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Py args practice : can anyone helps correcting my code plz

def my_min(x, *args): if x < args: return x elseļ¼š for arg in args: if arg < x: yield arg print(my_min(8, 13, 4, 42, 120, 7))

7th Jul 2021, 3:40 PM
jiejie wu
jiejie wu - avatar
3 Answers
+ 2
what's the purpose if your code? if you expect to return the min value from the arguments passed, you can simply do: def my_min(*args): return min(args) but why not using directly min function ;P if you want to make a recursive function (only for *args practice, as you use min built-in inside): def my_min(x, *args): if len(args) == 0 or x < min(args): return x else: return my_min(*args)
7th Jul 2021, 7:24 PM
visph
visph - avatar
+ 2
You're probably having issues because it's a generator function and your if statement isn't correct. Your if statement is equivalent to the following with the current given arguments; if 8 < [13, 4, 42, 120, 7]: You can make the code work like; def my_min(x, *args): if x < min(args): yield x else: for arg in args: if arg < x: yield arg [print(x) for x in my_min(8, 13, 4, 42, 120, 7)] Or def my_min(x, *args): if x < min(args): return [x] def nums_less_than_x(): for arg in args: if arg < x: yield arg return nums_less_than_x() [print(x) for x in my_min(2, 13, 4, 42, 120, 7)]
7th Jul 2021, 5:29 PM
ChaoticDawg
ChaoticDawg - avatar
0
ļ¼ˆfrom ā€œIpangā€ļ¼‰ def my_min( x, *args ): return min( x, min( args ) )
8th Jul 2021, 5:08 AM
jiejie wu
jiejie wu - avatar