On 5/30/07, Steve Howell <showell30@yahoo.com> wrote:
--- Steven Bethard <steven.bethard@gmail.com> wrote:
def rowexpr(expr): ... def evaluate(row): ... d = dict(row) ... d.update(globals()) ... exec '__result__ = ' + expr in d ... return d['__result__'] ... return evaluate ... [snip] I tried it out with a slightly more involved expression.
x = rowexpr( 'convert_to_euros(salary) ' 'if dept == "foo" else 0') rows = [dict(dept='foo', salary=i) for i in range(10000)] transform = [x(row) for row in rows]
Here were my findings:
1) Not horribly slow.
3.4 seconds on my box for 10000 calls
2) I introduced a syntax error, and it was more clear than I thought it would be. It happens at runtime, of course, which is less than ideal
You can get the syntax error a little earlier (at the time of the rowexpr() call) by using compile()::
def rowexpr(expr): ... code = compile('__result__ = ' + expr, '<rowexpr>', 'exec') ... def evaluate(row): ... d = dict(globals()) ... d.update(row) ... exec code in d ... return d['__result__'] ... return evaluate ... def convert_to_euros(amt): ... return amt / 2 ... x = rowexpr('convert_to_euros(salary)') row = dict(salary=50000) print x(row) 25000 rowexpr('convert_to_euros(salary) if dept = "foo" else 0') Traceback (most recent call last): File "<interactive input>", line 1, in <module> File "<interactive input>", line 2, in rowexpr File "<rowexpr>", line 1 __result__ = convert_to_euros(salary) if dept = "foo" else 0 ^ SyntaxError: invalid syntax
STeVe -- I'm not *in*-sane. Indeed, I am so far *out* of sane that you appear a tiny blip on the distant coast of sanity. --- Bucky Katt, Get Fuzzy