Related question here.
So I have a character vector with currency values that contain both dollar signs and commas. However, I want to try and remove both the commas and dollar signs in the same step.
This removes dollar signs =
d = c("$0.00", "$10,598.90", "$13,082.47")
gsub('\\$', '', d)
This removes commas =
library(stringr)
str_replace_all(c("10,0","tat,y"), fixed(c(","), "")
I'm wondering if I could remove both characters in one step.
I realize that I could just save the gsub results into a new variable, and then reapply that (or another function) on that variable. But I guess I'm wondering about a single step to do both.
Since answering in the comments is bad:
gsub('\\$|,', '', d)
replaces either $ or (|) , with an empty string.
take a look at ?regexp for additional special regex notation:
> gsub('[[:punct:]]', '', d)
[1] "000" "1059890" "1308247"
Related
I have a large dataset with two sorts of labels. The first is of the form 'numeric_alphanumeric_alpha' and another which is 'alphanumeric_alpha'. I need to strip the numeric prefix from the first label so that it matches the second label. I know how to remove numbers from alphanumeric data (as below) but this would remove numbers that I need.
gsub('[0-9]+', '', x)
Below is an example of the two different labels I am encountered with well as the prefer
c('12345_F24R2_ABC', 'r87R2_DEFG')
Below is the desired output
c('F24R2_ABC', 'r87R2_DEFG')
A simple regex can do it. ^ refers to the start of a string, \\d refers to any digits, + indicates one or more time it appears.
gsub("^\\d+_", "", c('12345_F24R2_ABC', 'r87R2_DEFG'), perl = T)
[1] "F24R2_ABC" "r87R2_DEFG"
Your code a litte modified:
^[0-9]*.....starts with number followed by numbers
\\_ .... matches underscore
gsub('^[0-9]*\\_', '', x)
[1] "F24R2_ABC" "r87R2_DEFG"
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
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"
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$", "Apartment")
# then do the rest of your replacements
There's probabably a better way to do the replacements than regular expressions (using switch(), maybe?)
Use paste0(fixedApts, collapse = "") to collapse the vector into a single string at the end if that's what you need to do.
After I collapse my rows and separate using a semicolon, I'd like to delete the semicolons at the front and back of my string. Multiple semicolons represent blanks in a cell. For example an observation may look as follows after the collapse:
;TX;PA;CA;;;;;;;
I'd like the cell to look like this:
TX;PA;CA
Here is my collapse code:
new_df <- group_by(old_df, unique_id) %>% summarize_each(funs(paste(., collapse = ';')))
If I try to gsub for semicolon it removes all of them. If if I remove the end character it just removes one of the semicolons. Any ideas on how to remove all at the beginning and end, but leaving the ones in between the observations? Thanks.
use the regular expression ^;+|;+$
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"