Succeded in using NLTK stop words and Lin thesaurus to find nearby meaning for certain words in the sentence
from nltk.corpus import stopwords, lin_thesaurus as thes
sentence = "The door is made of glass"
stop = set(stopwords.words('english'))
print [thes.synonyms(i,fileid="simN.lsp") for i in sentence.lower().split() if i not in stop]
Question?
1.How to "Find and replace" a random synonym from the list?
Lin thesaurus gives scored synonyms, Is it possible to calculate 1% band across identified word and randomly select a word within that band and replace it.
For example suppose door is 1 and to check for a word within variance of 0.09 plus or minus and replace it.
Related
I have a dataframe with text in one of its columns.
I have listed some predefined keywords which I need for analysis and words associated with it (and later make a wordcloud and counter of occurrences) to understand topics /context associated with such keywords.
Use case:
df.text_column()
keywordlist = [coca , food, soft, aerated, soda]
lets say one of the rows of the text column has text : ' coca cola is expanding its business in soft drinks and aerated water'.
another entry like : 'lime soda is the best selling item in fast food stores'
my objective is to get Bigram/trigram like:
'coca_cola','coca_cola_expanding', 'soft_drinks', 'aerated_water', 'business_soft_drinks', 'lime_soda', 'food_stores'
Kindly help me to do that [Python only]
First, you can optioanlly load the nltk's stop word list and remove any stop words from the text (such as "is", "its", "in", and "and"). Alternatively, you can define your own stop words list, as well as even extend the nltk's list with additional words. Following, you can use nltk.bigrams() and nltk.trigrams() methods to get bigrams and trigrams joined with an underscore _, as you asked. Also, have a look at Collocations.
Edit:
If you haven't already, you need to include the following once in your code, in order to download the stop words list.
nltk.download('stopwords')
Code:
import nltk
from nltk.tokenize import word_tokenize
from nltk.corpus import stopwords
word_data = "coca cola is expanding its business in soft drinks and aerated water"
#word_data = "lime soda is the best selling item in fast food stores"
# load nltk's stop word list
stop_words = list(stopwords.words('english'))
# extend the stop words list
#stop_words.extend(["best", "selling", "item", "fast"])
# tokenise the string and remove stop words
word_tokens = word_tokenize(word_data)
clean_word_data = [w for w in word_tokens if not w.lower() in stop_words]
# get bigrams
bigrams_list = ["_".join(item) for item in nltk.bigrams(clean_word_data)]
print(bigrams_list)
# get trigrams
trigrams_list = ["_".join(item) for item in nltk.trigrams(clean_word_data)]
print(trigrams_list)
Update
Once you get the bigram and trigram lists, you can check for matches against your keyword list to keep only the relevant ones.
keywordlist = ['coca' , 'food', 'soft', 'aerated', 'soda']
def find_matches(n_grams_list):
matches = []
for k in keywordlist:
matching_list = [s for s in n_grams_list if k in s]
[matches.append(m) for m in matching_list if m not in matches]
return matches
all_matching_bigrams = find_matches(bigrams_list) # find all mathcing bigrams
all_matching_trigrams = find_matches(trigrams_list) # find all mathcing trigrams
# join the two lists
all_matches = all_matching_bigrams + all_matching_trigrams
print(all_matches)
Output:
['coca_cola', 'business_soft', 'soft_drinks', 'drinks_aerated', 'aerated_water', 'coca_cola_expanding', 'expanding_business_soft', 'business_soft_drinks', 'soft_drinks_aerated', 'drinks_aerated_water']
I want to input a sentence, and output a sentence with hard words made simpler.
I'm using Nltk to tokenize sentences and tag words, but I'm having trouble using WordNet to find a synonym for the specific meaning of a word that I want.
For example:
Input:
"I refuse to pick up the refuse"
Maybe refuse #1 is the easiest word for rejecting, but the refuse #2 means garbage, and there are simpler words that could go there.
Nltk might be able to tag refuse #2 as a noun, but then how do I get synonyms for refuse (trash) from WordNet?
Sounds like you want word synonyms based upon the part of speech of the word (i.e. noun, verb, etc.)
Follows creates synonyms for each word in a sentence based upon part of speech.
