Hi I am trying to peform a specific pattern matching.
I want to stand. street names.
y <- c("Straße des 18 JAN.")
gsub("(.*)([1-3]?[0-9]\\.?)(JAN\\.?U?A?R?)(.*)","\\1 \\2 JANUAR \\4",y, perl=T)
What I want is that it keeps everything but rewrites the bracket 3 to JANUAR, so far i could not handle that.
Thanks in advance.
The regular expression has to be
gsub("(.*)([1-3]?[0-9]\\.?) (JAN\\.?U?A?R?)(.*)","\\1\\2 JANUAR\\4",y, perl=TRUE)
# [1] "Straße des 18 JANUAR"
I added a whitespace () before the term beginning with (JAN. Furthermore, I removed the whitespaces between \\1 and \\2 and between JANUAR and \\4.
Related
I got a lot of these strings in one txt-file:
X00NAP-0111-OG02Flur-A 2 AIR-CAP2702I-E-K9 00:b8:b8:b8:7d:b8 0111-HGS DE 10.100.100.100 8
X006NAP-0500-EG00Grossrau-A 2 AIR-CAP2702I-E-K9 50:0f:80:94:82:c0 HGS 0500 DE 10.100.100.100 1
Y008NAP-8399-OG04OE3020-A 2 AIR-CAP2702I-E-K9 00:b8:b8:b8:7d:b8 HGS Erfurter Hof DE 10.100.100.100 1
A1234NAP-4101-OG02Raum237-A 2 AIR-CAP2602I-E-K9 00:b8:b8:b8:7d:b8 AP 2 Anmeldung V DE 10.100.100.100 0
I am only interested in the first string and the number on the end of the lines. The number can be max. 99
So in the end I would like to have a output like this:
X00NAP-0111-OG02Flur-A 8
X006NAP-0500-EG00Grossrau-A 1
Y008NAP-8399-OG04OE3020-A 1
A1234NAP-4101-OG02Raum237-A 0
I tried a lot of things with regex, but nothing worked really.
Here is a general regex solution:
Find:
^([^\s]*).*(\d+)$
Replace:
$1 $2
The idea here is to match the first string and final number as capture groups, which are indicated by the two terms in the pattern surrounded by parentheses. These capture groups are made available in the replacement as $1 and $2 (sometimes \1 and \2, depending on the regex tool/engine). We can replace each line with these capture groups to leave you with the output you expect.
Note that this may "trash" the original file, but if you are using a tool like Notepad++, you can simply copy this result out, then undo the replacement, or just close the original file without saving.
Demo
The simplest way I can think of is:
Find: " .* "
Replace: " "
This replaces everything from the first space to the last space with a single space, achieving your goal.
Note: Quotes are only there to help show where spaces are in the regex.
Here's the thing:
test=" 2 15 3 23 12 0 0.18"
#I want to extract the 1st number separately
pattern="^ *(\\d+) +"
d=regmatches(test,gregexpr(pattern,test))
> d
[[1]]
[1] " 2 "
library(stringr)
f=str_extract(test,pattern)
> f
[1] " 2 "
They both bring whitespaces to the result despite usage of ()-brackets. Why? The brackets are for specifying which part of the matched pattern you want, am I wrong? I know I can trim them with trimws() or coerce them directly to numeric, but I wonder if I misunderstand some mechanics of patterns.
Using str_match (or str_match_all)
Since you want to extract a capture group, you can use str_match (or str_match_all). str_extract only extracts whole matches.
From R stringr help:
str_match Extract matched groups from a string.
and
str_extract to extract the complete match
R code:
library(stringr)
test=" 2 15 3 23 12 0 0.18"
pattern="^ *(\\d+) +"
f=str_match(test,pattern)
f[[2]]
## [1] "2"
The f[[2]] will output the 2nd item that is the first capture group value.
