I have a vector that is constructed with numbers and letters. I want to get all the characters before the LAST letter of each value (which is I guess, always the 2nd letter of the vector). Using stringr (preferably)...
Example :
x = c("1H23456789H10", "97845784584H2", "0H987654321H0", "0P45454545A3", "63A00000000000A91")
str_extract_all(string = x, pattern = ????????)
I tried some tricks here : https://evoldyn.gitlab.io/evomics-2018/ref-sheets/R_strings.pdf
The result I want is :
"1H23456789" instead of "1H23456789H10"
"97845784584" instead of "97845784584H2"
"0H987654321", instead of "0H987654321H0"
"0P45454545", instead of "0P45454545A3"
"63A00000000000" instead of "63A00000000000A91"
str_extract(string = x, pattern = "[^A-Z]*[A-Z][^A-Z]*")
# [1] "1H23456789" "0H987654321" "0P45454545" "63A00000000000"
Explanation: we want to extract 1 pattern match per input, so we use str_extract not str_extract_all. Our pattern [^A-Z]*, any number of non-letters, followed by [A-Z] exactly one letter, followed by [^A-Z]* any number of non-letters. I just used capital letters based on your input, but you could change A-Z to A-Za-z inside the brackets to include lower case letters.
Related
I wanted to extract certain words from a bigger word-list. One example of a desired extracted word-list is: extract all the words that contain /s/ followed by /r/. So this should give me words such as sər'ka:rəh, e:k'sa:r, səmʋitərəɳ, and so:'ha:rd. from the bigger word-list.
Consider the data (IPA transcription) to be the one given below:
sər'ka:rəh
sə'lᴔ:nija:
hã:ki:
pu:'dʒa:ẽ:
e:k'sa:r
mritko:
dʒʱã:sa:
pə'hũtʃ'ne:'ʋa:le:
kərəpʈ
tʃinhirit
tʃʰəʈʈʰi:
dʱũdʱ'la:pən
səmʋitərəɳ
so:'ha:rd
məl'ʈi:spe:'ʃijliʈi:
la:'pər'ʋa:i:
upləbɡʱ
Thanks much!
Here's an answer to the issue described in the first paragraph of your post. (To my mind, the examples in the second paragraph are inconsistent with the issue described in the first para, so I'll take the liberty of ignoring them here).
You say you want to "extract all the words that contain p followed by t". The word 'extract' implies that there are other characters in the same string than those you want to match and extract. The verb 'contain' implies that the words you want to extract need not necessarily have p in word-initial position. Based on these premises, here's some mock data and a solution to the task:
Data:
x <- c("pastry is to the pastor's appetite what pot is to the pupil's")
Solution:
libary(stringr)
unlist(str_extract_all(x, "\\b\\w*(?<=p)\\w*t\\w*\\b"))
This uses word boundaries \\b to extract the target words from the surrounding context; it further uses positive lookbehind (?<=...) to assert the condition that for there to be a matching t there needs to be a p character occurring prior to the match.
The regex in more detail:
\\b: the opening word boundary
\\w*: zero or more alphanumeric chars (or an underscore)
(?<=p): positive lookbehind: "if and only if you see a p char on
the left..."
\\w*: zero or more alphanumeric chars (or an underscore)
t: the literal character t
\\w*: zero or more alphanumeric chars (or an underscore)
\\b: the closing word boundary
Result:
[1] "pastry" "pastor" "appetite" "pot"
EDIT 1:
Now that the question has been updated, a more definitive answer is possible.
