I was using grep to do a case-insensitive search, but the problem is I get all values containg the pattern, not just the exact match, but if I use fixed=TRUE that invalidates the ignore.case=TRUE parameter.
g = c("PLD3","PLD2","PLD2ABC","DTPLD2a")
r = "pLd2"
grep(r,g,ignore.case=TRUE,value=TRUE)
>[1] "PLD2" "PLD2ABC" "DTPLD2a"
grep(r,g,ignore.case=TRUE,value=TRUE,fixed=TRUE)
>character(0)
EDIT
r is a user input, so basically it can be anything from a list of 30,000 genes, and it can be all lower-case, all upper-case, or a mixture of both.
And also in my list g the elements can be upper-case, lower-case or a mixture (it is a list of around 15,000 genes)
try
g = c("PLD3","PLD2","PLD2ABC","DTPLD2a")
r <- 'pLd2'
r2 <- paste('^', r, '$', sep = '')
grep(r2 , g ,ignore.case = T, value=TRUE)
[1] "PLD2"
basically the meta characters ^ and $ force grep to fix the regular expression at the start and the end.
Related
I am beginner programmer in R.
I have "cCt/cGt" and I want to extract C and G and write it like C>G.
test ="cCt/cGt"
str_extract(test, "[A-Z]+$")
Try this:
gsub(".*([A-Z]).*([A-Z]).*", "\\1>\\2", test )
[1] "C>G"
Here, we capture the two occurrences of the upper case letters in capturing groups given in parentheses (...). This enables us to refer to them (and only to them but not the rest of the string!) in gsub's replacement clause using backreferences \\1 and \\2. In the replacement clause we also include the desired >.
You seem to look for a mutation in two concatenated strings, this function should solve your problem:
extract_mutation <- function(text){
splitted <- strsplit(text, split = "/")[[1]]
pos <- regexpr("[[:upper:]]", splitted)
uppercases <- regmatches(splitted, pos)
mutation <- paste0(uppercases, collapse = ">")
return(mutation)
}
If the two base exchanges are always at the same index, you could also return the position if you're interested:
position <- pos[1]
return(list(mutation, position))
instead of the return(mutation)
You might also capture the 2 uppercase chars followed and preceded by optional lowercase characters and a / in between.
test ="cCt/cGt"
res = str_match(test, "([A-Z])[a-z]*/[a-z]*([A-Z])")
sprintf("%s>%s", res[2], res[3])
Output
[1] "C>G"
See an R demo.
An exact match for the whole string could be:
^[a-z]([A-Z])[a-z]/[a-z]([A-Z])[a-z]$
I need to split the following sequence of letters into distinct chunks
SCDKSFNRGECSCDKSFNRGECSCDKSFNRGEC
I have used the following code provided from a previous user to achieve what I initially wanted, which was to split the sequence after every C.
library(dplyr)
TestSequence <- "SCDKSFNRGECSCDKSFNRGECSCDKSFNRGEC"
Test <- strsplit(TestSequence, "(?<=[C])", perl = TRUE) %>% unlist
df <- data.frame(Fragment = Test) %>%
mutate("position" = cumsum(nchar(Test)))
This allowed me to split the sequence after every C and retain it's position in the sequence, for example C at position 2, 11 etc.
Now I need to split the same sequence at different locations, which I can do using the following to split after P,A,G or S:
Test2 <- strsplit(TestSequence, "(?<=[P,A,G,S])", perl = TRUE) %>% unlist
This is fine if I want it to split after a given character, but if I try to split it before a character for example D, I cannot seem to retain the D in the fragment. I can only have it retained if it is split after the D.
I have tried every combination of look behind or look ahead I can think of, the following cuts before and after every D which isn't that useful.
Test3 <- strsplit(TestSequence, "(?=[D])", perl = TRUE) %>% unlist
Also is there a way to retain the exact position of every C in the original sequence?
So if I were to split the test sequence after the initial K, I'd have a fragment that was SCDK, could I have a separate column that tells me where the C was in the original sequence. Just as a second example, the next fragment would be SFNRGECSCDK and in that separate column it would say the C was originally in position 11.
Zero-length matches that result from the use of lookahead only patterns used in strsplit are not handled properly.
