In numeric column old_code of dataframe df1 my integers have this structure:
head(df1$old_code)
[1] 12101 18201 13202 11301 13302 10401
In the column new_code I would like the same exact integers, minus the second to last 0 (i.e. 12101 I would like to see as 1011, 18201 as 1021). I am positive this is solved using regex but I can't crack the code.
Thank you for your help!
You can try to use gsub:
gsub("(.*)0(.)$", "\\1\\2", df$old_code)
# [1] "1211" "1821" "1322" "1131" "1332" "1041"
You could try converting it into a string:
df$old_code=as.character(df$old_code)
a= substr(df$old_code, 1, 1)
b=substr(df$old_code, 3,nchar(df$old_code))
df$old_code=paste0(a,b)
df$old_code=as.integer(df$old_code)
Related
I have:
'30Jun2021'
I want to skip/remove the first two digits of the four digit number (or any other way of doing this):
'30Jun21'
I have tried:
^.{0,5}
https://regex101.com/r/hAJcdE/1
I have the first 5 characters but I have not figured out how to skip/remove the '20'
Manipulating datetimes is better using the dedicated date/time functions.
You can convert the variable to date and use format to get the output in any format.
x <- '30Jun2021'
format(as.Date(x, '%d%b%Y'), '%d%b%y')
#[1] "30Jun21"
You can also use lubridate::dmy(x) to convert x to date.
You don't even need regex for this. Just use substring operations:
x <- '30Jun2021'
paste0(substr(x, 1, 5), substr(x, 8, 9))
[1] "30Jun21"
Use sub
sub('\\d{2}(\\d{2})$', "\\1", x)
[1] "30Jun21"
or with str_remove
library(stringr)
str_remove(x, "\\d{2}(?=\\d{2}$)")
[1] "30Jun21"
data
x <- '30Jun2021'
You could also match the format of the string with 2 capture groups, where you would match the part that you want to omit and capture what you want to keep.
\b(\d+[A-Z][a-z]+)\d\d(\d\d)\b
Regex demo
sub("\\b(\\d+[A-Z][a-z]+)\\d\\d(\\d\\d)\\b", "\\1\\2", "30Jun2021")
Output
[1] "30Jun21"
This question already has answers here:
How to add leading zeros?
(8 answers)
Closed 1 year ago.
I have a list of US postal zip codes of 5 digits, but some lost their leading zeros. How do I add those zeros back in, while keeping others without the leading 0s intact? I tried formatC, springf, str_pad, and none of them worked, because I am not adding 0s to all values.
We can use sprintf
sprintf('%05d', as.integer(zipcodes))
In which way did str_pad not work?
https://www.rdocumentation.org/packages/stringr/versions/1.4.0/topics/str_pad
df<-data.frame(zip=c(1,22,333,4444,55555))
df$zip <- stringr::str_pad(df$zip, width=5, pad = "0")
[1] "00001" "00022" "00333" "04444" "55555"
Update:
As of the valuable comment of r2evans:
My solution is not very efficient and to get leading 0 we have to modify the paste0 part slightly see here with a dataframe example:
sapply(df$zip, function(x){if(nchar(x)<5){paste0(0,x)}else{x}})
data:
df <- tribble(
~zip,
7889,
2345,
45567,
4394,
34566,
4392,
4599)
df
Output:
[1] "07889" "02345" "45567" "04394" "34566" "04392" "04599"
Fist answer:
This will add a trailing zero to each integer < 5 digits
Where zip is a vector:
sapply(zip, function(x){if(nchar(x)<5){paste0(x,0)}else{x}})
If they start as strings and you don't want to (or cannot) convert to integers first, then an alternative to sprintf is
vec <- c('1','11','11111')
paste0(strrep('0', pmax(0, 5 - nchar(vec))), vec)
# [1] "00001" "00011" "11111"
This will handle strings of any length, and is a no-op for strings of 5 or greater characters.
In a frame, that would be
dat$colname <- paste0(strrep('0', pmax(0, 5 - nchar(dat$colname))), dat$colname)
I have a dataset with different values in R. Some values are like 11.474 and others like 1.034.496 in the same column. I would like to change the values with two dots from 1.034.496 to 1034.496. Is there anyone who could help me please?
Thanks for the help!
Use gsub with Perl regexes:
df <- data.frame(a = c('11.474', '1.034.496', '1.234.034.496'))
df$a = gsub('[.](?=.*[.])', '', df$a, perl = TRUE)
print(df)
## a
## 1 11.474
## 2 1034.496
## 3 1234034.496
Here, [.](?=.*[.]) is a literal dot (has to be escaped like so \. or put into a character class like so: [.]), followed by a literal dot using positive lookahead: (?=PATTERN).
I guess there must be other smarter regex approaches than the below one, but here is my attempt
> ifelse(lengths(gregexpr("\\.",v))>1,sub("\\.","",v),v)
[1] "11.474" "1034.496"
where
v <- c("11.474","1.034.496")
I have a column as below.
9453, 55489, 4588, 18893, 4457, 2339, 45489HQ, 7833HQ
I would like to add leading zero if the number is less than 5 digits. However, some numbers have "HQ" in the end, some don't.(I did check other posts, they dont have similar problem in the "HQ" part)
so the finally desired output should be:
09453, 55489, 04588, 18893, 04457, 02339, 45489HQ, 07833HQ
any idea how to do this? Thank you so much for reading my post!
A one-liner using regular expressions:
my_strings <- c("9453", "55489", "4588",
"18893", "4457", "2339", "45489HQ", "7833HQ")
gsub("^([0-9]{1,4})(HQ|$)", "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")
I have string like
'abbb'
I need to understand how many times I can find substring 'bb'.
grep('bb','abbb')
returns 1. Therefore, the answer is 2 (a-bb and ab-bb). How can I count number of occurrences the way I need?
You can make the pattern non-consuming with '(?=bb)', as in:
length(gregexpr('(?=bb)', x, perl=TRUE)[[1]])
[1] 2
Here is an ugly approach using substr and sapply:
input <- "abbb"
search <- "bb"
res <- sum(sapply(1:(nchar(input)-nchar(search)+1),function(i){
substr(input,i,i+(nchar(search)-1))==search
}))
We can use stri_count
library(stringi)
stri_count_regex(input, '(?=bb)')
#[1] 2
stri_count_regex(x, '(?=bb)')
#[1] 0 1 0
data
input <- "abbb"
x <- c('aa','bb','ba')