Recursive list comprehension in Python?

Actually you can! This example with an explanation hopefully will illustrate how.

define recursive example to get a number only when it is 5 or more and if it isn't, increment it and call the 'check' function again. Repeat this process until it reaches 5 at which point return 5.

print [ (lambda f,v: v >= 5 and v or f(f,v+1))(lambda g,i: i >= 5 and i or g(g,i+1),i) for i in [1,2,3,4,5,6] ]

result:

[5, 5, 5, 5, 5, 6]
>>> 

essentially the two anonymous functions interact in this way:

let f(g,x) = {  
                 expression, terminal condition
                 g(g,x), non-terminal condition
             }

let g(f,x) = {  
                 expression, terminal condition
                 f(f,x), non-terminal condition
             }

make g,f the 'same' function except that in one or both add a clause where the parameter is modified so as to cause the terminal condition to be reached and then go f(g,x) in this way g becomes a copy of f making it like:

f(g,x) = {  
                 expression, terminal condition
                 {
                    expression, terminal condition,
                    g(g,x), non-terminal codition
                 }, non-terminal condition
             }

You need to do this because you can't access the the anonymous function itself upon being executed.

i.e

(lambda f,v: somehow call the function again inside itself )(_,_)

so in this example let A = the first function and B the second. We call A passing B as f and i as v. Now as B is essentially a copy of A and it's a parameter that has been passed you can now call B which is like calling A.

This generates the factorials in a list

print [ (lambda f,v: v == 0 and 1 or v*f(f,v-1))(lambda g,i: i == 0 and 1 or i*g(g,i-1),i) for i in [1,2,3,5,6,7] ]

[1, 2, 6, 120, 720, 5040]
>>> 

No, there's no (documented, solid, stable, ...;-) way to refer to "the current comprehension". You could just use a loop:

res = []
for x in nums:
  if x not in res:
    res.append(x)

of course this is very costly (O(N squared)), so you can optimize it with an auxiliary set (I'm assuming that keeping the order of items in res congruent to that of the items in nums, otherwise set(nums) would do you;-)...:

res = []
aux = set()
for x in nums:
  if x not in aux:
    res.append(x)
    aux.add(x)

this is enormously faster for very long lists (O(N) instead of N squared).

Edit: in Python 2.5 or 2.6, vars()['_[1]'] might actually work in the role you want for self (for a non-nested listcomp)... which is why I qualified my statement by clarifying there's no documented, solid, stable way to access "the list being built up" -- that peculiar, undocumented "name" '_[1]' (deliberately chosen not to be a valid identifier;-) is the apex of "implementation artifacts" and any code relying on it deserves to be put out of its misery;-).