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Using regex in R to find strings as whole words (but not strings as part of words)
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I might be missing something very obvious but how can I write efficient code to get all matches of a singular version of a noun but NOT its plural? for example, I want to match
angel investor
angel
BUT NOT
angels
try angels
If I try
grep("angel ", string)
Then a string with JUST the word
angel
won't match.
Please help!
Use word-boundary markers \\b:
x <- c("angel investor", "angel","angels", "try angels")
grep("\\bangel\\b", x, value = T)
[1] "angel investor" "angel"
You can try the following approach. It still believe there are other excellent ways to solve this problem.
df <- data.frame(obs = 1:4, words = c("angle", "try angles", "angle investor", "angles"))
df %>%
filter(!str_detect(words, "(?<=[ertkgwmnl])s\\b"))
# obs words
# 1 1 angle
# 2 3 angle investor
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I have a .csv-file with a column containing book descriptions scraped from the web which I import into R for further analysis. My goal is to extract the protagonists' ages from this column in R, so what I imagine is this:
Match strings like "age" and "-year-old" with a regex
Copy the sentences containing these strings into a new column (so that I can make sure that the sentence is not, for example "In the middle ages 50 people lived in xy"
Extract the numbers (and, if possible some number words) from this column into a new column.
The resulting table (or probably data.frame) would then hopefully look like this
|Description |Sentence |Age
|YY is a novel by Mr. X |The 12-year-old boy| 12
|about a boy. The 12-year|is named Dave. |
|-old boy is named Dave..| |
If you could me help out that would great since my R-skills are still very limited and I have not found a solution for this problem!
Another option if the string contains other numbers/descriptions besides just age, but you only want age.
library(stringr)
description <- "YY is a novel by Mr. X about a boy. The boy is 5 feet tall. The 12-year-old boy is named Dave. Dave is happy. Dave lives at 42 Washington street."
sentence <- str_split(description, "\\.")[[1]][which(grepl("-year-old", unlist(str_split(description, "\\."))))]
> sentence
[1] " The 12-year-old boy is named Dave"
age <- as.numeric(str_extract(description, "\\d+(?=-year-old)"))
> age
[1] 12
Here we use the string "-year-old" to tell us which sentence to pull and then we extract the age that is followed by that string.
You can try the following
library(stringr)
description <- "YY is a novel by Mr. X about a boy. The 12-year-old boy is named Dave. Dave is happy."
sentence <- str_extract(description, pattern = "\\.[^\\.]*[0-9]+[^\\.]*.") %>%
str_replace("^\\. ", "")
> sentence
[1] "The 12-year-old boy is named Dave."
age <- str_extract(sentence, pattern = "[0-9]+")
> age
[1] "12"
Edited to shorten and provide sample data.
I have text data consisting of 8 questions asked of a number of participants twice. I want to use text2vec to compare the similarity of their responses to these questions at the two points in time (duplicate detection). Here is how my initial data is structured (in this example there are just 3 participants, 4 questions instead of 8, and 2 quarters/time periods). I want to do similarity comparison for each participant's response in the first quarter vs. the second quarter. I intend to use package text2vec's psim command to do this.
df<-read.table(text="ID,Quarter,Question,Answertext
Joy,1,And another question,adsfjasljsdaf jkldfjkl
Joy,2,And another question,dsadsj jlijsad jkldf
Paul,1,And another question,adsfj aslj sd afs dfj ksdf
Paul,2,And another question,dsadsj jlijsad
Greg,1,And another question,adsfjasljsdaf
Greg,2,And another question, asddsf asdfasd sdfasfsdf
Joy,1,this is the first question that was asked,this is joys answer to this question
Joy,2,this is the first question that was asked,this is joys answer to this question
Paul,1,this is the first question that was asked,this is Pauls answer to this question
Paul,2,this is the first question that was asked,Pauls answer is different
Greg,1,this is the first question that was asked,this is Gregs answer to this question nearly the same
Greg,2,this is the first question that was asked,this is Gregs answer to this question
Joy,1,This is the text of another question,more random text
Joy,2,This is the text of another question, adkjjlj;ds sdafd
Paul,1,This is the text of another question,more random text
Paul,2,This is the text of another question, adkjjlj;ds sdafd
Greg,1,This is the text of another question,more random text
Greg,2,This is the text of another question,sdaf asdfasd asdff
Joy,1,this was asked second.,some random text
Joy,2,this was asked second.,some random text that doesn't quite match joy's response the first time around
Paul,1,this was asked second.,some random text
Paul,2,this was asked second.,some random text that doesn't quite match Paul's response the first time around
Greg,1,this was asked second.,some random text
Greg,2,this was asked second.,ada dasdffasdf asdf asdfa fasd sdfadsfasd fsdas asdffasd
", header=TRUE,sep=',')
I've done some more thinking and I believe the right approach is to split the dataframe into a list of dataframes, not separate items.
