I am reading a PDF file using R. I would like to transform the given text in such a way, that whenever multiple spaces are detected, I want to replace them by some value (for example "_"). I've come across questions where all spaces of 1 or more can be replaced using "\\s+" (Merge Multiple spaces to single space; remove trailing/leading spaces) but this will not work for me. I have a string that looks something like this;
"[1]This is the first address This is the second one
[2]This is the third one
[3]This is the fourth one This is the fifth"
When I apply the answers I found; replacing all spaces of 1 or more with a single space, I will not be able to recognise separate addresses anymore, because it would look like this;
gsub("\\s+", " ", str_trim(PDF))
"[1]This is the first address This is the second one
[2]This is the third one
[3]This is the fourth one This is the fifth"
So what I am looking for is something like this
"[1]This is the first address_This is the second one
[2]This is the third one_
[3]This is the fourth one_This is the fifth"
However if I rewrite the code used in the example, I get the following
gsub("\\s+", "_", str_trim(PDF))
"[1]This_is_the_first_address_This_is_the_second_one
[2]This_is_the_third_one_
[3]This_is_the_fourth_one_This_is_the_fifth"
Would anyone know a workaround for this? Any help will be greatly appreciated.
Whenever I come across string and reggex problems I like to refer to the stringr cheat sheet: https://raw.githubusercontent.com/rstudio/cheatsheets/master/strings.pdf
On the second page you can see a section titled "Quantifiers", which tells us how to solve this:
library(tidyverse)
s <- "This is the first address This is the second one"
str_replace(s, "\\s{2,}", "_")
(I am loading the complete tidyverse instead of just stringr here due to force of habit).
Any 2 or more whitespace characters will no be replaced with _.
Related
I'd like to replace the content of a column in a data frame with only a specific word in that column.
The column always looks like this:
Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')
Place(fullName='Iphofen, Deutschland', name='Iphofen', type='city', country='Germany', countryCode='DE')
I'd like to extract the city name (in this case Würzburg or Iphofen) into a new column, or replace the entire row with the name of the town. There are many different towns so having a gsub-command for every city name will be tough.
Is there a way to maybe just use a gsub and tell Rstudio to replace whatever it finds inside the first two ' '?
Might it be possible to tell it, "give me the word after "name=' until the next '?
I'm very new to using R so I'm kind of out of ideas.
Thanks a lot for any help!
I know of the gsub command, but I don't think it will be the most appropriate in this case.
Yes, with a regular expression you can do exactly that:
string <- "Place(fullName='Würzburg, Germany', name='Würzburg', type='city', country='Germany', countryCode='DE')"
city <- gsub(".*name='(.*?)'.*", "\\1", string)
The regular expression says "match any characters followed by name=', then capture any characters until the next ' and then match any additional characters". Then you replace all of that with just the captured characters ("\\1").
The parentheses mean "capture this part", and the value becomes "\\1". (You can do multiple captures, with subsequent captures being \\2, \\3, etc.
Note the question mark in (.*?). This means "match as little as possible while still satisfying the rest of the regex". If you don't include the question mark, the regular expression will match "greedily" and you will capture the entire rest of the line instead of just the city since that would also satisfy the regular expression.
More about regular expression (specific to R) can be found here
I have a question similar to this one but instead of having two specific characters to look between, I want to get the text between a space and a specific character. In my example, I have this string:
myString <- "This is my string I scraped from the web. I want to remove all instances of a picture. picture-file.jpg. The text continues here. picture-file2.jpg"
but if I were to do something like this: str_remove_all(myString, " .*jpg) I end up with
[1] "This"
I know that what's happening is R is finding the first instance of a space and removing everything between that space and ".jpg" but I want it to be the first space immediately before ".jpg". My final result I hope for looks like this:
[1] "This is my string I scraped from the web. I want to remove all instances of a picture. the text continues here.
NOTE: I know that a solution may arise which does what I want, but ends up putting two periods next to each other. I do not mind a solution like that because later in my analysis I am removing punctuation.
You can use
str_remove_all(myString, "\\S*\\.jpg")
Or, if you also want to remove optional whitespace before the "word":
str_remove_all(myString, "\\s*\\S*\\.jpg")
Details:
\s* - zero or more whitespaces
\S* - zero or more non-whitespaces
\.jpg - .jpg substring.
To make it case insensitive, add (?i) at the pattern part: "(?i)\\s*\\S*\\.jpg".
If you need to make sure there is no word char after jpg, add a word boundary: "(?i)\\s*\\S*\\.jpg\\b"
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 have this vector Target <- c( "tes_1123_SS1G_340T01", "tes_23_SS2G_340T021". I want to remove anything before SS and anything after T0 (including T0).
Result I want in one line of code:
SS1G_340 SS2G_340
Code I have tried:
gsub("^.*?SS|\\T0", "", Target)
We can use str_extract
library(stringr)
str_extract(Target, "SS[^T]*")
#[1] "SS1G_340" "SS2G_340"
Try this:
gsub(".*(SS.*)T0.*","\\1",Target)
[1] "SS1G_340" "SS2G_340"
Why it works:
With regex, we can choose to keep a pattern and remove everything outside of that pattern with a two-step process. Step 1 is to put the pattern we'd like to keep in parentheses. Step 2 is to reference the number of the parentheses-bound pattern we'd like to keep, as sometimes we might have multiple parentheses-bound elements. See the example below for example:
gsub(".*(SS.*)+(T0.*)","\\1",Target)
[1] "SS1G_340" "SS2G_340"
Note that I've put the T0.* in parentheses this time, but we still get the correct answer because I've told gsub to return the first of the two parentheses-bound patterns. But now see what happens if I use \\2 instead:
gsub(".*(SS.*)+(T0.*)","\\2",Target)
[1] "T01" "T021"
The .* are wild cards by the way. If you'd like to learn more about using regex in R, here's a reference that can get you started.
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)