I would like to lemmatize some Italian text in order to perform some frequency counting of words and further investigations on the output of this lemmatized content.
I am preferring lemmatizing than stemming because I could extract the word meaning from the context in the sentence (e.g. distinguish between a verb and a noun) and obtain words that exist in the language, rather than roots of those words that don't usually have a meaning.
I found out this library called pattern (pip2 install pattern) that should complement nltk in order to perform lemmatization of the Italian language, however I am not sure the approach below is correct because each word is lemmatized by itself, not in the context of a sentence.
Probably I should give pattern the responsibility to tokenize a sentence (so also annotating each word with the metadata regarding verbs/nouns/adjectives etc), then retrieving the lemmatized word, but I am not able to do this and I am not even sure it is possible at the moment?
Also: in Italian some articles are rendered with an apostrophe so for example "l'appartamento" (in English "the flat") is actually 2 words: "lo" and "appartamento". Right now I am not able to find a way to split these 2 words with a combination of nltk and pattern so then I am not able to count the frequency of the words in the correct way.
import nltk
import string
import pattern
# dictionary of Italian stop-words
it_stop_words = nltk.corpus.stopwords.words('italian')
# Snowball stemmer with rules for the Italian language
ita_stemmer = nltk.stem.snowball.ItalianStemmer()
# the following function is just to get the lemma
# out of the original input word (but right now
# it may be loosing the context about the sentence
# from where the word is coming from i.e.
# the same word could either be a noun/verb/adjective
# according to the context)
def lemmatize_word(input_word):
in_word = input_word#.decode('utf-8')
# print('Something: {}'.format(in_word))
word_it = pattern.it.parse(
in_word,
tokenize=False,
tag=False,
chunk=False,
lemmata=True
)
# print("Input: {} Output: {}".format(in_word, word_it))
the_lemmatized_word = word_it.split()[0][0][4]
# print("Returning: {}".format(the_lemmatized_word))
return the_lemmatized_word
it_string = "Ieri sono andato in due supermercati. Oggi volevo andare all'ippodromo. Stasera mangio la pizza con le verdure."
# 1st tokenize the sentence(s)
word_tokenized_list = nltk.tokenize.word_tokenize(it_string)
print("1) NLTK tokenizer, num words: {} for list: {}".format(len(word_tokenized_list), word_tokenized_list))
# 2nd remove punctuation and everything lower case
word_tokenized_no_punct = [string.lower(x) for x in word_tokenized_list if x not in string.punctuation]
print("2) Clean punctuation, num words: {} for list: {}".format(len(word_tokenized_no_punct), word_tokenized_no_punct))
# 3rd remove stop words (for the Italian language)
word_tokenized_no_punct_no_sw = [x for x in word_tokenized_no_punct if x not in it_stop_words]
print("3) Clean stop-words, num words: {} for list: {}".format(len(word_tokenized_no_punct_no_sw), word_tokenized_no_punct_no_sw))
# 4.1 lemmatize the words
word_tokenize_list_no_punct_lc_no_stowords_lemmatized = [lemmatize_word(x) for x in word_tokenized_no_punct_no_sw]
print("4.1) lemmatizer, num words: {} for list: {}".format(len(word_tokenize_list_no_punct_lc_no_stowords_lemmatized), word_tokenize_list_no_punct_lc_no_stowords_lemmatized))
# 4.2 snowball stemmer for Italian
word_tokenize_list_no_punct_lc_no_stowords_stem = [ita_stemmer.stem(i) for i in word_tokenized_no_punct_no_sw]
print("4.2) stemmer, num words: {} for list: {}".format(len(word_tokenize_list_no_punct_lc_no_stowords_stem), word_tokenize_list_no_punct_lc_no_stowords_stem))
# difference between stemmer and lemmatizer
print(
"For original word(s) '{}' and '{}' the stemmer: '{}' '{}' (count 1 each), the lemmatizer: '{}' '{}' (count 2)"
.format(
word_tokenized_no_punct_no_sw[1],
word_tokenized_no_punct_no_sw[6],
word_tokenize_list_no_punct_lc_no_stowords_stem[1],
word_tokenize_list_no_punct_lc_no_stowords_stem[6],
word_tokenize_list_no_punct_lc_no_stowords_lemmatized[1],
word_tokenize_list_no_punct_lc_no_stowords_lemmatized[1]
)
)
Gives this output:
1) NLTK tokenizer, num words: 20 for list: ['Ieri', 'sono', 'andato', 'in', 'due', 'supermercati', '.', 'Oggi', 'volevo', 'andare', "all'ippodromo", '.', 'Stasera', 'mangio', 'la', 'pizza', 'con', 'le', 'verdure', '.']
