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
Related
I just learnt R and was trying to clean data for analysis using R using string manipulation using the code given below for Amount_USD column of a table. I could not find why changes were not made. Please help.
Code:
csv_file2$Amount_USD <- ifelse(str_sub(csv_file$Amount_USD,1,10) == "\\\xc2\\\xa0",
str_sub(csv_file$Amount_USD,12,-1),csv_file2$Amount_USD)
Result:
\\xc2\\xa010,000,000
\\xc2\\xa016,200,000
\\xc2\\xa019,350,000
Expected Result:
10,000,000
16,200,000
19,350,000
You could use the following code, but maybe there is a more compact way:
vec <- c("\\xc2\\xa010,000,000", "\\xc2\\xa016,200,000", "\\xc2\\xa019,350,000")
gsub("(\\\\x[[:alpha:]]\\d\\\\x[[:alpha:]]0)([d,]*)", "\\2", vec)
[1] "10,000,000" "16,200,000" "19,350,000"
A compact way to extract the numbers is by using str_extract and negative lookahead:
library(stringr)
str_extract(vec, "(?!0)[\\d,]+$")
[1] "10,000,000" "16,200,000" "19,350,000"
How this works:
(?!0): this is negative lookahead to make sure that the next character is not 0
[\\d,]+$: a character class allowing only digits and commas to occur one or more times right up to the string end $
Alternatively:
str_sub(vec, start = 9)
There were a few minor issues with your code.
The main one being two unneeded backslashes in your matching statement. This also leads to a counting error in your first str_sub(), where you should be getting the first 8 characters not 10. Finally, you should be getting the substring from the next character after the text you want to match (i.e. position 9, not 12). The following should work:
csv_file2$Amount_USD <- ifelse(str_sub(csv_file$Amount_USD,1,8) == "\\xc2\\xa0", str_sub(csv_file$Amount_USD,9,-1),csv_file2$Amount_USD)
However, I would have done this with a more compact gsub than provided above. As long as the text at the start to remove is always going to be "\\xc2\\xa0", you can simply replace it with nothing. Note that for gsub you will need to escape all the backslashes, and hence you end up with:
csv_file2$Amount_USD <- gsub("\\\\xc2\\\\xa0", replacement = "", csv_file2$Amount_USD)
Personally, especially if you plan to do any sort of mathematics with this column, I would go the additional step and remove the commas, and then coerce the column to be numeric:
csv_file2$Amount_USD <- as.numeric(gsub("(\\\\xc2\\\\xa0)|,", replacement = "", csv_file2$Amount_USD))
I am trying to get to grips with the world of regular expressions in R.
I was wondering whether there was any simple way of combining the functionality of "grep" and "gsub"?
Specifically, I want to append some additional information to anything which matches a specific pattern.
For a generic example, lets say I have a character vector:
char_vec <- c("A","A B","123?")
Then lets say I want to append any letter within any element of char_vec with
append <- "_APPEND"
Such that the result would be:
[1] "A_APPEND" "A_APPEND B_APPEND" "123?"
Clearly a gsub can replace the letters with append, but this does not keep the original expression (while grep would return the letters but not append!).
Thanks in advance for any / all help!
It seems you are not familiar with backreferences that you may use in the replacement patterns in (g)sub. Once you wrap a part of the pattern with a capturing group, you can later put this value back into the result of the replacement.
So, a mere gsub solution is possible:
char_vec <- c("A","A B","123?")
append <- "_APPEND"
gsub("([[:alpha:]])", paste0("\\1", append), char_vec)
## => [1] "A_APPEND" "A_APPEND B_APPEND" "123?"
See this R demo.
Here, ([[:alpha:]]) matches and captures into Group 1 any letter and \1 in the replacement reinserts this value into the result.
Definatly not as slick as #Wiktor Stribiżew but here is what i developed for another method.
char_vars <- c('a', 'b', 'a b', '123')
grep('[A-Za-z]', char_vars)
gregexpr('[A-Za-z]', char_vars)
matches = regmatches(char_vars,gregexpr('[A-Za-z]', char_vars))
for(i in 1:length(matches)) {
for(found in matches[[i]]){
char_vars[i] = sub(pattern = found,
replacement = paste(found, "_append", sep=""),
x=char_vars[i])
}
}
Let's say I want a Regex expression that will only match numbers between 18 and 31. What is the right way to do this?
I have a set of strings that look like this:
"quiz.18.player.total_score"
"quiz.19.player.total_score"
"quiz.20.player.total_score"
"quiz.21.player.total_score"
I am trying to match only the strings that contain the numbers 18-31, and am currently trying something like this
(quiz.)[1-3]{1}[1-9]{1}.player.total_score
This obviously won't work because it will actually match all numbers between 11-39. What is the right way to do this?
