How can I in R extract the number that always comes after the string -{any single letter}, e.g. from the vector:
c("JFSDLKJ-H465", "FJSLKJHSD-Y5FSDLKJ", "DFSJLKJAAA-Z3216FJJ")
one should get:
(465, 5, 3216).
The -{any single letter} pattern occurs only once.
You could use gsub, e.g.:
x <- c("JFSDLKJ-H465", "FJSLKJHSD-Y5FSDLKJ", "DFSJLKJAAA-Z3216FJJ")
as.numeric(gsub("^.*-[A-Z]+([0-9]+).*$", "\\1", x))
# [1] 465 5 3216
library(stringr)
v <- c("JFSDLKJ-H465", "FJSLKJHSD-Y5FSDLKJ", "DFSJLKJAAA-Z3216FJJ")
as.numeric(sapply(str_match_all(v, "\\-[a-zA-Z]([0-9]+)"),"[")[2,])
## [1] 465 5 3216
> x <- c("JFSDLKJ-H465", "FJSLKJHSD-Y5FSDLKJ", "DFSJLKJAAA-Z3216FJJ")
> as.numeric(gsub("[A-Z]|-", "", x))
## [1] 465 5 3216
Related
I have a set of strings x for example:
[1] "0000000000000000000000000000000000000Y" "9000000000D00000000000000000000Y"
[3] "0000000000000D00000000000000000000X" "000000000000000000D00000000000000000000Y"
[5] "000000000000000000D00000000000000000000Y" "000000000000000000D00000000000000000000Y"
[6]"000000000000000000000000D0000000011011D1X"
I want to extract the last position of a particular character like 1. I am running this code:
ifelse(grepl("1",x),rev(gregexpr("1",x)[[1]])[1],50)
But this is returning -1 for all elements. How do I correct this?
We can use stri_locate_last from stringi. If there are no matches, it will return NA.
library(stringi)
r1 <- stri_locate_last(v1, fixed=1)[,1]
r1
#[1] NA NA NA NA NA NA 40
nchar(v1)
#[1] 38 32 35 40 40 40 41
If we need to replace the NA values with number of characters
ifelse(is.na(r1), nchar(v1), r1)
data
v1 <- c("0000000000000000000000000000000000000Y",
"9000000000D00000000000000000000Y",
"0000000000000D00000000000000000000X",
"000000000000000000D00000000000000000000Y",
"000000000000000000D00000000000000000000Y",
"000000000000000000D00000000000000000000Y",
"000000000000000000000000D0000000011011D1X")
In base R, the following returns the position of the last matched "1".
# Make some toy data
toydata <- c("001", "007", "00101111Y", "000AAAYY")
# Find last postion
last_pos <- sapply(gregexpr("1", toydata), function(m) m[length(m)])
print(last_pos)
#[1] 3 -1 8 -1
It returns -1 whenever the pattern is not matched.
I am trying to find an elegant way to find matches between the two following character columns in a data frame. The complicated part is that either string can contain a comma-separated list, and if a member of one list is a match for any member of the other list, then that whole entry would be considered a match. I'm not sure how well I've explained this, so here's sample data and output:
Alt1:
AT
A
G
CGTCC,AT
CGC
Alt2:
AA
A
GG
AT,GGT
CG
Expected Match per row:
Row 1 = none
Row 2 = A
Row 3 = none
Row 4 = AT
Row 5 = none
Non-working solutions:
First attempt: merge entire data frames by desired columns, then match up the alt columns shown above:
match1 = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
matches = unique(match1[unlist(sapply(match1$Alt1 grep, match1$Alt2, fixed=TRUE)),])
Second method, using findoverlaps feature from VariantAnnoatation/Granges:
findoverlaps(ranges(vcf1), ranges(vcf2))
Any suggestions would be greatly appreciated! Thank you!
