I have a character vector that looks like this:
z <- c("./.", "To/TO", "my/PRP$", "starved/VBN", ",/,", "wretched/JJ") # test input
[9992] "./."
[9993] "To/TO"
[9994] "my/PRP$"
[9995] "starved/VBN"
[9996] ",/,"
[9997] "wretched/JJ"
I want to remove all entries that consist of three consecutive punctuation marks, resulting in something like this:
[9993] "To/TO"
[9994] "my/PRP$"
[9995] "starved/VBN"
[9997] "wretched/JJ"
I've tried different regex expressions:
sub("[:punct:]/[:punct:]", "", z)
and
sub("[:punct:]{3}", "", z)
with either single/double brackets, both yield:
[9992] "./."
[9993] "To"
[9994] "my$"
[9995] "starved"
[9996] ",/,"
[9997] "wretched"
Any ideas? And I apologize in advance if the question is dumb; I'm not very good at this!
Try this:
x <- c("./.", "To/TO", "my/PRP$", "starved/VBN", ",/,", "wretched/JJ") # test input
grep("[[:punct:]]{3}", x, value = TRUE, invert = TRUE)
## [1] "To/TO" "my/PRP$" "starved/VBN" "wretched/JJ"
Related
I'd like to create a function in r using regular expressions that extracts hashtags (and one for #'s as well) but checks to see if its a part of a string and return those parts of that string. I'm still picking up hashtags (and #'s) and so I'm assuming that I'm not picking up pure hashtag strings (#word) because this is after using a function to remove URLs, emails, hashtags, and #'s via:
clean.text <- function(x){
x <- gsub("http[^[:space:]]+"," ", x)
x <- gsub("([_+A-Za-z0-9-]+(\\.[_+A-Za-z0-9-]+)*#[A-Za-z0-9-]+(\\.[A-Za-z0-9-]+)*(\\.[A-Za-z]{2,14}))","", x)
x <- gsub("\\s#[[:alnum:]_]+"," ", x)
x <- gsub("\\s#[^[:space:]]+"," ", x)
x
}
So I'd like to know what parts of the string are attached to the hashtags (and #'s) because I'm still getting hashtags (and #'s) when use the following on my cleaned text.
findHash2 <- function(x){
m <- gregexpr("(#\\w+)", x, perl=TRUE)
w <- unlist(regmatches(x,m))
op <- paste(w,collapse=" ")
return(op)
}
findAT2 <- function(x){
m <- gregexpr("#(\\w+)", x, perl=TRUE)
w <- unlist(regmatches(x,m))
op <- paste(w,collapse=" ")
return(op)
}
Note: again, this is after I apply my clean.text function to my text. Would it be something like this?
findHash1 <- function(x){
m <- gregexpr("(?<=^)#\\w+(?=$)", x, perl=TRUE)
w <- unlist(regmatches(x, m))
return(paste(w, collapse=" "))
}
UPDATE Example
x <- "yp#MonicaSarkar: RT #saultracey: Sun kissed .....#olmpicrings at #towerbridge #london2012 # Tower Bridge http://t.co/wgIutHUl
x <-I don'nt#know #It would %%%%#be #best if#you just.idk#provided/a fewexample#character! strings# #my#&^( 160,000+posts#in #of text #my) #data is#so huge!# (some# that #should match#and some that #shouldn't) and post# the desired#output.#We'll take it from there."
As for the desired output, I guess something like:
[1] yp#MonicaSarkar: #saultracey: .....#olmpicrings
Or in the second example:
[1] don'nt#know if#you 160,000+posts#in %%%%#be fewexample#character!
Ultimately, I'd like to see what's attached to the hash tags.
I'd like to use a function or functions that would extract a hashtag (another function or set of functions for #'s) if part of a string in three scenarios and presented my attempt at the first: one that says it must be preceded and followed by one or more characters, another that matches if only followed by one or more characters and a third that matches only if preceded by one or more characters. That is: one that would match the string hashtag only if it's at the middle not if it's present at the start or at the end of a string, one that would match the string only if it's present at the start, and one that would match the string if it's present at the end.
Would three functions like I discussed need to be created for that type of procedure or could it be combined into one?
I have an equation as a string where the variables in the string equation are variables in the R workspace. I would like to replace each variable with its numeric value in the R workspace. This is easy enough when the variable names don't contain punctuation.
Here is a simple example.
x <- 5
y <- 10
yy <- 15
z <- x*(y + yy)
zAsChar <- "z=x*(y+yy)"
vars <- unlist(strsplit(zAsChar, "[[:punct:]]"))
notVars <- unlist(strsplit(zAsChar, "[^[:punct:]]"))
varsValues <- sapply(vars[vars != ""], FUN=function(aaa) get(aaa))
notVarsValues <- notVars[notVars != ""]
paste(paste0(varsValues, notVarsValues), collapse="")
This yields "125=5*(10+15)", which is great.
