I have a string printed out like this:
"\"Jenna and Alex were making cupcakes.\", \"Jenna asked Alex whether all were ready to be frosted.\", \"Alex said that\", \" some of them \", \"were.\", \"He added\", \"that\", \"the rest\", \"would be\", \"ready\", \"soon.\", \"\""
(The "\" wasn't there. R just automatically prints it out.)
I would like to calculate how many non-empty segments there are in this string. In this case the answer should be 11.
I tried to convert it to a vector, but R ignores the quotation marks so I still ended up with a vector with length 1.
I don't know whether I need to extract those segments first and then count, or there're easier ways to do that.
If it's the former case, which regular expression function best suits my need?
Thank you very much.
You can use scan to convert your large string into a vector of individual ones, then use nchar to count the lengths. Assuming your large string is x:
y <- scan(text=x, what="character", sep=",", strip.white=TRUE)
Read 12 items
sum(nchar(y)>0)
[1] 11
I assume a segment is defined as anything between . or ,. An option using strsplit can be found as:
length(grep("\\w+", trimws(strsplit(str, split=",|\\.")[[1]])))
#[1] 11
Note: trimws is not mandatory in above statement. I have included so that one can get the value of each segment by just adding value = TRUE argument in grep.
Data:
str <- "\"Jenna and Alex were making cupcakes.\", \"Jenna asked Alex whether all were ready to be frosted.\", \"Alex said that\", \" some of them \", \"were.\", \"He added\", \"that\", \"the rest\", \"would be\", \"ready\", \"soon.\", \"\""
strsplit might be one possibility?
txt <- "Jenna and Alex were making cupcakes., Jenna asked Alex whether all were ready to be frosted.,
Alex said that, some of them , were., He added, that, the rest, would be, ready, soon.,"
a <- strsplit(txt, split=",")
length(a[[1]])
[1] 11
If the backslashes are part of the text it doesnt really change a lot, except for the last element which would have "\"" in it. By filtering that out, the result is the same:
txt <- "\"Jenna and Alex were making cupcakes.\", \"Jenna asked Alex whether all
were ready to be frosted.\", \"Alex said that\", \" some of them \",
\"were.\", \"He added\", \"that\", \"the rest\", \"would be\", \"ready\", \"soon.\", \"\""
a <- strsplit(txt, split=", \"")
length(a[[1]][a[[1]] != "\""])
[1] 11
This is an absurd idea, but it does work:
txt <- "\"Jenna and Alex were making cupcakes.\", \"Jenna asked Alex whether all were ready to be frosted.\", \"Alex said that\", \" some of them \", \"were.\", \"He added\", \"that\", \"the rest\", \"would be\", \"ready\", \"soon.\", \"\""
Txt <-
read.csv(text = txt,
header = FALSE,
colClasses = "character",
na.strings = c("", " "))
sum(!vapply(Txt, is.na, logical(1)))
Related
I am working with really long strings. How can I remove the first 4 words after a certain string pattern occurs? For example:
string <- "hello I am a user of stackoverflow and I am really happy with all the help the community usually offers when I'm in need of some coding expertise."
#remove the fist 4 words after and including "stackoverflow"
result
"hello I am a user of happy with all the help the community usually offers when I'm in need of some coding expertise."
Solution with base R
A one line solution:
pattern <- "stackoverflow"
string <- "hello I am a user of stackoverflow and I am really happy with all the help the community usually offers when I'm in need of some coding expertise."
gsub(paste0(pattern, "(\\W+\\w+){0,4}\\W*"), "", string)
#> [1] "hello I am a user of happy with all the help the community usually offers when I'm in need of some coding expertise."
How it works
Create the pattern you want with a regex:
"stackoverflow" followed by 4 words.
Definitely, check out ?regex for more info about it.
Words are identified by \\w+ and separators are identified by \\W+ (capital w, it includes spaces and special characters like the apostrophe that you have in the sentence)
(...){0,4} means that the combination of word and separator may repeat up to 4 times.
\\W* needs to identify a possible final separator, so that the remaining two pieces of the sentence won't have two separators dividing them. Try it without, you'll see what I mean.
gsub locates the pattern you want and replace it with "" (thus deliting it).