References:
Extract Word from Synset using Wordnet in NLTK 3.0
Printing the part of speech along with the synonyms of the word
Code
import nltk; nltk.download('popular')
from nltk.corpus import wordnet as wn
def get_synonyms(word, pos):
' Gets word synonyms for part of speech '
for synset in wn.synsets(word, pos=pos_to_wordnet_pos(pos)):
for lemma in synset.lemmas():
yield lemma.name()
def pos_to_wordnet_pos(penntag, returnNone=False):
' Mapping from POS tag word wordnet pos tag '
morphy_tag = {'NN':wn.NOUN, 'JJ':wn.ADJ,
'VB':wn.VERB, 'RB':wn.ADV}
try:
return morphy_tag[penntag[:2]]
except:
return None if returnNone else ''
Example Usage
# Tokenize text
text = nltk.word_tokenize("I refuse to pick up the refuse")
for word, tag in nltk.pos_tag(text):
print(f'word is {word}, POS is {tag}')
# Filter for unique synonyms not equal to word and sort.
unique = sorted(set(synonym for synonym in get_synonyms(word, tag) if synonym != word))
for synonym in unique:
print('\t', synonym)
Output
Note the different sets of synonyms for refuse based upon POS.
word is I, POS is PRP
word is refuse, POS is VBP
decline
defy
deny
pass_up
reject
resist
turn_away
turn_down
word is to, POS is TO
word is pick, POS is VB
beak
blame
break_up
clean
cull
find_fault
foot
nibble
peck
piece
pluck
plunk
word is up, POS is RP
word is the, POS is DT
word is refuse, POS is NN
food_waste
garbage
scraps
This program is to find similarities between the a sentences and words and how they are similar in synonyms I have downloaded the nltk when i first coded it was run and there were no errors but after some days when i run the program
import nltk
nltk.download('stopwords')
nltk.download('wordnet')
from nltk.tokenize import word_tokenize
from nltk.corpus import stopwords
from nltk.corpus import wordnet as wn
filtered_uploaded_sentences = []
uploaded_sentence_synset = []
database_word_synset = []
uploaded_doc_sentence=" The issue of text semantics, such as word semantics and sentence semantics has received increasing attentions in recent years. However, rare research focuses on the document-level semantic matching due to its complexity. Long documents usually have sophisticated structure and massive information, which causes hardship to measure their semantic similarity. The semantic similarity between words, sentences, texts, and documents is widely studied in various fields, including natural language processing, document semantic comparison, artificial intelligence, semantic web, and semantic search engines. "
database_word=["car","complete",'focus',"semantics"]
stopwords = stopwords.words('english')
uploaded_sentence_words_tokenized = word_tokenize(uploaded_doc_sentence)
#filtering the sentence and synset
for word in uploaded_sentence_words_tokenized:
if word not in stopwords:
filtered_uploaded_sentences.append(word)
print (filtered_uploaded_sentences)
for sentences_are in filtered_uploaded_sentences:
uploaded_sentence_synset.append(wn.synsets(sentences_are))
print(uploaded_sentence_synset)
#for finding similrity in the words
for databasewords in database_word:
database_word_synset.append(wn.synsets(databasewords)[0])
print(database_word_synset)
Index Error: list index out of range
this error comes when the uploaded_doc_sentence is short and long sentence is used the error accorded
check.append(wn.wup_similarity(data,sen[0]))
I want to compare the sentence and the words and the results are stored. this type
#the similarity main function for words
for data in database_word_synset:
for sen in uploaded_sentence_synset :
check.append(wn.wup_similarity(data,sen[0]))
print(check)
The problem is that there are empty lists contained in uploaded_sentence_synset. I'm not sure what you're trying to do, but modify the last block of code to:
for data in database_word_synset:
for sen in uploaded_sentence_synset:
if sen:
check.append(wn.wup_similarity(data, sen[0]))
Without the if-else block, you're essentially trying to index the first element of a list, giving you an IndexError.
by removing the empty [] list blocks from the list and making the multi dimension list into single dimension list the problem was solver
list2 = [x for x in main_sen if x != []]
print(list2)
result=list()
for t in list2:
for x in t:
result.append(x)
I have written the following (crude) code to find the association strengths among the words in a given piece of text.
