R splitting string on predefined location - r

I have string, which should be split into parts from "random" locations. Split occurs always from next comma after colon.
My idea was to find colons with
stringr::str_locate_all(test, ":") %>%
unlist()
then find commas
stringr::str_locate_all(test, ",") %>%
unlist()
and from there to figure out position where it should be split up, but could not find suitable way to do it. Feels like there is always 6 characters after colon before the comma, but I can't be sure about that for whole data.
Here is example string:
dput(test)
"AA,KK,QQ,JJ,TT,99,88:0.5083,66,55:0.8303,AK,AQ,AJs,AJo:0.9037,ATs:0.0024,ATo:0.5678"
Here is what result should be
dput(result)
c("AA,KK,QQ,JJ,TT,99,88:0.5083", "66,55:0.8303", "AK,AQ,AJs,AJo:0.9037",
"ATs:0.0024", "ATo:0.5678")

Perehaps we can use regmatches like below
> regmatches(test, gregexpr("(\\w+,?)+:[0-9.]+", test))[[1]]
[1] "AA,KK,QQ,JJ,TT,99,88:0.5083" "66,55:0.8303"
[3] "AK,AQ,AJs,AJo:0.9037" "ATs:0.0024"
[5] "ATo:0.5678"

here is one option with strsplit - replace the , after the digit followed by the . and one or more digits (\\d+) with a new delimiter using gsub and then split with strsplit in base R
result1 <- strsplit(gsub("([0-9]\\.[0-9]+),", "\\1;", test), ";")[[1]]
-checking
> identical(result, result1)
[1] TRUE
If the number of characters are fixed, use a regex lookaround
result1 <- strsplit(test, "(?<=:.{6}),", perl = TRUE)[[1]]

Related

Convert sign in column names if not at certain position in R [duplicate]

I have a character string of names which look like
"_6302_I-PAL_SPSY_000237_001"
I need to remove the first occurred underscore, so that it will be as
"6302_I-PAL_SPSY_000237_001"
I aware of gsub but it removes all of underscores. Thank you for any suggestions.
gsub function do the same, to remove starting of the string symbol ^ used
x <- "_6302_I-PAL_SPSY_000237_001"
x <- gsub("^\\_","",x)
[1] "6302_I-PAL_SPSY_000237_001"
We can use sub with pattern as _ and replacement as blanks (""). This will remove the first occurrence of '_'.
sub("_", "", str1)
#[1] "6302_I-PAL_SPSY_000237_001"
NOTE: This will remove the first occurence of _ and it will not limit based on the position i.e. at the start of the string.
For example, suppose we have string
str2 <- "6302_I-PAL_SPSY_000237_001"
sub("_", "", str2)
#[1] "6302I-PAL_SPSY_000237_001"
As the example have _ in the beginning, another option is substring
substring(str1, 2)
#[1] "6302_I-PAL_SPSY_000237_001"
data
str1 <- "_6302_I-PAL_SPSY_000237_001"
This can be done with base R's trimws() too
string1<-"_6302_I-PAL_SPSY_000237_001"
trimws(string1, which='left', whitespace = '_')
[1] "6302_I-PAL_SPSY_000237_001"
In case we have multiple words with leading underscores, we may have to include a word boundary (\\b) in our regex, and use either gsub or stringr::string_remove:
string2<-paste(string1, string1)
string2
[1] "_6302_I-PAL_SPSY_000237_001 _6302_I-PAL_SPSY_000237_001"
library(stringr)
str_remove_all(string2, "\\b_")
> str_remove_all(string2, "\\b_")
[1] "6302_I-PAL_SPSY_000237_001 6302_I-PAL_SPSY_000237_001"

Get substring before the second capital letter

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

How would I remove the text before the initial period, the initial period itself and text after final period in a string?

