I have been mucking around with regex strings and strsplit but can't figure out how to solve my problem.
I have a collection of html documents that will always contain the phrase "people own these". I want to extract the number immediately preceding this phrase. i.e. '732,234 people own these' - I'm hoping to capture the number 732,234 (including the comma, though I don't care if it's removed).
The number and phrase are always surrounded by a . I tried using Xpath but that seemed even harder than a regex expression. Any help or advice is greatly appreciated!
example string: >742,811 people own these<
-> 742,811
Could you please try following.
val <- "742,811 people own these"
gsub(' [a-zA-Z]+',"",val)
Output will be as follows.
[1] "742,811"
Explanation: using gsub(global substitution) function of R here. Putting condition here where it should replace all occurrences of space with small or capital alphabets with NULL for variable val.
Try using str_extract_all from the stringr library:
str_extract_all(data, "\\d{1,3}(?:,\\d{3})*(?:\\.\\d+)?(?= people own these)")
Related
I'm trying to eliminate apostrophes from my a column of words in my dataset using
str_replace(tidy_posts$word, "'", "")
But the vector it returns still contains apostrophes. The vector's class is character, so I can't understand why this won't work. My only guess is that because the words came from data I got from the reddit API, the encoding is funky or something.
edit: "[:punct:]" also doesn't work.
When you are having some "special" punctuation such as . or ' a way to select them in regex is by using \\ before the punctuation of interest. This example may help you:
ch = c("The vector's class is character, so I can't understand why this won't work")
> stringr::str_replace_all(ch, "\\'","")
[1] "The vectors class is character, so I cant understand why this wont work"
I have a column within a data frame with a series of identifiers in, a letter and 8 numbers, i.e. B15006788.
Is there a way to remove all instances of B15.... to make them empty cells (there’s thousands of variations of numbers within each category) but keep B16.... etc?
I know if there was just one thing I wanted to remove, like the B15, I could do;
sub(“B15”, ””, df$col)
But I’m not sure on the how to remove a set number of characters/numbers (or even all subsequent characters after B15).
Thanks in advance :)
Welcome to SO! This is a case of regex. You can use base R as I show here or look into the stringR package for handy tools that are easier to understand. You can also look for regex rules to help define what you want to look for. For what you ask you can use the following code example to help:
testStrings <- c("KEEPB15", "KEEPB15A", "KEEPB15ABCDE")
gsub("B15.{2}", "", testStrings)
gsub is the base R function to replace a pattern with something else in one or a series of inputs. To test our regex I created the testStrings vector for different examples.
Breaking down the regex code, "B15" is the pattern you're specifically looking for. The "." means any character and the "{2}" is saying what range of any character we want to grab after "B15". You can change it as you need. If you want to remove everything after "B15". replace the pattern with "B15.". the "" means everything till the end.
edit: If you want to specify that "B15" must be at the start of the string, you can add "^" to the start of the pattern as so: "^B15.{2}"
https://www.rstudio.com/wp-content/uploads/2016/09/RegExCheatsheet.pdf has a info on different regex's you can make to be more particular.
I'm studying the recent hashtag #BalanceTonPorc in one of my classes. I'm trying to get all the occurrences of this hashtag appearing in tweets, but of course nobody uses the same format.
Some people use #BalanceTonPorc, some #balancetonporc, and son on and so forth.
Using gsub, I've so far done this :
df$hashtags <- gsub(".alance.on.orc", "BalanceTonPorc", df$hashtags)
Which does what I want, and all variations of this hashtag are stored under the same one. But there are A LOT of other variations. Some people used #BalanceTonPorc... or #BalanceTonPorc.
Is there a way to have a RegEx that says I want everything that contains .alance.on.orc with every character possible after the hashtag, except , (because it separates hashtags)? Here is a screenshot to illustrate what I mean.
I'm also having another issue, in my frequency table I have twice #BalanceTonPorc, so I guess R must consider them to be different. Can you spot the difference?