Handle Exceptions
Note that it works even for particular cases:
# end of a sentence with fewer than 4 words after
string <- "hello I am a user of stackoverflow and I am"
gsub(paste0(pattern, "(\\W+\\w+){0,4}\\W*"), "", string)
#> [1] "hello I am a user of "
# beginning of a sentence
string <- "stackoverflow and I am really happy with all the help the community usually offers when I'm in need of some coding expertise."
gsub(paste0(pattern, "(\\W+\\w+){0,4}\\W*"), "", string)
#> [1] "happy with all the help the community usually offers when I'm in need of some coding expertise."
# pattern == string
string <- "stackoverflow and I am really"
gsub(paste0(pattern, "(\\W+\\w+){0,4}\\W*"), "", string)
#> [1] ""
A tidyverse solution
library(stringr)
# locate start and end position of pattern
tmp <- str_locate(string, paste0(pattern,"(\\W+\\w+){0,4}\\W*"))
# get positions: start_sentence-start_pattern and end_pattern-end_sentence
tmp <- invert_match(tmp)
# get the substrings
tmp <- str_sub(string, tmp[,1], tmp[,2])
# collapse substrings together
str_c(tmp, collapse = "")
#> [1] "hello I am a user of happy with all the help the community usually offers when I'm in need of some coding expertise."
Search for your pattern with additional spaces and words after it. Find the positions of the first last match, split the string and paste it back together. At the end gsub any double (or more) spaces.
string <- "hello I am a user of stackoverflow and I am really happy with all the help the community usually offers when I'm in need of some coding expertise."
pat="stackoverflow"
library(stringr)
tmp=str_locate(
string,
paste0(
pat,
paste0(
rep("\\s?[a-zA-Z]+",4),
collapse=""
)
)
)
gsub("\\s{2,}"," ",
paste0(
substring(string,1,tmp[1]-1),
substring(string,tmp[2]+1)
)
)
[1] "hello I am a user of happy with all the help the community usually offers when I'm in need of some coding expertise."
Quick answer, I am sure you can have better code thant that:
string <- "hello I am a user of stackoverflow and I am really happy with all the help the community usually offers when I'm in need of some coding expertise."
t<-read.table(textConnection(string))
string2<-''
i<-0
j<-0
for(i in 1:length(t)){
if(t[i]=="stackoverflow"){
j=i
}else if(j>0){
if(i-j>4){
string2=paste0(string2, " " , t[i])
}
}else if(j==0){
if(i>1){
string2=paste0(string2, " " , t[i])
}else{
string2=t[i]
}
}
}
print(string2)
I have some strings of text (example below). As you can see each string was split at a period or question mark.
[1]"I am a Mr."
[2]"asking for help."
[3]"Can you help?"
[4]"Thank you ms."
[5]"or mr."
I want to collapse where the string ends with an abbreviation like mr., mrs. so the end result would be the desired output below.
[1]"I am a Mr. asking for help."
[2]"Can you help?"
[3]"Thank you ms. or mr."
I already created a vector (called abbr) containing all my abbreviations in the following format:
> abbr
[1] "Mr|Mrs|Ms|Dr|Ave|Blvd|Rd|Mt|Capt|Maj"
but I can't figure out how to use it in paste function to collapse. I have also tried using gsub (didn't work) to replace \n following abbreviation with a period with a space like this:
lines<-gsub('(?<=abbr\\.\\n)(?=[A-Z])', ' ', lines, perl=FALSE)
We can use tapply to collapse string and grepl to create groups to collapse.
x <- c("I am a Mr.", "asking for help.","Can you help?","Thank you ms.", "or Mr.")
#Include all the abbreviations with proper cases
#Note that "." has a special meaning in regex so you need to escape it.
abbr <- 'Mr\\.|Mrs\\.|Ms\\.|Dr\\.|mr\\.|ms\\.'
unname(tapply(x, c(0, head(cumsum(!grepl(abbr, x)), -1)), paste, collapse = " "))
#[1] "I am a Mr. asking for help." "Can you help?" "Thank you ms. or mr."
R struggles. I am using the following to extract quotations from text, with multiple results on a large datset. I am trying to have the output be a character string within a dataframe, so I can easily share this as an csv with others.
