I have a column as below.
9453, 55489, 4588, 18893, 4457, 2339, 45489HQ, 7833HQ
I would like to add leading zero if the number is less than 5 digits. However, some numbers have "HQ" in the end, some don't.(I did check other posts, they dont have similar problem in the "HQ" part)
so the finally desired output should be:
09453, 55489, 04588, 18893, 04457, 02339, 45489HQ, 07833HQ
any idea how to do this? Thank you so much for reading my post!
A one-liner using regular expressions:
my_strings <- c("9453", "55489", "4588",
"18893", "4457", "2339", "45489HQ", "7833HQ")
gsub("^([0-9]{1,4})(HQ|$)", "0\\1\\2",my_strings)
[1] "09453" "55489" "04588" "18893"
"04457" "02339" "45489HQ" "07833HQ"
Explanation:
^ start of string
[0-9]{1,4} one to four numbers in a row
(HQ|$) the string "HQ" or the end of the string
Parentheses represent capture groups in order. So 0\\1\\2 means 0 followed by the first capture group [0-9]{1,4} and the second capture group HQ|$.
Of course if there is 5 numbers, then the regex isn't matched, so it doesn't change.
I was going to use the sprintf approach, but found the the stringr package provides a very easy solution.
library(stringr)
x <- c("9453", "55489", "4588", "18893", "4457", "2339", "45489HQ", "7833HQ")
[1] "9453" "55489" "4588" "18893" "4457" "2339" "45489HQ" "7833HQ"
This can be converted with one simple stringr::str_pad() function:
stringr::str_pad(x, 5, side="left", pad="0")
[1] "09453" "55489" "04588" "18893" "04457" "02339" "45489HQ" "7833HQ"
If the number needs to be padded even if the total string width is >5, then the number and text need to be separated with regex.
The following will work. It combines regex matching with the very helpful sprintf() function:
sprintf("%05.0f%s", # this encodes the format and recombines the number with padding (%05.0f) with text(%s)
as.numeric(gsub("^(\\d+).*", "\\1", x)), #get the number
gsub("[[:digit:]]+([a-zA-Z]*)$", "\\1", x)) #get just the text at the end
[1] "09453" "55489" "04588" "18893" "04457" "02339" "45489HQ" "07833HQ"
Another attempt, which will also work in cases like "123" or "1HQR":
x <- c("18893","4457","45489HQ","7833HQ","123", "1HQR")
regmatches(x, regexpr("^\\d+", x)) <- sprintf("%05d", as.numeric(sub("\\D+$","",x)))
x
#[1] "18893" "04457" "45489HQ" "07833HQ" "00123" "00001HQR"
This basically finds any numbers at the start of the string (^\\d+) and replaces them with a zero-padded (via sprintf) string that was subset out by removing any non-numeric characters (\\D+$) from the end of the string.
We can use only sprintf() and gsub() by splitting up the parts then putting them back together.
sprintf("%05d%s", as.numeric(gsub("[^0-9]+", "", x)), gsub("[0-9]+", "", x))
# [1] "18893" "04457" "45489HQ" "07833HQ" "00123" "00001HQR"
Using #thelatemail's data:
x <- c("18893", "4457", "45489HQ", "7833HQ", "123", "1HQR")
Related
Is there an R function to get only the part of a string before the 2nd capital character appears?
For example:
Example <- "MonkeysDogsCats"
Expected output should be:
"Monkeys"
Maybe something like
stringr::str_extract("MonkeysDogsCats", "[A-Z][a-z]*")
#[1] "Monkeys"
Here is an alternative approach:
Here we first put a space before all uppercase and then extract the first word:
library(stringr)
word(gsub("([a-z])([A-Z])","\\1 \\2", Example), 1)
[1] "Monkeys"
A base solution with sub():
x <- "MonkeysDogsCats"
sub("(?<=[a-z])[A-Z].*", "", x, perl = TRUE)
# [1] "Monkeys"
Another way using stringr::word():
stringr::word(x, 1, sep = "(?=[A-Z])\\B")
# [1] "Monkeys"
If the goal is strictly to capture any string before the 2nd capital character, one might want pick a solution it'll also work with all types of strings including numbers and special characters.
strings <- c("MonkeysDogsCats",
"M4DogsCats",
"M?DogsCats")
stringr::str_remove(strings, "(?<=.)[A-Z].*")
Output:
[1] "Monkeys" "M4" "M?"
