I'm trying to merge two strings based on their matching suffix/prefix. For example, given the two strings 'a' and 'b' below, I'm first using Biostrings::pairwiseAlignement to get their common suffix/prefix, which in this case is "cutie". I then need to merge the two strings. Concatenating would not be helpful because I would get repeats.
This is all I have for now:
a= "bahahahallocutie"
b = "cutiepalaohaha"
pairwiseAlignment(a, b, type = "overlap")
Which gives me:
Overlap PairwiseAlignmentsSingleSubject (1 of 1)
pattern: [12] cutie
subject: [1] cutie
score: 17.20587
What I want to get is the merging the two strings by the pattern that's the suffix of one and prefix of the other:
"bahahahallocutiepalaohaha"
You may extract the pattern from the result of pairwiseAlignment. Then using gsub to remove the pattern from the strings you may use paste0 to get the desired merged string. Note that in your final code you'll need to take account for the order of the original strings.
library(Biostrings)
pat <- pairwiseAlignment(a, b, type = "overlap")#pattern
paste0(gsub(pat, "", a), pat, gsub(pat, "", b))
# [1] "bahahahallocutiepalaohaha"
Related
I have a table where one column has data like this:
table$test_string<- "[projectname](https://somewebsite.com/projectname/Abc/xyz-09)"
1.) I am trying to extract the first part of this string within the square brackets in one column, i.e.
table$project_name <- "projectname"
using the regex:
project_name <- "^\\[|(?:[a-zA-Z]|[0-9])+|\\]$"
table$project_name <- str_extract(table$test_string, project_name)
If I test the regex on 1 value (1 row individually) of the table, the above regex works with using
str_extract_all(table$test_string, project_name[[1]][2]).
However, I get NA when I apply the regex pattern to the whole table and an error if I use str_extract_all.
2.) Second part of the string, which is a URL in another column,
table$url_link <- "https://somewebsite.com/projectname/Abc/xyz-09"
I am using the following regex expression for URL:
url_pattern <- "http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*\\(\\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+"
table$url_link <- str_extract(table$test_string, url_pattern)
and this works on the whole table, however, I still get the ')' last paranthesis in the url link.
What am I missing here? and why does the first regex work individually and not on the whole table?
and for the url, how do I not get the last paranthesis?
It feels like you could simplify things considerably by using parentheses to group capture. For example:
test_string<- "[projectname](https://somewebsite.com/projectname/Abc/xyz-09)"
regex <- "\\[(.*)\\]\\((.*)\\)"
gsub(regex, "\\1", test_string)
#> [1] "projectname"
gsub(regex, "\\2", test_string)
#> [1] "https://somewebsite.com/projectname/Abc/xyz-09"
We can make use of convenient functions from qdapRegex
library(qdapRegex)
rm_round(test_string, extract = TRUE)[[1]]
#[1] "https://somewebsite.com/projectname/Abc/xyz-09"
rm_square(test_string, extract = TRUE)[[1]]
#[1] "projectname"
data
test_string<- "[projectname](https://somewebsite.com/projectname/Abc/xyz-09)"
I'm trying to extract certain records from a dataframe with grepl.
This is based on the comparison between two columns Result and Names. This variable is build like this "WordNumber" but for the same word I have multiple numbers (more than 30), so when I use the grepl expression to get for instance Word1 I get also results that I would like to avoid, like Word12.
Any ideas on how to fix this?
Names <- c("Word1")
colnames(Names) <- name
Results <- c("Word1", "Word11", "Word12", "Word15")
Records <- c("ThisIsTheResultIWant", "notThis", "notThis", "notThis")
Relationships <- data.frame(Results, Records)
Relationships <- subset(Relationships, grepl(paste(Names$name, collapse = "|"), Relationships$Results))
This doesn't work, if I use fixed = TRUE than it doesn't return any result at all (which is weird). I have also tried concatenating the name part with other numbers like this, but with no success:
Relationships <- subset(Relationships, grepl(paste(paste(Names$name, '3', sep = ""), collapse = "|"), Relationships$Results))
Since I'm concatenating I'm not really sure of how to use the \b to enforce a full match.
Any suggestions?
In addition to #Richard's solution, there are multiple ways to enforce a full match.
\b
"\b" is an anchor to identify word before/after pattern
> grepl("\\bWord1\\b",c("Word1","Word2","Word12"))
[1] TRUE FALSE FALSE
\< & \>
"\<" is an escape sequence for the beginning of a word, and ">" is used for end
> grepl("\\<Word1\\>",c("Word1","Word2","Word12"))
[1] TRUE FALSE FALSE
Use ^ to match the start of the string and $ to match the end of the string
Names <-c('^Word1$')
Or, to apply to the entire names vector
Names <-paste0('^',Names,'$')
I think this is just:
Relationships[Relationships$Results==Names,]
If you end up doing ^Word1$ you're just doing a straight subset.
