This question already has answers here:
Extract a substring according to a pattern
(9 answers)
Closed 4 years ago.
I have a column of data that looks like this:
**varX**
Q1#_1
Q1#_5
Q1#_10
I would like to edit the data to look like this:
**varX**
1
5
10
Is there a command I could use to simply keep all information after the underscore?
If you want a tidyverse solution, you can use str_extract from the stringr package:
data %>%
mutate(varx = str_extract(varx, "[0-9]+$")) %>%
mutate(varx = as.numeric(varx)) # include this last line if you want a number and not character
In case you always have the Q1#_ string, you can do:
gsub("Q1#_", "", df$varX)
I think you're looking for sub, substitute a certain part of a string with something else. You can give it a regular expression if you want to go fancy, or just give it a literal:
VarX <- sub('Q1#_', '', VarX, fixed=T)
The fancy way ("remove everything before and including the underscore") would be
VarX <- sub('^.*_', '', VarX)
And you may want to convert it to a numeric or an integer:
VarX <- as.integer(sub('Q1#_', '', VarX, fixed=T)) # or as.numeric
You could you use regular expressions:
df[["varX"]] <- sub(".+_", "", df[["varX"]])
df
varX
1 1
2 5
3 10
Or regular expressions-free: with strsplit():
df[["varX"]] <- sapply(df[["varX"]], function(x) strsplit(x, "_")[[c(1,2)]])
Related
I would like to substitute the strings in the list by cutting each string after the first 3-digit number.
a <- c("MTH314PHY410","LB471LB472","PHY472CHM141")
I would like for it to look something like
a <- c("MTH314","LB471","PHY472")
I have tried something like
b <- gsub("[100-999].*","",a)
but it returns c("MTH","LB","PHY") without the first number
A possible solution, based on stringr::str_remove:
library(stringr)
a <- c("MTH314PHY410","LB471LB472","PHY472CHM141")
str_remove(a, "(?<=\\d{3}).*")
#> [1] "MTH314" "LB471" "PHY472"
c("MTH314PHY410","LB471LB472","PHY472CHM141") %>%
stringr::str_extract('.+?\\d{3}')
[1] "MTH314" "LB471" "PHY472"
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 tibble and the vectors within the tibble are character strings with a mix of English and Mandarin characters. I want to split the tibble into two, with one column returning the English, the other column returning the Mandarin. However, I had to resort to the following code in order to accomplish this:
tb <- tibble(x = c("I我", "love愛", "you你")) #create tibble
en <- str_split(tb[[1]], "[^A-Za-z]+", simplify = T) #split string when R reads a character that is not a-z
ch <- str_split(tb[[1]], "[A-Za-z]+", simplify = T) #split string after R reads all the a-z characters
tb <- tb %>%
mutate(EN = en[,1],
CH = ch[,2]) %>%
select(-x)#subset the matrices created above, because the matrices create a column of blank/"" values and also remove x column
tb
I'm guessing there's something wrong with my RegEx that's causing this to occur. Ideally, I would like to write one str_split line that would return both of the columns.
We can use strsplit from base R
do.call(rbind, strsplit(tb$x, "(?<=[A-Za-z])(?=[^A-Za-z])", perl = TRUE))
Or we can use
library(stringr)
tb$en <- str_extract(tb$x,"[[:alpha:]]+")
tb$ch <- str_extract(tb$x,"[^[:alpha:]]+")
We can use str_match and get data for English and rest of the characters separately.
stringr::str_match(tb$x, "([A-Za-z]+)(.*)")[, -1]
# [,1] [,2]
#[1,] "I" "我"
#[2,] "love" "愛"
#[3,] "you" "你"
A simple solution using str_extract from package stringr:
library(stringr)
tb$en <- str_extract(tb$x,"[A-z]+")
tb$ch <- str_extract(tb$x,"[^A-z]")
In case there's more than one Chinese character, just add +to [^A-z].
Alternatively, use gsuband backreference:
tb$en <- gsub("(\\w+).$", "\\1", tb$x)
tb$ch <- gsub("\\w+(.$)", "\\1", tb$x)
Yet another solution macthes unicode characters with [ -~]+ and excludes them with [^ -~]+:
tb$en <- str_extract(tb$x, "[ -~]+")
tb$ch <- str_extract(tb$x, "[^ -~]+")
Result:
tb
# A tibble: 3 x 3
x en ch
<chr> <chr> <chr>
1 I我 I 我
2 love愛 love 愛
3 you你 you 你
I have a date value as follows
"'2015-10-24'"
class Character
I am trying to format this value such that it looks like this '10/24/2015'
I know how to use noquote function and strip the quotes and gsub function to replace the - with / but I am not sure how to switch the year, date and month such that it looks like this '10/24/2015'
Any help is much appreciated.
We can convert to Date class after removing the ' with gsub, and then use format to get the expected output
format(as.Date(gsub("'", '', v1)), "'%m/%d/%Y'")
#[1] "'10/24/2015'" "'10/25/2015'"
Or without using the gsub to remove ', we can specify the ' also in the format within as.Date
format(as.Date(v1, "'%Y-%m-%d'"), "'%m/%d/%Y'")
#[1] "'10/24/2015'" "'10/25/2015'"
This can be made more compact if we are using library(lubridate)
library(lubridate)
format(ymd(v1), "'%m/%d/%Y'")
#[1] "'10/24/2015'" "'10/25/2015'"
If we don't need the ' in the output, we don't have to specify that in the format,
format(ymd(v1), "%m/%d/%Y")
#[1] "10/24/2015" "10/25/2015"
Or we can do this using only gsub by capturing the characters as a group. In the below code, we capture the first 4 characters (.{4}) as a group by wrapping with parentheses followed by matching the -, then capturing the next two characters, followed by -, and capturing the last two characters. In the replacement, we can shuffle the capture groups as per the requirement. In this case, the second capture group should come first (\\2) followed by /, then the third (\\3) and so on...
gsub('(.{4})-(.{2})-(.{2})', '\\2/\\3/\\1', v1)
#[1] "'10/24/2015'" "'10/25/2015'"
To avoid the quotes,
gsub('.(.{4})-(.{2})-(.{2}).', '\\2/\\3/\\1', v1)
#[1] "10/24/2015" "10/25/2015"
In addition, there are other ways such as splitting the string
vapply(strsplit(v1, "['-]"), function(x) paste(x[c(3,4,2)], collapse='/'), character(1))
#[1] "10/24/2015" "10/25/2015"
or extracting the numeric part with str_extract_all and pasteing as before.
library(stringr)
vapply(str_extract_all(v1, '\\d+'), function(x)
paste(x[c(2,3,1)], collapse='/'), character(1))
#[1] "10/24/2015" "10/25/2015"
data
v1 <- c("'2015-10-24'", "'2015-10-25'")
You can also use the function strftime to get the result
d <- "'2015-10-24'"
strftime(as.Date(gsub("'", "", d)), "%m/%d/%Y")
# [1] "10/24/2015"