It depends on what you want to allow to match. You can for example match an uppercase char [A-Z] optionally followed by any character that is not an uppercase character [^A-Z]*
If you don't want to allow whitespace chars, you can exclude them [^A-Z\\s]*
library(stringr)
str_extract("MonkeysDogsCats", "[A-Z][^A-Z]*")
Output
[1] "Monkeys"
R demo
If there should be an uppercase character following, and there are only lowercase characters allowed:
str <- "MonkeysDogsCats"
regmatches(str, regexpr("[A-Z][a-z]*(?=[A-Z])", str, perl = TRUE))
Output
[1] "Monkeys"
R demo
I have:
'30Jun2021'
I want to skip/remove the first two digits of the four digit number (or any other way of doing this):
'30Jun21'
I have tried:
^.{0,5}
https://regex101.com/r/hAJcdE/1
I have the first 5 characters but I have not figured out how to skip/remove the '20'
Manipulating datetimes is better using the dedicated date/time functions.
You can convert the variable to date and use format to get the output in any format.
x <- '30Jun2021'
format(as.Date(x, '%d%b%Y'), '%d%b%y')
#[1] "30Jun21"
You can also use lubridate::dmy(x) to convert x to date.
You don't even need regex for this. Just use substring operations:
x <- '30Jun2021'
paste0(substr(x, 1, 5), substr(x, 8, 9))
[1] "30Jun21"
Use sub
sub('\\d{2}(\\d{2})$', "\\1", x)
[1] "30Jun21"
or with str_remove
library(stringr)
str_remove(x, "\\d{2}(?=\\d{2}$)")
[1] "30Jun21"
data
x <- '30Jun2021'
You could also match the format of the string with 2 capture groups, where you would match the part that you want to omit and capture what you want to keep.
\b(\d+[A-Z][a-z]+)\d\d(\d\d)\b
Regex demo
sub("\\b(\\d+[A-Z][a-z]+)\\d\\d(\\d\\d)\\b", "\\1\\2", "30Jun2021")
Output
[1] "30Jun21"
I have the following data
GT_BUC-01_BUCST-19
ADT_BURC-1_BUCST-09
BT_BUDDC-1_BUDSCST-29
CAST_BUC-31_BUCST-9
CAST_BUC-1_BUCST-9
How do I use R to make the numbers after both hyphens to pad leading zeros so it will have Two digits? The resulting format should look like this:
GT_BUC-01_BUCST-19
ADT_BURC-01_BUCST-09
BT_BUDDC-01_BUDSCST-29
CAST_BUC-31_BUCST-09
CAST_BUC-01_BUCST-09
One option would be to use stringr::str_replace_all
x <- c('GT_BUC-01_BUCST-19', 'ADT_BURC-1_BUCST-09',
'BT_BUDDC-1_BUDSCST-29', 'CAST_BUC-31_BUCST-9', 'CAST_BUC-1_BUCST-9')
stringr::str_replace_all(x, '\\d+', function(m) sprintf('%02s', m))
#[1] "GT_BUC-01_BUCST-19" "ADT_BURC-01_BUCST-09"
#[3] "BT_BUDDC-01_BUDSCST-29" "CAST_BUC-31_BUCST-09"
#[5] "CAST_BUC-01_BUCST-09"
You could try using gsub as follows:
x <- gsub("-(\\d)(?!\\d)", "-0\\1", x, perl=TRUE)
x
[1] "GT_BUC-01_BUCST-19" "ADT_BURC-01_BUCST-09" "BT_BUDDC-01_BUDSCST-29"
[4] "CAST_BUC-31_BUCST-09" "CAST_BUC-01_BUCST-09"
Data:
x <- c("GT_BUC-01_BUCST-19",
"ADT_BURC-1_BUCST-09",
"BT_BUDDC-1_BUDSCST-29",
"CAST_BUC-31_BUCST-9",
"CAST_BUC-1_BUCST-9")
The regex pattern used here matches dash followed by a single number only. In this case, we then replace by prepending a zero to this single number.
