This question already has answers here:
Remove Rows From Data Frame where a Row matches a String
(6 answers)
Delete rows containing specific strings in R
(7 answers)
Closed 4 years ago.
I'd have hundreds of observations and I'd like to remove the ones that contain the string "english basement". I can't seem to find the right syntax to do so. I can only figure out how to keep observations with the that string. For instance, I used the code below to get only observations containing the string, and it worked perfectly:
eng_base <- zdata %>%
filter(str_detect(zdata$ListingDescription, “english basement”))
Now I want a data set,top_10mpEB, that excludes observations containing "english basement". Your help is greatly appreciated.
I do not know how your data looks like, but maybe this example helps you - I think you just need to negate the logical vector returned by str_detect:
library(dplyr)
library(stringr)
zdata <- data.frame(ListingDescription = c(rep("english basement, etc",3), letters[1:2] ))
zdata
# ListingDescription
#1 english basement, etc
#2 english basement, etc
#3 english basement, etc
#4 a
#5 b
zdata %>%
filter(!str_detect(ListingDescription, "english basement"))
# ListingDescription
#1: a
#2: b
Or using data.table package (no need of stringr::str_detect):
library(data.table)
setDT(zdata)
zdata[! ListingDescription %like% "english basement"]
# ListingDescription
#1: a
#2: b
You can do this using grepl():
x <- data.frame(ListingDescription = c('english basement other words description continued',
'great fireplace and an english basement',
'no basement',
'a house with a sauna!',
'the pool is great... and wait till you see the english basement!',
'new listing...will go fast'),
rent = c(3444, 23444, 346, 9000, 1250, 599))
x_english_basement <- x[grepl('english basement',
x$ListingDescription)==FALSE, ]
You can use dplyr to easily filter your dataframe.
library(dplyr)
new_data <- data %>%
filter(!ListingDescription=="english basement")
The ! became my best friend once I realized it meant "doesnt equal"
Related
This question already has answers here:
Using regex in R to find strings as whole words (but not strings as part of words)
(2 answers)
Closed 2 years ago.
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
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
I apologize if this question has been answered. I have searched this for way too long.
I have coded data that has a prefix of a letter and suffix of numbers.
ex:
A01, A02,...A99 ### (for each letter A-Z)
I need R code that mirrors this SAS code:
Proc SQL;
Create table NEW as
Select *
From DATA
Where VAR contains 'D';
Quit;
EDIT
Sorry y'all, I'm new! (also, mediocre in R at best.) I thought posting the SAS/SQL code would help make it easier.
Anyway, the data is manufacturing data. I have a variable whose values are the A01...A99, etc. values.
(rough) example of the dataframe:
OBS PRODUCT PRICE PLANT
1 phone 8.55 A87
2 paper 105.97 X67
3 cord .59 D24
4 monitor 98.65 D99
The scale of the data is massive, and I'm only wanting to focus on the observations that come from the plant 'D', so I'm trying to subset the data based on the 'PLANT' variable that contains (or starts with) 'D'. I know how to filter the data with a specific value (ie. ==, >=, != , etc.). I just can't figure out how to do it when only part of the value is known and I have yet to find anything about a 'contains' operator in R. I hope that clarifies things more.
Assuming DATA is your data.frame and VAR is your column value,
DATA <- data.frame(
VAR=apply(expand.grid(LETTERS[1:4], 1:3), 1, paste0, collapse=""),
VAL = runif(3*4)
)
then you can do
subset(DATA, grepl("D", VAR))
A slight alternative to MrFlick's solution: use a vector of row-indices:
DATA[grep('D', DATA$VAR), ]
VAR VAL
4 D1 0.31001091
8 D2 0.71562382
12 D3 0.00981055
where we defined:
DATA <- data.frame(
VAR=apply(expand.grid(LETTERS[1:4], 1:3), 1, paste0, collapse=""),
VAL = runif(3*4)
)
I have the following data:
> head(bigdata)
type text
1 neutral The week in 32 photos
2 neutral Look at me! 22 selfies of the week
3 neutral Inside rebel tunnels in Homs
4 neutral Voices from Ukraine
5 neutral Water dries up ahead of World Cup
6 positive Who's your hero? Nominate them
My duplicates will look like this (with empty $type):
7 Who's your hero? Nominate them
8 Water dries up ahead of World Cup
I remove duplicates like this:
bigdata <- bigdata[!duplicated(bigdata$text),]
The problem is, it removes the wrong duplicate. I want to remove the one where $type is empty, not the one that has a value for $type.
How can I remove a specific duplicate in R?
So here's a solution that does not use duplicated(...).
# creates an example - you have this already...
set.seed(1) # for reproducible example
bigdata <- data.frame(type=rep(c("positive","negative"),5),
text=sample(letters[1:10],10),
stringsAsFactors=F)
# add some duplicates
bigdata <- rbind(bigdata,data.frame(type="",text=bigdata$text[1:5]))
# you start here...
newdf <- with(bigdata,bigdata[order(text,type,decreasing=T),])
result <- aggregate(newdf,by=list(text=newdf$text),head,1)[2:3]
This sorts bigdata by text and type, in decreasing order, so that for a given text, the empty type will appear after any non-empty type. Then we extract only the first occurrence of each type for every text.
If your data really is "big", then a data.table solution will probably be faster.
library(data.table)
DT <- as.data.table(bigdata)
setkey(DT, text, type)
DT.result <- DT[, list(type = type[.N]), by = text]
This does basically the same thing, but since setkey sorts only in increasing order, we use type[.N] to get the last occurrence of type for a every text. .N is a special variable that holds the number of elements for that group.
Note that the current development version implements a function setorder(), which orders a data.table by reference, and can order in both increasing and decreasing order. So, using the devel version, it'd be:
require(data.table) # 1.9.3
setorder(DT, text, -type)
DT[, list(type = type[1L]), by = text]
You should keep rows that are either not duplicated or not missing a type value. The duplicated function only returns the second and later duplicates of each value (check out duplicated(c(1, 1, 2))), so we need to use both that value and the value of duplicated called with fromLast=TRUE.
bigdata <- bigdata[!(duplicated(bigdata$text) |
duplicated(bigdata$text, fromLast=TRUE)) |
!is.na(bigdata$type),]
foo = function(x){
x == ""
}
bigdata <- bigdata[-(!duplicated(bigdata$text)&sapply(bigdata$type, foo)),]