I have a vector a.
I want to extract the numbers between PUBMED and \nREFERENCE, which means the number is 32634600
I don't know how to code it using str_extract().
a = "234 4dfd 123PUBMED 32634600\nREFERENCE"
# expected output is 32634600
Using a lookbehind and stringr:
library(stringr)
str_extract_all(a, "(?<=PUBMED )[0-9]+")
[[1]]
[1] "32634600"
We can use sub() here with a capture group:
a <- "234 4dfd 123PUBMED 32634600\nREFERENCE"
num <- sub(".*PUBMED\\s*(\\d+)\\s*\\bREFERENCE\\b.*", "\\1", a)
num
[1] "32634600"
Related
I am in R and would like to extract a two digit number 38y from the following string:
"/Users/files/folder/file_number_23a_version_38y_Control.txt"
I know that _Control always comes after the 38y and that 38y is preceded by an underscore. How can I use strsplit or other R commands to extract the 38y?
You could use
regmatches(x, regexpr("[^_]+(?=_Control)", x, perl = TRUE))
# [1] "38y"
or equivalently
stringr::str_extract(x, "[^_]+(?=_Control)")
# [1] "38y"
Using gsub.
gsub('.*_(.*)_Control.*', '\\1', x)
# [1] "38y"
See demo with detailed explanation.
A possible solution:
library(stringr)
text <- "/Users/files/folder/file_number_23a_version_38y_Control.txt"
str_extract(text, "(?<=_)\\d+\\D(?=_Control)")
#> [1] "38y"
You can find an explanation of the regex part at:
https://regex101.com/r/PQSZHX/1
text <- c('d__Viruses|f__Closteroviridae|g__Closterovirus|s__Citrus_tristeza_virus',
'd__Viruses|o__Tymovirales|f__Alphaflexiviridae|g__Mandarivirus|s__Citrus_yellow_vein_clearing_virus',
'd__Viruses|o__Ortervirales|f__Retroviridae|s__Columba_palumbus_retrovirus')
I have tried but failed:
str_extract(text, pattern = 'f.*\\|')
How can I get
f__Closteroviridae
f__Alphaflexiviridae
f__Retroviridae
Any help will be high appreciated!
Make the regex non-greedy and since you don't want "|" in final output use positive lookahead.
stringr::str_extract(text, 'f.*?(?=\\|)')
#[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
In base R, we can use sub :
sub('.*(f_.*?)\\|.*', '\\1', text)
#[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
For a base R solution, I would use regmatches along with gregexpr:
m <- gregexpr("\\bf__[^|]+", text)
as.character(regmatches(text, m))
[1] "f__Closteroviridae" "f__Alphaflexiviridae" "f__Retroviridae"
The advantage of using gregexpr as above is that should an input contain more than one f__ matching term, we could also capture it. For example:
x <- 'd__Viruses|f__Closteroviridae|g__Closterovirus|f__some_virus'
m <- gregexpr("\\bf__[^|]+", x)
regmatches(x, m)[[1]]
[1] "f__Closteroviridae" "f__some_virus"
Data:
text <- c('d__Viruses|f__Closteroviridae|g__Closterovirus|s__Citrus_tristeza_virus',
'd__Viruses|o__Tymovirales|f__Alphaflexiviridae|g__Mandarivirus|s__Citrus_yellow_vein_clearing_virus',
'd__Viruses|o__Ortervirales|f__Retroviridae|s__Columba_palumbus_retrovirus')
I have a vector of URLs and need to extract a certain part of it. I've tried using a regex tester to see if my attempts worked, but they were no good.
The URLs I have are in this format: https://www.baseball-reference.com/teams/MIL/1976.shtml
I ned to extract the three letters after "teams/" (so for the example above, I need "MIL")
Does anyone have any idea how to get the correct regular expression to get this working? Thanks.
1) basename/dirname Try this:
u <- "https://www.baseball-reference.com/teams/MIL/1976.shtml" # input data
basename(dirname(u))
## [1] "MIL"
2) sub or with a regular expression:
sub(".*teams/(.*?)/.*", "\\1", u)
## [1] "MIL"
3) strsplit Split the string on / and take the second last component.
s <- strsplit(u, "/")[[1]]
s[length(s) - 1]
## [1] "MIL"
4) gsub Since the required substring is all upper case and no other characters in the input are this gsub which removes all characters that are not upper case letters would work:
gsub("[^A-Z]", "", u)
## [1] "MIL"
Many different ways to achieve this using regexp's. Here's one:
url <- "https://www.baseball-reference.com/teams/MIL/1976.shtml"
gsub(".+teams/(\\w{3}).+$", "\\1", url);
#[1] "MIL"
Or
x <- c('https://www.baseball-reference.com/teams/MIL/1976.shtml')
pattern <- "/teams/([^/]+)"
m <- regexec(pattern, x)
res = regmatches(x, m)[[1]]
res[2]
which yields
[1] "MIL"
Consider using the stringr package to simplify your code when handling strings.
Use a regular expression with positive lookbehind to catch alphanumeric codes following the string "teams\":
stringr::str_extract(url, "(?<=teams\\/)[A-Z]*")
In your case, if the URLs literally all begin with the same string https://www.baseball-reference.com/teams/ then you can avoid regex entirely and use a simple substring to get the three-letter code which follows:
stringr::str_sub(url, 42, 44)
Here are the results:
> url <- "https://www.baseball-reference.com/teams/MIL/1976.shtml"
>
> stringr::str_extract(url, "(?<=teams\\/)[A-Z]*")
[1] "MIL"
>
> stringr::str_sub(url, 42, 44)
[1] "MIL"
I have an string called PATTERN:
PATTERN <- "MODEL_Name.model-OUTCOME_any.outcome-IMP_number"
and I would like to parse the string using a pattern matching function, like grep, sub, ... to obtain a string variable MODEL equal to "Name.model", a string variable OUTCOME equal to "any.outcome" and an integer variable IMP equal to number.