Regex: 1[89]|2\d|3[01]
For matching add additional text and escape the dots:
quiz\.(?:1[89]|2\d|3[01])\.player\.total_score
Details:
(?:) non-capturing group
[] match a single character present in the list
| or
\d matches a digit (equal to [0-9])
\. dot
. matches any character
!) If s is the character vector read the fields into a data frame picking off the second field and check whether it is in the desired range. Put the result in logical vector ok and get those elements from s. This uses no regular expressions and only base R.
digits <- read.table(text = s, sep = ".")$V2
s[digits %in% 18:31]
2) Another approach based on the pattern "\\D" matching any non-digit is to remove all such characters and then check if what is left is in the desired range:
digits <- gsub("\\D", "", s)
s[digits %in% 18:31]
2a) In the development version of R (to be 3.6.0) we could alternately use the new whitespace argument of trimws like this:
digits <- trimws(s, whitespace = "\\D")
s[digits %in% 18:31]
3) Another alternative is to simply construct the boundary strings and compare s to them. This will work only if all the number parts in s are exactly the same number of digits (which for the sample shown in the question is the case).
ok <- s >= "quiz.18.player.total_score" & s <= "quiz.31.player.total_score"
s[ok]
This is done using character ranges and alternations. For your range
3[10]|[2][0-9]|1[8-9]
Demo
I have the following codes
"ABC.A.SVN.10.10.390.10.UDGGL"
"XYZ.Z.SVN.11.12.111.99.ASDDL"
and I need to replace the characters that exist between the 2nd and the 3rd dot. In this case it is SVN but it may well be any combination of between A and ZZZ, so really the only way to make this work is by using the dots.
The required outcome would be:
"ABC.A..10.10.390.10.UDGGL"
"XYZ.Z..11.12.111.99.ASDDL"
I tried variants of grep("^.+(\\.\\).$", "ABC.A.SVN.10.10.390.10.UDGGL") but I get an error.
Some examples of what I have tried with no success :
Link 1
Link 2
EDIT
I tried #Onyambu 's first method and I ran into a variant which I had not accounted for: "ABC.A.AB11.1.12.112.1123.UDGGL". In the replacement part, I also have numeric values. The desired outcome is "ABC.A..1.12.112.1123.UDGGL" and I get it using sub("\\.\\w+.\\B.",".",x) per the second part of his answer!
See code in use here
x <- c("ABC.A.SVN.10.10.390.10.UDGGL", "XYZ.Z.SVN.11.12.111.99.ASDDL")
sub("^(?:[^.]*\\.){2}\\K[^.]*", "", x, perl=T)
^ Assert position at the start of the line
(?:[^.]*\.){2} Match the following exactly twice
[^.]*\. Match any character except . any number of times, followed by .
\K Resets the starting point of the pattern. Any previously consumed characters are no longer included in the final match
[^.]* Match any character except . any number of times
Results in [1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
x= "ABC.A.SVN.10.10.390.10.UDGGL" "XYZ.Z.SVN.11.12.111.99.ASDDL"
sub("([A-Z]+)(\\.\\d+)","\\2",x)
[1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
([A-Z]+) Capture any word that has the characters A-Z
(\\.\\d+) The captured word above, must be followed with a dot ie\\..This dot is then followed by numbers ie \\d+. This completes the capture.
so far the captured part of the string "ABC.A.SVN.10.10.390.10.UDGGL" is SVN.10 since this is the part that matches the regular expression. But this part was captured as SVN and .10. we do a backreference ie replace the whole SVN.10 with the 2nd part .10
Another logic that will work:
sub("\\.\\w+.\\B.",".",x)
[1] "ABC.A..10.10.390.10.UDGGL" "XYZ.Z..11.12.111.99.ASDDL"
Not exactly regex but here is one more approach
#DATA
S = c("ABC.A.SVN.10.10.390.10.UDGGL", "XYZ.Z.SVN.11.12.111.99.ASDDL")
sapply(X = S,
FUN = function(str){
ind = unlist(gregexpr("\\.", str))[2:3]
paste(c(substring(str, 1, ind[1]),
"SUBSTITUTION",
substring(str, ind[2], )), collapse = "")
},
USE.NAMES = FALSE)
#[1] "ABC.A.SUBSTITUTION.10.10.390.10.UDGGL" "XYZ.Z.SUBSTITUTION.11.12.111.99.ASDDL"
I am trying to get to grips with the world of regular expressions in R.
I was wondering whether there was any simple way of combining the functionality of "grep" and "gsub"?
Specifically, I want to append some additional information to anything which matches a specific pattern.
For a generic example, lets say I have a character vector:
char_vec <- c("A","A B","123?")
Then lets say I want to append any letter within any element of char_vec with
append <- "_APPEND"
Such that the result would be:
[1] "A_APPEND" "A_APPEND B_APPEND" "123?"
Clearly a gsub can replace the letters with append, but this does not keep the original expression (while grep would return the letters but not append!).
Thanks in advance for any / all help!
It seems you are not familiar with backreferences that you may use in the replacement patterns in (g)sub. Once you wrap a part of the pattern with a capturing group, you can later put this value back into the result of the replacement.
So, a mere gsub solution is possible:
char_vec <- c("A","A B","123?")
append <- "_APPEND"
gsub("([[:alpha:]])", paste0("\\1", append), char_vec)
## => [1] "A_APPEND" "A_APPEND B_APPEND" "123?"
See this R demo.
Here, ([[:alpha:]]) matches and captures into Group 1 any letter and \1 in the replacement reinserts this value into the result.
Definatly not as slick as #Wiktor Stribiżew but here is what i developed for another method.
char_vars <- c('a', 'b', 'a b', '123')
grep('[A-Za-z]', char_vars)
gregexpr('[A-Za-z]', char_vars)
matches = regmatches(char_vars,gregexpr('[A-Za-z]', char_vars))
for(i in 1:length(matches)) {
for(found in matches[[i]]){
char_vars[i] = sub(pattern = found,
replacement = paste(found, "_append", sep=""),
x=char_vars[i])
}
}