Solution
Thanks to #Marat Talipov's answer below, the following solution works to compare two comma-separated strings:
> ##read in edited kaviar vcf and human ref
> ref <- readVcfAsVRanges("ref.vcf.gz", humie_ref)
Warning message:
In .vcf_usertag(map, tag, ...) :
ScanVcfParam ‘geno’ fields not present: ‘AD’
> ##rename chromosomes to match with vcf files
> ref <- renameSeqlevels(ref, c("1"="chr1"))
> ##################################
> ## Gather VCF files to process ##
> ##################################
> ##data frame *.vcf.gz files in directory path
> vcf_path <- data.frame(path=list.files(vcf_dir, pattern="*.vcf.gz$", full=TRUE))
> ##read in everything but sample data for speediness
> vcf_param = ScanVcfParam(samples=NA)
> vcf <- readVcfAsVRanges("test.vcf.gz", humie_ref, param=vcf_param)
> #################
> ## Match SNP's ##
> #################
> ##create data frames of info to match on
> vcf.df = data.frame(chr =as.character(seqnames(vcf)), start = start(vcf), end = end(vcf), ref = as.character(ref(vcf)),
+ alt=alt(vcf), stringsAsFactors=FALSE)
> ref.df = data.frame(chr =as.character(seqnames(ref)), start = start(ref), end = end(ref),
+ ref = as.character(ref(ref)), alt=alt(ref), stringsAsFactors=FALSE)
>
> ##merge based on all positional fields except vcf
> col_match = data.frame(merge(vcf.df, ref.df, by=c("chr", "start", "end", "ref")))
> library(stringi)
> ##split each alt column by comma and bind together
> M1 <- stri_list2matrix(sapply(col_match$alt.x,strsplit,','))
> M2 <- stri_list2matrix(sapply(col_match$alt.y,strsplit,','))
> M <- rbind(M1,M2)
> ##compare results
> result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
> ##add results column to col_match df for checking/subsetting
> col_match$match = result
> head(col_match)
chr start end ref alt.x alt.y match
1 chr1 39998059 39998059 A G G G
2 chr1 39998059 39998059 A G G G
3 chr1 39998084 39998084 C A A A
4 chr1 39998084 39998084 C A A A
5 chr1 39998085 39998085 G A A A
6 chr1 39998085 39998085 G A A A
In the case that input lists are of equal length and you'd like to compare list elements in the pairwise manner, you could use this solution:
library(stringi)
M1 <- stri_list2matrix(sapply(Alt1,strsplit,','))
M2 <- stri_list2matrix(sapply(Alt2,strsplit,','))
M <- rbind(M1,M2)
result <- apply(M,2,function(z) unique(na.omit(z[duplicated(z)])))
Sample input:
Alt1 <- list('AT','A','G','CGTCC,AT','CGC','GG,CC')
Alt2 <- list('AA','A','GG','AT,GGT','CG','GG,CC')
Output:
# [[1]]
# character(0)
#
# [[2]]
# [1] "A"
#
# [[3]]
# character(0)
#
# [[4]]
# [1] "AT"
#
# [[5]]
# character(0)
#
# [[6]]
# [1] "GG" "CC"
Sticking with the stringi package, you could do something like this, using the Alt1 and Alt2 data from Marat's answer.
library(stringi)
f <- function(x, y) {
ssf <- stri_split_fixed(c(x, y), ",", simplify = TRUE)
if(any(sd <- stri_duplicated(ssf))) ssf[sd] else NA_character_
}
Map(f, Alt1, Alt2)
# [[1]]
# [1] NA
#
# [[2]]
# [1] "A"
#
# [[3]]
# [1] NA
#
# [[4]]
# [1] "AT"
#
# [[5]]
# [1] NA
#
# [[6]]
# [1] "GG" "CC"
Or in base R, we can use scan() to separate the strings with commas.
g <- function(x, y, sep = ",") {
s <- scan(text = c(x, y), what = "", sep = sep, quiet = TRUE)
s[duplicated(s)]
}
Map(g, Alt1, Alt2)
you could do something like this:
Alt1 <- list('AT','A','G',c('CGTCC','AT'),'CGC')
Alt2 <- list('AA','A','GG',c('AT','GGT'),'CG')
# make sure you change the lists within in the lists into vectors
matchlist <- list()
for (i in 1:length(Alt1)){
matchlist[[i]] <- ifelse(Alt1[[i]] %in% Alt2[[i]],
paste("Row",i,"=",c(Alt1[[i]],Alt2[[i]])[duplicated(c(Alt1[[i]],Alt2[[i]]))],sep=" "),
paste("Row",i,"= none",sep=" "))
}
print(matchlist)
Shouldn't this code work for repeating number detection in R?
> grep(pattern = "\\d{2}", x = 1223)
[1] 1
> grep(pattern = "\\d{3}", x = 1223)
[1] 1
If we have 988 we should get true and if 123 we should get false.
Sounds like it isn't.
> grep(pattern = "\\d{2}", x = "1223")
[1] 1
> grep(pattern = "\\d{2}", x = "13")
[1] 1
You need to use backreferences:
> grep(pattern = "(\\d)\\1", x = "1224")
[1] 1
> grep(pattern = "(\\d)\\1{1,}", x = "1224")
[1] 1
> grep(pattern = "(\\d)\\1", x = "1234")
integer(0)
EDIT: Seems like you need to figure how it works: (\\d) creates a capture group for the \\d, which can be referred to using a backreference \\1. For example, if you have numbers like x2y and you want to find those where x is the same as y, then:
> grep(pattern = "(\\d)2\\1", x = "121")
[1] 1
> grep(pattern = "(\\d)2\\1", x = "124")
integer(0)
I'd strongly recommend that you read a basic tutorial on regular expressions.
