Both creative solution for Min and Max by veky - python coding challenges

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Both solution in Creative category for Min and Max by veky

min, max = (lambda *args, key=None, r=r: sorted(args[0] if len(args) == 1
    else args, key=key, reverse=r)[0] for r in range(2))

May 10, 2014

Comments:

Sim0000 on May 10, 2014, 8:29 p.m.
Wao, tuple of lambda!

veky on May 10, 2014, 9:02 p.m.
Yes. With defaultification of closed-over vars (r=r). ;-) And range(2) as bool domain. I find it cute... I could have had [-r] instead of [0], but it wouldn't be so cute. ;-)

Cjkjvfnby on May 12, 2014, 1:11 p.m.
<blockquote> range(2) </blockquote> why not (0, 1)?

veky on May 12, 2014, 1:57 p.m.
Because I like range. :-) Here, of course, I wasn't trying to golf.

Cjkjvfnby on May 12, 2014, 2:55 p.m.
it is not golf it is not production(I hope :)) how did you categorize it?

veky on May 12, 2014, 5 p.m.
Why do I have to categorize it? I'm just playing. :-) Seriously, "educational" might be a good category. For example, can you explain what ", r=r" is for? And what would happen if it weren't there? BTW, if someone told me I have to write production code for min and max (in Python), I'd kill him. Justifiably. :-P

Cjkjvfnby on May 12, 2014, 5:38 p.m.
<blockquote> And what would happen if it weren't there? </blockquote> double max

veky on May 12, 2014, 5:55 p.m.
I guess you find it too trivial to provide an explanation. :-)

Janoo12 on Sept. 6, 2015, 3:03 p.m.
if there's no r=r, then reverse = r produces double max, and reverse = 0 produces double min. I tried it ;-)

veky on Sept. 6, 2015, 4:12 p.m.
<blockquote> I tried it </blockquote> That's great, but do you have any explanation? Or at least a curiosity? :-)

Janoo12 on Sept. 6, 2015, 4:28 p.m.
I guess if r is not named as an argument of the lambda function, it can't be changed by later assigning r a 0 and 1 value. By reverse = r we default the reverse to be true and this way it will be declining, so the first element will be the biggest, so it will be both times a max. Reverse = 0 does the opposite.

veky on Sept. 6, 2015, 6:08 p.m.
You're very close, but not quite there. The point is not that r was named as an argument: functions can easily access the variables in outer scopes, because they are closures: <pre class='brush: python'>def f(): def g(): print(i) i = 2; g() i = 3; g() >>> f() 2 3 </pre> What is the important point, is that r is a default value of an argument: since defaults are calculated when function is defined, every function will have its own r. Otherwise, you'd have <pre class='brush: python'>def f(): i = 2 def g(): print(i) i = 3 def h(): print(i) g(); h() >>> f() 3 3 </pre> Of course, since g and h are for all practical functions equal functions, and i is evaluated when they are called (when i is of course 3). So, if you do <pre class='brush: python'>def f(): i = 2 def g(i=i): print(i) i = 3 def h(i=i): print(i) g(); h() >>> f() 2 3 </pre> then g and h are still equal when called with an argument, but when called without an argument, g is called with 2 and h with 3.

Janoo12 on Sept. 6, 2015, 9:20 p.m.
So you say that it should work also without r=r and neither you don't know why it isn't? Other question: lambda here serves only for returng the two value? is it like: min, max = return (sorted(args[0] if len(args) == 1 else args, key=key, reverse=r)[0] for r in range(2)) ? Thank you for the explanations :)