questlist<-split(df,f=df$Question)
then write a function to create the vocabulary for each question.
library(text2vec)
vocabmkr<-function(x) {
itoken(x$AnswerText, ids=x$ID) %>% create_vocabulary()%>% prune_vocabulary(term_count_min = 2) %>% vocab_vectorizer()
}
test<-lapply(questlist, vocabmkr)
But then I think I need to split the original dataframe into question-quarter combinations and apply the vocab from the other list to it and am not sure how to go about that.
Ultimately, I want a similarity score telling me if the participants are duplicating some or all of their responses from the first and second quarters.
EDIT: Here is how I would do this for a single question starting with the above dataframe.
quest1 <- filter(df,Question=="this is the first question that was asked")
quest1vocab <- itoken(as.character(quest1$Answertext), ids=quest1$ID) %>% create_vocabulary()%>% prune_vocabulary(term_count_min = 1) %>% vocab_vectorizer()
quest1q1<-filter(quest1,Quarter==1)
quest1q1<-itoken(as.character(quest1q1$Answertext),ids=quest1q1$ID) # tokenize question1 quarter 1
quest1q2<-filter(quest1,Quarter==2)
quest1q2<-itoken(as.character(quest1q2$Answertext),ids=quest1q2$ID) # tokenize question1 quarter 2
#now apply the vocabulary to the two matrices
quest1q1<-create_dtm(quest1q1,quest1vocab)
quest1q2<-create_dtm(quest1q2,quest1vocab)
similarity<-psim2(quest1q1, quest1q2, method="jaccard", norm="none") #row by row similarity.
b<-data.frame(ID=names(similarity),Similarity=similarity,row.names=NULL) #make dataframe of similarity scores
endproduct<-full_join(b,quest1)
Edit:
Ok, I have worked with the lapply some more.
df1<-split.data.frame(df,df$Question) #now we have 4 dataframes in the list, 1 for each question
vocabmkr<-function(x) {
itoken(as.character(x$Answertext), ids=x$ID) %>% create_vocabulary()%>% prune_vocabulary(term_count_min = 1) %>% vocab_vectorizer()
}
vocab<-lapply(df1,vocabmkr) #this gets us another list and in it are the 4 vocabularies.
dfqq<-split.data.frame(df,list(df$Question,df$Quarter)) #and now we have 8 items in the list - each list is a combination of question and quarter (4 questions over 2 quarters)
How do I apply the vocab list (consisting of 4 elements) to the dfqq list (consisting of 8)?
I'm sorry, that sounds frustrating. In case you have more to do and did want a more automatic way to do it, here's one approach that might work for you:
First, convert your example code for a single dataframe into a function:
analyze_vocab <- function(df_) {
quest1vocab =
itoken(as.character(df_$Answertext), ids = df_$ID) %>%
create_vocabulary() %>%
prune_vocabulary(term_count_min = 1) %>%
vocab_vectorizer()
quarter1 = filter(df_, Quarter == 1)
quarter1 = itoken(as.character(quarter1$Answertext),
ids = quarter1$ID)
quarter2 = filter(df_, Quarter == 2)
quarter2 = itoken(as.character(quarter2$Answertext),
ids = quarter2$ID)
q1mat = create_dtm(quarter1, quest1vocab)
q2mat = create_dtm(quarter2, quest1vocab)
similarity = psim2(q1mat, q2mat, method = "jaccard", norm = "none")
b = data.frame(
ID = names(similarity),
Similarity = similarity)
output <- full_join(b, df_)
return(output)
}
Now, you can split if you want and then use lapply like this: lapply(split(df, df$Question), analyze_vocab). However, you already seem comfortable with piping so you might as well go with that approach:
similarity_df <- df %>%
group_by(Question) %>%
do(analyze_vocab(.))