import re
## The first paragraph of Wikipedia's article on itself - you can try with other pieces of text with preferably more words (to produce more meaningful word pairs)
text = "Wikipedia was launched on January 15, 2001, by Jimmy Wales and Larry Sanger.[10] Sanger coined its name,[11][12] as a portmanteau of wiki[notes 3] and 'encyclopedia'. Initially an English-language encyclopedia, versions in other languages were quickly developed. With 5,748,461 articles,[notes 4] the English Wikipedia is the largest of the more than 290 Wikipedia encyclopedias. Overall, Wikipedia comprises more than 40 million articles in 301 different languages[14] and by February 2014 it had reached 18 billion page views and nearly 500 million unique visitors per month.[15] In 2005, Nature published a peer review comparing 42 science articles from Encyclopadia Britannica and Wikipedia and found that Wikipedia's level of accuracy approached that of Britannica.[16] Time magazine stated that the open-door policy of allowing anyone to edit had made Wikipedia the biggest and possibly the best encyclopedia in the world and it was testament to the vision of Jimmy Wales.[17] Wikipedia has been criticized for exhibiting systemic bias, for presenting a mixture of 'truths, half truths, and some falsehoods',[18] and for being subject to manipulation and spin in controversial topics.[19] In 2017, Facebook announced that it would help readers detect fake news by suitable links to Wikipedia articles. YouTube announced a similar plan in 2018."
text = re.sub("[\[].*?[\]]", "", text) ## Remove brackets and anything inside it.
text=re.sub(r"[^a-zA-Z0-9.]+", ' ', text) ## Remove special characters except spaces and dots
text=str(text).lower() ## Convert everything to lowercase
## Can add other preprocessing steps, depending on the input text, if needed.
from nltk.corpus import stopwords
import nltk
stop_words = stopwords.words('english')
desirable_tags = ['NN'] # We want only nouns - can also add 'NNP', 'NNS', 'NNPS' if needed, depending on the results
word_list = []
for sent in text.split('.'):
for word in sent.split():
'''
Extract the unique, non-stopword nouns only
'''
if word not in word_list and word not in stop_words and nltk.pos_tag([word])[0][1] in desirable_tags:
word_list.append(word)
'''
Construct the association matrix, where we count 2 words as being associated
if they appear in the same sentence.
Later, I'm going to define associations more properly by introducing a
window size (say, if 2 words seperated by at most 5 words in a sentence,
then we consider them to be associated)
'''
table = np.zeros((len(word_list),len(word_list)), dtype=int)
for sent in text.split('.'):
for i in range(len(word_list)):
for j in range(len(word_list)):
if word_list[i] in sent and word_list[j] in sent:
table[i,j]+=1
df = pd.DataFrame(table, columns=word_list, index=word_list)
# Count the number of occurrences of each word from word_list in the text
all_words = pd.DataFrame(np.zeros((len(df), 2)), columns=['Word', 'Count'])
all_words.Word = df.index
for sent in text.split('.'):
count=0
for word in sent.split():
if word in word_list:
all_words.loc[all_words.Word==word,'Count'] += 1
# Sort the word pairs in decreasing order of their association strengths
df.values[np.triu_indices_from(df, 0)] = 0 # Make the upper triangle values 0
assoc_df = pd.DataFrame(columns=['Word 1', 'Word 2', 'Association Strength (Word 1 -> Word 2)'])
for row_word in df:
for col_word in df:
'''
If Word1 occurs 10 times in the text, and Word1 & Word2 occur in the same sentence 3 times,
the association strength of Word1 and Word2 is 3/10 - Please correct me if this is wrong.
'''
assoc_df = assoc_df.append({'Word 1': row_word, 'Word 2': col_word,
'Association Strength (Word 1 -> Word 2)': df[row_word][col_word]/all_words[all_words.Word==row_word]['Count'].values[0]}, ignore_index=True)
assoc_df.sort_values(by='Association Strength (Word 1 -> Word 2)', ascending=False)
This produces the word associations like so:
Word 1 Word 2 Association Strength (Word 1 -> Word 2)
330 wiki encyclopedia 3.0
895 encyclopadia found 1.0
1317 anyone edit 1.0
754 peer science 1.0
755 peer encyclopadia 1.0
756 peer britannica 1.0
...
...
...
However, the code contains a lot of for loops which hampers its running time. Specially the last part (sort the word pairs in decreasing order of their association strengths) consumes a lot of time as it computes the association strengths of n^2 word pairs/combinations, where n is the number of words we are interested in (those in word_list in my code above).
So, the following are what I would like some help on:
How do I vectorize the code, or otherwise make it more efficient?
Instead of producing n^2 combinations/pairs of words in the last step, is there any way to prune some of them before producing them? I am going to prune some of the useless/meaningless pairs by inspection after they are produced anyway.