I know the question explicitly says "using regex" in the title, but here is a non-regex method that could work, depending on what you want to do.
strings <- c("1223","1233","1234","113")
# detect consecutive repeat digits, or characters
(strings.rle <- lapply(strings, function(x)rle(unlist(strsplit(x,"")))))
[[1]]
Run Length Encoding
lengths: int [1:3] 1 2 1
values : chr [1:3] "1" "2" "3"
[[2]]
Run Length Encoding
lengths: int [1:3] 1 1 2
values : chr [1:3] "1" "2" "3"
[[3]]
Run Length Encoding
lengths: int [1:4] 1 1 1 1
values : chr [1:4] "1" "2" "3" "4"
[[4]]
Run Length Encoding
lengths: int [1:2] 2 1
values : chr [1:2] "1" "3"
Now you can work with strings.rle to do what you want
# which entries have consecutive repeat digits, or characters
strings[sapply(strings.rle, function(x) any(x$lengths > 1))]
[1] "1223" "1233" "113"
or
# which digits or characters are consecutively repeated?
lapply(strings.rle, function(x) x$values[which(x$lengths > 1)])
[[1]]
[1] "2"
[[2]]
[1] "3"
[[3]]
character(0)
[[4]]
[1] "1"
I would like to find the location of a character in a string.
Say: string = "the2quickbrownfoxeswere2tired"
I would like the function to return 4 and 24 -- the character location of the 2s in string.
You can use gregexpr
gregexpr(pattern ='2',"the2quickbrownfoxeswere2tired")
[[1]]
[1] 4 24
attr(,"match.length")
[1] 1 1
attr(,"useBytes")
[1] TRUE
or perhaps str_locate_all from package stringr which is a wrapper for gregexpr stringi::stri_locate_all (as of stringr version 1.0)
library(stringr)
str_locate_all(pattern ='2', "the2quickbrownfoxeswere2tired")
[[1]]
start end
[1,] 4 4
[2,] 24 24
note that you could simply use stringi
library(stringi)
stri_locate_all(pattern = '2', "the2quickbrownfoxeswere2tired", fixed = TRUE)
Another option in base R would be something like
lapply(strsplit(x, ''), function(x) which(x == '2'))
should work (given a character vector x)
Here's another straightforward alternative.
> which(strsplit(string, "")[[1]]=="2")
[1] 4 24
You can make the output just 4 and 24 using unlist:
unlist(gregexpr(pattern ='2',"the2quickbrownfoxeswere2tired"))
[1] 4 24
find the position of the nth occurrence of str2 in str1(same order of parameters as Oracle SQL INSTR), returns 0 if not found
instr <- function(str1,str2,startpos=1,n=1){
aa=unlist(strsplit(substring(str1,startpos),str2))
if(length(aa) < n+1 ) return(0);
return(sum(nchar(aa[1:n])) + startpos+(n-1)*nchar(str2) )
}
instr('xxabcdefabdddfabx','ab')
[1] 3
instr('xxabcdefabdddfabx','ab',1,3)
[1] 15
instr('xxabcdefabdddfabx','xx',2,1)
[1] 0
To only find the first locations, use lapply() with min():
my_string <- c("test1", "test1test1", "test1test1test1")
unlist(lapply(gregexpr(pattern = '1', my_string), min))
#> [1] 5 5 5
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(min) %>%
unlist()
#> [1] 5 5 5
To only find the last locations, use lapply() with max():
unlist(lapply(gregexpr(pattern = '1', my_string), max))
#> [1] 5 10 15
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(max) %>%
unlist()
#> [1] 5 10 15
You could use grep as well:
grep('2', strsplit(string, '')[[1]])
#4 24
Given a list whose length <= N, what is the best / most efficient way to fill it up with trailing NULLs up to length (so that it has length N).
This is something which is a one-liner in any decent language, but I don't have a clue how to do it (efficiently) in a few lines in R so that it works for every corner case (zero length list etc.).
Let's keep it really simple:
tst<-1:10 #whatever, to get a vector of length 10
tst<-tst[1:15]
Try this :
> l = list("a",1:3)
> N = 5
> l[N+1]=NULL
> l
[[1]]
[1] "a"
[[2]]
[1] 1 2 3
[[3]]
NULL
[[4]]
NULL
[[5]]
NULL
>
How about this ?
> l = list("a",1:3)
> length(l)=5
> l
[[1]]
[1] "a"
[[2]]
[1] 1 2 3
[[3]]
NULL
[[4]]
NULL
[[5]]
NULL
Directly editing the list's length appears to be the fastest as far as I can tell:
tmp <- vector("list",5000)
sol1 <- function(x){
x <- x[1:10000]
}
sol2 <- function(x){
x[10001] <- NULL
}
sol3 <- function(x){
length(x) <- 10000
}
library(rbenchmark)
benchmark(sol1(tmp),sol2(tmp),sol3(tmp),replications = 5000)
test replications elapsed relative user.self sys.self user.child sys.child
1 sol1(tmp) 5000 2.045 1.394952 1.327 0.727 0 0
2 sol2(tmp) 5000 2.849 1.943383 1.804 1.075 0 0
3 sol3(tmp) 5000 1.466 1.000000 0.937 0.548 0 0
But the differences aren't huge, unless you're doing this a lot on very long lists, I suppose.
I'm sure there are shorter ways, but I would be inclined to do:
l <- as.list(1:10)
N <- 15
l <- c(l, as.list(rep(NA, N - length(l) )))
Hi: I'm not sure if you were talking about an actual list but, if you were, below will work. It works because, once you access the element of a vector ( which is a list is ) that is not there, R expands the vector to that length.
length <- 10
temp <- list("a","b")
print(temp)
temp[length] <- NULL
print(temp)