列表中最小的数字-Python

Question

I am trying to write a function that takes a list input and returns the index of the smallest number in that list. For example,

minPos( [5,4,3,2,1] ) → 4

When I run my function, I get a List Index error, can someone please help? Thanks. I cannot use the built in function min().

def MinPos(L):
     Subscript = 0
     Hydrogen = 1
     SmallestNumber = L[Subscript]

    while L[Subscript] < len(L):
          while  L[Subscript] < L[Subscript + Hydrogen]:
                Subscript += 1
                return SmallestNumber

          while L[Subscript] > L[Subscript + Hydrogen]:
                Subscript += 1

    return SmallestNumber


def main():
    print MinPos( [-5,-4] )

Answer 1

Maybe something like this:

>>> def min_pos(L):
...    min = None
...    for i,v in enumerate(L):
...        if min is None or min[1] > v:
...            min = (i,v)
...    return min[0] if min else None


>>> min_pos([1,3,4,5])
0

>>> min_pos([1,3,4,0,5])
3

Edit: Return None if empty list

Answer 2

Since you already know how to find the minimum value, you simply feed that value to the index() function to get the index of this value in the list. Ie,

>>> n = ([5,4,3,2,1])
>>> n.index(min(n))
4

This will return the index of the minimum value in the list. Note that if there are several minima it will return the first.

Answer 3

I would recommend use of for ... and enumarate() :

data = [6, 3, 2, 4, 2, 5]

try:
    index, minimum = 0, data[0]
    for i, value in enumerate(data):
        if value < minimum:
            index, minimum = i, value
except IndexError:
    index = None
print index
# Out[49]: 2

EDIT added guard against empty data