Sample data:
normalCase <- 'He said, "I am a test," very quickly.'
endCase <- 'This is a long quote, which we said, "Would never happen."'
shortCase <- 'A "quote" yo';
beginningCase <- '"I said this," he said quickly';
multipleCase <- 'When asked, "No," said Sam "I do not like green eggs and ham."'
testdata = c(normalCase,endCase,shortCase,beginningCase,multipleCase)
Using the following to extract quotations and a buffer of characters:
result <-function(testdata) {
str_extract_all(testdata, '[^\"]?{15}"[^\"]+"[^\"]?{15}')
}
extract <- sapply(testdata, FUN=result)
The extract is a list within a matrix. However, I want the extract to be a character string that I can later merge to a dataframe as a column. How do I convert this?
Code
normalCase <- 'He said, "I am a test," very quickly.'
endCase <- 'This is a long quote, which we said, "Would never happen."'
shortCase <- 'A "quote" yo';
beginningCase <- '"I said this," he said quickly';
multipleCase <- 'When asked, "No," said Sam "I do not like green eggs and ham."'
testdata = c(normalCase,endCase,shortCase,beginningCase,multipleCase)
# extract quotations
gsub(pattern = "[^\"]*((?:\"[^\"]*\")|$)", replacement = "\\1 ", x = testdata)
Output
[1] "\"I am a test,\" "
[2] "\"Would never happen.\" "
[3] "\"quote\" "
[4] "\"I said this,\" "
[5] "\"No,\" \"I do not like green eggs and ham.\" "
Explanation
pattern = "[^\"]" will match with any character except a double quote
pattern = "[^\"]*" will match with any character except a double quote 0 or more times
pattern = "\"[^\"]*\"" will match with a double quote, then any
character except a double quote 0 or more times, then another double
quote (i.e.) quotations
pattern = "(?:\"[^\"]*\")" will match with quotations, but wont capture
it
pattern = "((?:\"[^\"]*\")|$)" will match with quotations or endOfString,
and capture it. Note that this is the first group we capture
replacement = "\\1 " will replace with the first group we captured followed by a space
For a multiple pattern matches (present in a character vector), i tried to apply grep(paste(States,collapse="|), Description). It works fine, but the problem here is that
Consider,
Descritpion=C("helloWorld Washington DC","Hello Stackoverflow////Newyork RBC")
States=C("DC","RBC","WA")
if the multiple pattern match for "WA" in the Description Vector. My function works for "helloWorld **Wa**shington DC" because "WA" is present. But i need a suggestion regarding the search pattern not in the whole String but at the end of String here with DC,RBC.
Thanks in advance
I guess you want something like the following. I've taken the liberty to clean up your example a bit.
Description <- c("helloWorld Washington DC", "Hello Stackoverflow", "Newyork RBC")
States <- c("DC","RBC","WA")
search.string <- paste0(States, "$", collapse = "|") # Construct the reg. exprs.
grep(search.string, Description, value = TRUE)
#[1] "helloWorld Washington DC" "Newyork RBC"
Note, we use $ to signify end-of-string match.
Referring the title, I'm figuring how to convert space between words to be %20 .
For example,
> y <- "I Love You"
How to make y = I%20Love%20You
> y
[1] "I%20Love%20You"
Thanks a lot.
Another option would be URLencode():
y <- "I love you"
URLencode(y)
[1] "I%20love%20you"
gsub() is one option:
R> gsub(pattern = " ", replacement = "%20", x = y)
[1] "I%20Love%20You"
The function curlEscape() from the package RCurl gets the job done.
library('RCurl')
y <- "I love you"
curlEscape(urls=y)
[1] "I%20love%20you"
I like URLencode() but be aware that sometimes it does not work as expected if your url already contains a %20 together with a real space, in which case not even the repeated option of URLencode() is doing what you want.
In my case, I needed to run both URLencode() and gsub consecutively to get exactly what I needed, like so:
a = "already%20encoded%space/a real space.csv"
URLencode(a)
#returns: "encoded%20space/real space.csv"
#note the spaces that are not transformed
URLencode(a, repeated=TRUE)
#returns: "encoded%2520space/real%20space.csv"
#note the %2520 in the first part
gsub(" ", "%20", URLencode(a))
#returns: "encoded%20space/real%20space.csv"
In this particular example, gsub() alone would have been enough, but URLencode() is of course doing more than just replacing spaces.