Here is some data.
library(stringr)
library(dplyr)
df <- tibble(sentences)
I want to identify all sentences with the word "her." But this, of course, also returns sentences with words like "there" and "here."
df %>% filter(str_detect(sentences, "her"))
# A tibble: 43 x 1
sentences
<chr>
1 The boy was there when the sun rose.
2 Help the woman get back to her feet.
3 What joy there is in living.
4 There are more than two factors here.
5 Cats and dogs each hate the other.
6 The wharf could be seen at the farther shore.
7 The tiny girl took off her hat.
8 Write a fond note to the friend you cherish.
9 There was a sound of dry leaves outside.
10 Add the column and put the sum here.
The documentation for stringr::str_detect says, "Match character, word, line and sentence boundaries with boundary()." I can't figure out how to do this, nor can I find an example anywhere. All of the documentation examples involve the str_split or str_count functions.
My question is related to this question, but I would specifically like to understand how to use the stringr::boundary function.
We can specify the word boundary (\\b) at the start and end to avoid any partial matches
library(stringr)
library(dplyr)
df %>%
filter(str_detect(sentences, "\\bher\\b"))
# sentences
#1 Help the woman get back to her feet.
#2 The tiny girl took off her hat.
Or use boundary to wrap
df %>%
filter(str_detect(sentences, boundary("her")))
Related
I'm trying to create a filter for large dataframe. I'm trying to use grepl to search for a series of text within a specific column. I've done this for single words/combinations, but now I want to search for two words in close proximity (ie the word tumo(u)r within 3 words of the word colon).
I've checked my regular expression on https://www.regextester.com/109207 and my search works there, but it doesn't work within R.
The error I get is
Error: '\W' is an unrecognized escape in character string starting ""\btumor|tumour)\W"
Example below - trying to search for tumo(u)r within 3 words of cancer.
Can anyone help?
library(tibble)
example.df <- tibble(number = 1:4, AB = c('tumor of the colon is a very hard disease to cure', 'breast cancer is also known as a neoplasia of the breast', 'tumour of the colon is bad', 'colon cancer is also bad'))
filtered.df <- example.df %>%
filter(grepl(("\btumor|tumour)\W|\w+(\w+\W+){0,3}colon\b"), AB, ignore.case=T)
R uses backslashes as escapes and the regex engine does,too. Need to double your backslashes. This is explained in multiple prior questions on StackOverflow as well as in the help page brought up at ?regex. You should try to use the escaped operators in a more simple set of tests before attempting complex operations. And you should pay better attention to the proper placement of parentheses and quotes in the pattern argument.
filtered.df <- example.df %>%
#filter(grepl(("\btumor|tumour)\W|\w+(\w+\W+){0,3}colon\b"), AB,
# errors here ....^.^..............^..^...^..^.............^.^
filter(grepl( "(\\btumor|tumour)\\W|\\w+(\\w+\\W+){0,3}colon\\b", AB,
ignore.case=T) )
> filtered.df
# A tibble: 2 × 2
number AB
<int> <chr>
1 1 tumor of the colon is a very hard disease to cure
2 3 tumour of the colon is bad
Hello I have a data frame that looks something like this
dataframe <- data_frame(text = c('WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12',
'WUFF;other stuff to keep;WIFF2;yes yes IGWIFF'))
print(dataframe)
# A tibble: 2 × 1
text
<chr>
1 WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12
2 WUFF;other stuff to keep;WIFF2;yes yes IGWIFF
I want to extract the segment of the strings containing the word "keep". Note that these segments can be separated from other parts by different symbols for example , and ;.
the final dataset should look something like this.
final_dataframe <- data_frame(text = c('some words to keep',
'other stuff to keep'))
print(final_dataframe)
# A tibble: 2 × 1
text
<chr>
1 some words to keep
2 other stuff to keep
Does anyone know how I could do this?
With stringr ...
library(stringr)
library(dplyr)
dataframe %>%
mutate(text = trimws(str_extract(text, "(?<=[,;]).*keep")))
# A tibble: 2 × 1
text
<chr>
1 some words to keep
2 other stuff to keep
Created on 2022-02-01 by the reprex package (v2.0.1)
I've made great use of the positive lookbehind and positive lookahead group constructs -- check this out: https://regex101.com/r/Sc7h8O/1
If you want to assert that the text you're looking for comes after a character/group -- in your first case the apostrophe, use (?<=').
