My task is to split and extract the part from a string until the occurrence of the fourth underscore.
I am working with R right now but I am kind of a beginner with programming and stuff.
The input looks like this:
6_10_36_0_1
6_10_38_16_15
6_100_76_16_18.1
My required result would look like this:
6_10_36_0
6_10_38_16
6_100_76_16
My idea is the following:
substr(data$x, 0, XXX)
While XXX defines the position before the fourth underscore, maybe using grep or strsplit?
Sorry, if I asked a stupid and easy-to-answer question. However I didn't find a fitting to answers already posted.
edit:
> bestand$ID<-sub("(_[0-9.]+$)", "", bestand$x)
Fehler in `$<-.data.frame`(`*tmp*`, "ID", value = character(0)) :
replacement has 0 rows, data has 36513
> gsub("(_[0-9.]+$)", "", "6_100_63_8_2")
[1] "6_100_63_8"
>
apparently the command works, however it doesnot work with the matrix..
You can use regular expression to replace with null, in php we do
$string = '6_10_36_0_1';
$newstring =preg_replace('/(_[0-9.]+$)/', '', $string);
Edit (I dono exactly about r but roughly it would be like this)
sub("(_[0-9.]+$)", "", 'your strings or array of strings')
gsub("(_[0-9.]+$)", "", 'your strings or array of strings')
and the tutorial is here
The stringr package has lots of handy shortcuts for this kind of work:
# input data
data <- read.table(text = "6_10_36_0_1
6_10_38_16_15
6_100_76_16_18.1")
# load library
library(stringr)
# prepare regular expression
regexp <- "([[:digit:]]+_){3}[[:digit:]]+"
# process string
(str_extract(data$V1, regexp))
Which gives the desired result:
[1] "6_10_36_0" "6_10_38_16" "6_100_76_16"
To explain the regexp a little:
[[:digit:]] is any number 0 to 9
+ means the preceding item (in this case, a digit) will be matched one or more times
_ is the underscore, as is
{3} means repeat the previous string three times
This page is also very useful for this kind of string processing: http://en.wikibooks.org/wiki/R_Programming/Text_Processing
Related
A field from my dataset has some of it's observations to start with a "." e.g ".TN34AB1336"instead of "TN34AB1336".
I've tried
truck_log <- truck_log %>%
filter(BookingID != "WDSBKTP49392") %>%
filter(BookingID != "WDSBKTP44502") %>%
mutate(GpsProvider = str_replace(GpsProvider, "NULL", "UnknownGPS")) %>%
mutate(vehicle_no = str_replace(vehicle_no, ".TN34AB1336", "TN34AB1336"))
the last mutate command in my code worked but there are more of such issues in another field e.g "###TN34AB1336" instead of "TN34AB1336".
So I need a way of automating the process such that all observations that doesn't start with a character is trimmed from the left by a single command in R.
I've attached a picture of a filtered field from spreadsheet to make the question clearer.
We can use regular expressions to replace anything up to the first alphanumeric character with ""to remove everything that is not a Number/Character from the beginning of a string:
names <- c("###TN34AB1336",
".TN34AB1336",
",TN654835A34",
":+?%TN735345")
stringr::str_replace(names,"^.+?(?=[[:alnum:]])","") # Matches up to, but not including the first alphanumeric character and replaces with ""
[1] "TN34AB1336" "TN34AB1336" "TN654835A34" "TN735345"
``
You can use sub.
s <- c(".TN34AB1336", "###TN34AB1336")
sub("^[^A-Z]*", "", s)
#[1] "TN34AB1336" "TN34AB1336"
Where ^ is the start of the string, [^A-Z] matches everything what is not A, B, C, ... , Z and * matches it 0 to n times.
Ideally, in base R I need some kind of string manipulation that will let me detect a pattern and change the string 3 positions after the pattern.
example <- "when string says SOMETHING = #c792ea"
desired output:
when string says SOMETHING = #001628
I have tried gsub but I am not sure how I can get it to replace the characters after a pattern.
If it based on the position of character, then we can use substring assignment
substring(example, 30) <- "#001628"
example
#[1] "when string says SOMETHING = #001628"
Or if we need to find the position of the word that starts with #
library(stringr)
posvec <- c(str_locate(example, "#\\w+"))
substring(example, posvec[1], posvec[2]) <- "#001628"
# // or with
# str_sub(example, posvec[1], posvec[2]) <- "#001628"
Another option is sub to change the substring after the = and one or more space (\\s*)
sub("=\\s*.*", "= #001628", example)
#[1] "when string says SOMETHING = #001628"
I need to extract substrings from some strings,for example:
My data is a vector: c("Shigella dysenteriae","PREDICTED: Ceratitis")
a = "Shigella dysenteriae"
b = "PREDICTED: Ceratitis"
I hope that if the string starts with "PREDICTED:", it can be extracted to the subsequent word(maybe "Ceratitis"), and if the string doesn't start with "PREDICTED", it can be extracted to the first word(maybe Shigella);
In this example, the result would be:
result_of_a = "Shigella"
result_of_b = "Ceratitis"
Well,it is a typical conditional regular expression.I tried,but always failed;
I used R which can compatible perl's regular expression.
I know R supports perl's regular expression so I tried to use regexpr and regmatches, two functions to extract the substrings that I want.
