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"
Related
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.
I am trying to add a - between letter S and any number in a column of a data frame. So, this is an example:
VariableA
TRS34
MMH22
GFSR104
GS23
RRTM55
P3
S4
My desired output is:
VariableA
TRS-34
MMH22
GFSR104
GS-23
RRTM55
P3
S-4
I was trying yo use gsub:
gsub('^([a-z])-([0-9]+)$','\\1d\\2',myDF$VariableA)
but this is not working.
How can I solve this?
Thanks!
Your ^([a-z])-([0-9]+)$ regex attempts to match strings that start with a letter, then have a - and then one or more digits. This can't work as there are no hyphens in the strings, you want to introduce it into the strings.
You can use
gsub('(S)([0-9])', '\\1-\\2', myDF$VariableA)
The (S)([0-9]) regex matches and captures S into Group 1 (\1) and then any digit is captured into Group 2 (\2) and the replacement pattern is a concatenation of group values with a hyphen in between.
If there is only one substitution expected, replace gsub with sub.
See the regex demo and the online R demo.
Other variations:
gsub('(S)(\\d)', '\\1-\\2', myDF$VariableA) # \d also matches digits
gsub('(?<=S)(?=\\d)', '-', myDF$VariableA, perl=TRUE) # Lookarounds make backreferences redundant
Here is the version I like using sub:
myDF$VariableA <- gsub('S(\\d)', 'S-\\1', myDF$VariableA)
This requires using only one capture group.
Using stringr package
library(stringr)
str_replace_all(myDF$VariableA, 'S(\\d)', 'S-\\1')
You could also use lookbehinds if you set perl=TRUE:
> gsub('(?<=S)([0-9]+)', '-\\1', myDF$VariableA, perl=TRUE)
[1] "TRS-34" "MMH22" "GFSR104" "GS-23" "RRTM55" "P3" "S-4"
>
I have a vector of strings and I want to remove -es from all strings (words) ending in either -ses or -ces at the same time. The reason I want to do it at the same time and not consequitively is that sometimes it happens that after removing one ending, the other ending appears while I don't want to apply this pattern to a single word twice.
I have no idea how to use two patterns at the same time, but this is the best I could:
text <- gsub("[sc]+s$", "[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.
I am trying to extract usernames tagged in a text-chat, such as "#Jack #Marie Hi there!"
I am trying to do it on the combination of # and whitespace but I cannot get the regex to match non-greedy (or at least this is what I think is wrong):
library(stringr)
str_extract(string = '#This is what I want to extract', pattern = "(?<=#)(.*)(?=\\s+)")
[1] "This is what I want to"
What I would like to extract instead is only This.
You could make your regex non greedy:
(?<=#)(.*?)(?=\s+)
Or if you want to capture only "This" after the # sign, you could try it like this using only a positive lookbehind:
(?<=#)\w+
Explanation
A positive lookbehind (?<=
That asserts that what is behind is an #
Close positive lookbehind )
Match one or more word characters \w+
The central part of your regex ((.*)) is a sequence of any chars.
Instead you shoud look for a sequence of chars other than white space
(\S+) or word chars (\w+).
Note also that I changed * to +, as you are probably not interested
in any empty sequence of chars.
To capture also a name which has "last" position in the source
string, the last part of your regex should match not only a sequence
of whitespace chars, but also the end of the string, so change
(?=\\s+) to (?=\\s+|$).
And the last remark: Actually you don't need the parentheses around
the "central" part.
So to sum up, the whole regex can be like this:
(?<=#)\w+(?=\s+|$)
(with global oprion).
Here is a non-regex approach or rather a minimal-regex approach since grep takes the detection of # through the regex engine
grep('#', strsplit(x, ' ')[[1]], value = TRUE)
#[1] "#This"
Or to avoid strsplit, we can use scan (taken from this answer), i.e.
grep('#', scan(textConnection(x), " "), value=TRUE)
#Read 7 items
#[1] "#This"