If you want to do the same but match something before ' then use (?=')
And you want to match between 0 and unlimited characters surrounding "keep" so use .* on either side, and you wind up with (?<=').*keep.*(?=')
I did find in my test that a string like text =' c('WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12', will also match the c(, which I didn't intend. But I assume your strings are all captured by pairs of apostrophes.
My dataframe column looks like this:
head(tweets_date$Tweet)
[1] b"It is #DineshKarthik's birthday and here's a rare image of the captain of #KKRiders. Have you seen him do this before? Happy birthday, DK\\xf0\\x9f\\x98\\xac
[2] b'The awesome #IPL officials do a wide range of duties to ensure smooth execution of work! Here\\xe2\\x80\\x99s #prabhakaran285 engaging with the #ChennaiIPL kid-squad that wanted to meet their daddies while the presentation was on :) #cutenessoverload #lineofduty \\xf0\\x9f\\x98\\x81
[3] b'\\xf0\\x9f\\x8e\\x89\\xf0\\x9f\\x8e\\x89\\n\\nCHAMPIONS!!
[4] b'CHAMPIONS - 2018 #IPLFinal
[5] b'Chennai are Super Kings. A fairytale comeback as #ChennaiIPL beat #SRH by 8 wickets to seal their third #VIVOIPL Trophy \\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86. This is their moment to cherish, a moment to savour.
[6] b"Final. It's all over! Chennai Super Kings won by 8 wickets
These are tweets which have mentions starting with '#', I need to extract all of them and save each mention in that particular tweet as "#mention1 #mention2". Currently my code just extracts them as lists.
My code:
tweets_date$Mentions<-str_extract_all(tweets_date$Tweet, "#\\w+")
How do I collapse those lists in each row to a form a string separated by spaces as mentioned earlier.
Thanks in advance.
I trust it would be best if you used an asis column in this case:
extract words:
library(stringr)
Mentions <- str_extract_all(lis, "#\\w+")
some data frame:
df <- data.frame(col = 1:6, lett = LETTERS[1:6])
create a list column:
df$Mentions <- I(Mentions)
df
#output
col lett Mentions
1 1 A #DineshK....
2 2 B #IPL, #p....
3 3 C
4 4 D
5 5 E #ChennaiIPL
6 6 F
I think this is better since it allows for quite easy sub setting:
df$Mentions[[1]]
#output
[1] "#DineshKarthik" "#KKRiders"
df$Mentions[[1]][1]
#output
[1] "#DineshKarthik"
and it succinctly shows whats inside the column when printing the df.
data:
lis <- c("b'It is #DineshKarthik's birthday and here's a rare image of the captain of #KKRiders. Have you seen him do this before? Happy birthday, DK\\xf0\\x9f\\x98\\xac",
"b'The awesome #IPL officials do a wide range of duties to ensure smooth execution of work! Here\\xe2\\x80\\x99s #prabhakaran285 engaging with the #ChennaiIPL kid-squad that wanted to meet their daddies while the presentation was on :) #cutenessoverload #lineofduty \\xf0\\x9f\\x98\\x81",
"b'\\xf0\\x9f\\x8e\\x89\\xf0\\x9f\\x8e\\x89\\n\\nCHAMPIONS!!",
"b'CHAMPIONS - 2018 #IPLFinal",
"b'Chennai are Super Kings. A fairytale comeback as #ChennaiIPL beat #SRH by 8 wickets to seal their third #VIVOIPL Trophy \\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86. This is their moment to cherish, a moment to savour.",
"b'Final. It's all over! Chennai Super Kings won by 8 wickets")
The str_extract_all function from the stringr package returns a list of character vectors. So, if you instead want a list of single CSV terms, then you may try using sapply for a base R option:
tweets <- str_extract_all(tweets_date$Tweet, "#\\w+")
tweets_date$Mentions <- sapply(tweets, function(x) paste(x, collapse=", "))
Demo
Via Twitter's help site: "Your username cannot be longer than 15 characters. Your real name can be longer (20 characters), but usernames are kept shorter for the sake of ease. A username can only contain alphanumeric characters (letters A-Z, numbers 0-9) with the exception of underscores, as noted above. Check to make sure your desired username doesn't contain any symbols, dashes, or spaces."
