I have a list of filenames in python and I would want to construct a set out of all the filenames.

for filename in filelist:

This does not seem to work. How can do this?

If you have a list of hashable objects (filenames would probably be strings, so they should count):

lst = ['foo.py', 'bar.py', 'baz.py', 'qux.py', Ellipsis]

you can construct the set directly:

s = set(lst)

In fact, set will work this way with any iterable object! (Isn’t duck typing great?)

If you want to do it iteratively:

s = set()
for item in iterable:

But there’s rarely a need to do it this way. I only mention it because the set.add method is quite useful.

The most direct solution is this:

s = set(filelist)

The issue in your original code is that the values weren’t being assigned to the set. Here’s the fixed-up version of your code:

s = set()
for filename in filelist:

You can do

my_set = set(my_list)

or, in Python 3,

my_set = {*my_list}

to create a set from a list. Conversely, you can also do

my_list = list(my_set)

or, in Python 3,

my_list = [*my_set]

to create a list from a set.

Just note that the order of the elements in a list is generally lost when converting the list to a set since a set is inherently unordered. (One exception in CPython, though, seems to be if the list consists only of non-negative integers, but I assume this is a consequence of the implementation of sets in CPython and that this behavior can vary between different Python implementations.)

Here is another solution:

set(['E:\\', 'D:\\', 'C:\\'])

In this code I have used the set method in order to turn it into a set and then it removed all duplicate values from the list

Simply put the line:

new_list = set(your_list)

One general way to construct set in iterative way like this:

aset = {e for e in alist}

You can also use list comprehension to create set.

s = {i for i in range(5)}

All the above answers are correct but if you want to preserve the order of your list you’ll need to proceed as follow