Transforming nested Python loops into list comprehensions

I've started working on some Project Euler problems, and have solved number 4 with a simple brute force solution:

def mprods(a,b):
 c = range(a,b)
 f = []
 for d in c:
  for e in c:
   f.append(d*e)
 return f

max([z for z in mprods(100,1000) if str(z)==(''.join([str(z)[-i] for i in range(1,len(str(z))+1)]))])

After solving, I tried to make it as compact as possible, and came up with that horrible bottom line!

Not to leave something half-done, I am trying to condense the mprods function into a list comprehension. So far, I've come up with these attempts:

• [d*e for d,e in (range(a,b), range(a,b))]

  Obviously completely on the wrong track. :-)

• [d*e for x in [e for e in range(1,5)] for d in range(1,5)]

  This gives me [4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16], where I expect [1, 2, 3, 4, 2, 4, 6, 8, 3, 6, 9, 12, 4, 8, 12, 16]开发者_如何学运维 or similar.

Any Pythonistas out there that can help? :)

c = range(a, b)
print [d * e for d in c for e in c]

from itertools import product

def palindrome(i):
  return str(i) == str(i)[::-1]

x = xrange(900,1000)

max(a*b for (a,b) in (product(x,x)) if palindrome(a*b))

• xrange(900,1000) is like range(900,1000) but instead of returning a list it returns an object that generates the numbers in the range on demand. For looping, this is slightly faster than range() and more memory efficient.

• product(xrange(900,1000),xrange(900,1000)) gives the Cartesian product of the input iterables. It is equivalent to nested for-loops. For example, product(A, B) returns the same as: ((x,y) for x in A for y in B). The leftmost iterators are in the outermost for-loop, so the output tuples cycle in a manner similar to an odometer (with the rightmost element changing on every iteration).

  product('ab', range(3)) --> ('a',0) ('a',1) ('a',2) ('b',0) ('b',1) ('b',2) product((0,1), (0,1), (0,1)) --> (0,0,0) (0,0,1) (0,1,0) (0,1,1) (1,0,0) ...

• str(i)[::-1] is list slicing shorthand to reverse a list.

• Note how everything is wrapped in a generator expression, a high performance, memory efficient generalization of list comprehensions and generators.

• Also note that the largest palindrome made from the product of two 2-digit numbers is made from the numbers 91 99, two numbers in the range(90,100). Extrapolating to 3-digit numbers you can use range(900,1000).

I think you'll like this one-liner (formatted for readability):

max(z for z in (d*e
                for d in xrange(100, 1000)
                for e in xrange(100, 1000))
            if str(z) == str(z)[::-1])

Or slightly changed:

c = range(100, 1000)
max(z for z in (d*e for d in c for e in c) if str(z) == str(z)[::-1])

Wonder how many parens that would be in Lisp...