Using regmatches
As it is mentioned in the comment above, it is also possible with regmatches and regexec:
test=" 2 15 3 23 12 0 0.18"
pattern="^ *(\\d+) +"
res <- regmatches(test,regexec(pattern,test))
res[[1]][2] // The res list contains all matches and submatches
## [1] "2" // We get the item[2] from the first match to get "2"
See regexec help page that says:
regexec returns a list of the same length as text each element of which is either -1 if there is no match, or a sequence of integers with the starting positions of the match and all substrings corresponding to parenthesized subexpressions of pattern, with attribute "match.length" a vector giving the lengths of the matches (or -1 for no match).
OP task specific solution
Actually, since you only are interested in 1 integer number in the beginning of a string, you could achieve what you want with a mere gsub:
> gsub("^ *(\\d+) +.*", "\\1", test)
[1] "2"
This regex: (.*?)(?:I[0-9]-)*I3(?:-I[0-9])* matches an expression using multiple groups. The point of the regex is that it captures patterns in pairs of two, where the first part of the regex has to be followed by the second part of the regex.
How can I extract each of these two groups?
library(stringr)
data <- c("A-B-C-I1-I2-D-E-F-I1-I3-D-D-D-D-I1-I1-I2-I1-I1-I3-I3-I7")
str_extract_all(data, "(.*?)(?:I[0-9]-)*I3(?:-I[0-9])*")
Gives me:
[[1]]
[1] "A-B-C-I1-I2-D-E-F-I1-I3" "-D-D-D-D-I1-I1-I2-I1-I1-I3-I3-I7"
However, I would want something along the lines of:
[[1]]
[1] "A-B-C-I1-I2-D-E-F" [2] "I1-I3"
[[2]]
[1] "D-D-D-D" [2] "I1-I1-I2-I1-I1-I3-I3-I7"
The key here is that regex matches twice, each time containing 2 groups. I want every match to have a list of it's own, and that list to contain 2 elements, one for each group.
You need to wrap a capturing group around the second part of your expression and if you're using stringr for this task, I would use str_match_all instead to return the captured matches ...
library(stringr)
data <- c('A-B-C-I1-I2-D-E-F-I1-I3-D-D-D-D-I1-I1-I2-I1-I1-I3-I3-I7')
mat <- str_match_all(data, '-?(.*?)-((?:I[0-9]-)*I3(?:-I[0-9])*)')[[1]][,2:3]
colnames(mat) <- c('Group 1', 'Group 2')
# Group 1 Group 2
# [1,] "A-B-C-I1-I2-D-E-F" "I1-I3"
# [2,] "D-D-D-D" "I1-I1-I2-I1-I1-I3-I3-I7"
I want to have a regular expression that match anything that is not a correct mathematical number. the list below is a sample list as input for regex:
1
1.7654
-2.5
2-
2.
m
2..3
2....233..6
2.2.8
2--5
6-4-9
So the first three (in Bold) should not get selected and the rest should.
This is a close topic to another post but because of it's negative nature, it is different.
I'm using R but any regular expression will do I guess.
The following is the best shot in the mentioned post:
a <- c("1", "1.7654", "-2.5", "2-", "2.", "m", "2..3", "2....233..6", "2.2.8", "2--5", "6-4-9")
grep(pattern="(-?0[.]\\d+)|(-?[1-9]+\\d*([.]\\d+)?)|0$", x=a)
which outputs:
\[1\] 1 2 3 4 5 7 8 9 10 11
You can use following regex :
^(?:((\d+(?=[^.]+|\.{2,})).)+|(\d\.){2,}).*|[^\d]+$
See demo https://regex101.com/r/tZ3uH0/6
Note that your regex engine should support look-ahead with variable length.and you need to use multi-line flag and as mentioned in comment you can use perl=T to active look-ahead in R.
this regex is contains 2 part that have been concatenated with an OR.first part is :
(?:((\d+(?=[^.]+|\.{2,})).)+|(\d\.){2,}).*
which will match a combination of digits that followed by anything except dot or by 2 or more dot.which the whole of this is within a capture group that can be repeat and instead of this group you can have a digit which followed by dot 2 or more time (for matching some strings like 2.3.4.) .
and at the second part we have [^\d]+ which will match anything except digit.
Debuggex Demo
a[grep("^-?\\d*(\\.?\\d*)$", a, invert=T)]
With a suggested edit from #Frank.
