (Scala, Haskell, etc) allow you to simply assign names to expressions calculated from comprehension variables, which lets you do things like the following:

# NOT REAL PYTHON
[result for x in xrange(5) for result = expensive_function(x) if result % 2 == 0]

But you can easily emulate this by turning the assignment result = expensive_function(x) into another for iteration over a sequence of one element:

[result for x in xrange(5) for result in (expensive_function(x),) if result % 2 == 0]

And the proof:

>>> def expensive_function(x):
        print 'expensive_function({})'.format(x)
        return x + 10
>>> [expensive_function(x) for x in xrange(5) if expensive_function(x) % 2 == 0]
expensive_function(0)
expensive_function(0)
expensive_function(1)
expensive_function(2)
expensive_function(2)
expensive_function(3)
expensive_function(4)
expensive_function(4)
[10, 12, 14]
>>> [result for x in xrange(5) for result in (expensive_function(x),) if result % 2 == 0]
expensive_function(0)
expensive_function(1)
expensive_function(2)
expensive_function(3)
expensive_function(4)
[10, 12, 14]