Removing the smaller value from a list using function min()

user_8835166 0
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In short, this is the plan: I want to remove the smallest value from a Python list. My approach to that was creating a new list trimmed , which is basically a copy of the original one but with a condition like:

add it to trimmed only if the value is not equal to min(). Like this:

lis=[]
trimmed =[]

n = int(input("Enter number of elements : "))
for i in range(0, n):
    elements = int(input("Type a number: "))
    lis.append(elements) # adding the elements

# "trimmed" appends a new copy of "lis" removing the smaller value

trimmed = [x for i,x in enumerate(lis) if i != min(lis)]

print(trimmed)

The problem is, it sometimes seems to build trimmed by removing a value in a specific index rather than removing the actual smaller one. E.g. I have just inputted 5 ,2 ,6 and 9. It should ramv removed number two. Instead, it prints out [5, 2, 9]

min python
2 Answers

It's because your if i != min(lis) checks the minimum value against the index variable i instead of the value itself x. enumerate gives back (index, val) pairs i.e, (i, x) here. It happened that min(lis) was 2, so whenever i hits 2, that value will be trimmed, hence you saw 6 disappearing (which was at 2nd index). Fix is to compare against x:

trimmed = [x for i,x in enumerate(lis) if x != min(lis)]

should do.

But better yet, let us compute the minimum only once (above computes many times):

minimum = min(lis)
trimmed = [x for i,x in enumerate(lis) if x != minimum]

and further, you don't seem to need i at all:

minimum = min(lis)
trimmed = [x for x in list if x != minimum]
user_93321870

The index doesn't have anything to do with the minimum value. No need to enumerate.

minval = min(lis)
trimmed = [x for x in lis if x != minval]
user_90051140
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