If you have multiple names, then instead use:
Relationships[Relationships$Results %in% Names,]
Minimal Reprex
Suppose I have the string as1das2das3D. I want to extract everything from the letter a to the letter D. There are three different substrings that match this - I want the shortest / right-most match, i.e. as3D.
One solution I know to make this work is stringr::str_extract("as1das2das3D", "a[^a]+D")
Real Example
Unfortunately, I can't get this to work on my real data. In my real data I have string with (potentially) two URLs and I'm trying to extract the one that's immediately followed by rel=\"next\". So, in the below example string, I'd like to extract the URL https://abc.myshopify.com/ZifQ.
foo <- "<https://abc.myshopify.com/YifQ>; rel=\"previous\", <https://abc.myshopify.com/ZifQ>; rel=\"next\""
# what I've tried
stringr::str_extract(foo, '(?<=\\<)https://.*(?=\\>; rel\\="next)') # wrong output
stringr::str_extract(foo, '(?<=\\<)https://(?!https)+(?=\\>; rel\\="next)') # error
You could do:
stringr::str_extract(foo,"https:[^;]+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
or even
stringr::str_extract(foo,"https(?:(?!https).)+(?=>; rel=\"next)")
[1] "https://abc.myshopify.com/ZifQ"
Would this be an option?
Splitting string on ; or , comparing it with target string and take url from its previous index.
urls <- strsplit(foo, ";\\s+|,\\s+")[[1]]
urls[which(urls == "rel=\"next\"") - 1]
#[1] "<https://abc.myshopify.com/ZifQ>"
Here may be an option.
gsub(".+\\, <(.+)>; rel=\"next\"", "\\1", foo, perl = T)
#[1] "https://abc.myshopify.com/ZifQ"
I would like to convert this:
AIR-GEN-SUM-UD-ELA-NH-COMBINED-3-SEG1
to this:
ELA-3
I tried this function:
str_extract(.,pattern = ":?(ELA).*(\\d\\-)"))
it printed this:
"ELA-NH-COMBINED-3-"
I need to get rid of the text or anything between the two extracts. The number will be a number between 3 and 9. How should I modify my expression in pattern =?
Thanks!
1) Match everything up to -ELA followed by anything (.*) up to - followed by captured digits (\\d+)followed by - followed by anything. Then replace that with ELA- followed by the captured digits. No packages are used.
x <- "AIR-GEN-SUM-UD-ELA-NH-COMBINED-3-SEG1"
sub(".*-ELA.*-(\\d+)-.*", "ELA-\\1", x)
## [1] "ELA-3"
2) Another approach if there is only one numeric field is that we can read in the fields, grep out the numeric one and preface it with ELA- . No packages are used.
s <- scan(text = x, what = "", quiet = TRUE, sep = "-")
paste("ELA", grep("^\\d+$", s, value = TRUE), sep = "-")
## [1] "ELA-3"
TL;DR;
You can't do that with a single call to str_extract because you cannot match discontinuous portions of texts within a single match operation.
Again, it is impossible to match texts that are separated with other text into one group.
Work-arounds/Solutions
There are two solutions:
Capture parts of text you need and then join them (2 operations: match + join)
Capture parts of text you need and then replace with backreferences to the groups needed (1 replace operation)
Capturing groups only keep parts of text you match in separate memory buffers, but you also need a method or function that is capable of accessing these chunks.
Here, in R, str_extract drops them, but str_match keeps them in the result.
s <- "AIR-GEN-SUM-UD-ELA-NH-COMBINED-3-SEG1"
m <- str_match(s, ":?(ELA).*-(\\d+)")
paste0(m[,2], "-", m[,3])
This prints ELA-3. See R demo online.
Another way is to replace while capturing the parts you need to keep and then using backreferences to those parts in the replacement pattern:
x <- "AIR-GEN-SUM-UD-ELA-NH-COMBINED-3-SEG1"
sub("^.*-ELA.*?-([^-]+)-[^-]+$", "ELA-\\1", x)
See this R demo
I would like to use R to compare written text and extract sections which differ between the elements.
Consider a and b two text paragraphs. One is a modified version of the other:
a <- "This part is the same. This part is old."
b <- "This string is updated. This part is the same."
I want to compare the two strings and receive the part of the string which is unique to either of the two as output, preferably separate for both input strings.
Expected output:
stringdiff <- list(a = " This part is old.", b = "This string is updated. ")
> stringdiff
$a
[1] " This part is old."
$b
[1] "This string is updated. "
I've tried a solution from Extract characters that differ between two strings, but this only compares unique characters. The answer in Simple Comparing of two texts in R comes closer, but still only compares unique words.
Is there any way to get the expected output without too much of a hassle?
We concatenate both the strings, split at the space after the . to create a list of sentences ('lst'), get the unique elements from unlisting the 'lst' ('un1'), using setdiff we get the elements that are not in 'un1'
lst <- strsplit(c(a= a, b = b), "(?<=[.])\\s", perl = TRUE)
un1 <- unique(unlist(lst))
lapply(lst, setdiff, x= un1)