+ 3

*args in python

The given code defined a function called my_min(), which takes two arguments and returns the smaller one. You need to improve the function, so that it can take any number of variables, so that the function call works. def my_min(x, y): if x < y: return x else: return y print(my_min(8, 13, 4, 42, 120, 7))

14th Mar 2021, 2:49 PM
pardeep
pardeep - avatar
18 Antworten
+ 6
faster approach def myfunc(*args): return min(*args) print(myfunc(1,2,3,4)) print(myfunc([1,2,3,4]))
15th Mar 2021, 10:50 AM
iTech
iTech - avatar
+ 5
def my_min(*args): minval = args[0] for num in args: if num<minval: minval=num return minval print(my_min(8, 13, 4, 42, 120, 7))
14th Mar 2021, 3:00 PM
deleted
+ 3
how about this it worked; def my- min(x, y, *args): return min (args) if x < y: return x else: return y print(my-min( 8, 13, 4, 42, 120, 7))
18th Oct 2022, 4:50 PM
kevin friend
kevin friend - avatar
+ 2
Look I have following solution it may help you. def my_min(x, *args): if len(args) == 1: if x < args[0]: return x else: return args[0] y, *args = args if x < y: return my_min(x, *args) else: return my_min(y, *args) print(my_min(8, 13, 4, 42, 120, 7)) It's a recursive way of finding minimum value DHANANJAY
15th Mar 2021, 8:31 PM
DHANANJAY PATEL
DHANANJAY PATEL - avatar
+ 1
You should add the language name in tags while asking a question.
14th Mar 2021, 2:51 PM
Abhay
Abhay - avatar
+ 1
def myFunc(*args): ... will populate args with a list containing all non keywords args. you should just find min value of this list and return it ;)
14th Mar 2021, 2:51 PM
visph
visph - avatar
+ 1
AssasinStudent if OP have to implement a my_min() function, he's probably not allowed to use min() built-in... and if he could, shorter would be: my_min = min anyway, if you give complete solution to OP, he will learn nothing appart copy-pasting without understanding ^^
14th Mar 2021, 3:07 PM
visph
visph - avatar
14th Mar 2021, 2:51 PM
pardeep
pardeep - avatar
0
ya thanks
14th Mar 2021, 2:51 PM
pardeep
pardeep - avatar
0
Agree. Giving hints to an OP rather than a complete solution is recommended in this section.
15th Mar 2021, 4:53 AM
Sonic
Sonic - avatar
0
Slution: def my_min(*args): return min(args) print(my_min(8, 13, 4, 42, 120, 7))
10th Nov 2022, 8:34 AM
Pooja Patel
Pooja Patel - avatar
0
Why it doesn't wok ? def my_min(x, *args): for y in args: if x < y: return x else: return y print(my_min(8, 13, 4, 42, 120, 7))
25th Dec 2022, 3:53 PM
Ouail ElMesoudy
Ouail ElMesoudy - avatar
0
def my_min(*args): return min(args) print(my_min(8, 13, 4, 42, 120, 7))
20th Mar 2023, 8:01 PM
Mateusz Kos
Mateusz Kos - avatar
0
def my_min(x,*args): for i in args: if x < i: res = x else: res = i return res print(my_min(8, 13, 4, 42, 120, 7)) Who can explain why it's not work?
11th Apr 2023, 8:39 AM
Tran Thuy Nha Truc
Tran Thuy Nha Truc - avatar
0
Truc Tran Above code will not work to find minimum value It will always compare with first value (means x) during the loop Here, it is 8, so it will always return minimum value between first value and last value in case of your code it will return 7 Since, between 8 and 7 minimum value is 7 Thanks for your concern DHANANJAY PATEL
11th Apr 2023, 9:49 AM
DHANANJAY PATEL
DHANANJAY PATEL - avatar
0
Oh I understand now. Thank you so much
11th Apr 2023, 10:29 AM
Tran Thuy Nha Truc
Tran Thuy Nha Truc - avatar
0
@DHANANJAY PATEL I understand your comment why it wouldn't work, because it will only compare with the first digit... However, that was the question of the assignment. If not, the assignment was poorly set up!! (... which takes two arguments and returns the smaller one....) Punctuation is key!!
16th Mar 2024, 10:48 AM
Abdi
- 1
which language?
14th Mar 2021, 2:50 PM
deleted