2) Clean punctuation, num words: 17 for list: ['ieri', 'sono', 'andato', 'in', 'due', 'supermercati', 'oggi', 'volevo', 'andare', "all'ippodromo", 'stasera', 'mangio', 'la', 'pizza', 'con', 'le', 'verdure']
3) Clean stop-words, num words: 12 for list: ['ieri', 'andato', 'due', 'supermercati', 'oggi', 'volevo', 'andare', "all'ippodromo", 'stasera', 'mangio', 'pizza', 'verdure']
4.1) lemmatizer, num words: 12 for list: [u'ieri', u'andarsene', u'due', u'supermercato', u'oggi', u'volere', u'andare', u"all'ippodromo", u'stasera', u'mangiare', u'pizza', u'verdura']
4.2) stemmer, num words: 12 for list: [u'ier', u'andat', u'due', u'supermerc', u'oggi', u'vol', u'andar', u"all'ippodrom", u'staser', u'mang', u'pizz', u'verdur']
For original word(s) 'andato' and 'andare' the stemmer: 'andat' 'andar' (count 1 each), the lemmatizer: 'andarsene' 'andarsene' (count 2)
How to effectively lemmatize some sentences with pattern using their tokenizer? (assuming lemmas are recognized as nouns/verbs/adjectives etc.)
Is there a python alternative to pattern to use for Italian lemmatization with nltk?
How to split articles that are bound to the next word using apostrophes?
I'll try to answer your question, knowing that I don't know a lot about italian!
1) As far as I know, the main responsibility for removing apostrophe is the tokenizer, and as such the nltk italian tokenizer seems to have failed.
3) A simple thing you can do about it is call the replace method (although you probably will have to use the re package for more complicated pattern), an example:
word_tokenized_no_punct_no_sw_no_apostrophe = [x.split("'") for x in word_tokenized_no_punct_no_sw]
word_tokenized_no_punct_no_sw_no_apostrophe = [y for x in word_tokenized_no_punct_no_sw_no_apostrophe for y in x]
It yields:
['ieri', 'andato', 'due', 'supermercati', 'oggi', 'volevo', 'andare', 'all', 'ippodromo', 'stasera', 'mangio', 'pizza', 'verdure']
2) An alternative to pattern would be treetagger, granted it is not the easiest install of all (you need the python package and the tool itself, however after this part it works on windows and Linux).
A simple example with your example above:
import treetaggerwrapper
from pprint import pprint
it_string = "Ieri sono andato in due supermercati. Oggi volevo andare all'ippodromo. Stasera mangio la pizza con le verdure."
tagger = treetaggerwrapper.TreeTagger(TAGLANG="it")
tags = tagger.tag_text(it_string)
pprint(treetaggerwrapper.make_tags(tags))
The pprint yields:
[Tag(word=u'Ieri', pos=u'ADV', lemma=u'ieri'),
Tag(word=u'sono', pos=u'VER:pres', lemma=u'essere'),
Tag(word=u'andato', pos=u'VER:pper', lemma=u'andare'),
Tag(word=u'in', pos=u'PRE', lemma=u'in'),
Tag(word=u'due', pos=u'ADJ', lemma=u'due'),
Tag(word=u'supermercati', pos=u'NOM', lemma=u'supermercato'),
Tag(word=u'.', pos=u'SENT', lemma=u'.'),
Tag(word=u'Oggi', pos=u'ADV', lemma=u'oggi'),
Tag(word=u'volevo', pos=u'VER:impf', lemma=u'volere'),
Tag(word=u'andare', pos=u'VER:infi', lemma=u'andare'),
Tag(word=u"all'", pos=u'PRE:det', lemma=u'al'),
Tag(word=u'ippodromo', pos=u'NOM', lemma=u'ippodromo'),
Tag(word=u'.', pos=u'SENT', lemma=u'.'),
Tag(word=u'Stasera', pos=u'ADV', lemma=u'stasera'),
Tag(word=u'mangio', pos=u'VER:pres', lemma=u'mangiare'),
Tag(word=u'la', pos=u'DET:def', lemma=u'il'),
Tag(word=u'pizza', pos=u'NOM', lemma=u'pizza'),
Tag(word=u'con', pos=u'PRE', lemma=u'con'),
Tag(word=u'le', pos=u'DET:def', lemma=u'il'),
Tag(word=u'verdure', pos=u'NOM', lemma=u'verdura'),
Tag(word=u'.', pos=u'SENT', lemma=u'.')]
It also tokenized pretty nicely the all'ippodromo to al and ippodromo (which is hopefully correct) under the hood before lemmatizing. Now we just need to apply the removal of stop words and punctuation and it will be fine.
The doc for installing the TreeTaggerWrapper library for python
I know this issue has been solved few years ago, but I am facing the same problem with nltk tokenization and Python 3 in regards to parsing words like all'ippodromo or dall'Italia. So I want to share my experience and give a partial, although late, answer.
The first action/rule that an NLP must take into account is to prepare the corpus. So I discovered that by replacing the ' character with a proper accent ’ by using accurate regex replacing during text parsing (or just a propedeutic replace all at once in basic text editor), then the tokenization works correctly and I am having the proper splitting with just nltk.tokenize.word_tokenize(text)
I'm using the R (3.2.3) tm-package (0.6-2) and would like to subset my corpus according to partial string matches contained with the metadatum "id".
For example, I would like to filter all documents that contain the string "US" within the "id" column. The string "US" would be preceded and followed by various characters and numbers.
I have found a similar example here. It is recommended to download the quanteda package but I think this should also be possible with the tm package.
Another more relevant answer to a similar problem is found here. I have tried to adapt that sample code to my context. However, I don't manage to incorporate the partial string matching.
I imagine there might be multiple things wrong with my code so far.
What I have so far looks like this:
US <- tm_filter(corpus, FUN = function(corpus, filter) any(meta(corpus)["id"] == filter), grep(".*US.*", corpus))
And I receive the following error message:
Error in structure(as.character(x), names = names(x)) :
'names' attribute [3811] must be the same length as the vector [3]
I'm also not sure how to come up with a reproducible example simulating my problem for this post.
It could work like this:
library(tm)
reut21578 <- system.file("texts", "crude", package = "tm")
(corp <- VCorpus(DirSource(reut21578), list(reader = readReut21578XMLasPlain)))
# <<VCorpus>>
# Metadata: corpus specific: 0, document level (indexed): 0
# Content: documents: 20
(idx <- grep("0", sapply(meta(corp, "id"), paste0), value=TRUE))
# 502 704 708
# "502" "704" "708"
(corpsubset <- corp[idx] )
# <<VCorpus>>
# Metadata: corpus specific: 0, document level (indexed): 0
# Content: documents: 3
You are looking for "US" instead of "0". Have a look at ?grep for details (e.g. fixed=TRUE).
I need to remove defined strings from sentences in data frame:
sent1 = data.frame(Sentences=c("bad printer for the money wireless setup was surprisingly easy",
"love my samsung galaxy tabinch gb whitethis is the first"), user = c(1,2))
Sentences User
bad printer for the money wireless setup was surprisingly easy 1
love my samsung galaxy tabinch gb whitethis is the first 2
Defined strings for excluding, e.g.:
stop_words <- c("bad", "money", "love", "is", "the")
I was wondering about something like this:
library(stringr)
words1 <- (str_split(unlist(sent1$Sentences)," "))
ddd = which(words1[[1]] %in% stop_words)
words1[[1]][-ddd]
But I need it for all items in the list. Then I need to have output table in the same structure as input table sent1, but without defined strings.
Please, I very appreciate any of help or advice.
You can combine the stop words and create a regex pattern. Therefore, you only need a single gsub command.
# create regex pattern
pattern <- paste0("\\b(?:", paste(stop_words, collapse = "|"), ")\\b ?")
# [1] "\\b(?:bad|money|love|is|the)\\b ?"
# remove stop words
res <- gsub(pattern, "", sent1$Sentences)
# [1] "printer for wireless setup was surprisingly easy"
# [2] "my samsung galaxy tabinch gb whitethis first"
# store result in a data frame
data.frame(Sentences = res)
# Sentences
# 1 printer for wireless setup was surprisingly easy
# 2 my samsung galaxy tabinch gb whitethis first
I want to parse an access.log in R. It has the following form and I want to get it into a data.frame:
TIME="2013-07-25T06:28:38+0200" MOBILE_AGENT="0" HTTP_REFERER="-" REQUEST_HOST="www.example.com" APP_ENV="envvar" APP_COUNTRY="US" APP_DEFAULT_LOCATION="New York" REMOTE_ADDR="11.222.33.444" SESSION_ID="rstg35tsdf56tdg3" REQUEST_URI="/get/me/something" HTTP_USER_AGENT="Mozilla/5.0 (compatible; Googlebot/2.1; +http://www.google.com/bot.html)" REQUEST_METHOD="GET" REWRITTEN_REQUEST_URI="/index.php?url=/get/me/something" STATUS="200" RESPONSE_TIME="155,860ms" PEAK_MEMORY="18965" CPU="99,99"
The logs are 400MB per file and currently I have about 4GB logs so size matters.
