I'm trying to use stringr/dplyr to extract a pathway name from a table cell containing excess information. All cells in this table follow the same general format. Some examples are:
(R)-lactate from methylglyoxal: step 1/2. {ECO:0000256|ARBA:ARBA00005008, ECO:0000256|RuleBase:RU361179}.
(S)-dihydroorotate from bicarbonate: step 3/3. {ECO:0000256|ARBA:ARBA00004880}.
3,4',5-trihydroxystilbene biosynthesis
From these examples, I want to extract "(R)-lactate from methylglyoxal", "(S)-dihydroorotate from bicarbonate", and "3,4',5-trihydroxystilbene biosynthesis" respectively. I'm struggling to figure out which combination of regular expressions to use in order to accomplish this. I've been trying to use the positive look behind assertion ?<=... along with str_extract to extract all information preceding the first ":", but I can't get it to work. Any help would be appreciated!
please try the following pattern:
(?<=^)(.+?)(:|$)
(?<=^) the first part is looking exclusively at the beginning of the sentence
(.+?)(:|$) the second part is looking for at least one character before first ":" or end of sentence
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You don't need any lookarounds, you can match the values using:
^[^\r\n:]+
The pattern matches:
^ Start of string
[^\r\n:]+ Match 1+ chars other than newlines or :
Regex demo
library(stringr)
s <- c("(R)-lactate from methylglyoxal: step 1/2. {ECO:0000256|ARBA:ARBA00005008, ECO:0000256|RuleBase:RU361179}.",
"(S)-dihydroorotate from bicarbonate: step 3/3. {ECO:0000256|ARBA:ARBA00004880}.",
"3,4',5-trihydroxystilbene biosynthesis")
str_extract(s, "^[^\\r\\n:]+")
Output
[1] "(R)-lactate from methylglyoxal"
[2] "(S)-dihydroorotate from bicarbonate"
[3] "3,4',5-trihydroxystilbene biosynthesis"
Related
I am trying to remove the decimal points in decimal numbers in R. Please note I want to keep the full stop of strings.
Example:
data= c("It's 6.00pm, and is late.")
I know that I have to use regex for this, but I am struggling. My desired output is:
"It's 6 00pm, and is late."
Thank you in advance.
Try this:
sub("(?<=\\d)\\.(?=\\d)", " ", data, perl = TRUE)
This solution uses lookbehind (?<=...) and lookahead (?=...)to assert that the period you wish to remove be enclosed by digits (thus avoiding matching the period at the sentence end). If you have several such cases within strings, then use gsubinstead of sub.
I suggest using a simple pattern to find the target text, then adding parenthesis to identify the parts of the matching text that you want to retain.
# Test data
data <- c("It's 6.00pm, and is late.")
The target pattern is a literal dot with a string of digits before and after it. \\d+ matches one or more digits and \\. matches a literal dot. Testing the pattern to see if it works:
grepl("\\d+\\.\\d+", data)
Result
TRUE
If we wanted too eliminate the whole thing we could do a simple replacement with an empty string. Testing if this targets the correct text:
sub("\\d+\\.\\d+", "", data)
Result
"It's pm, and is late."
Instead, to discard only a section of matched text we can identify the parts we want to keep, which is done by surrounding them with parenthesis. Once done we can refer to the captured text in the replacement. \\1 refers to the first chunk of text captured and \\2 refers to the second chunk of text, corresponding to the first and second sets of parenthesis
# pattern replacement
sub("(\\d+)\\.(\\d+)", "\\1\\2", data)
Result
[1] "It's 600pm, and is late."
This effectively removes the dot by omitting it from the replacement text.
This may be a very simple question but I have not much experience with regex expressions. This page is a good source of regex expressions but could not figure out how to include them into my following code:
data %>% filter(grepl("^A01H1", icl))
Question
I would like to extract the values in one column of my data frame starting with this A01H1 up to 2 more digits, for example A01H100, A01H140, A01H110. I could not find a solution despite my few attempts:
Attempts
I looked at this question from which I used ^A01H1[0-9].{2} to select up tot two more digits.
I tried with adding any character ^A01H1[0-9][0-9][x-y] to stop after two digits.
Any help would be much appreciated :)
You can use "^A01H1\\d{1,2}$".
The first part ("^A01H1"), you figured out yourself, so what are we doing in the second part ("\\d{1,2}$")?
\d includes all digits and is equivalent to [0-9], since we are working in R you need to escape \ and thus we use \\d
{1,2} indicates we want to have 1 or 2 matches of \\d
$ specifies the end of the string, so nothing should come afterwards and this prevents to match more than 2 digits
It looks as if you want to match a part of a string that starts with A01H1, then contains 1 or 2 digits and then is not followed with any digit.
