Just as a quck note, I thought I'd post the cleanest python I could come up with to solve the roman numerals problem I discussed earlier. It tries to use a functional style while actually avoiding recursion. To do so, I wrote an iterative python unfold:

```
def unfold(f, x):
res = []
while 1:
try:
w, x = f(x)
res.append(w)
except TypeError:
return res
```

And then the answer becomes:

```
numerals = [("M", 1000), ("CM", 900), ("D", 500), ("CD", 400),
("C", 100), ("XC", 90), ("L", 50), ("XL", 40),
("X", 10), ("IX", 9), ("V", 5), ("IV", 4),
("I", 1)]
def next(x):
for n in numerals:
if n[1] <= x: return (n[0], x-n[1])
def romanize(n):
return "".join(unfold(next, n))
```

Compare to the haskell I posted before:

```
romanize = concat . unfoldr next
where next 0 = Nothing
next x = Just $ second (x-) $ head $ filter ((<=x) . snd) numerals
numerals = [("M", 1000), ("CM", 900), ("D", 500), ("CD", 400),
("C", 100), ("XC", 90), ("L", 50), ("XL", 40),
("X", 10), ("IX", 9), ("V", 5), ("IV", 4),
("I", 1)]
```

The "where" idiom allows you to group helper functions and constants underneath the main function, and the unimportance of function order means you can highlight the high-level logic. The haskell code is much shorter and tighter than the python.

Having powerful iterative functions like unfoldr in the standard library is another clear win for Haskell; it took me more lines to define unfold than it did to define both next() and romanize(). (By the by, some googling failed to turn up a previous python implementation of unfold(); I think this code gets a lot uglier without it. I also don't see any obvious way to do it with itertools.)

On the other hand, I find the simplicity of the python next() appealing, with its constructive instead of declarative approach. It's a less generic solution that's intuitively simpler (for me, still an imperative thinker).

You could match the haskell more exactly:

```
from itertools import ifilter
def second(f, (a, b)): return (a, f(b))
def next(x):
return second(lambda a: x-a, ifilter(lambda (y, z): z <= x, numerals).next()
```

but the result is pretty ugly. The inability to compose functions compactly results in a lot of boilerplate, and in having to pick a lot of meaningless variable names that obscure what's going on. We could move the lambdas out of the call and give them names:

```
def next(x):
subx = lambda a: x-a
ltx = lambda (y, z): z <= x
return second(subx, ifilter(ltx, numerals).next())
```

And it's a little better, but I think we're now pretty far from idiomatic python.

The python translation of the haskell isn't bad, but I do think it loses something. The unfold() trick works really well, and translates easily into an imperative, pythonic implementation; I'll look for ways to use it in the future. And, finally, I'm pretty jealous of Haskell's function combination abilities.

In the comments at reddit, nostrademons posts a nicer unfold function:

```
def unfold(f, x):
while True:
w, x = f(x)
yield w
```

And Kay Schluehr prefers the iterative solution:

```
numerals = (("M", 1000), ("CM", 900), ("D", 500), ("CD", 400),
("C", 100),("XC", 90),("L", 50),("XL", 40), ("X", 10), ("IX", 9), ("V", 5),
("IV", 4), ("I", 1))
def romanize(n):
roman = []
for ltr, num in numerals:
(k,n) = divmod(n, num)
roman.append(ltr*k)
return "".join(roman)
```

David Pollak contributes a Scala unfold and romanize at his blog.