If MODEL, OUTCOME and IMP were all integers, I could get the values using function sub:
PATTERN <- "MODEL_002-OUTCOME_007-IMP_001"
pattern_build <- "MODEL_([0-9]+)-OUTCOME_([0-9]+)-IMP_([0-9]+)"
MODEL <- as.integer(sub(pattern_build, "\\1", PATTERN))
OUTCOME <- as.integer(sub(pattern_build, "\\2", PATTERN))
IMP <- as.integer(sub(pattern_build, "\\3", PATTERN))
Do you have any idea of how to match the string contained in variable PATTERN?
Possible tricky patterns are:
PATTERN <- "MODEL_PS2-OUTCOME_stroke_i-IMP_001"
PATTERN <- "MODEL_linear-model-OUTCOME_stroke_i-IMP_001"
A solution which is also able to deal with the 'tricky' patterns:
PATTERN <- "MODEL_linear-model-OUTCOME_stroke_i-IMP_001"
lst <- strsplit(PATTERN, '([A-Z]+_)')[[1]][2:4]
lst <- sub('-$','',lst)
which gives:
> lst
[1] "linear-model" "stroke_i" "001"
And if you want that in a dataframe:
df <- as.data.frame.list(lst)
names(df) <- c('MODEL','OUTCOME','IMP')
which gives:
> df
MODEL OUTCOME IMP
1 linear-model stroke_i 001
A minimal-regex approach,
sapply(strsplit(PATTERN, '-'), function(i) sub('(.*?_){1}', '', i))
# [,1]
#[1,] "PS2"
#[2,] "stroke_i"
#[3,] "001"
You may use a pattern with capturing groups matching any chars, as few as possible between known delimiting substrings:
MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)
See the regex demo. Note that the last .* is greedy since you get all the rest of the string into this capture.
You may precise this pattern to only allow matching expected characters (say, to match digits into the last capturing group, use ([0-9]+) rather than (.*).
Use it with, say, str_match from stringr:
> library(stringr)
> x <- "MODEL_Name.model-OUTCOME_any.outcome-IMP_number"
> res <- str_match(x, "MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)")
> res[,2]
[1] "Name.model"
> res[,3]
[1] "any.outcome"
> res[,4]
[1] "number"
>
A base R solution using the same regex will involve a regmatches / regexec:
> res <- regmatches(x, regexec("MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)", x))[[1]]
> res[2]
[1] "Name.model"
> res[3]
[1] "any.outcome"
> res[4]
[1] "number"
>
I would like to remove specific characters from strings within a vector, similar to the Find and Replace feature in Excel.
Here are the data I start with:
group <- data.frame(c("12357e", "12575e", "197e18", "e18947")
I start with just the first column; I want to produce the second column by removing the e's:
group group.no.e
12357e 12357
12575e 12575
197e18 19718
e18947 18947
With a regular expression and the function gsub():
group <- c("12357e", "12575e", "197e18", "e18947")
group
[1] "12357e" "12575e" "197e18" "e18947"
gsub("e", "", group)
[1] "12357" "12575" "19718" "18947"
What gsub does here is to replace each occurrence of "e" with an empty string "".
See ?regexp or gsub for more help.
Regular expressions are your friends:
R> ## also adds missing ')' and sets column name
R> group<-data.frame(group=c("12357e", "12575e", "197e18", "e18947")) )
R> group
group
1 12357e
2 12575e
3 197e18
4 e18947
Now use gsub() with the simplest possible replacement pattern: empty string:
R> group$groupNoE <- gsub("e", "", group$group)
R> group
group groupNoE
1 12357e 12357
2 12575e 12575
3 197e18 19718
4 e18947 18947
R>
Summarizing 2 ways to replace strings:
group<-data.frame(group=c("12357e", "12575e", "197e18", "e18947"))
1) Use gsub
group$group.no.e <- gsub("e", "", group$group)
2) Use the stringr package
group$group.no.e <- str_replace_all(group$group, "e", "")
Both will produce the desire output:
group group.no.e
1 12357e 12357
2 12575e 12575
3 197e18 19718
4 e18947 18947
You do not need to create data frame from vector of strings, if you want to replace some characters in it. Regular expressions is good choice for it as it has been already mentioned by #Andrie and #Dirk Eddelbuettel.
Pay attention, if you want to replace special characters, like dots, you should employ full regular expression syntax, as shown in example below:
ctr_names <- c("Czech.Republic","New.Zealand","Great.Britain")
gsub("[.]", " ", ctr_names)
this will produce
[1] "Czech Republic" "New Zealand" "Great Britain"
Use the stringi package:
require(stringi)
group<-data.frame(c("12357e", "12575e", "197e18", "e18947"))
stri_replace_all(group[,1], "", fixed="e")
[1] "12357" "12575" "19718" "18947"
> library(stringi)
> group <- c('12357e', '12575e', '12575e', ' 197e18', 'e18947')
> pattern <- "e"
> replacement <- ""
> group <- str_replace(group, pattern, replacement)
> group
[1] "12357" "12575" "12575" " 19718" "18947"
You can use chartr as well:
group$group.no.e <- chartr("e", "", group$group)