I have this variable:
names<-c("Sophia Abbe", "Olivia Abbett", "Emma Abbey", "Ava Abbitt", "Isabella Abbot", "Mia Abbott", "Aria Abbs")
I want to abbreviate the first names and place them into a vector.
I want to obtain a vector like ("S. Abbe", "O. Abbett", ... , "A. Abbs)
What would be an efficient way of doing this with the stringr functions str_c(), str_split() and str_sub() ?
An option with sub by matching the lower case letters and replace with a . in base R
sub("[a-z]+", ".", names)
#[1] "S. Abbe" "O. Abbett" "E. Abbey" "A. Abbitt" "I. Abbot" "M. Abbott" "A. Abbs"
In this the [a-z]+ matching one or more lower blocks of characters i.e. those in the first word (because we are using sub) and replace with ""
Or using str_replace
library(stringr)
str_replace(names, "[a-z]+", ".")
Related
I've got the following data.
df <- data.frame(Name = c("TOMTom Catch",
"BIBill Ronald",
"JEFJeffrey Wilson",
"GEOGeorge Sic",
"DADavid Irris"))
How do I clean the data in names column?
I've tried nchar and substring however some names need the first two characters removed where as other need the first three?
We can use regex lookaround patterns.
gsub("^[A-Z]+(?=[A-Z])", "", df$Name, perl = T)
#> [1] "Tom Catch" "Bill Ronald" "Jeffrey Wilson" "George Sic"
#> [5] "David Irris"
I am trying to extract a string between two commas with gsub. If I have the following
xz<- "1620 Honeylocust Drive, 60210 IL, USA"
and I want to extract everything between the two commas, (60120 IL), is it possible to use gsub?
I have tried
gsub(".*,","",xz)
The result is USA. How can I do it?
We can match zero or more characters that are not a , ([^,]*) followed by a , followed by zero or more space from the start (^) of the string or | a , followed by zero or more characters that are not a , ([^,]*) at the end ($) of string and replace with blank ("")
gsub("^[^,]*,\\s*|,[^,]*$", "", xz)
#[1] "60210 IL"
Or another option is using sub and capture as a group
sub("^[^,]+,\\s+([^,]+).*", "\\1", xz)
#[1] "60210 IL"
Or another option is regexpr/regmatches
regmatches(xz, regexpr("(?<=,\\s)[^,]*(?=,)", xz, perl = TRUE))
#[1] "60210 IL"
Or with str_extract from stringr
library(stringr)
str_extract(xz, "(?<=,\\s)[^,]*(?=,)")
#[1] "60210 IL"
Update
With the new string,
xz1 <- "1620, Honeylocust Drive, 60210 IL, USA"
sub(".*,\\s+(+[0-9]+[^,]+).*", "\\1", xz1)
#[1] "60210 IL"
You could also do this using strsplit and grep (here I did it in 2 lines for readability):
xz1 <- "1620, Honeylocust Drive, 60210 IL, USA"
a1 <- strsplit(xz1, "[ ]*,[ ]*")[[1]]
grep("^[0-9]+[ ]+[A-Z]+", a1, value=TRUE)
#[1] "60210 IL"
It's not using gsub, and in the present case it is not better, but maybe it is easier to adapt to other situations.
I have a vector including certain strings, and I would like remove other parts in each string except the word including certain patter (here is mir).
s <- c("a mir-96 line (kk27)", "mir-133a cell",
"d mir-14-3p in", "m mir133 (sas)", "mir_23_5p r 27")
I want to obtain:
mir-96, mir-133a, mir-14-3p, mir133, mir_23_5p
I know the idea: use the gsub() and pattern is: a word beginning with (or including) mir.
But I have no idea how to construct such patter.
Or other idea?
Any help will be appreciated!
One way in base R would be splitting every string into words and then extracting only those with mir in it
unlist(lapply(strsplit(s, " "), function(x) grep("mir", x, value = TRUE)))
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p"
We can save the unlist step in lapply by using sapply as suggested by #Rich Scriven in comments
sapply(strsplit(s, " "), function(x) grep("mir", x, value = TRUE))
We can use sub to match zero or more characters (.*) followed by a word boundary (\\b) followed by the string (mir and one or more characters that are not a white space (\\S+), capture it as a group by placing inside the (...) followed by other characters, and in the replacement use the backreference of the captured group (\\1)
sub(".*\\b(mir\\S+).*", "\\1", s)
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p"
Update
If there are multiple 'mir.*' substring, then we want to extract strings having some numeric part
sub(".*\\b(mir[^0-9]*[0-9]+\\S*).*", "\\1", s1)
#[1] "mir-96" "mir-133a" "mir-14-3p" "mir133" "mir_23_5p" "mir_23-5p"
data
s1 <- c("a mir-96 line (kk27)", "mir-133a cell", "d mir-14-3p in", "m mir133 (sas)",
"mir_23_5p r 27", "a mir_23-5p 1 mir-net")
I want to use gsub to correct some names that are in my data. I want names such as "R. J." and "A. J." to have no space between the letters.
For example:
x <- "A. J. Burnett"
I want to use gsub to match the pattern of his first name, and then remove the space:
gsub("[A-Z]\\.\\s[A-Z]\\.", "[A-Z]\\.[A-Z]\\.", x)
But I get:
[1] "[A-Z].[A-Z]. Burnett"
Obviously, instead of the [A-Z]'s I want the actual letters in the original name. How can I do this?
Use capture groups by enclosing patterns in (...), and refer to the captured patterns with \\1, \\2, and so on. In this example:
x <- "A. J. Burnett"
gsub("([A-Z])\\.\\s([A-Z])\\.", "\\1.\\2.", x)
[1] "A.J. Burnett"
Also note that in the replacement you don't need to escape the . characters, as they don't have a special meaning there.
You can use a look-ahead ((?=\\w\\.)) and a look-behind ((?<=\\b\\w\\.)) to target such spaces and replace them with "".
x <- c("A. J. Burnett", "Dr. R. J. Regex")
gsub("(?<=\\b\\w\\.) (?=\\w\\.)", "", x, perl = TRUE)
# [1] "A.J. Burnett" "Dr. R.J. Regex"
The look-ahead matches a word character (\\w) followed by a period (\\.), and the look-behind matches a word-boundary (\\b) followed by a word character and a period.
Rearranging simpsons names with R to follow first name, last name format but there are large spaces between the names, is it possible to remove spaces outside the quoted names?
library(stringr)
simpsons <- c("Moe Syzlak", "Burns, C. Montgomery", "Rev. Timothy Lovejoy", "Ned Flanders", "Simpson, Homer", "Dr. Julius Hibbert")
reorder <- sapply(sapply(str_split(simpsons, ","), str_trim),rev)
for (i in 1:length(name) ) {
splitname[i]<- paste(unlist(splitname[i]), collapse = " ")
}
splitname <- unlist(splitname)
If we need to rearrange the first name followed by last name, we could use sub. We capture one or more than character which is not a , in a group, followed by , followed by 0 or more space (\\s*), capture one or more characters that are not a , as the 2nd group, and in the replacement reverse the backreference to get the output.
sub("([^,]+),\\s*([^,]+)", "\\2 \\1", simpsons)
#[1] "Moe Syzlak" "C. Montgomery Burns" "Rev. Timothy Lovejoy" "Ned Flanders" "Homer Simpson" "Dr. Julius Hibbert"