In this case, you need to "anchor" the matches on the left, too. Either use a non-word boundary, or a lookbehind that disallows the match at the start of string:
TestSequence <- "SCDKSFNRGECSCDKSFNRGECSCDKSFNRGEC"
strsplit(TestSequence, "\\B(?=D)", perl = TRUE)
# => [[1]]
# => [1] "SC" "DKSFNRGECSC" "DKSFNRGECSC" "DKSFNRGEC"
strsplit(TestSequence, "(?<!^)(?=D)", perl = TRUE)
# => [[1]]
# => [1] "SC" "DKSFNRGECSC" "DKSFNRGECSC" "DKSFNRGEC"
See the online R demo.
The \B(?=D) pattern matches a location that is immediately preceded with a word char and is immediately followed with D.
The (?<!^)(?=D) pattern matches a location that is not immediately preceded with a start of string location (i.e. not at the start of string) and is immediately followed with D.
Also, note that [P,A,G,S] matches P, A, G, S and a comma. You should use [PAGS] to match one of the letters.
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+")
how can I encode a url as this
http://www.chemspider.com/inchi.asmx/InChIToSMILES?inchi=InchI=1S/C21H30O9/c1-11(5-6-21(28)12(2)8-13(23)9-20(21,3)4)7-15(24)30-19-18(27)17(26)16(25)14(10-22)29-19/h5-8,14,16-19,22,25-28H,9-10H2,1-4H3/b6-5+,11-7-/t14-,16-,17+,18-,19+,21-/m1/s1&token=e4a6d6fb-ae07-4cf6-bae8-c0e6115bc681
to make this
http://www.chemspider.com/inchi.asmx/InChIToSMILES?inchi=InChI%3D1S%2FC21H30O9%2Fc1-11(5-6-21(28)12(2)8-13(23)9-20(21%2C3)4)7-15(24)30-19-18(27)17(26)16(25)14(10-22)29-19%2Fh5-8%2C14%2C16-19%2C22%2C25-28H%2C9-10H2%2C1-4H3%2Fb6-5%2B%2C11-7-%2Ft14-%2C16-%2C17%2B%2C18-%2C19%2B%2C21-%2Fm1%2Fs1
on R?
I tried
URLencode
but it does not work.
Thanks
It seems that you want to get rid of all but first URL GET data specifier and then to encode the associated data.
url <- "..."
library(stringi)
(addr <- stri_replace_all_regex(url, "\\?.*", ""))
## [1] "http://www.chemspider.com/inchi.asmx/InChIToSMILES"
args <- stri_match_first_regex(url, "[?&](.*?)=([^&]+)")
(data <- stri_replace_all_regex(
stri_trans_general(args[,3], "[^a-zA-Z0-9\\-()]Any-Hex/XML"),
"&#x([0-9a-fA-F]{2});", "%$1"))
## [1] "InchI%3D1S%2FC21H30O9%2Fc1-11(5-6-21(28)12(2)8-13(23)9-20(21%2C3)4)7-15(24)30-19-18(27)17(26)16(25)14(10-22)29-19%2Fh5-8%2C14%2C16-19%2C22%2C25-28H%2C9-10H2%2C1-4H3%2Fb6-5%2B%2C11-7-%2Ft14-%2C16-%2C17%2B%2C18-%2C19%2B%2C21-%2Fm1%2Fs1"
(addr <- stri_c(addr, "?", args[,2], "=", data))
## [1] "http://www.chemspider.com/inchi.asmx/InChIToSMILES?inchi=InchI%3D1S%2FC21H30O9%2Fc1-11(5-6-21(28)12(2)8-13(23)9-20(21%2C3)4)7-15(24)30-19-18(27)17(26)16(25)14(10-22)29-19%2Fh5-8%2C14%2C16-19%2C22%2C25-28H%2C9-10H2%2C1-4H3%2Fb6-5%2B%2C11-7-%2Ft14-%2C16-%2C17%2B%2C18-%2C19%2B%2C21-%2Fm1%2Fs1"
Here I made use of the ICU's transliterator (via stri_trans_general). All characters but A..Z, a..z, 0..9, (, ), and - have been converted to hexadecimal representation
(it seems that URLencode does not handle , even with reserved=TRUE) of the form &#xNN;. Then, each &#xNN; was converted to %NN with stri_replace_all_regex.