You may use [^,]* to match any char but ,, 0+ occurrences:
gsub(".alance.on.orc[^,]*", "BalanceTonPorc", df$hashtags)
Or, to exactly match balancetonporc,
gsub("balancetonporc[^,]*", "BalanceTonPorc", df$hashtags, ignore.case=TRUE)
See a regex demo and an R online test:
x <- c("#balancetonPorc#%$%#$%^","#balancetonporc#%$%, text")
gsub("balancetonporc[^,]*", "BalanceTonPorc", x, ignore.case=TRUE)
# => [1] "#BalanceTonPorc" "#BalanceTonPorc, text"
I looked around both here and elsewhere, I found many similar questions but none which exactly answer mine. I need to clean up naming conventions, specifically replace/remove certain words and phrases from a specific column/variable, not the entire dataset. I am migrating from SPSS to R, I have an example of the code to do this in SPSS below, but I am not sure how to do it in R.
EG:
"Acadia Parish" --> "Acadia" (removes Parish and space before Parish)
"Fifth District" --> "Fifth" (removes District and space before District)
SPSS syntax:
COMPUTE county=REPLACE(county,' Parish','').
There are only a few instances of this issue in the column with 32,000 cases, and what needs replacing/removing varies and the cases can repeat (there are dozens of instances of a phrase containing 'Parish'), meaning it's much faster to code what needs to be removed/replaced, it's not as simple or clean as a regular expression to remove all spaces, all characters after a specific word or character, all special characters, etc. And it must include leading spaces.
I have looked at the replace() gsub() and other similar commands in R, but they all involve creating vectors, or it seems like they do. What I'd like is syntax that looks for characters I specify, which can include leading or trailing spaces, and replaces them with something I specify, which can include nothing at all, and if it does not find the specific characters, the case is unchanged.
Yes, I will end up repeating the same syntax many times, it's probably easier to create a vector but if possible I'd like to get the syntax I described, as there are other similar operations I need to do as well.
Thank you for looking.
> x <- c("Acadia Parish", "Fifth District")
> x2 <- gsub("^(\\w*).*$", "\\1", x)
> x2
[1] "Acadia" "Fifth"
Legend:
^ Start of pattern.
() Group (or token).
\w* One or more occurrences of word character more than 1 times.
.* one or more occurrences of any character except new line \n.
$ end of pattern.
\1 Returns group from regexp
Maybe I'm missing something but I don't see why you can't simply use conditionals in your regex expression, then trim out the annoying white space.
string <- c("Arcadia Parish", "Fifth District")
bad_words <- c("Parish", "District") # Write all the words you want removed here!
bad_regex <- paste(bad_words, collapse = "|")
trimws( sub(bad_regex, "", string) )
# [1] "Arcadia" "Fifth"
dataframename$varname <- gsub(" Parish","", dataframename$varname)
I am using the following code for finding number of occurrences of a word memory in a file and I am getting the wrong result. Can you please help me to know what I am missing?
NOTE1: The question is looking for exact occurrence of word "memory"!
NOTE2: What I have realized they are exactly looking for "memory" and even something like "memory," is not accepted! That was the part which has brought up the confusion I guess. I tried it for word "action" and the correct answer is 7! You can try as well.
#names=scan("hamlet.txt", what=character())
names <- scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character())
Read 28230 items
> length(grep("memory",names))
[1] 9
Here's the file
The problem is really Shakespeare's use of punctuation. There are a lot of apostrophes (') in the text. When the R function scan encounters an apostrophe it assumes it is the start of a quoted string and reads all characters up until the next apostrophe into a single entry of your names array. One of these long entries happens to include two instances of the word "memory" and so reduces the total number of matches by one.
You can fix the problem by telling scan to regard all quotation marks as normal characters and not treat them specially:
names <- scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character(), quote=NULL )
Be careful when using the R implementation of grep. It does not behave in exactly the same way as the usual GNU/Linux program. In particular, the way you have used it here WILL find the number of matching words and not just the total number of matching lines as some people have suggested.
As pointed by #andrew, my previous answer would give wrong results if a word repeats on the same line. Based on other answers/comments, this one seems ok:
names = scan('http://pastebin.com/raw.php?i=kC9aRvfB', what=character(), quote=NULL )
idxs = grep("memory", names, ignore.case = TRUE)
length(idxs)
# [1] 10