Select an item from a set in Python

Question

I am looking to select one item from a set. It does not matter which item it is, but I want it to be the same item every time the function is called.

For example, if I had the set:

my_set = set(["ABC","ABC inc", "ABC ltd"])

Then I want to return any one of those, but the same one whenever the function is run. [i.e. so I would always get "ABC", or always get "ABC inc" whenever I ran with the above set.] If I was using lists, I would just take the first item, but sets don't have a first per se.

This is in contrast to selecting randomly [e.g. How do I pick 2 random items from a Python set? ] where it will change every time it is run.

your call to `set` gives an error since it expects an iterable, not three of them — Pynchia, Sep 10 '15 at 21:53
You can't, since a set is unordered. The only way I can think of is to compare the hash of each element until you find the element with the same hash as you had before. — , Sep 10 '15 at 21:54
I think [this answer][1] can help you. [1]: http://stackoverflow.com/questions/59825/how-to-retrieve-an-element-from-a-set-without-removing-it — jlnabais, Sep 10 '15 at 21:54
Why are you using a set for this? If there's a good reason, use a set backed by a tree or heap. — Marcin, Sep 10 '15 at 21:58
this doesn't sound like the right way to approach whatever problem you're trying to solve... — acushner, Sep 10 '15 at 22:07
@acushner - working on old code - I already have the data I want to extract from in a set format - they are all alternative ways of writing the same company name - currently just need to consistently select the same value every time i run. — kyrenia, Sep 10 '15 at 22:14
The *right* way to do it is to define an an ordering scheme and then use a form of ordered set as in [this recipe](http://code.activestate.com/recipes/576694/) — dawg, Sep 10 '15 at 22:18

score 7 · Accepted Answer · edited Sep 11 '15 at 00:30

7

What about converting to list and sorting?

my_list = list(my_set)
my_list.sort()
chosen_element = my_list[0]

edited Sep 11 '15 at 00:30

dawg

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answered Sep 10 '15 at 21:56

dorverbin

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You can combine this into a single step: sorted(my_set) returns a list in both Python2 and Python3. – Alex Huszagh Sep 10 '15 at 22:36

Joran Beasley · Answer 2 · 2015-09-10T22:46:49.837

3

you could use a function with memoization

def get_random(my_set,memo={}):
    if id(my_set) not in memo:
       memo[id(my_set)] = random.choice(list(my_set))
    return memo[id(my_set)]

a_set = set([1,2,3,4,5])
print get_random(a_set)
print get_random(a_set)

this would always give you the same value as long as you passed in a_set ... (a different set would give a different answer)

if you wanted to make sure the item was still in the set you could change the memo if check

def get_random(my_set,memo={}):
    if id(my_set) not in memo or memo[id(my_set)] not in my_set:
       memo[id(my_set)] = random.choice(list(my_set))
    return memo[id(my_set)]

edited Sep 10 '15 at 22:46

answered Sep 10 '15 at 21:56

Joran Beasley

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Dangerous in so many ways. E.g., remove the memoized item from the set, and it will still show up as the answer. (You'll also need to make a list of `my_set` before applying `random.choice`.) – Alan Sep 10 '15 at 22:06
1

meh its what OP asked for ... you could always do something like `or memo[id(my_set)] not in my_set` – Joran Beasley Sep 10 '15 at 22:09
... I very quickly ran into issues using this [e.g. when re-initialized the set with new value, the prior result was returned] – kyrenia Sep 10 '15 at 22:42
1

if the memory location is the same then its the same set and it returns to you the first value it returned ... ... which is what was requested ... anyway I altered it to check for the items existance still in the set ... – Joran Beasley Sep 10 '15 at 22:45
If memory is also an issue and you don't want to duplicate the set in memory, you could grab the first value from an iterator rather than random.choice(list(my_set)), which is an advantage of this approach rather than generating a list. – Alex Huszagh Sep 10 '15 at 23:08

score 1 · Answer 3 · answered Sep 10 '15 at 22:04

I may have misunderstood your question, but a check for membership in the set should work, if you are looking for anything specific.

letter_set = set(['abc','ABC','xyz','XYZ'])
check_string = 'ABC'
if check_string in letter_set:
    output = check_string

This should only give an output value if the desired string is in the set.

score 0 · Answer 4 · answered Dec 18 '20 at 09:22

0

Use min().

>>> my_set = set(["ABC","ABC inc", "ABC ltd"])
>>> min(my_set)
'ABC'

Raises ValueError if empty. From Python 3.4 a default value can be returned instead.

answered Dec 18 '20 at 09:22

Mattias Wallin

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score 0 · Answer 5 · answered Mar 28 '21 at 16:23

0

Another answer I found for this is to use min:

elem = min(mySet)

answered Mar 28 '21 at 16:23

cumin

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Practical · Answer 6 · 2021-12-17T18:36:22.650

Maybe this can help you:

from iteration_utilities import first
from simple_benchmark import benchmark

def forLoop(s):
    value = len(s)/2
    for elem in s:
        if elem == value:
            return elem


def RemoveAndSet(s: set):
    value = len(s)/2
    s.remove(value)
    s.add(value)


argumente = {2 ** i: set(range(2 ** i)) for i in range(10, 13)}
a = benchmark([forLoop, RemoveAndSet],
              argumente,
              function_aliases={first: 'First'})

print(a)

Output:

         forLoop       RemoveAndSet
   32    8.139000e-07  4.400154e-07
   64    1.403343e-06  5.584737e-07
   128   2.584880e-06  5.288382e-07
   256   4.903100e-06  4.909571e-07
   512   9.628500e-06  5.034354e-07
   1024  1.899860e-05  4.541565e-07
   2048  3.760350e-05  5.803184e-07

Or that: https://pypi.org/project/ordered-set/

score -1 · Answer 7 · answered Sep 10 '15 at 22:04

-1

Sets are implemented as hash tables. So the order is deterministic unless you insert/remove an element.

say x = set (['1','123','2','10'])

this set will always have an order '1', '2', '123', '10' (lexicographic I guess)

but if you are inserting or removing an element then you will have to use some sort of sorting function to make the order deterministic

answered Sep 10 '15 at 22:04

hackbot89

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*Never* count on a set iterating through its elements in a certain order. – Alan Sep 10 '15 at 22:22
I think you should be good as long as you are using pop to pull the elements out, because after a pop the order still remains the same. Do you have an example that explains why I should "Never count ... " ? : ) – hackbot89 Sep 10 '15 at 22:27
Sort of ... http://stackoverflow.com/questions/3848091/python-set-iteration-order-varies-from-run-to-run – Alan Sep 11 '15 at 02:12

Select an item from a set in Python

7 Answers7