I need to remove the text before the leading period (as well as the leading period) and the text following the last period from a string.
Given this string for example:
"ABCD.EF.GH.IJKL.MN"
I'd like to get the output:
[1] "IJKL"
I have tried the following:
split_string <- sub("^.*?\\.","", string)
split_string <- sub("^\\.+|\\.[^.]*$", "", string)
I believe I have it working for the period and text after for that string output I want. However, the first line needs to be executed multiple times to remove the text before that period in question e.g. '.I'.
One option in base R is to capture as a group ((...)) the word followed by the dot (\\.) and the word (\\w+) till the end ($) of the string. In the replacement, use the backreference (\\1) of the captured word
sub(".*\\.(\\w+)\\.\\w+$", "\\1", str1)
#[1] "IJKL"
Here, we match characters (.*) till the . (\\. - escaped to get the literal value because . is a metacharacter that will match any character if not escaped), followed by the word captured ((\\w+)), followed by a dot and another word at the end ($)of the string. The replacement part is mentioned above
Or another option is regmatches/regexpr from base R
regmatches(str1, regexpr("\\w+(?=\\.\\w+$)", str1, perl = TRUE))
#[1] "IJKL"
Or another option is word from stringr
library(stringr)
word(str1, -2, sep="[.]")
#[1] "IJKL"
data
str1 <- "ABCD.EF.GH.IJKL.MN"
Here is a janky dplyr version in case the other values are of importance and you want to select them later on, just include them in the "select".
df<- data.frame(x=c("ABCD.EF.GH.IJKL.MN"))
df2<-df %>%
separate(x, into=c("var1", "var2","var3","var4","var5")) %>%
select("var4")
Split into groups at period and take the second one from last.
sapply(strsplit(str1, "\\."), function(x) x[length(x) - 1])
#[1] "IJKL"
Get indices of the periods and use substr to extract the relevant portion
sapply(str1, function(x){
ind = gregexpr("\\.", x)[[1]]
substr(x, ind[length(ind) - 1] + 1, ind[length(ind)] - 1)
}, USE.NAMES = FALSE)
#[1] "IJKL"
These alternatives all use no packages or regular expressions.
1) basename/dirname Assuming the test input s shown in the Note at the end convert the dots to slashes and then use dirname and basename.
basename(dirname(chartr(".", "/", s)))
## [1] "IJKL" "IJKL"
2) strsplit Using strsplit split the strings at dot creating a list of character vectors, one vector per input string, and then for each such vector take the last 2 elements using tail and the first of those using indexing.
sapply(strsplit(s, ".", fixed = TRUE), function(x) tail(x, 2)[1])
## [1] "IJKL" "IJKL"
3) read.table It is not clear from the question what the general case is but if all the components of s have the same number of dot separated fields then we can use read.table to create a data.frame with one row per input string and one column per dot-separated component. Then take the column just before the last.
dd <- read.table(text = s, sep = ".", as.is = TRUE)
dd[[ncol(dd)-1]]
## [1] "IJKL" "IJKL"
4) substr Again, the general case is not clear but if the string of interest is always at character positions 12-15 then a simple solution is:
substr(s, 12, 15)
## [1] "IJKL" "IJKL"
Note
s <- c("ABCD.EF.GH.IJKL.MN", "ABCD.EF.GH.IJKL.MN")

Apply a regex only to the first word of a phrase (defined with spaces)

I have this regex to separate letters from numbers (and symbols) of a word: (?<=[a-zA-Z])(?=([[0-9]|[:punct:]])). My test string is: "CALLE15 CRA22".
I want to apply this regex only to the first word of that sentence (the word is defined with spaces). Namely, I want apply that only to "CALLE15".
One solution is split the string (sentence) into words and then apply the regex to the first word, but I want to do all in one regex. Other solution is to use r stringr::str_replace() (or sub()) that replace only the first match, but I need stringr::str_replace_all (or gsub()) for other reasons.
What I need is to insert a space between the two that I do with the replacement function. The outcome I want is "CALLE 15 CRA22" and with the posibility of "CALLE15 CRA 22". I try a lot of positions for the space and nothing, neither the ^ at the beginning.
https://rubular.com/r/7dxsHdOA3avTdX
Thanks for your help!!!!
I am unsure about your problem statement (see my comment above), but the following reproduces your expected output and uses str_replace_all
ss <- "CALLE15 CRA22"
library(stringr)
str_replace_all(ss, "^([A-Za-z]+)(\\d+)(\\s.+)$", "\\1 \\2\\3")
#[1] "CALLE 15 CRA22"
Update
To reproduce the output of the sample string from the comment above
ss <- "CLL.6 N 5-74NORTE"
pat <- c(
"(?<=[A-Za-z])(?![A-Za-z])",
"(?<![A-Za-z])(?=[A-Za-z])",
"(?<=[0-9])(?![0-9])",
"(?<![0-9])(?=[0-9])")
library(stringr)
str_split(ss, sprintf("(%s)", paste(pat, collapse = "|"))) %>%
unlist() %>%
.[nchar(trimws(.)) > 0] %>%
paste(collapse = " ")
#[1] "CLL . 6 N 5 - 74 NORTE"

