I have strings like this: "X96HE6.10nMBI_1_2", "X96HE6.10nMBI_2_2", "X96HE6.10nMBI_3_2" and I would like to match only numbers 1, 2 and 3 in between underscores but without them(underscores). The best solution I could come up with is this str_match(sample_names, "_+[1-3]?") I would really appreciate the help.
The simplest method is by using suband backreference:
Data:
d <- c("X96HE6.10nMBI_1_2", "X96HE6.10nMBI_2_2", "X96HE6.10nMBI_3_2")
Solution:
sub(".*_(\\d)_.*", "\\1", d)
Here, (\\d) defines the capturing group for a single number (if the number in question can be more than one digit, use \\d+) that is 'recalled' by the backreference \\1in subs replacement argument
Alternatively use str_extract and positive lookaround:
library(stringr)
str_extract(d, "(?<=_)\\d(?=_)")
(?<=_) is positive lookbehind which can be glossed as "If you see _ on the left..."
\\d is the number to be matched
(?=_) is positive lookahead, which can be glossed as "If you see _ on the right..."
Result:
[1] "1" "2" "3"
You can use Look Arounds, I personally rely heavily on the stringr Cheatsheets for these kind of regex, the syntax is a bit hard to remember, here is the rstudio page for Cheatsheets look for stringr ->LOOK AROUNDS
library(tidyverse)
codes <- c("X96HE6.10nMBI_1_2", "X96HE6.10nMBI_2_2", "X96HE6.10nMBI_3_2")
codes %>%
str_extract("(?<=_)[:digit:]+(?=_)")
#> [1] "1" "2" "3"
Created on 2020-06-14 by the reprex package (v0.3.0)
Using x in the Note at the end, read it in using read.table and pick off the second field. No packages or regular expressions are used.
read.table(text = x, sep = "_")[[2]]
## [1] 1 2 3
Note
x <- c("X96HE6.10nMBI_1_2", "X96HE6.10nMBI_2_2", "X96HE6.10nMBI_3_2")
No need for any third-party module:
strings <- c("X96HE6.10nMBI_1_2", "X96HE6.10nMBI_2_2", "X96HE6.10nMBI_3_2")
pattern <- "(?<=_)(\\d+)(?=_)"
unlist(regmatches(strings, gregexpr(pattern, strings, perl = TRUE)))
Which yields:
[1] "1" "2" "3"
I'm trying to extract values from a vector of strings. Each string in the vector, (there are about 2300 in the vector), follows the pattern of the example below:
"733|Overall (-2 to 2): _________2________________|How controversial is each sentence (1-5)?|Sent. 1 (ANALYSIS BY...): ________1__________|Sent. 2 (Bail is...): ____3______________|Sent. 3 (2) A...): _______1___________|Sent. 4 (3) A...): _______1___________|Sent. 5 (Proposition 100...): _______5___________|Sent. 6 (In 2006,...): _______3___________|Sent. 7 (That legislation...): ________1__________|Types of bias (check all that apply):|Pro Anti|X O Word use (bold, add alternate)|X O Examples (italicize)|O O Extra information (underline)|X O Any other bias (explain below)|Last sentence makes it sound like an urgent matter.|____________________________________________|NA|undocumented, without a visa|NA|NA|NA|NA|NA|NA|NA|NA|"
What I'd like is to extract the numbers following the pattern "Sent. " and place them into a separate vector. For the example, I'd like to extract "1311531".
I'm having trouble using gsub to accomplish this.
library(tidyverse)
Data <- c("PASTE YOUR WHOLE STRING")
str_locate(Data, "Sent. ")
Reference <- str_locate_all(Data, "Sent. ") %>% as.data.frame()
Reference %>% names() #Returns [1] "start" "end"
Reference <- Reference %>% mutate(end = end +1)
YourNumbers <- substr(Data,start = Reference$end[1], stop = Reference$end[1])
for (i in 2:dim(Reference)[1]){
Temp <- substr(Data,start = Reference$end[i], stop = Reference$end[i])
YourNumbers <- paste(YourNumbers, Temp, sep = "")
}
YourNumbers #Returns "1234567"
We can use str_match_all from stringr to get all the numbers followed by "Sent".
str_match_all(text, "Sent.*?_+(\\d+)")[[1]][, 2]
#[1] "1" "3" "1" "1" "5" "3" "1"
A base R option using strsplit and sub
lapply(strsplit(ss, "\\|"), function(x)
sub("Sent.+: _+(\\d+)_+", "\\1", x[grepl("^Sent", x)]))
#[[1]]
#[1] "1" "3" "1" "1" "5" "3" "1"
Sample data
ss <- "733|Overall (-2 to 2): _________2________________|How controversial is each sentence (1-5)?|Sent. 1 (ANALYSIS BY...): ________1__________|Sent. 2 (Bail is...): ____3______________|Sent. 3 (2) A...): _______1___________|Sent. 4 (3) A...): _______1___________|Sent. 5 (Proposition 100...): _______5___________|Sent. 6 (In 2006,...): _______3___________|Sent. 7 (That legislation...): ________1__________|Types of bias (check all that apply):|Pro Anti|X O Word use (bold, add alternate)|X O Examples (italicize)|O O Extra information (underline)|X O Any other bias (explain below)|Last sentence makes it sound like an urgent matter.|____________________________________________|NA|undocumented, without a visa|NA|NA|NA|NA|NA|NA|NA|NA|"
I am new to R and would like to know how to remove leading 0s from a determinate column in a database.
