I have strings like Sacher, Franz Xaver or Nishikawa, Kiyoko.
Using R, I want to change them to Sacher, F. X. or Nishikawa, K..
In other words, the first letter of each word after the comma should be retained with a dot (and a whitespace if another word follows).
Here is a related response, but it cannot be applied to my case 1:1 as it does not have a comma in its strings; it seems that the simple addition of (<?=, ) does not work.
E.g. in the following attempts, gsub() replaces everything, while my str_replace_all()-attempt leads to an error:
TEST <- c("Sacher, Franz Xaver", "Nishikawa, Kiyoko", "Al-Assam, Muhammad")
# first attempt
# (resembles the response from the other thread)
gsub('\\b(\\pL)\\pL{2,}|.','\\U\\1', TEST, perl = TRUE)
# second attempt
# error: "Incorrect unicode property"
stringr::str_replace_all(TEST, '(?<=, )\\b(\\pL)\\pL{2,}|.','\\U\\1')
I would be grateful for your help!
You can use
gsub("(*UCP)^[^,]+(*SKIP)(*F)|\\b(\\p{L})\\p{L}*", "\\U\\1.", TEST, perl=TRUE)
See the regex demo. Details:
(*UCP) - the PCRE verb that will make \b Unicode aware
^[^,]+(*SKIP)(*F) - start of string and then any zero or more chars other than a comma, and then the match is failed and skipped, the next match starts at the location where the failure occurred
| - or
\b - word boundary
(\p{L}) - Group 1: any Unicode letter
\p{L}* - zero or more Unicode letters
See the R demo:
TEST <- c("Sacher, Franz Xaver", "Nishikawa, Kiyoko", "Al-Assam, Muhammad")
gsub("(*UCP)^[^,]+(*SKIP)(*F)|\\b(\\p{L})\\p{L}*", "\\U\\1.", TEST, perl=TRUE)
## => [1] "Sacher, F. X." "Nishikawa, K." "Al-Assam, M."
A crude approach splitting the string :
TEST <- c("Sacher, Franz Xaver", "Nishikawa, Kiyoko", "Al-Assam, Muhammad")
sapply(strsplit(TEST, '\\s+'), function(x)
paste0(x[1], paste0(substr(x[-1], 1, 1), collapse = '.'), '.'))
#[1] "Sacher,F.X." "Nishikawa,K." "Al-Assam,M."
An approach using multiple backreference:
gsub("(\\b\\w+,\\s)(\\b\\w).*(\\b\\w)*", "\\1\\2.\\3", TEST)
[1] "Sacher, F." "Nishikawa, K." "Al-Assam, M."
Here, we use three capturing groups to refer back to in gsub's replacment argument via backreference:
(\\b\\w+,\\s): this, first, group captures the last name plus the comma followed by whitespace
(\\b\\w): this, second, group captures the initial of the first name
(\\b\\w): this, third, group captures the initial of the middle name
Related
I need to capture the numbers out of a string that come after a certain parameter name.
I have it working for most, but there is one parameter that is sometimes at the end of the string, but not always. When using the regular expression, it seems to matter.
I've tried different things, but nothing seems to work in both cases.
# Regular expression to capture the digit after the phrase "AppliedWhenID="
p <- ".*&AppliedWhenID=(.\\d*)"
# Tried this, but when at end, it just grabs a blank
#p <- ".*&AppliedWhenID=(.\\d*)&.*|.*&AppliedWhenID=(.\\d*)$"
testAtEnd <- "ReportType=233&ReportConID=171&MonthQuarterYear=0TimePeriodLabel=Year%202020&AppliedWhenID=2"
testNotAtEnd <- "ReportType=233&ReportConID=171&MonthQuarterYear=0TimePeriodLabel=Year%202020&AppliedWhenID=2&AgDateTypeID=1"
# What should be returned is "2"
gsub(p, "\\1", testAtEnd) # works
gsub(p, "\\1", testNotAtEnd) # doesn't work, it captures 2 + &AgDateTypeID=1
Note that sub and gsub replace the found text(s), thus, in order to extract a part of the input string with a capturing group + a backreference, you need to actually match (and consume) the whole string.