Note that email addresses can be in tweets as can URLs with #'s in them (and not just the silly URLs with username/password in the host component). Thus, something like:
(^|[^[[:alnum:]_]#/\\!?=&])#([[:alnum:]_]{1,15})\\b
is likely a better, safer choice
This is the vector I have:
posts = c("originally by: cearainmy only concern with csm is they seem a bit insulated from players. they have private message boards where it appears most of their work goes on. i would bet they are posting more there than in jita speakers corner. i think that is unfortunate because its hard to know who to vote for if you never really see what positions they hold. its sort of like ccp used to post here on the forums then they stopped. so they got a csm to represent players and use jita park forum to interact. now the csm no longer posts there as they have their internal forums where they hash things out. perhaps we need a csm to the csm to find out what they are up to.i don't think you need to worry too much. the csm has had an internal forum for over 2 years, although it is getting used a lot more now than it was. a lot of what goes on in there is nda stuff that we couldn't discuss anyway.i am quite happy to give my opinion on any topic, to the extent that the nda allows, and i" , "fot those of you bleating about imagined nda scandals as you attempt to cast yourselves as the julian assange of eve, here's a quote from the winter summit thread:originally by: sokrateszday 3post dominion 0.0 (3hrs!)if i had to fly to iceland only for this session i would have done it. we had gathered a list of items and prepared it a bit. important things we went over were supercaps, force projection, empire building, profitability of 0.0, objectives for small gangs and of course sovereingty.the csm spent 3 hours talking to ccp about how dominion had changed 0.0, and the first thing on sokratesz's list is supercaps. its not hard to figure out the nature of the discussion.on the other hand, maybe you're right, and the csm's priority for this discussion was to talk about how underpowered and useless supercarriers are and how they needed triple the ehp and dps from their current levels?(it wasn't)"
I want a data frame as a result, that would contain words and the frequecy of times they occur.
So result should look something like:
word count
a 300
and 260
be 200
... ...
... ...
What I tried to do, was use tm
corpus <- VCorpus(VectorSource(posts))
corpus <-tm_map(corpus, removeNumbers)
corpus <-tm_map(corpus, removePunctuation)
m <- DocumentTermMatrix(corpus)
Running findFreqTerms(m, lowfreq =0, highfreq =Inf ) just gives me the words, so I understand its a sparse matrix, how do I extract the words and their frequency?
Is there a easier way to do this, maybe by not using tm at all?
posts = c("originally by: cearainmy only concern with csm is they seem a bit insulated from players. they have private message boards where it appears most of their work goes on. i would bet they are posting more there than in jita speakers corner. i think that is unfortunate because its hard to know who to vote for if you never really see what positions they hold. its sort of like ccp used to post here on the forums then they stopped. so they got a csm to represent players and use jita park forum to interact. now the csm no longer posts there as they have their internal forums where they hash things out. perhaps we need a csm to the csm to find out what they are up to.i don't think you need to worry too much. the csm has had an internal forum for over 2 years, although it is getting used a lot more now than it was. a lot of what goes on in there is nda stuff that we couldn't discuss anyway.i am quite happy to give my opinion on any topic, to the extent that the nda allows, and i" , "fot those of you bleating about imagined nda scandals as you attempt to cast yourselves as the julian assange of eve, here's a quote from the winter summit thread:originally by: sokrateszday 3post dominion 0.0 (3hrs!)if i had to fly to iceland only for this session i would have done it. we had gathered a list of items and prepared it a bit. important things we went over were supercaps, force projection, empire building, profitability of 0.0, objectives for small gangs and of course sovereingty.the csm spent 3 hours talking to ccp about how dominion had changed 0.0, and the first thing on sokratesz's list is supercaps. its not hard to figure out the nature of the discussion.on the other hand, maybe you're right, and the csm's priority for this discussion was to talk about how underpowered and useless supercarriers are and how they needed triple the ehp and dps from their current levels?(it wasn't)")
posts <- gsub("[[:punct:]]", '', posts) # remove punctuations
posts <- gsub("[[:digit:]]", '', posts) # remove numbers
word_counts <- as.data.frame(table(unlist( strsplit(posts, "\ ") ))) # split vector by space
word_counts <- with(word_counts, word_counts[ Var1 != "", ] ) # remove empty characters
head(word_counts)
# Var1 Freq
# 2 a 8
# 3 about 3
# 4 allows 1
# 5 although 1
# 6 am 1
# 7 an 1
Plain R solution, assuming all words are separated by space:
words <- strsplit(posts, " ", fixed = T)
words <- unlist(words)
counts <- table(words)
The names(counts) holds words, and values are the counts.