Speed Test
a <- rep(a, 1e4)
all.equal(a[is.na(as.numeric(a))], a[grep("^-?\\d+(\\.?\\d+)?$|^\\d+\\.$", a, invert=T)])
[1] TRUE
library(microbenchmark)
microbenchmark(dosc = a[is.na(as.numeric(a))],
plafort = a[grep("^-?\\d*(\\.?\\d*)$", a, invert=T)])
# Unit: milliseconds
# expr min lq mean median uq max neval
# dosc 27.83477 28.32346 28.69970 28.51254 28.76202 31.24695 100
# plafort 31.92118 32.14915 32.62036 32.33349 32.71107 35.12258 100
I think this should do the job:
re <- "^-?[0-9]+$|^-?[0-9]+\\.[0-9]+$"
R> a[!grepl(re, a)]
#[1] "2-" "2." "m" "2..3" "2....233..6" "2.2.8" "2--5"
#[8] "6-4-9"
The solution here is good. You only have to add the negative case [-] and invert the selection!
a <- c("1", "1.7654", "-2.5", "2-", "2.", "m", "2..3", "2....233..6", "2.2.8", "2--5", "6-4-9")
a[grep(pattern="(^[1-9]\\d*(\\.\\d+)?$)|(^[-][1-9]\\d*(\\.\\d+)?$)",invert=TRUE, x=a)]
[1] "2-" "2." "m" "2..3" "2....233..6"
[6] "2.2.8" "2--5" "6-4-9"
Try this:
a[!grepl("^\\-?\\d?\\.?\\d+$", a)]
I like the simplicity of as.numeric(). This would be my suggestion:
require(stringr)
a <- c("1", "1.7654", "-2.5", "2-", "2.", "m", "2..3", "2....233..6", "2.2.8", "2--5", "6-4-9")
a
a1 <- ifelse(str_sub(a, -1) == ".", "string filler", a)
a1
outvect <- is.na(as.numeric(a1))
outvect
I match and replace 4-digit numbers preceded and followed by white space with:
str12 <- "coihr 1234 &/()= jngm 34 ljd"
sub("\\s\\d{4}\\s", "", str12)
[1] "coihr&/()= jngm 34 ljd"
but, every try to invert this and extract the number instead fails.
I want:
[1] 1234
does someone has a clue?
ps: I know how to do it with {stringr} but am wondering if it's possible with {base} only..
require(stringr)
gsub("\\s", "", str_extract(str12, "\\s\\d{4}\\s"))
[1] "1234"
regmatches(), only available since R-2.14.0, allows you to "extract or replace matched substrings from match data obtained by regexpr, gregexpr or regexec"
Here are examples of how you could use regmatches() to extract either the first whitespace-cushioned 4-digit substring in your input character string, or all such substrings.
## Example strings and pattern
x <- "coihr 1234 &/()= jngm 34 ljd" # string with 1 matching substring
xx <- "coihr 1234 &/()= jngm 3444 6789 ljd" # string with >1 matching substring
pat <- "(?<=\\s)(\\d{4})(?=\\s)"
## Use regexpr() to extract *1st* matching substring
as.numeric(regmatches(x, regexpr(pat, x, perl=TRUE)))
# [1] 1234
as.numeric(regmatches(xx, regexpr(pat, xx, perl=TRUE)))
# [1] 1234
## Use gregexpr() to extract *all* matching substrings
as.numeric(regmatches(xx, gregexpr(pat, xx, perl=TRUE))[[1]])
# [1] 1234 3444 6789
(Note that this will return numeric(0) for character strings not containing a substring matching your criteria).
It's possible to capture group in regex using (). Taking the same example
str12 <- "coihr 1234 &/()= jngm 34 ljd"
gsub(".*\\s(\\d{4})\\s.*", "\\1", str12)
[1] "1234"
I'm pretty naive about regex in general, but here's an ugly way to do it in base:
# if it's always in the same spot as in your example
unlist(strsplit(str12, split = " "))[2]
# or if it can occur in various places
str13 <- unlist(strsplit(str12, split = " "))
str13[!is.na(as.integer(str13)) & nchar(str13) == 4] # issues warning