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Find NLTK Wordnet Synsets for combined items in a list

I am new to NLTK. I want to use nltk to extract hyponyms for a given list of words, specifically, for some combined words
my code:
import nltk
from nltk.corpus import wordnet as wn
list = ["real_time", 'Big_data', "Healthcare",
'Fuzzy_logic', 'Computer_vision']
def get_synset(a_list):
synset_list = []
for word in a_list:
a = wn.synsets(word)[:1] #The index is to ensure each word gets assigned 1st synset only
synset_list.append(a)
return synset_list
lst_synsets = get_synset(list)
lst_synsets
Here is the output:
[[Synset('real_time.n.01')],
[],
[Synset('healthcare.n.01')],
[Synset('fuzzy_logic.n.01')],
[]]
How could I find NLTK Wordnet Synsets for combined items? if no, any suggestion to use one of these methods for combined terms?

How to extract all words in a noun food category in WordNet?

I am trying to get all the words in Wordnet dictionary that are of type noun and category food.
I have found a way to check if a word is noun.food but I need the reverse method:
import nltk
nltk.download('wordnet')
from nltk.corpus import wordnet as wn
def if_food(word):
syns = wn.synsets(word, pos = wn.NOUN)
for syn in syns:
print(syn.lexname())
if 'food' in syn.lexname():
return 1
return 0

So I think I have found a solution:
# Using the NLTK WordNet dictionary check if the word is noun and a food.
import nltk
nltk.download('wordnet')
from nltk.corpus import wordnet as wn
def if_food(word):
syns = wn.synsets(str(word), pos = wn.NOUN)
for syn in syns:
if 'food' in syn.lexname():
return 1
return 0
Then using the qdapDictionaries::GradyAugmented R English words dictionary I have checked each word if it's a noun.food:
en_dict = pd.read_csv("GradyAugmentedENDict.csv")
en_dict['is_food'] = en_dict.word.apply(if_food)
en_dict[en_dict.is_food == 1].to_csv("en_dict_is_food.csv")
It it actually did the job.
Hope it will help others.

Looping through Lemmas in NLTK Wordnet

Have a script for getting italian synonyms from Wordnet like this:
from nltk.corpus import wordnet as wn
it_lemmas = wn.lemmas("problema", lang="ita")
hypernyms = it_lemmas[0].synset().hypernyms()
print(hypernyms[0].lemmas(lang="ita"))
When I do the looping I get message that list indices must be integers or slices, not Lemma
How should I do the looping to get not only one value ([0]) but all the values in this dictionary (the amount can be different) and print them all?

from nltk.corpus import wordnet as wn
it_lemmas = wn.lemmas("problema", lang="ita")
for i in range(len(it_lemmas)):
hypernyms = it_lemmas[i].synset().hypernyms()
for i in range(len(hypernyms)):
syn = hypernyms[i].lemmas(lang="ita")
print (syn)

How to generate a list of antonyms for adjectives in WordNet using Python

I want to do the following in Python (I have the NLTK library, but I'm not great with Python, so I've written the following in a weird pseudocode):
from nltk.corpus import wordnet as wn #Import the WordNet library
for each adjective as adj in wn #Get all adjectives from the wordnet dictionary
print adj & antonym #List all antonyms for each adjective
once list is complete then export to txt file
This is so I can generate a complete dictionary of antonyms for adjectives. I think it should be doable, but I don't know how to create the Python script. I'd like to do it in Python as that's the NLTK's native language.

from nltk.corpus import wordnet as wn
for i in wn.all_synsets():
if i.pos() in ['a', 's']: # If synset is adj or satelite-adj.
for j in i.lemmas(): # Iterating through lemmas for each synset.
if j.antonyms(): # If adj has antonym.
# Prints the adj-antonym pair.
print j.name(), j.antonyms()[0].name()
Note that there will be reverse duplicates.
[out]:
able unable
unable able
abaxial adaxial
adaxial abaxial
acroscopic basiscopic
basiscopic acroscopic
abducent adducent
adducent abducent
nascent dying
dying nascent
abridged unabridged
unabridged abridged
absolute relative
relative absolute
absorbent nonabsorbent
nonabsorbent absorbent
adsorbent nonadsorbent
nonadsorbent adsorbent
absorbable adsorbable
adsorbable absorbable
abstemious gluttonous
gluttonous abstemious
abstract concrete
...

The following function uses WordNet to return a set of adjective-only antonyms for a given word:
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
Usage:
print(antonyms_for("good"))

Getting adjective from an adverb in nltk or other NLP library

Is there a way to get an adjective corresponding to a given adverb in NLTK or other python library.
For example, for the adverb "terribly", I need to get "terrible".
Thanks.

There is a relation in wordnet that connects the adjectives to adverbs and vice versa.
>>> from itertools import chain
>>> from nltk.corpus import wordnet as wn
>>> from difflib import get_close_matches as gcm
>>> possible_adjectives = [k.name for k in chain(*[j.pertainyms() for j in chain(*[i.lemmas for i in wn.synsets('terribly')])])]
['terrible', 'atrocious', 'awful', 'rotten']
>>> gcm('terribly',possible_adjectives)
['terrible']
A more human readable way to computepossible_adjective is as followed:
possible_adj = []
for ss in wn.synsets('terribly'):
for lemmas in ss.lemmas: # all possible lemmas.
for lemma in lemmas:
for ps in lemma.pertainyms(): # all possible pertainyms.
for p in ps:
for ln in p.name: # all possible lemma names.
possible_adj.append(ln)
EDIT: In the newer version of NLTK:
possible_adj = []
for ss in wn.synsets('terribly'):
for lemmas in ss.lemmas(): # all possible lemmas
for ps in lemmas.pertainyms(): # all possible pertainyms
possible_adj.append(ps.name())

As MKoosej mentioned, nltk's lemmas is no longer an attribute but a method. I also made a little simplification to get the most possible word. Hope someone else can use it also:
wordtoinv = 'unduly'
s = []
winner = ""
for ss in wn.synsets(wordtoinv):
for lemmas in ss.lemmas(): # all possible lemmas.
s.append(lemmas)
for pers in s:
posword = pers.pertainyms()[0].name()
if posword[0:3] == wordtoinv[0:3]:
winner = posword
break
print winner # undue