Also, and I know this does not fall into the purview of a coding question, but I would love to know if there's any mistake in my logic, specially when calculating the word association strengths.
Since you asked about your specific code, I will not go into alternate libraries. I will be mostly concentrating on points 1) and 2) of your question:
Instead of iterating through the whole word ist twice (i and j) you can already cut the processing time by ~ half by just iterating j between i + i and the end of the list. This removes duplicate pairs (index 24 and 42 as well as index 42 and 24) as well as the identical pair (index 42 and 42).
for sent in text.split('.'):
for i in range(len(word_list)):
for j in range(i+1, len(word_list)):
if word_list[i] in sent and word_list[j] in sent:
table[i,j]+=1
Be careful with this, though. the in operator will also match partial words (like and in hand)
Of course, you could also remove the j iteration completely by first filtering for all words in your word list and then pairing them afterward:
word_list = set() # Using set instead of list makes lookups faster since this is a hashed structure
for sent in text.split('.'):
for word in sent.split():
'''
Extract the unique, non-stopword nouns only
'''
if word not in word_list and word not in stop_words and nltk.pos_tag([word])[0][1] in desirable_tags:
word_list.add(word)
(...)
for sent in text.split('.'):
found_words = [word for word in sent.split() if word in word_list] # list comprehensions are usually faster than pure for loops
# If you want to count duplicate words, then leave the whole line below out.
found_words = tuple(frozenset(found_words)) # make every word unique using a set and then iterable by index again by converting it into a tuple.
for i in range(len(found_words):
for j in range(i+1, len(found_words):
table[i, j] += 1
In general, you should really think about using external libraries for most of this, though. As some of the comments on your question already pointed out, splitting on . may get you wrong results, the same counts for splitting on whitespace, for example with words that are separated with a - or words that are followed by a ,.
Assuming I have two small dictionaries
posList=['interesting','novel','creative','state-of-the-art']
negList=['outdated','straightforward','trivial']
I have a new word, say "innovative", which is out of my knowledge and I am trying to figure out its sentiment via finding out its synonyms via NLTK function, if the synonyms fall out my small dictionaries, then I recursively call the NLTK function to find the synonyms of the synonyms from last time
The start input could be like this:
from nltk.corpus import wordnet
innovative = wordnet.synsets('innovative')
for synset in innovative:
print synset
print synset.lemmas
It produces the output like this
Synset('advanced.s.03')
[Lemma('advanced.s.03.advanced'), Lemma('advanced.s.03.forward-looking'), Lemma('advanced.s.03.innovative'), Lemma('advanced.s.03.modern')]
Synset('innovative.s.02')
[Lemma('innovative.s.02.innovative'), Lemma('innovative.s.02.innovational'), Lemma('innovative.s.02.groundbreaking')]
Clearly new words include 'advanced','forward-looking','modern','innovational','groundbreaking' are the new words and not in my dictionary, so now I should use these words as start to call synsets function again until no new lemma word appearing.
Anyone can give me a demo code how to extract these lemma words from Synset and keep them in a set strcutre?
It involves dealing with re module in Python I think but I am quite new to Python. Another point I need to address is that I need to get adjective only, so only 's' and 'a' symbol in the Lemma('advanced.s.03.modern'), not 'v' (verb) or 'n' (noun).
Later I would try to calculate the similarity score for a new word with any dictionary word, I need to define the measure. This problem is difficult since adj words are not arranged in hierarchy way and no available measure according to my knowledge. Anyone can advise?
You can get the synonyms of the synonyms as follows.
(Please note that the code uses the WordNet functions of the NodeBox Linguistics library because it offers an easier access to WordNet).
def get_remote_synonyms(s, pos):
if pos == 'a':
syns = en.adjective.senses(s)
if syns:
allsyns = sum(syns, [])
# if there are multiple senses, take only the most frequent two
if len(syns) >= 2:
syns = syns[0] + syns[1]
else:
syns = syns[0]
else:
return []
remote = []
for syn in syns:
newsyns = en.adjective.senses(syn)
remote.extend([r for r in newsyns[0] if r not in allsyns])
return [unicode(i) for i in list(set(remote))]
As far as I know, all semantic measurement functions of the NLTK are based on the hypernym / hyponym hierarchy, so that they cannot be applied to adjectives. Besides, I found a lot of synonyms to be missing in WordNet if you compare its results with the results from a thesaurus like thesaurus.com.