Here are two approaches:
1) gsubfn/URLencode If u is an R character string containing the URL then try this. This inputs everything after ? to URLencode replacing the input with the output of that function. Note that "\\K" kills everything in the buffer up to that point so that the ? itself does not get encoded:
library(gsubfn)
gsubfn("\\?\\K(.*)", ~ URLencode(x, TRUE), u, perl = TRUE)
It gives the following (which is not identical to the output in the question but may be sufficient):
http://www.chemspider.com/inchi.asmx/InChIToSMILES?inchi%3dInchI%3d1S%2fC21H30O9%2fc1-11(5-6-21(28)12(2)8-13(23)9-20(21,3)4)7-15(24)30-19-18(27)17(26)16(25)14(10-22)29-19%2fh5-8,14,16-19,22,25-28H,9-10H2,1-4H3%2fb6-5+,11-7-%2ft14-,16-,17+,18-,19+,21-%2fm1%2fs1%26token%3de4a6d6fb-ae07-4cf6-bae8-c0e6115bc681
2) gsubfn/curlEscape For a somewhat different output continuing to use gsubfn try:
library(RCurl)
gsubfn("\\?\\K(.*)", curlEscape, u, perl = TRUE)
giving:
http://www.chemspider.com/inchi.asmx/InChIToSMILES?inchi%3DInchI%3D1S%2FC21H30O9%2Fc1%2D11%285%2D6%2D21%2828%2912%282%298%2D13%2823%299%2D20%2821%2C3%294%297%2D15%2824%2930%2D19%2D18%2827%2917%2826%2916%2825%2914%2810%2D22%2929%2D19%2Fh5%2D8%2C14%2C16%2D19%2C22%2C25%2D28H%2C9%2D10H2%2C1%2D4H3%2Fb6%2D5%2B%2C11%2D7%2D%2Ft14%2D%2C16%2D%2C17%2B%2C18%2D%2C19%2B%2C21%2D%2Fm1%2Fs1%26token%3De4a6d6fb%2Dae07%2D4cf6%2Dbae8%2Dc0e6115bc681
ADDED curlEscape approach
I am using R to do some data pre-processing, and here is the problem that I am faced with: I input the data using read.csv(filename,header=TRUE), and then the space in variable names became ".", for example, a variable named Full Code became Full.Code in the generated dataframe. After the processing, I use write.xlsx(filename) to export the results, while the variable names are changed. How to address this problem?
Besides, in the output .xlsx file, the first column become indices(i.e., 1 to N), which is not what I am expecting.
If your set check.names=FALSE in read.csv when you read the data in then the names will not be changed and you will not need to edit them before writing the data back out. This of course means that you would need quote the column names (back quotes in some cases) or refer to the columns by location rather than name while editing.
To get spaces back in the names, do this (right before you export - R does let you have spaces in variable names, but it's a pain):
# A simple regular expression to replace dots with spaces
# This might have unintended consequences, so be sure to check the results
names(yourdata) <- gsub(x = names(yourdata),
pattern = "\\.",
replacement = " ")
To drop the first-column index, just add row.names = FALSE to your write.xlsx(). That's a common argument for functions that write out data in tabular format (write.csv() has it, too).
Here's a function (sorry, I know it could be refactored) that makes nice column names even if there are multiple consecutive dots and trailing dots:
makeColNamesUserFriendly <- function(ds) {
# FIXME: Repetitive.
# Convert any number of consecutive dots to a single space.
names(ds) <- gsub(x = names(ds),
pattern = "(\\.)+",
replacement = " ")
# Drop the trailing spaces.
names(ds) <- gsub(x = names(ds),
pattern = "( )+$",
replacement = "")
ds
}
Example usage:
ds <- makeColNamesUserFriendly(ds)
Just to add to the answers already provided, here is another way of replacing the “.” or any other kind of punctation in column names by using a regex with the stringr package in the way like:
require(“stringr”)
colnames(data) <- str_replace_all(colnames(data), "[:punct:]", " ")
For example try:
data <- data.frame(variable.x = 1:10, variable.y = 21:30, variable.z = "const")
colnames(data) <- str_replace_all(colnames(data), "[:punct:]", " ")
and
colnames(data)
will give you
[1] "variable x" "variable y" "variable z"