Python min function with *args

Question

How can i find the minimum number in a list by passing them to my own function.
def myfunc(a,*b):
..........

Answer 1

sid ,

as we don't know what your function will receive as arguments and what the function is doing exactly (? only finding the min nunbers ?), please post the complete code, and also a complete task description.

▪︎in general: for finding the min number of a list, you don't need to use *args. just use:

def myfunc(the_list)
.....

Answer 2

No need to declare x in the function. Just declare *y and then use a for loop to iterate and return the minimum value. Maybe your want like this:
https://code.sololearn.com/cYzdRU49Gb0U

Answer 3

Do you mean something like this?
https://code.sololearn.com/c63RYk5iMq3q

Answer 4

If you want the function to return <x> when there's nothing in <y> having value less than <x>, without using built-in function

def my_min(x, *y):
for v in y:
if v < x:
x = v
return x

print(my_min(8, 13, 4, 42, 120, 7))

Answer 5

Lothar
def my_min(x, *y):
if x < y:
return x
else:
return y

print(my_min(8, 13, 4, 42, 120, 7))

The given code defined a function called my_min(), which takes two arguments and returns the smaller one.
I need to improve the function, so that it can take any number of variables, so that the function call works.

Answer 6

The future is now thanks to science[In a break] i have to find the minimum using my own,not with inbuilt min()

Answer 7

def get_min(*args):
if all([type(i) is int for i in args]):
return min(args)
else:
return "wrong type of item in list"