First creative solution for Min and Max by samulih - python coding challenges

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First solution in Creative category for Min and Max by samulih

min = lambda *a, **k: sorted(a if len(a)-1 else a[0], **k)[0]
max = lambda *a, **k: min(*a, **dict(k, reverse=True))

May 14, 2014

Comments:

Mycroft0121 on May 16, 2014, 6:01 p.m.
Brilliant...

MackM on June 18, 2014, 5:29 p.m.
Clever reuse :-)

cmantas on Dec. 23, 2015, 12:39 p.m.
I do not like using "sorted" for this, but this is seriously cool

veky on Oct. 17, 2016, 5:48 p.m.
If you used the true python, you'd even be able to write it more easily. :-) <pre class='brush: python'>min(*a, **k, reverse=True) </pre>

DanyC97 on Oct. 30, 2016, 7:40 p.m.
@veky note the above works for python v > 3.5 (before we can't have any arguments after **). However you can have and that will work <pre class='brush: python'>min(*a, reverse = True, **k) </pre>

veky on Nov. 1, 2016, 2:46 a.m.
You're right (with a nit that it should be >=, I think.) By "true Python" I meant the Platonic ideal, towards which the "Python n" implementations converge as n tends to infinity. :-D

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