Data:
x <- c("sər'ka:rəh","sə'lᴔ:nija:","hã:ki:","pu:'dʒa:ẽ:","e:k'sa:r",
"mritko:","dʒʱã:sa:","pə'hũtʃ'ne:'ʋa:le:","kərəpʈ","tʃinhirit",
"tʃʰəʈʈʰi:","dʱũdʱ'la:pən","səmʋitərəɳ","so:'ha:rd",
"məl'ʈi:spe:'ʃijliʈi:", "la:'pər'ʋa:i:","upləbɡʱ")
If you want to match (rather than extract) words that "contain /s/ followed by /r/", you can use grepin various ways. Here are two ways:
grep("s.*r", x, value = T)
or:
grep("(?<=s).*r", x, value = T, perl = T) # with lookbehind
The result is the same in either case:
[1] "sər'ka:rəh" "e:k'sa:r" "səmʋitərəɳ" "so:'ha:rd"
EDIT 2:
If the aim is to match words that "contain /s/ or /p/ followed by /r/ or /t/", you can use the metacharacter | to allow for alternatives:
grep("s.*r|s.*t|p.*r|p.*t", x, value = T)
# or, more succinctly:
grep("(s|p).*(r|t)", x, value = T)
[1] "sər'ka:rəh" "e:k'sa:r" "pə'hũtʃ'ne:'ʋa:le:" "səmʋitərəɳ" "so:'ha:rd"
[6] "la:'pər'ʋa:i:"
You can use grep function. Assuming your list is called list:
grep("p[a-z]+t", list, value=TRUE)
I need to put a dot before a letter in this type of strings
name of data set: V2
6K102
62D102
627Z102
I would like to get this:
6.K102
62.D102
627.Z102
I am using this regex:
mutate(V2 = gsub("^[A-Z]",'\\.', V2))
If the string has to start with 1 or more digits followed by a char A-Z, you could use 2 capturing groups
^(\d+)([A-Z])
In the replacement use "\\1.\\2"
sub("^([0-9]+)([A-Z])", "\\1.\\2", V2)
you could use sub("([A-Z])",".\\1", V2)
This apply to the question before it was updated.
Usually \u2022 prints a bullet in text items. So if your question is regarding a label, you may just inserted it there "\u2022 ..."
Otherwise, for text items in datasets such as V2, you can work around by applying paste0 in combination with ifelse. In this case, your text items the you need a black dot infront if, is stored in V2$name
V2$name <- ifelse(V2$name==1,1,paste0("\u2022", V2$name))
Your regex lacks a capturing group around the letter pattern (so that you could keep it after replacement) and contains a redundant ^ anchor that matches the string start location. Also, you are using a gsub function while you just need a sub, since only one replacement is expected.
Use
sub("([[:upper:]])", ".\\1", V2)
With stringr (see demo):
stringr::str_replace(V2, "[[:upper:]]", "\\.\\0")
Details
sub - only the first match is replaced
([[:upper:]]) - matches and captures any uppercase letter into Group 1 (later referenced to with \1 from the replacement pattern)
\1 - the value of Group 1 (the uppercase letter matched)
Note that stingr solution uses \0, the placeholder for the whole match value, so no need to capture the uppercase letter in the regex pattern.
See the R demo:
V2 <- c("6K102","62D102","627Z102")
sub("([[:upper:]])", ".\\1", V2)
# => [1] "6.K102" "62.D102" "627.Z102"
I have a vector of strings and I want to remove -es from all strings (words) ending in either -ses or -ces at the same time. The reason I want to do it at the same time and not consequitively is that sometimes it happens that after removing one ending, the other ending appears while I don't want to apply this pattern to a single word twice.
I have no idea how to use two patterns at the same time, but this is the best I could:
text <- gsub("[sc]+s$", "[sc]", text)
I know the replacement is not correct, but I wonder how can I show that I want to replace it with the letter I just detected (c or s in this case). Thank you in advance.
To remove es at the end of words, that is preceded with s or c, you may use
gsub("([sc])es\\b", "\\1", text)
gsub("(?<=[sc])es\\b", "", text, perl=TRUE)
To remove them at the end of strings, you can go on using your $ anchor:
gsub("([sc])es$", "\\1", text)
gsub("(?<=[sc])es$", "", text, perl=TRUE)
The first gsub TRE pattern is ([sc])es\b: a capturing group #1 that matches either s or c, and then es is matched, and then \b makes sure the next char is not a letter, digit or _. The \1 in the replacement is the backreference to the value stored in the capturing group #1 memory buffer.
In the second example with the PCRE regex (due to perl=TRUE), (?<=[sc]) positive lookbehind is used instead of the ([sc]) capturing group. Lookbehinds are not consuming text, the text they match does not land in the match value, and thus, there is no need to restore it anyhow. The replacement is an empty string.
