I realize this is a rather simple question and I have searched throughout this site, but just can't seem to get my syntax right for the following regex challenges. I'm looking to do two things. First have the regex to pick up the first three characters and stop at a semicolon. For example, my string might look as follows:
Apt;House;Condo;Apts;
I'd like to go here
Apartment;House;Condo;Apartment
I'd also like to create a regex to substitute a word in between delimiters, while keep others unchanged. For example, I'd like to go from this:
feline;labrador;bird;labrador retriever;labrador dog; lab dog;
To this:
feline;dog;bird;dog;dog;dog;
Below is the regex I'm working with. I know ^ denotes the beginning of the string and $ the end. I've tried many variations, and am making substitutions, but am not achieving my desired out put. I'm also guessing one regex could work for both? Thanks for your help everyone.
df$variable <- gsub("^apt$;", "Apartment;", df$variable, ignore.case = TRUE)
Here is an approach that uses look behind (so you need perl=TRUE):
> tmp <- c("feline;labrador;bird;labrador retriever;labrador dog; lab dog;",
+ "lab;feline;labrador;bird;labrador retriever;labrador dog; lab dog")
> gsub( "(?<=;|^) *lab[^;]*", "dog", tmp, perl=TRUE)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
The (?<=;|^) is the look behind, it says that any match must be preceded by either a semi-colon or the beginning of the string, but what is matched is not included in the part to be replaced. The * will match 0 or more spaces (since your example string had one case where there was space between the semi-colon and the lab. It then matches a literal lab followed by 0 or more characters other than a semi-colon. Since * is by default greedy, this will match everything up to, but not including' the next semi-colon or the end of the string. You could also include a positive look ahead (?=;|$) to make sure it goes all the way to the next semi-colon or end of string, but in this case the greediness of * will take care of that.
You could also use the non-greedy modifier, then force to match to end of string or semi-colon:
> gsub( "(?<=;|^) *lab.*?(?=;|$)", "dog", tmp, perl=TRUE)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
The .*? will match 0 or more characters, but as few as it can get away with, stretching just until the next semi-colon or end of line.
You can skip the look behind (and perl=TRUE) if you match the delimiter, then include it in the replacement:
> gsub("(;|^) *lab[^;]*", "\\1dog", tmp)
[1] "feline;dog;bird;dog;dog;dog;"
[2] "dog;feline;dog;bird;dog;dog;dog"
With this method you need to be careful that you only match the delimiter on one side (the first in my example) since the match consumes the delimiter (not with the look-ahead or look-behind), if you consume both delimiters, then the next will be skipped and only every other field will be considered for replacement.
I'd recommend doing this in two steps:
Split the string by the delimiters
Do the replacements
(optional, if that's what you gotta do) Smash the strings back together.
To split the string, I'd use the stringr library. But you can use base R too:
myString <- "Apt;House;Condo;Apts;"
# base R
splitString <- unlist(strsplit(myString, ";", fixed = T))
# with stringr
library(stringr)
splitString <- as.vector(str_split(myString, ";", simplify = T))
Once you've done that, THEN you can do the text substitution:
# base R
fixedApts <- gsub("^Apt$|^Apts$", "Apartment", splitString)
# with stringr
fixedApts <- str_replace(splitString, "^Apt$|^Apts$", "Apartment")
# then do the rest of your replacements
There's probabably a better way to do the replacements than regular expressions (using switch(), maybe?)
Use paste0(fixedApts, collapse = "") to collapse the vector into a single string at the end if that's what you need to do.
Related
I have a vector of strings and I want to remove -es from all strings (words) ending in either -ses or -ces at the same time. The reason I want to do it at the same time and not consequitively is that sometimes it happens that after removing one ending, the other ending appears while I don't want to apply this pattern to a single word twice.
I have no idea how to use two patterns at the same time, but this is the best I could:
text <- gsub("[sc]+s$", "[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
In R, I am having trouble replacing a substring that has punctuation. Ie within the string "r.Export", I am trying to replace "r." with "Report.". I've used gsub and below is my code:
string <- "r.Export"
short <- "r."
replacement <- "Report."
gsub(short,replacement,string)
The desired output is: "Report.Export" however gsub seems to replace the second r such that the output is:
Report.ExpoReport.
Using sub() instead is not a solution either because I am doing multiple gsubs where sometimes the string to be replaced is:
short <- "o."
So, then the o's in r.Export are replaced anyway and it becomes a complete mess.
string <- "r.Export"
short <- "r\\."
replacement <- "Report."
gsub(short,replacement,string)
Returns:
[1] "Report.Export"
Or, using fixed=TRUE:
string <- "r.Export"
short <- "r."
replacement <- "Report."
gsub(short,replacement,string, fixed=TRUE)
Returns:
[1] "Report.Export"
Explanation: Without the fixed=TRUE argument, gsub expects a regular expression as first argument. And with regular expressions . is a placeholder for 'any character'. If you want the literal . (period) you have to use either \\. (i.e. escaping the period) or the aforementioned argument fixed=TRUE
Since you have characters in your pattern (.) which has a special meaning in regex use fixed = TRUE which matches the string as is.
gsub(short,replacement,string, fixed = TRUE)
#[1] "Report.Export"
I might actually add word boundaries and lookaheads to the mix here, to ensure as targeted a match as possible:
string <- "r.Export"
replacement <- "Report."
output <- gsub("\\br\\.(?=\\w)", replacement, string, perl=TRUE)
output
[1] "Report.Export"
This approach ensures that we only match r. when the r is preceded by whitespace or is the start of the string, and also when what follows the dot is another word. Consider the sentence The project r.Export needed a programmer. We wouldn't want to replace the final r. in this case.
We can use sub
sub(short,replacement,string, fixed = TRUE)
#[1] "Report.Export"
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"
After I collapse my rows and separate using a semicolon, I'd like to delete the semicolons at the front and back of my string. Multiple semicolons represent blanks in a cell. For example an observation may look as follows after the collapse:
;TX;PA;CA;;;;;;;
I'd like the cell to look like this:
TX;PA;CA
Here is my collapse code:
new_df <- group_by(old_df, unique_id) %>% summarize_each(funs(paste(., collapse = ';')))
If I try to gsub for semicolon it removes all of them. If if I remove the end character it just removes one of the semicolons. Any ideas on how to remove all at the beginning and end, but leaving the ones in between the observations? Thanks.
use the regular expression ^;+|;+$
x <- ";TX;PA;CA;;;;;;;"
gsub("^;+|;+$", "", x)
The ^ indicates the start of the string, the + indicates multiple matches, and $ indicates the end of the string. The | states "OR". So, combined, it's searching for any number of ; at the start of a string OR any number of ; at the end of the string, and replace those with an empty space.
The stringi package allows you to specify patterns which you wish to preserve and trim everything else. If you only have letters there (though you could specify other pattern too), you could simply do
stringi::stri_trim_both(";TX;PA;CA;;;;;;;", "\\p{L}")
## [1] "TX;PA;CA"