Output:
> head(similarity_df, 12)
# A tibble: 12 x 5
# Groups: Question [2]
ID Similarity Quarter Question Answertext
<fct> <dbl> <int> <fct> <fct>
1 Joy 0 1 And another question adsfjasljsdaf jkldfjkl
2 Joy 0 2 And another question "dsadsj jlijsad jkldf "
3 Paul 0 1 And another question adsfj aslj sd afs dfj ksdf
4 Paul 0 2 And another question dsadsj jlijsad
5 Greg 0 1 And another question adsfjasljsdaf
6 Greg 0 2 And another question " asddsf asdfasd sdfasfsdf"
7 Joy 1 1 this is the first question that was asked this is joys answer to this question
8 Joy 1 2 this is the first question that was asked this is joys answer to this question
9 Paul 0.429 1 this is the first question that was asked this is Pauls answer to this question
10 Paul 0.429 2 this is the first question that was asked "Pauls answer is different "
11 Greg 0.667 1 this is the first question that was asked this is Gregs answer to this question nearly the same
12 Greg 0.667 2 this is the first question that was asked this is Gregs answer to this question
The values in similarity match the ones shown in your example endproduct (note that values shown are rounded for tibble display), so it seems to be working as intended.
I gave up and did this manually one dataframe at a time. I'm sure there's a simple way to do it as a list but I can't for the life of me figure out how to apply a list of functions (the vocab vectorizers) to the "Answertext" column in the list of dataframes.
As powerful as R is, a simple for loop that allows text swapping into the command (a la Stata's "foreach") is grossly lacking. I get that there is a different workflow involving breaking a dataframe into a list and iterating over that but for some activities this complicates matters grossly, necessitating complex indexes to refer not just to the list but also to the specific vectors contained in the list. I also recognize that the Stata-like behavior can be achieved using assign and paste0 but this, like most code in R, is terribly clunky and obtuse. sigh.
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
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I have a dataset with unstructured text data.
From the text I want to extract sentences that have the following words:
education_vector <- c("university", "academy", "school", "college")
For example, from the text I am a student at the University of Wyoming. My major is biology. I want to get I am a student at the University of Wyoming.
From the text I love statistics and I enjoy working with numbers. I graduated from Walla Wall Community College I want to get I graduated from Walla Wall Community College. and so on
I tried using grep function but it returned wrong results
Answer modified to select first match.
texts = c("I am a student at the University of Wyoming. My major is biology.",
"I love statistics and I enjoy working with numbers. I graduated from Walla Wall Community College",
"First, I went to the Bowdoin College. Then I went to the University of California.")
gsub(".*?([^\\.]*(university|academy|school|college)[^\\.]*).*",
"\\1", texts, ignore.case=TRUE)
[1] "I am a student at the University of Wyoming"
[2] " I graduated from Walla Wall Community College"
[3] "First, I went to the Bowdoin College"
Explanation:
.*? is a non-greedy match up to the rest of the pattern. This is there to remove any sentences before the relevant sentence.
([^\\.]*(university|academy|school|college)[^\\.]*) matches any string of characters other than a period immediately before and after one of the key words.
.* handles anything after the relevant sentence.
This replaces the entire string with only the relevant part.
Here is a solution using grep
education <- c("university", "academy", "school", "college")
str1 <- "I am a student at the University of Wyoming. My major is biology."
str2 <- "I love statistics and I enjoy working with numbers. I graduated from Walla Wall Community College"
str1 <- tolower(str1) # we use tolower because "university" != "University"
str2 <- tolower(str2)
grep(paste(education, collapse = "|"), unlist(strsplit(str1, "(?<=\\.)\\s+",
perl = TRUE)),
value = TRUE)
grep(paste(education, collapse = "|"), unlist(strsplit(str2, "(?<=\\.)\\s+",
perl = TRUE)),
value = TRUE)