You might want to use gsub to get rid of (),.?: and 's, 't or 're as in your example. As in:
posts <- gsub("'t|'s|'t|'re", "", posts)
posts <- gsub("[(),.?:]", " ", posts)
You've got two options. Depends if you want word count per document, or for all documents.
All Documents
library(dplyr)
count <- as.data.frame(t(inspect(m)))
sel_cols <- colnames(count)
count$word <- rownames(count)
rownames(count) <- seq(length = nrow(count))
count$count <- rowSums(count[,sel_cols])
count <- count %>% select(word,count)
count <- count[order(count$count, decreasing=TRUE), ]
### RESULT of head(count)
# word count
# 140 the 14
# 144 they 10
# 4 and 9
# 25 csm 7
# 43 for 5
# 55 had 4
This should capture occurrences across all documents (by use of rowSums).
Per Document
I would suggesting using the tidytext package, if you want word frequency per document.
library(tidytext)
m_td <- tidy(m)
The tidytext package allows fairly intuitive text mining, including tokenization. It is designed to work in a tidyverse pipeline, so it supplies a list of stop words ("a", "the", "to", etc.) to exclude with dplyr::anti_join. Here, you might do
library(dplyr) # or if you want it all, `library(tidyverse)`
library(tidytext)
data_frame(posts) %>%
unnest_tokens(word, posts) %>%
anti_join(stop_words) %>%
count(word, sort = TRUE)
## # A tibble: 101 × 2
## word n
## <chr> <int>
## 1 csm 7
## 2 0.0 3
## 3 nda 3
## 4 bit 2
## 5 ccp 2
## 6 dominion 2
## 7 forum 2
## 8 forums 2
## 9 hard 2
## 10 internal 2
## # ... with 91 more rows
Can someone help me with how to find the most frequently used two and three words in a text using R?
My text is...
text <- c("There is a difference between the common use of the term phrase and its technical use in linguistics. In common usage, a phrase is usually a group of words with some special idiomatic meaning or other significance, such as \"all rights reserved\", \"economical with the truth\", \"kick the bucket\", and the like. It may be a euphemism, a saying or proverb, a fixed expression, a figure of speech, etc. In grammatical analysis, particularly in theories of syntax, a phrase is any group of words, or sometimes a single word, which plays a particular role within the grammatical structure of a sentence. It does not have to have any special meaning or significance, or even exist anywhere outside of the sentence being analyzed, but it must function there as a complete grammatical unit. For example, in the sentence Yesterday I saw an orange bird with a white neck, the words an orange bird with a white neck form what is called a noun phrase, or a determiner phrase in some theories, which functions as the object of the sentence. Theorists of syntax differ in exactly what they regard as a phrase; however, it is usually required to be a constituent of a sentence, in that it must include all the dependents of the units that it contains. This means that some expressions that may be called phrases in everyday language are not phrases in the technical sense. For example, in the sentence I can't put up with Alex, the words put up with (meaning \'tolerate\') may be referred to in common language as a phrase (English expressions like this are frequently called phrasal verbs\ but technically they do not form a complete phrase, since they do not include Alex, which is the complement of the preposition with.")
The tidytext package makes this sort of thing pretty simple:
library(tidytext)
library(dplyr)
data_frame(text = text) %>%
unnest_tokens(word, text) %>% # split words
anti_join(stop_words) %>% # take out "a", "an", "the", etc.
