Tadeusz Pietraszek

© 2004-2007 Tadeusz Pietraszek

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Implementing ?: in Python

Python (at least up to version 2.4) doesn’t have a much needed ?: operator. Here’s how you can hack it yourself (I’m not sure if I would use it, except for very special situations, though).

# if-then-else function taking functions as arguments.
def ite(condition, true, false):
    if condition:
        return true()
    else:
        return false()

# Call it like this.
ite(condition, lambda:arg_true, lambda:arg_false);


# Just to check if it really works
def fun(arg):
    print "Evaluated "+str(arg)
    return arg

print ite(0, lambda:fun(1), lambda:fun(0));
print ite(1, lambda:fun(1), lambda:fun(0));

This entry was posted on Friday, November 24th, 2006 at 5:42 pm and is filed under Tips&Tricks. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Implementing ?: in Python”

  1. James Says:
    November 28th, 2006 at 11:12 pm

    Fun to note that

    print ite(0, lambda:fun(fun(1)), lambda:fun(fun(0))); print ite(1, lambda:fun(fun(1)), lambda:fun(fun(0)));

    is also lazy. For that matter…

    #!/usr/bin/ruby

    def ife(condition,iftrue,iffalse) if condition return iftrue.call else return iffalse.call end end

    def debug(x) puts “Debug “+x.to_s return x end

    ife(true, lambda {debug(debug(true))},lambda{debug(debug(false))}) ife(false, lambda {debug(debug(true))},lambda{debug(debug(false))})

    works.

    Or the less sugary (no need for lambda wrapper) and generalized version in scheme:

    (define (debug x) (let () (display “debug: “) (display x) (display “\n”) x))

    (debug (list (debug “one”) (debug “two”) (debug “three”) (debug “four”) (debug “five”)))

    (display “\nversus…\n\n”)

    (define-syntax nth (syntax-rules () ((nth n var) var) ((nth n var rest …) (if (= n 0) var (nth (- n 1) rest …)))))

    (nth 0 (debug “one”) (debug “two”) (debug “three”) (debug “four”) (debug “five”)) (display “\nor\n\n”) (nth 1 (debug “one”) (debug “two”) (debug “three”) (debug “four”) (debug “five”))

    yields

    $ scheme ife.scm debug: one debug: two debug: three debug: four debug: five debug: (one two three four five)

    versus…

    debug: one

    or

    debug: two $

    How I miss real macros. ;^(

  2. tadekp Says:
    November 29th, 2006 at 8:17 pm

    Cool. Apparently Python 2.5 offers conditional expressions with a very weird syntax.

    Conditional expressions bring the equivalent of C/C++’s ternary operator to Python:

    ratio = x/y if x > y else y/x

    The condition is in the middle:

    result = true_value if condition else false_value

    The benefit of this syntax is that you can read it just like English, which is a quality I like in a language. The official excuse/explanation is that in the standard library most conditional expressions evaluate 90 percent of the time to |True| so the |true_value| should dominate.

    I am not sure if I like it though…

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