Split and re-concatenate a string

I am trying to get the host of an IP address from a list of strings.
ips <- c('140.112.204.42', '132.212.14.139', '31.2.47.93', '7.112.221.238')
I want to get the first 2 digits from the ips. output:
ips <- c('140.112', '132.212', '31.2', '7.112')
This is the code that I wrote to convert them:
cat(unlist(strsplit(ips, "\\.", fixed = FALSE))[1:2], sep = ".")
When I check the type of individual ips in the end I get something like this:
140.112 NULL
Not sure what I am doing wrong. If you have some other ideas completely different from this that is completely fine too.
With sub:
ips <- c('140.112.204.42', '132.212.14.139', '31.2.47.93', '7.112.221.238')
sub('\\.\\d+\\.\\d+$', '', ips)
# [1] "140.112" "132.212" "31.2" "7.112"
With str_extract from stringr:
library(stringr)
str_extract(ips, '^\\d+\\.\\d+')
# [1] "140.112" "132.212" "31.2" "7.112"
With strsplit + sapply:
sapply(strsplit(ips, '\\.'), function(x) paste(x[1:2], collapse = '.'))
# [1] "140.112" "132.212" "31.2" "7.112"
With read.table + apply:
apply(read.table(textConnection(ips), sep='.')[1:2], 1, paste, collapse = '.')
#[1] "140.112" "132.212" "31.2" "7.112"
Notes:
sub('\\.\\d+\\.\\d+$', '', ips):
i. \\.\\d+\\.\\d+$ matches a literal dot, a digit one or more times, a literal dot again, and a digit one or more times at the end of the string
ii. sub removes the above match from the string
str_extract(ips, '^\\d+\\.\\d+'):
i. ^\\d+\\.\\d+ matches a digit one or more times, a literal dot and a digit one or more times in the beginning of the string
ii. str_extract extracts the above match from the string
sapply(strsplit(ips, '\\.'), function(x) paste(x[1:2], collapse = '.')):
i. strsplit(ips, '\\.') splits each ip using a literal dot as the delimiter. This returns a list of vectors after the split
ii. With sapply, paste(x[1:2], collapse = '.') is applied to every element of the list, thus taking only the first two numbers from each vector, and collapsing them with a dot as the separator. sapply then coerces the list to a vector, thus returning a vector of the desired ips.
apply(read.table(textConnection(ips), sep='.')[1:2], 1, paste, collapse = '.'):
i. read.table(textConnection(ips), sep='.')[1:2] treats ips as text input and reads it in with dot as a delimiter. Only taking the first two columns.
ii. apply enables paste to be operated on each row, and collapses with a dot.
Could you please try following.
gsub("([0-9]+.[0-9]+)(.*)","\\1",ips)
Explanation: Using gsub function and putting regex there to match digits then DOT then digits in memory's 1st place holder and keeping .* everything after it in 2nd place holder of memory. Then substituting these with \\1 with first regex's value which will be first 2 fields.
One solution is the following:
vapply(strsplit(ips, ".", fixed = TRUE),
function(x) paste(x[1:2], collapse = "."),
character(1L))
vapply applies function(x) to each element of the output of strsplit
strsplit produces a list where each element of the list is the components of the IP addresses separated by "."; setting fixed = TRUE requests to split using the exact value of the splitting string (i.e., "."), not using regex
function(x) takes the first two elements (x[1:2]) of each item coming out of strsplit and pastes them together, seperated by "."
character(1L) tells vapply that each element of the output (i.e., returned from function(x) should be a string of length 1.
Edit: #useR posted this solution right before me (using sapply).
substr is vectorised on the stop argument, so you can use this with a vector of positions before the second dot. regexpr gives the positions of the first match, so if you sub out the first one you can match on the second - which will be conveniently one before it's true position as needed (since you removed the first one).
substr(ips,1,regexpr("\\.",sub("\\.","",ips)))
[1] "140.112" "132.212" "31.2" "7.112"
We can convert the ip addresses to numeric_version class and then format using this base R one-liner that employs no regular expressions:
format(numeric_version(ips)[, 1:2])
[1] "140.112" "132.212" "31.2" "7.112"

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