count(word, sort = TRUE) # count occurrences
# Source: local data frame [73 x 2]
#
# word n
# (chr) (int)
# 1 phrase 8
# 2 sentence 6
# 3 words 4
# 4 called 3
# 5 common 3
# 6 grammatical 3
# 7 meaning 3
# 8 alex 2
# 9 bird 2
# 10 complete 2
# .. ... ...
If the question is asking for counts of bigrams and trigrams, tokenizers::tokenize_ngrams is useful:
library(tokenizers)
tokenize_ngrams(text, n = 3L, n_min = 2L, simplify = TRUE) %>% # tokenize bigrams and trigrams
as_data_frame() %>% # structure
count(value, sort = TRUE) # count
# Source: local data frame [531 x 2]
#
# value n
# (fctr) (int)
# 1 of the 5
# 2 a phrase 4
# 3 the sentence 4
# 4 as a 3
# 5 in the 3
# 6 may be 3
# 7 a complete 2
# 8 a phrase is 2
# 9 a sentence 2
# 10 a white 2
# .. ... ...
Your text is:
text <- c("There is a difference between the common use of the term phrase and its technical use in linguistics. In common usage, a phrase is usually a group of words with some special idiomatic meaning or other significance, such as \"all rights reserved\", \"economical with the truth\", \"kick the bucket\", and the like. It may be a euphemism, a saying or proverb, a fixed expression, a figure of speech, etc. In grammatical analysis, particularly in theories of syntax, a phrase is any group of words, or sometimes a single word, which plays a particular role within the grammatical structure of a sentence. It does not have to have any special meaning or significance, or even exist anywhere outside of the sentence being analyzed, but it must function there as a complete grammatical unit. For example, in the sentence Yesterday I saw an orange bird with a white neck, the words an orange bird with a white neck form what is called a noun phrase, or a determiner phrase in some theories, which functions as the object of the sentence. Theorists of syntax differ in exactly what they regard as a phrase; however, it is usually required to be a constituent of a sentence, in that it must include all the dependents of the units that it contains. This means that some expressions that may be called phrases in everyday language are not phrases in the technical sense. For example, in the sentence I can't put up with Alex, the words put up with (meaning \'tolerate\') may be referred to in common language as a phrase (English expressions like this are frequently called phrasal verbs\ but technically they do not form a complete phrase, since they do not include Alex, which is the complement of the preposition with.")
In Natural Language Processing, 2-word phrases are referred to as "bi-gram", and 3-word phrases are referred to as "tri-gram", and so forth. Generally, a given combination of n-words is called an "n-gram".
First, we install the ngram package (available on CRAN)
# Install package "ngram"
install.packages("ngram")
Then, we will find the most frequent two-word and three-word phrases
library(ngram)
# To find all two-word phrases in the test "text":
ng2 <- ngram(text, n = 2)
# To find all three-word phrases in the test "text":
ng3 <- ngram(text, n = 3)
Finally, we will print the objects (ngrams) using various methods as below:
print(ng, output="truncated")
print(ngram(x), output="full")
get.phrasetable(ng)
ngram::ngram_asweka(text, min=2, max=3)
We can also use Markov Chains to babble new sequences:
# if we are using ng2 (bi-gram)
lnth = 2
babble(ng = ng2, genlen = lnth)
# if we are using ng3 (tri-gram)
lnth = 3
babble(ng = ng3, genlen = lnth)
We can split the words and use table to summarize the frequency:
words <- strsplit(text, "[ ,.\\(\\)\"]")
sort(table(words, exclude = ""), decreasing = T)
Simplest?
require(quanteda)
# bi-grams
topfeatures(dfm(text, ngrams = 2, verbose = FALSE))
## of_the a_phrase the_sentence may_be as_a in_the in_common phrase_is
## 5 4 4 3 3 3 2 2
## is_usually group_of
## 2 2
# for tri-grams
topfeatures(dfm(text, ngrams = 3, verbose = FALSE))
## a_phrase_is group_of_words of_a_sentence of_the_sentence for_example_in example_in_the
## 2 2 2 2 2 2
## in_the_sentence an_orange_bird orange_bird_with bird_with_a
# 2 2 2 2
Here's a simple base R approach for the 5 most frequent words:
head(sort(table(strsplit(gsub("[[:punct:]]", "", text), " ")), decreasing = TRUE), 5)
# a the of in phrase
# 21 18 12 10 8
What it returns is an integer vector with the frequency count and the names of the vector correspond to the words that were counted.
gsub("[[:punct:]]", "", text) to remove punctuation since you don't want to count that, I guess
strsplit(gsub("[[:punct:]]", "", text), " ") to split the string on spaces
table() to count unique elements' frequency
sort(..., decreasing = TRUE) to sort them in decreasing order
head(..., 5) to select only the top 5 most frequent words