veky on Sept. 7, 2015, 7:08 a.m.
<blockquote> So you say that it should work also without r=r and neither you don’t know why it isn’t? </blockquote> Either I am particularly bad at explaining this, or you're not reading very carefully. Let's try again. Call the three snippets of code in my previous post A,B,C in order. If I read you right, you say that the code in my solution without defaulting r is like A. That's not true. It is like B. In A, you define one function (and call it two times). In B, you define two functions (and call them one time each). In this task, you obviously need to define two functions (and name them min and max). Python doesn't support "implicit currying" like some functional bullies do, where you never know if you've called a function completely, or just half-way. :-) [It does support explicit currying, through functools.partial objects, but that's obviously not used here.] <hr /> If it surprises you how it can define two functions when there is only one lambda, the answer is: comprehensions. Executing literals creates new objects, not encountering them in code text. <pre class='brush: python'>>>> a, b = ([] for _ in range(2)) >>> a, b, a is b ([], [], False) </pre> There is only one [], but it's executed twice. (That's BTW the reason why `[[]]2` doesn't work: operators, like ``, are ordinary methods, and evaluate their operands exactly once.) I think that also answers your second question. It's not lambda that returns anything (except concrete minimal or maximal elements when resulting functions are called in some testing module:). lambda just makes functions. The interesting thing is that since for loop is gone through twice, it makes two functions, and since they have different default for the third parameter, they are different functions. One is <pre class='brush: python'>min = lambda *args, key=None, r=0: sorted(args[0] if len(args) == 1 else args, key=key, reverse=r)[0] </pre> and the other is <pre class='brush: python'>max = lambda *args, key=None, r=1: sorted(args[0] if len(args) == 1 else args, key=key, reverse=r)[0] </pre> <hr /> The point is, you can easily factor out those two functions into one literal in code, since they have the same code. You cannot easily factor out these two functions: <pre class='brush: python'>min = lambda *args, key=None: sorted(args[0] if len(args) == 1 else args, key=key, reverse=True)[0] max = lambda *args, key=None: sorted(args[0] if len(args) == 1 else args, key=key, reverse=False)[0] </pre> into one literal, since they have different code. If you try the usual "name whatever changes, and set that name to different values": <pre class='brush: python'>min, max = (lambda *args, key=None: sorted(args[0] if len(args) == 1 else args, key=key, reverse=r)[0] for r in range(2)) </pre> then it doesn't work, since there is only one r, and by the time those functions are called, it's 1. Variable substitution doesn't work as you think it does, since lambda delays it. <pre class='brush: python'>>>> a = 5 >>> f = lambda x: x + a </pre> doesn't (despite appearances:) mean: make me a function that takes x and adds 5 to it. It means: make me a function that takes x and adds the then-actual (at the moment of call) value of a to it. <pre class='brush: python'>>>> a = 3 >>> f(6) 9 </pre> If you want to express "make me a function that really takes x and adds current (at the moment of definition) value of a (that is, 5) to it", you must either construct function directly <pre class='brush: python'>f = a.__add__ </pre> [if you want to construct something more complicated, functools.partial will do it for you] or invent some way to save the current value of a till the moment of calling f. If you're using `class`, you can store a into some attribute of object that acts like f, <pre class='brush: python'>class adder: def __init__(self, a): self.a = a def __call__(self, x): return x + self.a </pre> or if you're using `def`, you can store a into an attribute of f itself, <pre class='brush: python'>def f(x): return x + f.a f.a = a </pre> but if you're using `lambda`, then you cannot use that approach since you don't know what you're called. But you can always use default arguments. <pre class='brush: python'>f = lambda x, a=a: x + a </pre> It's syntactically neat, and keeps the freezing of a close to the function expression. It does have a disadvantage that you can accidentally call f with two arguments, but even that can be helped: <pre class='brush: python'>f = lambda x, *, a=a: x + a </pre> Now you must explicitly say the name of parameter you want to set, and then you probably know what you're doing (and it can be considered a nice feature, to temporarily override the default a that f calculates with).

Janoo12 on Sept. 7, 2015, 11:42 a.m.
Thank you very much! Now everything is clear. I learnd a lot! :-)

AntoniGoldstein on Dec. 4, 2014, 9:35 a.m.
brilliant! it's O(n log n) instead of O(n) though.

veky on Dec. 4, 2014, 10:29 a.m.
Of course. But in any setting where speed is important, anyone who would forbid me to use a builtin written natively and expected me to write it from scratch in Python, would be shot. Justifiably. ;-P

AntoniGoldstein on Dec. 4, 2014, 11 a.m.
Unarguably ;)

kryksyh on April 16, 2015, 6:39 p.m.
50 shades of python...

veky on April 16, 2015, 7:21 p.m.
I'm in the Land of Weird Compliments again... :-D

vananabu on Sept. 29, 2015, 10:45 a.m.
range(2) as True and False! subtly as always)

veky on Sept. 29, 2015, 10:59 a.m.
Just "twice repeated". :-)

Mellivoran on Oct. 2, 2016, 6:29 a.m.
Thanks for the informative explanations. Archiving this publication in my personal dictionary for future reference.

Georgyphyton on Nov. 13, 2023, 8:22 a.m.
Hey! So smart solution. I'm sorry in advance for my English. Can you help me understand, in thies example, <pre class='brush: python'>max(2.2, 5.6, 5.9, key=int) </pre> if we use <pre class='brush: python'>sorted(args[0] if len(args) == 1 else args, key=key, reverse=r) </pre> it will give as <pre class='brush: python'>[5.6, 5.9, 2.2] </pre> so we can take index 0, and will have right answer, but if we use <pre class='brush: python'>sorted(args[0] if len(args) == 1 else args, key=key) </pre> we will have <pre class='brush: python'>[2.2, 5.6, 5.9] </pre> why thies happend? why when we use reverse, answer is not [5.9, 5.6, 2.2]?

veky on Nov. 13, 2023, 10:07 a.m.
Because you're asking it to sort with int as key. int(5.9) is the same as int(5.6) (both are 5), so they remain in the same order as in the original sequence.

Georgyphyton on Nov. 13, 2023, 10:10 a.m.
thx, sorry for stupid question ))

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