I have the next vector of strings
[1] "/players/playerpage.htm?ilkidn=BRYANPHI01"
[2] "/players/playerpage.htm?ilkidhh=WILLIROB027"
[3] "/players/playerpage.htm?ilkid=THOMPWIL01"
I am looking for a way to retrieve the part of the string that is placed after the equal sign meaning I would like to get a vector like this
[1] "BRYANPHI01"
[2] "WILLIROB027"
[3] "THOMPWIL01"
I tried using substr but for it to work I have to know exactly where the equal sign is placed in the string and where the part i want to retrieve ends
We can use sub to match the zero or more characters that are not a = ([^=]*) followed by a = and replace it with ''.
sub("[^=]*=", "", str1)
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
data
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
Using stringr,
library(stringr)
word(str1, 2, sep = '=')
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
Using strsplit,
strsplit(str1, "=")[[1]][2]
# [1] "BRYANPHI01"
With Sotos comment to get results as vector:
sapply(str1, function(x){
strsplit(x, "=")[[1]][2]
})
Another solution based on regex, but extracting instead of substituting, which may be more efficient.
I use the stringi package which provides a more powerful regex engine than base R (in particular, supporting look-behind).
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
stri_extract_all_regex(str1, pattern="(?<==).+$", simplify=T)
(?<==) is a look-behind: regex will match only if preceded by an equal sign, but the equal sign will not be part of the match.
.+$ matches everything until the end. You could replace the dot with a more precise symbol if you are confident about the format of what you match. For example, '\w' matches any alphanumeric character, so you could use "(?<==)\\w+$" (the \ must be escaped so you end up with \\w).
I have strings that looks like this.
x <- c("P2134.asfsafasfs","P0983.safdasfhdskjaf","8723.safhakjlfds")
I need to end up with:
"2134", "0983", and "8723"
Essentially, I need to extract the first four characters that are numbers from each element. Some begin with a letter (disallowing me from using a simple substring() function).
I guess technically, I could do something like:
x <- gsub("^P","",x)
x <- substr(x,1,4)
But I want to know how I would do this with regex!
You could use str_match from the stringr package:
library(stringr)
print(c(str_match(x, "\\d\\d\\d\\d")))
# [1] "2134" "0983" "8723"
You can do this with gsub too.
> sub('.?([0-9]{4}).*', '\\1', x)
[1] "2134" "0983" "8723"
>
I used sub instead of gsub to assure I only got the first match. .? says any single character and its optional (similar to just . but then it wouldn't match the case without the leading P). The () signify a group that I reference in the replacement '\\1'. If there were multiple sets of () I could reference them too with '\\2'. Inside the group, and you had the syntax correct, I want only numbers and I want exactly 4 of them. The final piece says zero or more trailing characters of any type.
Your syntax was working, but you were replacing something with itself so you wind up with the same output.
This will get you the first four digits of a string, regardless of where in the string they appear.
mapply(function(x, m) paste0(x[m], collapse=""),
strsplit(x, ""),
lapply(gregexpr("\\d", x), "[", 1:4))
Breaking it down into pieces:
What's going on in the above line is as follows:
# this will get you a list of matches of digits, and their location in each x
matches <- gregexpr("\\d", x)
# this gets you each individual digit
matches <- lapply(matches, "[", 1:4)
# individual characters of x
splits <- strsplit(x, "")
# get the appropriate string
mapply(function(x, m) paste0(x[m], collapse=""), splits, matches)
Another group capturing approach that doesn't assume 4 numbers.
x <- c("P2134.asfsafasfs","P0983.safdasfhdskjaf","8723.safhakjlfds")
gsub("(^[^0-9]*)(\\d+)([^0-9].*)", "\\2", x)
## [1] "2134" "0983" "8723"