heapq creative solution for Min and Max by Sim0000 - python coding challenges

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heapq solution in Creative category for Min and Max by Sim0000

def min(*args, key=lambda x: x):
    return __import__('heapq').nsmallest(1, args[0] if len(args) == 1 else args, key=key)[0]
    
def max(*args, key=lambda x: x):
    return __import__('heapq').nlargest(1, args[0] if len(args) == 1 else args, key=key)[0]

May 8, 2014

Comments:

veky on May 10, 2014, 7:49 p.m.
Nice and complicated. :-D Though you don't need such enormous duplication of code. ;-)

Sim0000 on May 10, 2014, 8:33 p.m.
I watch your code. Thank you, I can rewrite my code as follows. <pre class='brush: python'>min, max = [lambda *args, key=lambda x: x, func=f: func(1, args[0] if len(args) == 1 else args, key=key)[0] for f in (__import__('heapq').nsmallest, __import__('heapq').nlargest)] </pre>

veky on May 10, 2014, 8:57 p.m.
You can factor out even more. See getattr builtin. (Yes, it requires stringification, but that shouldn't be a problem for someone using __import__.;) <pre class='brush: python'>getattr(__import__("heapq"), "n{}est".format(adj)) ... ... for adj in ("small", "larg") </pre> ;-9

Sim0000 on May 10, 2014, 9:13 p.m.
Thank you again.

Jiyoqisv on June 5, 2014, 11:23 p.m.
if you aim to cheat, why not use __builtins__

sup_13200 on Dec. 3, 2025, 10:16 p.m.
Nice and short

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