This is the column I have in my df.
questionn
SI001
SI002
SI003
SI010
and I would like to get something like
questionn
1
2
3
10
I have tried something like this but it doesn't work because of the SI010
library(stringr)
df$questionn <- str_replace_all(df$questionn, 'SQ0', '')
data
df <- data.frame(questionn=c("SI001","SI002","SI003","SI010"),stringsAsFactors = FALSE)
Try:
as.numeric(str_replace_all(df$questionn,"SI0",""))
You can remove all characters that are not digits then convert as numeric:
as.numeric(gsub("\\D","",df$questionn))
[1] 1 2 3 10
or as.numeric(str_replace_all(df$questionn,"\\D","")) for same output.
substr(gsub("SI", "", question$question),
regexpr("[^0]",gsub("SI", "", question$question)),
nchar(gsub("SI", "", question$question)))
Produces:
"1" "2" "3" "10"
The first thing you do is strip out the SI, to get the data in a format of having leading zeros.
I have the following value (and similar formatting in hundreds of thousands of fields):
61.00.62.1
that I would like to use a stringr or stringi and (likely) a regex to turn into
61.0
0.6
2.1
I have been unsuccessfully using the stringi::stri_split_regex command of the form
"[0-9]\\d{1,2}(\\.[0-9]\\d{1})"
Thank you.
I would try stri_extract_all_regex:
stringi::stri_extract_all_regex('61.00.62.1', '\\d{1,2}\\.\\d')
# [[1]]
# [1] "61.0" "0.6" "2.1"
\\d{1,2}\\.\\d will match 1 to 2 digits followed by one point and one digit.
We can use strsplit
strsplit(str1, "(?<=\\..)", perl = TRUE)[[1]]
#[1] "61.0" "0.6" "2.1"
data
str1 <- "61.00.62.1"
Make list of given string and separate list by each character by checking '.'
I've got some problems deleting duplicate elements in a string.
My data look similar to this:
idvisit path
1 1,16,23,59
2 2,14,14,19
3 5,19,23,19
4 10,10
5 23,23,27,29,23
I have a column containing an unique ID and a column containing a path for web page navigation.
The right column contains some cases, where pages just were reloaded and the page were tracked twice or even more.
The pages are separated with commas and are saved as factors.
My problem is, that I don't want to have multiple pages in a row, so the data should look like this.
idvisit path
1 1,16,23,59
2 2,14,19
3 5,19,23,19
4 10
5 23,27,29,23
The multiple pages next to each other should be removed. I know how to delete a specific multiple number using regexpressions, but I have about 20.000 different pages and can't do this for all of them.
Does anyone have a solution or a hint, for my problem?
Thanks
Sebastian
We can use tidyverse. Use the separate_rows to split the 'path' variable by the delimiter (,) to convert to a long format, then grouped by 'idvisit', we paste the run-length-encoding values
library(tidyverse)
separate_rows(df1, path) %>%
group_by(idvisit) %>%
summarise(path = paste(rle(path)$values, collapse=","))
# A tibble: 5 × 2
# idvisit path
# <int> <chr>
#1 1 1,16,23,59
#2 2 2,14,19
#3 3 5,19,23,19
#4 4 10
#5 5 23,27,29,23
Or a base R option is
df1$path <- sapply(strsplit(df1$path, ","), function(x) paste(rle(x)$values, collapse=","))
NOTE: If the 'path' column is factor class, convert to character before passing as argument to strsplit i.e. strsplit(as.character(df1$path), ",")
Using stringr package, with function: str_replace_all, I think it gets what you want using the following regular expression: ([0-9]+),\\1and then replace it with \\1 (we need to scape the \ special character):
library(stringr)
> str_replace_all("5,19,23,19", "([0-9]+),\\1", "\\1")
[1] "5,19,23,19"
> str_replace_all("10,10", "([0-9]+),\\1", "\\1")
[1] "10"
> str_replace_all("2,14,14,19", "([0-9]+),\\1", "\\1")
[1] "2,14,19"
You can use it in a array form: x <- c("5,19,23,19", "10,10", "2,14,14,19") then:
str_replace_all(x, "([0-9]+),\\1", "\\1")
[1] "5,19,23,19" "10" "2,14,19"
or using sapply:
result <- sapply(x, function(x) str_replace_all(x, "([0-9]+),\\1", "\\1"))
Then:
> result
5,19,23,19 10,10 2,14,14,19
"5,19,23,19" "10" "2,14,19"
Notes:
The first line is the attribute information:
> str(result)
Named chr [1:3] "5,19,23,19" "10" "2,14,19"
- attr(*, "names")= chr [1:3] "5,19,23,19" "10,10" "2,14,14,19"
If you don't want to see them (it does not affect the result), just do:
attributes(result) <- NULL
Then,
> result
[1] "5,19,23,19" "10" "2,14,19"
Explanation about the regular expression used: ([0-9]+),\\1
([0-9]+): Starts with a group 1 delimited by () and finds any digit (at least one)
,: Then comes a punctuation sign: , (we can include spaces here, but the original example only uses this character as delimiter)
\\1: Then comes an identical string to the group 1, i.e.: the repeated number. If that doesn't happen, then the pattern doesn't match.
Then if the pattern matches, it replaces it, with the value of the variable \\1, i.e. the first time the number appears in the pattern matched.
How to handle more than one duplicated number, for example 2,14,14,14,19?:
Just use this regular expression instead: ([0-9]+)(,\\1)+, then it matches when at least there is one repetition of the delimiter (right) and the number. You can try other possibilities using this regex101.com (in MHO it more user friendly than other online regular expression checkers).
I hope this would work for you, it is a flexible solution, you just need to adapt it with the pattern you need.