How to retrieve an element from a set without removing it?

Two options that don't require copying the whole set:

for e in s:
    break
# e is now an element from s

Or...

e = next(iter(s))

But in general, sets don't support indexing or slicing.

Least code would be:

>>> s = set([1, 2, 3])
>>> list(s)[0]
1

Obviously this would create a new list which contains each member of the set, so not great if your set is very large.

tl;dr

for first_item in muh_set: break remains the optimal approach in Python 3.x. ^{Curse you, Guido.}

y u do this

Welcome to yet another set of Python 3.x timings, extrapolated from wr.'s excellent Python 2.x-specific response. Unlike AChampion's equally helpful Python 3.x-specific response, the timings below also time outlier solutions suggested above – including:

• list(s)[0], John's novel sequence-based solution.
• random.sample(s, 1), dF.'s eclectic RNG-based solution.

Code Snippets for Great Joy

Turn on, tune in, time it:

from timeit import Timer

stats = [
    "for i in range(1000): \n\tfor x in s: \n\t\tbreak",
    "for i in range(1000): next(iter(s))",
    "for i in range(1000): s.add(s.pop())",
    "for i in range(1000): list(s)[0]",
    "for i in range(1000): random.sample(s, 1)",
]

for stat in stats:
    t = Timer(stat, setup="import random\ns=set(range(100))")
    try:
        print("Time for %s:\t %f"%(stat, t.timeit(number=1000)))
    except:
        t.print_exc()

Quickly Obsoleted Timeless Timings

Behold! Ordered by fastest to slowest snippets:

$ ./test_get.py
Time for for i in range(1000): 
    for x in s: 
        break:   0.249871
Time for for i in range(1000): next(iter(s)):    0.526266
Time for for i in range(1000): s.add(s.pop()):   0.658832
Time for for i in range(1000): list(s)[0]:   4.117106
Time for for i in range(1000): random.sample(s, 1):  21.851104

Faceplants for the Whole Family

Unsurprisingly, manual iteration remains at least twice as fast as the next fastest solution. Although the gap has decreased from the Bad Old Python 2.x days (in which manual iteration was at least four times as fast), it disappoints the PEP 20 zealot in me that the most verbose solution is the best. At least converting a set into a list just to extract the first element of the set is as horrible as expected. Thank Guido, may his light continue to guide us.

Surprisingly, the RNG-based solution is absolutely horrible. List conversion is bad, but random really takes the awful-sauce cake. So much for the Random Number God.

I just wish the amorphous They would PEP up a set.get_first() method for us already. If you're reading this, They: "Please. Do something."

Since you want a random element, this will also work:

>>> import random
>>> s = set([1,2,3])
>>> random.sample(s, 1)
[2]

The documentation doesn't seem to mention performance of random.sample. From a really quick empirical test with a huge list and a huge set, it seems to be constant time for a list but not for the set. Also, iteration over a set isn't random; the order is undefined but predictable:

>>> list(set(range(10))) == range(10)
True 

If randomness is important and you need a bunch of elements in constant time (large sets), I'd use random.sample and convert to a list first:

>>> lst = list(s) # once, O(len(s))?
...
>>> e = random.sample(lst, 1)[0] # constant time

Yet another way in Python 3:

next(iter(s))

or

s.__iter__().__next__()

I use a utility function I wrote. Its name is somewhat misleading because it kind of implies it might be a random item or something like that.

def anyitem(iterable):
    try:
        return iter(iterable).next()
    except StopIteration:
        return None

What I usually do for small collections is to create kind of parser/converter method like this

def convertSetToList(setName):
return list(setName)

Then I can use the new list and access by index number

userFields = convertSetToList(user)
name = request.json[userFields[0]]

As a list you will have all the other methods that you may need to work with

You can unpack the values to access the elements:

s = set([1, 2, 3])

v1, v2, v3 = s

print(v1,v2,v3)
#1 2 3

I f you want just the first element try this: b = (a-set()).pop()

Another option is to use a dictionary with values you don't care about. E.g.,

poor_man_set = {}
poor_man_set[1] = None
poor_man_set[2] = None
poor_man_set[3] = None
...

You can treat the keys as a set except that they're just an array:

keys = poor_man_set.keys()
print "Some key = %s" % keys[0]

A side effect of this choice is that your code will be backwards compatible with older, pre-set versions of Python. It's maybe not the best answer but it's another option.

Edit: You can even do something like this to hide the fact that you used a dict instead of an array or set:

poor_man_set = {}
poor_man_set[1] = None
poor_man_set[2] = None
poor_man_set[3] = None
poor_man_set = poor_man_set.keys()