You may use
^A01H1\d{1,2}(?!\d)
See the regex demo. If there can be no text after two digits at all, replace (?!\d) with $.
Details
^ - start of strinmg
A01H1 - literal string
\d{1,2} - one to two digits
(?!\d) - no digit allowed immediately to the right
$ - end of string
In R, you could use it like
grepl("^A01H1\\d{1,2}(?!\\d)", icl, perl=TRUE)
Or, with the string end anchor,
grepl("^A01H1\\d{1,2}$", icl)
Note the perl=TRUE is only necessary when using PCRE specific syntax like (?!\d), a negative lookahead.
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)")
I am scraping PDFs for data and am trying to search for a numeric character (1:9) that is either of length 1 or 2. Unfortunately the value I am after changes position across the PDFs so I cannot simply call the index of the value and assign it to a variable.
I have tried many regex functions and can get numbers out of the list, but cannot seem to implement the argument to only pull numbers of the specific length.
# Data comes in as a long string
Test<-("82026-424 82026-424 1 CSX10 Store Room 75.74 75.74")
# Seperate data into individual pieces with str_split
Split_Test<-str_split(Test[1],"\\s+")
# We can easily unlist it with the following code (Not sure if needed)
Test_Unlisted<-unlist(Split_Test)
> Test_Unlisted
[1] "82026-424" "82026-424" "1" "CSX10" "Store" "Room"
[8] "75.74" "75.74"
My desired outcome would be to get the "1" out of the character list, and then if the value was "20" also be able to recognize that.
The best logic I can think of in code exists below, but this does not work.:
Test_Final<-str_match(Test_Unlisted, "\\d|\\d\\d")
Using this code I can grab anything of length=1, but it is not guaranteed to be a character:
Test_Final<-which(sapply(Test_Unlisted, nchar)==1)
Thanks for all the help!
You need to use
Test<-("82026-424 82026-424 1 CSX10 Store Room 75.74 75.74, 20")
regmatches(Test, gregexpr("\\b(?<!\\d\\.)\\d{1,2}\\b(?!\\.\\d)", Test, perl=TRUE))
See the regex demo and the regex demo.
Details
\b - a word boundary
(?<!\d\.) - a negative lookbehind that fails the match if, immediately to the left of the current location, there is a digit and a dot
\d{1,2} - 1 or 2 digits
\b - a word boundary
(?!\.\d) - a negative lookahead that fails the match if, immediately to the right of the current location, there is a dot and a digit.
Note that due to the lookarounds used in the pattern, the regex should be passed to the PCRE regex engine, hence the perl=TRUE argument is required.
With stringr that is ICU regex engine powered, you may use
library(stringr)
str_extract_all(Test, "\\b(?<!\\d\\.)\\d{1,2}\\b(?!\\.\\d)")
I have a question.
My text file contains lines such as:
1.1 Description.
This is the description.
1.1.1 Quality Assurance
Random sentence.
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
I'm trying to find out how to get:
1.1 Description
1.1.1 Quality Assurance
1.6.1 Quality Control
Right now, I have:
txt1 <- readLines("text1.txt")
txt2<-grep("^[0-9.]+", txt1, value = TRUE)
file<-write(txt2, "text3.txt")
which results in:
1.1 Description.
1.1.1 Quality Assurance
1.6.1 Quality Control. Quality Control is the responsibility of the contractor.
You are using grep with value=TRUE, which
returns a character vector containing the selected elements of x
(after coercion, preserving names but no other attributes).
This means, that if your regular expression matches anything in the line, the all line will be returned. You managed to build your regular expression to match numbers in the begining of the line. So all the lines which begin with numbers get selected.
It seems that your goal is not to select the all line, but to select only until there is a line break or a period.
So, you need to adjust the regular expression to be more specific, and you need to extract only the matching portion of the line.
A regular expression that matches what you want can be:
"^([0-9]\\.?)+ .+?(\\.|$)"
It selects numbers with dots, followed by a space, followed by anything, and stops matching things when a . comes or the line ends. I recommend the following website to better understand what the regex does: https://regexr.com/
The next step is extracting from the given lines only the matching portion, and not the all line where the regex has a match. For this we'll use the function regexpr, which tells us where the matches are, and the function regmatches, which helps us extract those matches:
txt1 <- readLines("text.txt")
regmatches(txt1, regexpr("^([0-9]\\.?)+ .+?(\\.|$)", txt1))