Hence, you need to match the string to the end by adding .* at the end of the pattern:
p <- ".*&AppliedWhenID=(\\d+).*"
sub(p, "\\1", testNotAtEnd)
# => [1] "2"
sub(p, "\\1", testAtEnd)
# => [1] "2"
See the regex demo and the R online demo.
Note that gsub matches multiple occurrences, you need a single one, so it makes sense to replace gsub with sub.
Regex details
.* - any zero or more chars as many as possible
&AppliedWhenID= - a &AppliedWhenID= string
(\d+) - Group 1 (\1): one or more digits
.* - any zero or more chars as many as possible.
You could try using the string look behind conditional "(?<=)" and str_extract() from the stringr library.
testAtEnd <- "ReportType=233&ReportConID=171&MonthQuarterYear=0TimePeriodLabel=Year%202020&AppliedWhenID=2"
testNotAtEnd <- "ReportType=233&ReportConID=171&MonthQuarterYear=0TimePeriodLabel=Year%202020&AppliedWhenID=2&AgDateTypeID=1"
p <- "(?<=AppliedWhenID=)\\d+"
# What should be returned is "2"
library(stringr)
str_extract(testAtEnd, p)
str_extract(testNotAtEnd, p)
Or in base R
p <- ".*((?<=AppliedWhenID=)\\d+).*"
gsub(p, "\\1", testAtEnd, perl=TRUE)
gsub(p, "\\1", testNotAtEnd, perl=TRUE)
I want to extract strings using rm_between function from the library(qdapRegex)
I need to extract the string between the second "|" and the word "_HUMAN".
I cant figure out how to select the second "|" and not the first.
example <- c("sp|B5ME19|EIFCL_HUMAN", "sp|Q99613|EIF3C_HUMAN")
prots <- rm_between(example, '|', 'HUMAN', extract=TRUE)
Thank you!!
Another alternative using regmatches, regexpr and using perl=TRUE to make use of \K
^(?:[^|]*\|){2}\K[^|_]+(?=_HUMAN)
Regex demo
For example
regmatches(example, regexpr("^(?:[^|]*\\|){2}\\K[^|_]+(?=_HUMAN)", example, perl=TRUE))
Output
[1] "EIFCL" "EIF3C"
In your rm_between(example, '|', 'HUMAN', extract=TRUE) command, the | is used to match the leftmost | and HUMAN is used to match the left most HUMAN right after.
Note the default value for the FIXED argument is TRUE, so | and HUMAN are treated as literal chars.
You need to make the pattern a regex pattern, by setting fixed=FALSE. However, the ^(?:[^|]*\|){2} as the left argument regex will not work because the qdap package creates an ICU regex with lookarounds (since you use extract=TRUE that sets include.markers to FALSE), which is (?<=^(?:[^|]*\|){2}).*?(?=HUMAN).
As a workaround, you could use a constrained-width lookbehind, by replacing * with a limiting quantifier with a reasonably large max parameter. Say, if you do not expect more than a 1000 chars between each pipe, you may use {0,1000}:
rm_between(example, '^(?:[^|]{0,1000}\\|){2}', '_HUMAN', extract=TRUE, fixed=FALSE)
# => [[1]]
# [1] "EIFCL"
#
# [[2]]
# [1] "EIF3C"
However, you really should think of using simpler approaches, like those described in other answers. Here is another variation with sub:
sub("^(?:[^|]*\\|){2}(.*?)_HUMAN.*", "\\1", example)
# => [1] "EIFCL" "EIF3C"
Details
^ - startof strig
(?:[^|]*\\|){2} - two occurrences of any 0 or more non-pipe chars followed with a pipe char (so, matching up to and including the second |)
(.*?) - Group 1: any 0 or more chars, as few as possible
_HUMAN.* - _HUMAN and the rest of the string.
\1 keeps only Group 1 value in the result.