However, I would love the option to use underscores in the variable names so that I can use "subscripts" for variable names. I am using these strings in math mode in R markdown.
So I need a [:punct:] that excludes _. I tried using [\\+\\-\\*\\/\\(\\)\\=] rather than [:punct:], but with this approach I couldn't split on the minus sign. Is there a way to preserve the _?
Instead of [:punct:] use the unicode character class \pP (shortcut for \p{P}) and its negation \PP to do that:
[^\\PP_]
(It works with perl=TRUE option)
Are you sure you need to do all this string manipulation? The substitute() function can help you out
substitute(z==x*(y+yy), list(x=x, y=y, yy=yy,z=z))
Or if you really need to start with a character value
do.call("substitute", list(parse(text=zAsChar)[[1]],list(x=x, y=y, yy=yy,z=z)))
# 125 = 5 * (10 + 15)
You can deparse() the result to turn it back into a character.
I have a character vector as:
x<- "\t\t<taxon id=\"TOT_F50\"/>"
and
y<- "TOT_A01"
and I want replace TOT_F50 with the text in y ("TOT_A01").
Do you know how to replace the text between " and \ (i.e. "TOT_F50) ?
Try
sub('(?<=").*(?=")', y, x, perl=TRUE)
#[1] "\t\t<taxon id=\"TOT_A01\"/>"
I would use something like
gsub("\".*\"", paste0("\"", y, "\""), x)
It just means "find text within two quotation marks in x and replace it with y inside two quotation marks"
I think this is what you want, your example is wrong though
I have seen a few questions concerning returning the position of a character with a String in R, but maybe I cannot seem to figure it out for my case. I think this is because I'm trying to do it for a whole column rather than a single string, but it could just be my struggles with regex.
Right now, I have a data.frame with a column, df$id that looks something like 13.23-45-6A. The number of digits before the period is variable, but I would like to retain just the part of the string after the period for each row in the column. I would like to do something like:
df$new <- substring(df$id, 1 + indexOf(".", df$id))
So 12.23-45-6A would become 23-45-6A, 0.1B would become 1B, 4.A-A would become A-A and so on for an entire column.
Right now I have:
df$new <- substr(df$id, 1 + regexpr("\\\.", data.count$id),99)
Thanks for any advice.
As #AnandaMahto mentioned his comment, you would probably be better simplifying things and using gsub:
> x <- c("13.23-45-6A", "0.1B", "4.A-A")
> gsub("[0-9]*\\.(.*)", "\\1", x, perl = T, )
[1] "23-45-6A" "1B" "A-A"
To make this work with your existing data frame you can try:
df$id <- gsub("[0-9]*\\.(.*)", "\\1", df$id, perl = T, )
another way is to use strsplit. Using #Tims example
x <- c("13.23-45-6A", "0.1B", "4.A-A")
sapply(strsplit(x, "\\."), "[", -1)
"23-45-6A" "1B" "A-A"
You could remove the characters including the . using
sub('[^.]*\\.', '', x)
#[1] "23-45-6A" "1B" "A-A"
data
x <- c("13.23-45-6A", "0.1B", "4.A-A")
I have a list of alphanumeric characters that looks like:
x <-c('ACO2', 'BCKDHB456', 'CD444')
I would like the following output:
x <-c('ACO', 'BCKDHB', 'CD')
Any suggestions?
# dput(tmp2)
structure(c(432L, 326L, 217L, 371L, 179L, 182L, 188L, 268L, 255L,...,
), class = "factor")
You can use gsub for this:
gsub('[[:digit:]]+', '', x)
or
gsub('[0-9]+', '', x)
# [1] "ACO" "BCKDHB" "CD"
If your goal is just to remove numbers, then the removeNumbers() function removes numbers from a text. Using it reduces the risk of mistakes.
library(tm)
x <-c('ACO2', 'BCKDHB456', 'CD444')
x <- removeNumbers(x)
x
[1] "ACO" "BCKDHB" "CD"
Using stringr
Most stringr functions handle regex
str_replace_all will do what you need
str_replace_all(c('ACO2', 'BCKDHB456', 'CD444'), "[:digit:]", "")
A solution using stringi:
# your data
x <-c('ACO2', 'BCKDHB456', 'CD444')
# extract capital letters
x <- stri_extract_all_regex(x, "[A-Z]+")
# unlist, so that you have a vector
x <- unlist(x)
Solution in one line: