from nltk.corpus import wordnet as wn


def antonyms_for(word):
    antonyms = set()
    for ss in wn.synsets(word):
        for lemma in ss.lemmas():
            any_pos_antonyms = [ antonym.name() for antonym in lemma.antonyms() ]
            for antonym in any_pos_antonyms:
                antonym_synsets = wn.synsets(antonym)
                if wn.ADJ not in [ ss.pos() for ss in antonym_synsets ]:
                    continue
                antonyms.add(antonym)
    return antonyms

print(antonyms_for("bad"))

prints: {'unregretful', 'good'}

is there a way to print only the first word {'unregretful', 'good'} -> {'unregretful'}

CodePudding user response：

If you don't care about the state of the set that's returned, you could use the method pop() on it:

print(antonyms_for("bad").pop())

This will return ONE of the set items. There is no definitive notion of first in a set as they are unordered.

If you expect the first value in alphabetical order then you can use built-in function min():

print(min(antonyms_for("bad")))

CodePudding user response：

You can use iter next:

next(iter(antonyms_for("bad")))

output: 'good' or 'unregretful'

Note that sets are unordered, so there is no way to reliably obtain a given value, it is not random either, you will just get what the hashing algorithm puts first in the python implementation that you are using.

CodePudding user response：

Sets are unordered collections (not sequences), so there is no first or last.

You can transform the iterable set to a sequence, e.g. one of the following data structures

• list, using constructor list(iterable)
• tuple, using constructor tuple(iterable)

and then use indexing to obtain the first:

words_set = antonyms_for("bad")
if words_set:
   first_word = tuple(words_set)[0]
   print(f'First antonym: {first_word}')  # expected to print 'unregretful'
else:
   print('No antonym found!')

Since there may be cases with an empty set returned, you may test for the presence of any elements first using if words_set:.