Strings ending with "ces" and "ses" follow the same pattern, i.e. "*es$"
If I understand it correctly than you don't need two patterns.
Example:
x = c("ces", "ses", "mes)
gsub( pattern = "*([cs])es$", replacement = "\\1", x)
[1] "c" "s" "mes"
Hope it helps.
M
I need to remove a parenthesis after a number in a string:
"dl_CONH_r = a0cons+a2cons*(CONH_r_lag_1)-a3cons*HGDI_r_lag_1)-(1-a3cons)*HNW_r_lag_2)+a4cons*rate_90_r_lag_1))+a5cons*dl_HCOE_r+a6cons*dl_HOY_r_lag_2)+a7cons*dl_HNW_r_lag_1)+a8cons*d_rate_UNE_lag_2)+(1-a5cons-a6cons-a7cons)*(dl_TREND_PROD+dl_TREND_AVEH+dl_TREND_WAP)"
The resulting string should look like this:
"dl_CONH_r = a0cons+a2cons*(CONH_r_lag_1-a3cons*HGDI_r_lag_1-(1-a3cons)*HNW_r_lag_2+a4cons*rate_90_r_lag_1)+a5cons*dl_HCOE_r+a6cons*dl_HOY_r_lag_2+a7cons*dl_HNW_r_lag_1+a8cons*d_rate_UNE_lag_2+(1-a5cons-a6cons-a7cons)*(dl_TREND_PROD+dl_TREND_AVEH+dl_TREND_WAP)"
The regular expression I am trying to capture here is the first parenthesis after the string "lag_" followed by some number. Note, that in places there are two parenthesis:
rate_90_r_lag_1))
And I only want to remove the first one.
I've tried a simple regex in gsub
a <- "dl_CONH_r = a0cons+a2cons*(CONH_r_lag_1)-a3cons*HGDI_r_lag_1)-(1-a3cons)*HNW_r_lag_2)+a4cons*rate_90_r_lag_1))+a5cons*dl_HCOE_r+a6cons*dl_HOY_r_lag_2)+a7cons*dl_HNW_r_lag_1)+a8cons*d_rate_UNE_lag_2)+(1-a5cons-a6cons-a7cons)*(dl_TREND_PROD+dl_TREND_AVEH+dl_TREND_WAP)"
gsub("[0-9]\\)","[0-9]",a)
But I the resulting string removes the number and replaces it with [0-9]:
"dl_CONH_r = a0cons+a2cons*(CONH_r_lag_[0-9]-a3cons*HGDI_r_lag_[0-9]-(1-a3cons)*HNW_r_lag_[0-9]+a4cons*rate_90_r_lag_[0-9])+a5cons*dl_HCOE_r+a6cons*dl_HOY_r_lag_[0-9]+a7cons*dl_HNW_r_lag_[0-9]+a8cons*d_rate_UNE_lag_[0-9]+(1-a5cons-a6cons-a7cons)*(dl_TREND_PROD+dl_TREND_AVEH+dl_TREND_WAP)"
I understand that the gsub is doing what it is intended to do. What I don't know is how to keep the number before the parenthesis?
You need to use a look around (in this case the preceded by) so that it will match just the parentheses as the matching group instead of the numbers and the parentheses. Then you can just remove the parentheses.
gsub("(?<=[0-9])\\)","", a, perl = TRUE)
You can do this using capture groups:
Lets just try it on the string my_string <- " = a0cons+a2cons*(CONH_r_lag_1)-a3cons*"
reg_expression <- "(.*[0-9])\\)(.*)" #two capture groups, with the parenthesis not in a group
my_sub_string <- sub(reg_expression,"\\1\\2", my_string)
Notice "\\1" reads like \1 to the regex engine, and so is a special character referring to the first capture group. (These can also be named)
Another way of doing this is lookarounds:
There are two basic kinds of lookarounds, a lookahead (?=) and a lookbehind (?<=). Since we want to match a pattern, but not capture, something behind our matched expression we need a lookbehind.
reg_expression <- "(?<=[0-9])\\)" #lookbehind
my_sub_string <- sub(reg_expression,"", my_string)
Which will match the pattern, but only replace the parenthesis.
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