On Sat, 20 Jun 2009 09:45:58 am Lie Ryan wrote:
How about this syntax which would solve your concern for the semantic change:
[x**x as F for x in lst if F() < 100]
Let's look at what that would be equivalent to. L = [] for x in lst: F = lambda x=x: x**x # Need to use default value in the # lambda otherwise all elements will have the same value. tmp = F() if tmp < 100: L.append(tmp) It's not clear why you think this is an improvement over: [x**x as F for x in lst if F < 100] # note the missing ()s which would be equivalent to: L = [] for x in lst: F = x**x if F < 100: L.append(F) Despite what you say here:
The advantage of F being callable is that it does not need semantic change, the filtering part will be done before expression just like it is right now.
That's not true. It can't be true. If you want to filter on x**x being greater than 100, you need to calculate x**x first. In theory, a sufficiently clever compiler could recognise that, say, x**x < 100 implies 0 <= x < 3.59728 (approx), but in general, you can't predict the value of f(x) without actually calculating f(x). What happens if you accidentally forget to put brackets after the F expression? Do you get a syntax error? Undefined behaviour? A runtime exception? -- Steven D'Aprano