The code is :
pattern = "(?<=PREDICTED:)?(?(1)(\\s+\\w+\\b)|(\\w+\\b))"
a = c("Shigella dysenteriae")
m_a = regexpr(pattern,a,perl = TRUE)
result_a = regmatches(a,m_a)
b = c("PREDICTED: Ceratitis")
m_b = regexpr(pattern,a,perl = TRUE)
result_b = regmatches(b,m_b)
Finaly,the result is :
# result_a = "Shigella"
# result_b = "PREDICTED"
It is not the result I expect,result_a is right,result_b is wrong.
WHY???Its seem that the condition didn't work...
PS:
I tried to read some details of conditional reg-expresstion. this is the web I tried to read : https://www.regular-expressions.info/conditional.html and I try to imitate "pattern" from this web ,and also tried to use "RegexBuddy" software to find the reason.
EDIT:
To use the function below on a vector, one can do:
Vector: myvec<-c("Shigella dysenteriae","PREDICTED: Ceratitis")
lapply(myvec,extractor)
[[1]]
[1] "Shigella"
[[2]]
[1] "Ceratitis"
Or:
unlist(lapply(myvec,extractor))
[1] "Shigella" "Ceratitis"
This assumes that the strings are always in the format shown above:
extractor<- function(string){
if(grepl("^PREDICTED",string)){
strsplit(string,": ")[[1]][2]
}
else{
strsplit(string," ")[[1]][1]
}
}
extractor(b)
#[1] "Ceratitis"
extractor(a)
#[1] "Shigella"
I think the reason it does not work is because (1) checks if a numbered capture group has been set but there is no first capturing group set yet, also not in the positive lookbehind (?<=PREDICTED:)?.
There are a first and second capturing group in the parts that follow. The if clause will check for group 1, it is not set so it will match group 2.
If you would make it the only capturing group (?<=(PREDICTED: )?) and omit the other 2 then the if clause will be true but you will get an error because the lookbehind assertion is not fixed length.
Instead of using a conditional pattern, to get both words you might use a capturing group and make PREDICTED: optional:
^(?:PREDICTED: )?(\w+)
Regex demo | R demo
If I understand correctly, the OP wants to extract
the first word after "PREDICTED:" if the strings starts with "PREDICTED:"
the first word of the string if the string does not start with "PREDICTED:".
So, if there is no specific requirement to use only one regex, this is what I would do:
Remove any leading "PREDICTED:" (if any)
Extract the first word from the intermediate result.
For working with regex, I prefer to use Hadley Wickham's stringr package:
inp <- c("Shigella dysenteriae", "PREDICTED: Ceratitis")
library(magrittr) # piping used to improve readability
inp %>%
stringr::str_replace("^PREDICTED:\\s*", "") %>%
stringr::str_extract("^\\w+")
[1] "Shigella" "Ceratitis"
To be on the safe side, I would remove any leading spaces beforehand:
inp %>%
stringr::str_trim() %>%
stringr::str_replace("^PREDICTED:\\s*", "") %>%
stringr::str_extract("^\\w+")
I am trying to parse strings with hyphens and/or numbers to call specific rows.
gene_name <- c("EP-CAM")
Genename=paste0("RNA$",gene_name)
Gene=eval(parse(text = paste0(Genename)))
This is the error:
Error in eval(parse(text = paste0(Genename))) :
object 'CAM' not found
I would need to get RNA$EP-CAM parsed for example. Backquotes will not give me the output and only show me the string.
With numbers the same would happen. I guess this is just a problem of the parse command. Is there an alternative to it?
This is in analogy to this problem: Unexpected symbol error in parse(text = str) with hyphen after a digit
Thank you so much for you support.
D
Adding back ticks to the call works for me. The problem, here, is that "EP-CAM" isn't really a valid name.
RNA <- list(`EP-CAM` = 0)
gene <- c("EP-CAM")
geneName <- paste0("RNA$`", gene, "`")
eval(parse(text = geneName))
# [1] 0
In fact, the following renames the column as EP.CAM.
data.frame(`EP-CAM` = 0)
# EP.CAM
# 1 0
I recently started using R and a newbie for data analysis.
Is it possible in R to find the number of repetitions in a single main string of data when a string of data is used for searching through it?
Example:
Main string: 'abcdefghikllabcdefgllabcd'
and search string: 'lla'
Desired output: 'abcdefghik lla bcdefg lla bcd'
[I tried using grep() function of R, but It is not working in the desired way and only gives the number of repetitions of search string in multiple main strings.]
Thank you in advance.
This works too using regex capture groups:
gsub("(lla)"," \\1 ","abcdefghikllabcdefgllabcd")
Try the gsub() method like this:
main_string <- 'abcdefghikllabcdefgllabcd'
search_string <- 'lla'
output_string <- gsub(search_string, paste(' ', search_string, ' ', sep = ''), main_string)
Your question says that you might want to just COUNT the number of occurrences of the search tring in the main string. If that is the case, try this one liner:
string = "abcdefghikllabcdefgllabcd"
search = 'lla'
( nchar(string) - nchar( gsub(search, "", string)) ) / nchar(search)
#returns 2
string2 = "llaabcdefghikllabcdefgllabcdlla"
( nchar(string2) - nchar( gsub(search, "", string2)) ) / nchar(search)
#returns 4
NOTE: Unit-test your solution for matches at the beginning and end of the string (i.e. make sure it works on 'llaabcdefghikllabcdefgllabcdlla'). I have seen several solutions elsewhere that rely on strsplit() to split on 'lla', but these solutions skip the final 'lla' at the end of the word.