Another thing.. There are two different log structures (different columns are included) so you can not assume to have the same columns always, but you can assume that only one kind of structure is parsed at a time.
What I have up to now is a regex for this structure:
(\\w+)[=][\"](.*?)[\"][ ]{0,1}
I can read the data in and somehow fit it into a dataframe using readlines, gsub and read.table but it is slow and messy.
Any ideas? Tnx!
You can do this for example:
text <- readLines(textConnection(text))
## since we can't use = as splitter (used in url) I create a new splitter
dd <- read.table(text=gsub('="','|"',text),sep=' ')
## use data.table since it is faster to apply operation by columns and bind them again
library(data.table)
DT <- as.data.table(dd)
DT.split <- DT[,lapply(.SD,function(x)
unlist(strsplit(as.character(x) ,"|",fixed=TRUE)))]
DT.split[c(F,T)]
I want to use the Regex by John Gruber (http://daringfireball.net/2010/07/improved_regex_for_matching_urls) to match complex URLs in text blocks. The Regex is quite complex (as is the task, see regex to find url in a text).
My problem is that I don't get it work with R:
x <- c("http://foo.com/blah_blah",
"http://foo.com/blah_blah/",
"(Something like http://foo.com/blah_blah)",
"http://foo.com/blah_blah_(wikipedia)",
"http://foo.com/more_(than)_one_(parens)",
"(Something like http://foo.com/blah_blah_(wikipedia))",
"http://foo.com/blah_(wikipedia)#cite-1",
"http://foo.com/blah_(wikipedia)_blah#cite-1",
"http://foo.com/unicode_(✪)_in_parens",
"http://foo.com/(something)?after=parens",
"http://foo.com/blah_blah.",
"http://foo.com/blah_blah/.",
"<http://foo.com/blah_blah>",
"<http://foo.com/blah_blah/>",
"http://foo.com/blah_blah,",
"http://www.extinguishedscholar.com/wpglob/?p=364.",
"http://✪df.ws/1234",
"rdar://1234",
"rdar:/1234",
"x-yojimbo-item://6303E4C1-6A6E-45A6-AB9D-3A908F59AE0E",
"message://%3c330e7f840905021726r6a4ba78dkf1fd71420c1bf6ff#mail.gmail.com%3e",
"http://➡.ws/䨹",
"www.c.ws/䨹",
"<tag>http://example.com</tag>",
"Just a www.example.com link.",
"http://example.com/something?with,commas,in,url, but not at end",
"What about <mailto:gruber#daringfireball.net?subject=TEST> (including brokets).",
"mailto:name#example.com",
"bit.ly/foo",
"“is.gd/foo/”",
"WWW.EXAMPLE.COM",
"http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55))/Web_ENG/View_DetailPhoto.aspx?PicId=752",
"http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55))",
"http://lcweb2.loc.gov/cgi-bin/query/h?pp/horyd:#field(NUMBER+#band(thc+5a46634))")
t <- regexec("\\b((?:[a-z][\\w-]+:(?:/{1,3}|[a-z0-9%])|www\\d{0,3}[.]|[a-z0-9.\\-]+[.][a-z]{2,4}/)(?:[^\\s()<>]+|\\(([^\\s()<>]+|(\\([^\\s()<>]+\\)))*\\))+(?:\\(([^\\s()<>]+|(\\([^\\s()<>]+\\)))*\\)|[^\\s`!()\\[\\]{};:'".,<>?«»“”‘’]))", x)
regmatches(x,t)
I appreciate your help.
I ended up using gregexpr as this supports perl=TRUE. After adapting the Regex for R, I came up with the following solution (use data above).
findURL <- function(x){
t <- gregexpr("(?xi)\\b(
(?:[a-z][\\w-]+:(?:/{1,3}|[a-z0-9%])|www\\d{0,3}[.]|[a-z0-9.\\-]+[.][a-z]{2,4}/)
(?:[^\\s\\(\\)<>]+|\\(([^\\s\\(\\)<>]+|(\\([^\\s\\(\\)<>]+\\)))*\\))+
(?:\\(([^\\s\\(\\)<>]+|(\\([^\\s\\(\\)<>]+\\)))*\\)|[^\\s`!\\(\\)\\[\\]{};:'\\\"\\.,<>\\?«»“”‘’])
)",x, perl=TRUE, fixed=FALSE)
regmatches(x,t)
}
# Find URLs
urls <- findURL(x)
# Count URLs
count.urls.temp <- lapply(urls, length)
count.urls <- sum(unlist(count.urls.temp))
I hope this is helpful for others.