A stringr variation:
stringr::str_match(example, "^(?:[^|]*\\|){2}(.*?)_HUMAN")[,2]
# => [1] "EIFCL" "EIF3C"
With str_match, the captures can be accessed easily, we do it with [,2] to get Group 1 value.
this is not exactly what you asked for, but you can achieve the result with base R:
sub("^.*\\|([^\\|]+)_HUMAN.*$", "\\1", example)
This solution is an application of regular expression.
"^.*\\|([^\\|]+)_HUMAN.*$" matches the entire character string.
\\1 matches whatever was matched inside the first parenthesis.
Using regular gsub:
example <- c("sp|B5ME19|EIFCL_HUMAN", "sp|Q99613|EIF3C_HUMAN")
gsub(".*?\\|.*?\\|(.*?)_HUMAN", "\\1", example)
#> [1] "EIFCL" "EIF3C"
The part (.*?) is replaced by itself as the replacement contains the back-reference \\1.
If you absolutely prefer qdapRegex you can try:
rm_between(example, '.{0,100}\\|.{0,100}\\|', '_HUMAN', fixed = FALSE, extract = TRUE)
The reason why we have to use .{0,100} instead of .*? is that the underlying stringi needs a mamixmum length for the look-behind pattern (i.e. the left argument in rm_between).
Just saying that you could easily just use sapply()/strsplit():
example <- c("sp|B5ME19|EIFCL_HUMAN", "sp|Q99613|EIF3C_HUMAN")
unlist(sapply(strsplit(example, "|", fixed = T),
function(item) strsplit(item[3], "_HUMAN", fixed = T)))
# [1] "EIFCL" "EIF3C"
It just splits on | in the first list and on _HUMAN on every third element within that list.
everyone.
I am completely new to regex in r, and i run into a problem when trying to retrieve a smaller set of pattern in the middle of a larger pattern using tagged xml file.
Here, i have a three-word sequence "reinforce the advantage" tagged by BNC (British National Corpus) Basic (C5) Tagset system. In specific, i want to only retrieve the three lemmatized words immediately after every "hw=" in this long sequence.
<w c5=VVI hw=reinforce pos=VERB>reinforce </w><w c5=AT0 hw=the pos=ART>the </w><w c5=NN2 hw=advantage pos=SUBST>advantages </w>
Can anyone please offer a possible solution with gsub or other functions in r? Many thanks in advance!
NF
vec <- "<w c5=VVI hw=reinforce pos=VERB>reinforce </w><w c5=AT0 hw=the pos=ART>the </w><w c5=NN2 hw=advantage pos=SUBST>advantages </w>"
m <- gregexpr("(?<=hw=)\\S+", vec, perl = T)
regmatches(vec, m)
# [[1]]
# [1] "reinforce" "the" "advantage"
copied from regex101.com
/
(?<=hw=)\S+
/
Positive Lookbehind (?<=hw=)
Assert that the Regex below matches
hw= matches the characters hw= literally (case sensitive)
\S+ matches any non-whitespace character (equal to [^\r\n\t\f\v ])
+ Quantifier — Matches between one and unlimited times, as many times as possible,
giving back as needed (greedy)
first ?unlist then collapse (?paste0)
paste0(unlist(
regmatches(vec, m)
), collapse = " ")
# [1] "reinforce the advantage"
Col
WBU-ARGU*06:03:04
WBU-ARDU*08:01:01
WBU-ARFU*11:03:05
WBU-ARFU*03:456
I have a column which has 75 rows of variables such as the col above. I am not quite sure how to use gsub or sub in order to get up until the integers after the first colon.
Expected output:
Col
WBU-ARGU*06:03
WBU-ARDU*08:01
WBU-ARFU*11:03
WBU-ARFU*03:456
I tried this but it doesn't seem to work:
gsub("*..:","", df$col)
Following may help you here too.
sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)
Output will be as follows.
> sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)
[1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456b"
Where Input for data frame is as follows.
dat <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456b")
df <- data.frame(dat)
Explanation: Following is only for explanation purposes.
sub(" ##using sub for global subtitution function of R here.
([^:]*) ##By mentioning () we are keeping the matched values from vector's element into 1st place of memory(which we could use later), which is till next colon comes it will match everything.
: ##Mentioning letter colon(:) here.
([^:]*) ##By mentioning () making 2nd place in memory for matched values in vector's values which is till next colon comes it will match everything.
.*" ##Mentioning .* to match everything else now after 2nd colon comes in value.
,"\\1:\\2" ##Now mentioning the values of memory holds with whom we want to substitute the element values \\1 means 1st memory place \\2 is second memory place's value.
,df$dat) ##Mentioning df$dat dataframe's dat value.
You may use
df$col <- sub("(\\d:\\d+):\\d+$", "\\1", df$col)
See the regex demo
Details
(\\d:\\d+) - Capturing group 1 (its value will be accessible via \1 in the replacement pattern): a digit, a colon and 1+ digits.
: - a colon
\\d+ - 1+ digits
$ - end of string.
R Demo:
col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("(\\d:\\d+):\\d+$", "\\1", col)
## => [1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
Alternative approach:
df$col <- sub("^(.*?:\\d+).*", "\\1", df$col)
See the regex demo
Here,
^ - start of string
(.*?:\\d+) - Group 1: any 0+ chars, as few as possible (due to the lazy *? quantifier), then : and 1+ digits
.* - the rest of the string.
However, it should be used with the PCRE regex engine, pass perl=TRUE:
col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("^(.*?:\\d+).*", "\\1", col, perl=TRUE)
## => [1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
See the R online demo.
sub("(\\d+:\\d+):\\d+$", "\\1", df$Col)
[1] "WBU-ARGU*06:03" "WBU-ARDU*08:01" "WBU-ARFU*11:03" "WBU-ARFU*03:456"
Alternatively match what you want (instead of subbing out what you don't want) with stringi:
stringi::stri_extract_first(df$Col, regex = "[A-Z-\\*]+\\d+:\\d+")
Slightly more concise stringr:
stringr::str_extract(df$Col, "[A-Z-\\*]+\\d+:\\d+")
# or
stringr::str_extract(df$Col, "[\\w-*]+\\d+:\\d+")
Suppose I have the following two strings and want to use grep to see which match:
business_metric_one
business_metric_one_dk
business_metric_one_none
business_metric_two
business_metric_two_dk
business_metric_two_none
And so on for various other metrics. I want to only match the first one of each group (business_metric_one and business_metric_two and so on). They are not in an ordered list so I can't index and have to use grep. At first I thought to do:
.*metric.*[^_dk|^_none]$
But this doesn't seem to work. Any ideas?
You need to use a PCRE pattern to filter the character vector:
x <- c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")
grep("metric(?!.*_(?:dk|none))", x, value=TRUE, perl=TRUE)
## => [1] "business_metric_one" "business_metric_two"
See the R demo
The metric(?!.*(?:_dk|_none)) pattern matches
metric - a metric substring
(?!.*_(?:dk|none)) - that is not followed with any 0+ chars other than line break chars followed with _ and then either dk or none.
See the regex demo.
NOTE: if you need to match only such values that contain metric and do not end with _dk or _none, use a variation, metric.*$(?<!_dk|_none) where the (?<!_dk|_none) negative lookbehind fails the match if the string ends with either _dk or _none.
You can also do something like this:
grep("^([[:alpha:]]+_){2}[[:alpha:]]+$", string, value = TRUE)
# [1] "business_metric_one" "business_metric_two"
or use grepl to match dk and none, then negate the logical when you're indexing the original string:
string[!grepl("(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
more concisely:
string[!grepl("business_metric_[[:alpha:]]+_(dk|none)", string)]
# [1] "business_metric_one" "business_metric_two"
Data:
string = c("business_metric_one","business_metric_one_dk","business_metric_one_none","business_metric_two","business_metric_two_dk","business_metric_two_none")