This question already has answers here:
How to delete everything after nth delimiter in R?
(2 answers)
Closed 2 years ago.
I currently have a dataframe with colnames that I'm trying to truncate the end after the 2nd period.
Example below:
GTEX.W5WGY.1726.SM.4LMI5 GTEX.WEY5.1226.SM.4LMIQ
23 20
0 32
Ideal output:
GTEX.W5WGY GTEX.WEY5
23 20
0 32
I'm trying to get it to this output instead and have tried sub but it isn't working.
colnames(x) <- sub("..*.SM..*", "", colnames(x))
Any help would be appreciated!
We can change the pattern to capture the characters that are not a . ([^.]+) from the start (^) of the string followed by a . and the second of no dots, replace with the backreference of the captured group
colnames(x) <- sub("^([^.]+\\.[^.]+)\\..*", "\\1", colnames(x))
Related
This question already has answers here:
Trim a string to a specific number of characters in R
(3 answers)
Using gsub in R to remove values in Zip Code field
(1 answer)
Closed 2 years ago.
I'm trying to use gsub on the df$Zipcode in the following data frame:
#Sample
df <-data.frame(ID = c(1,2,3,4,5,6,7),
Zipcode =c("10001-2838", "95011", "95011", "100028018", "84321", "84321", "94011"))
df
I want to take everything after the "-" (hyphen) out and replace it with nothing. Something like:
df$Zipcode <- gsub("\-", "", df$Zipcode)
But I don't think that is quite right. I also want to take the first 5 digits of all Zipcodes that are longer than 5 digits, like observation 4. Which should just be 10002. Maybe this is correct:
df$Zipcode <- gsub("[:6:]", "", df$Zipcode)
We can capture the first 5 characters that are not a - as a group and replace with the backreference (\\1) of the captured group
df$Zipcode <- sub("^([^-]{5}).*", "\\1", df$Zipcode)
df$Zipcode
#[1] "10001" "95011" "95011" "10002" "84321" "84321" "94011"
I think what you're looking for is this:
sub("(\\d{5}).*", "\\1", df$Zipcode)
[1] "10001" "95011" "95011" "10002" "84321" "84321" "94011"
This matches the first 5 digits, puts them into a capturing group, and 'remembers' them (but not the rest) via backreference \\1 in the replacement argument to sub.
This question already has answers here:
remove leading 0s with stringr in R
(3 answers)
Closed 2 years ago.
I'm trying to remove the 0 that appears at the beginning of some observations for Zipcode in the following table:
I think the sub function is probably my best choice but I only want to do the replacement for observations that begin with 0, not all observations like the following does:
data_individual$Zipcode <-sub(".", "", data_individual$Zipcode)
Is there a way to condition this so it only removes the first character if the Zipcode starts with 0? Maybe grepl for those that begin with 0 and generate a dummy variable to use?
We can specify the ^0+ as pattern i.e. one or more 0s at the start (^) of the string instead of . (. in regex matches any character)
data_individual$Zipcode <- sub("^0+", "", data_individual$Zipcode)
Or with tidyverse
library(stringr)
data_individual$Zipcode <- str_remove(data_individual$Zipcode, "^0+")
Another option without regex would be to convert to numeric as numeric values doesn't support prefix 0 (assuming all zipcodes include only digits)
data_individual$Zipcode <- as.numeric(data_individual$Zipcode)
This question already has answers here:
remove repeated character between words
(4 answers)
Closed 3 years ago.
I have this text:
F <- "hhhappy birthhhhhhdayyy"
and I want to remove the repeat characters, I tried this code
https://stackoverflow.com/a/11165145/10718214
and it works, but I need to remove repeat characters if it repeats more than 2, and if it repeated 2 times keep it.
so the output that I expect is
"happy birthday"
any help?
Try using sub, with the pattern (.)\\1{2,}:
F <- ("hhhappy birthhhhhhdayyy")
gsub("(.)\\1{2,}", "\\1", F)
[1] "happy birthday"
Explanation of regex:
(.) match and capture any single character
\\1{2,} then match the same character two or more times
We replace with just the single matching character. The quantity \\1 represents the first capture group in sub.
This question already has answers here:
How to delete everything after nth delimiter in R?
(2 answers)
Remove text after second colon
(3 answers)
Remove all characters after the 2nd occurrence of "-" in each element of a vector
(1 answer)
Closed 3 years ago.
How could I remove everything before the second pattern occurence in a dataframe using R?
I used:
for (i in 1:length(df1)){
df1[, i]<- gsub(".*_", "",df1[, i])
}
But I guess there is a better way to apply that for all the dataframe?
Here is an exemple of a value in the dataframe:
name_000004_A_B_C
name_00003_C_D
and get
A_B_C
C_D
Thank you for your help.
x <- c("name_000004_A_B_C", "name_00003_C_D")
gsub("(name_[0-9]*_)(.*)", "\\2", x)
##[1] "A_B_C" "C_D"
More generalised:
gsub("([a-z0-9]*_[a-z0-9]*_)(.*)", "\\2", x)
#[1] "A_B_C" "C_D"
The global substitution takes two matching group patterns into consideration, first is the pattern (name_[0-9]*_) and the second is whatever comes after. It keeps the second matching group. Hope this hepls!
This question already has answers here:
Trying to return a specified number of characters from a gene sequence in R
(3 answers)
Extracting the last n characters from a string in R
(15 answers)
Closed 5 years ago.
Is there a function in R that I can cut a value in vector.
for example i got this vec:
40754831597
64278107602
64212163451
and each vale in the vec i want to cut so from the number pos 3 to 6 for example and get a new vector look like this
7548
2781
2121
and so on
I don't really get why you would like to do this, but here you go:
# assuming it's a character vector
substring(vec,3,6)
# if it's numeric
substring(as.character(vec),3,6)
#output
#[1] "7548" "2781" "2121"
We can use sub
sub(".{2}(.{4}).*", "\\1", v1)
#[1] "7548" "2781" "2121"
data
v1 <- c(40754831597, 64278107602, 64212163451)