I have a string (x, see below) that has many different formats. They are all positions on a genome but have different names. These names were given to me and belong to a list of about 6 million so it's not easy for me to change manually. This is a subset, however there are others like X1 or chr 13 that are part of this list too.:
x <- c("rs62224609.16051249.T.C", "rs376238049.16052962.C.T","rs62224614.16053862.C.T","X22.17028719.G.A", "rs4535153", "X22.17028719.G.A", "kgp3171179", "rs375850426.17029070.GCAGTGGC.G" , "chr22.17030620.G.A")
I'd like all the string to look like this:
y <- c("rs62224609", "rs376238049", "rs62224614", "chr22:17028719", "rs4535153", "chr22:17028719", "kgp3171179", "rs375850426", "chr22:17030620")
I've tried the following, but everything after the first "." is removed... which isn't exactly what I want.
x.test = gsub(pattern = "\\.\\S+$", replacement = "", x = x)
Any help would be greatly appreciated!
If all your data corresponds to the examples you've given:
x = c("rs62224609.16051249.T.C", "rs376238049.16052962.C.T","rs62224614.16053862.C.T","X22.17028719.G.A", "rs4535153", "X22.17028719.G.A", "kgp3171179", "rs375850426.17029070.GCAGTGGC.G" , "chr22.17030620.G.A")
There are two types of ids, the ones with SNP ids (starting with rs or kgp), and the ones giving a chromosomal position (starting with the chromosome name).
You could start off by identifying your SNP ids, with something like:
x1 = gsub("((rs|kgp)\\d+).*","\\1",x)
This returns:
[1] "rs62224609" "rs376238049" "rs62224614" "X22.17028719.G.A" "rs4535153" "X22.17028719.G.A" "kgp3171179" "rs375850426" "chr22.17030620.G.A"
Then format the chromosome positions with (I've assumed that you had chromosomes from 1 to 22, X,Y and M, but this depends on your data):
## We look for [(chr OR X) (1 or 2 digits or X or Y or M) 1 or more punctuation marks (1 or more digits) anything] and
## we transform it into: [chr (the second captured element) : (the third captured element)]
x2 = gsub("(chr|X)(\\d{1,2}|X|Y|M)[[:punct:]]+(\\d+).*","chr\\2:\\3",x1)
This returns:
[1] "rs62224609" "rs376238049" "rs62224614" "chr22:17028719" "rs4535153" "chr22:17028719" "kgp3171179" "rs375850426" "chr22:17030620"
Related
So I have a vector the represents start times (trt.start) and end times (trt.end). Each of the start times represents one treatment (which takes 400 days) and the corresponding position in the vector trt.end represents the end.
trt.start = c(10000,10090,10180,10270, 10360)
trt.end = c(trt.start + 400)
Is there a way to (with out hard coding it), to create a new vector that will represent the duration of each treatment? So it will result in:
c(10000:10400, 10090:10490, 10180:10580, 10270:10670, 10360:10760)
I would like to be able to do this without hard coding because the trt.start vector will change values.
Thank you!
In R, we can use Map to get the sequence of corresponding vectors in a list and then unlist the list to create the single vector
v1 <- unlist(Map(`:`, trt.start, trt.end))
length(v1)
#[1] 2005
If we need as a string
v1 <- sprintf("%d:%d", trt.start, trt.end)
v1
#[1] "10000:10400" "10090:10490" "10180:10580" "10270:10670" "10360:10760"
or with paste
v1 <- paste0(trt.start, ":", trt.end)
v1
#[1] "10000:10400" "10090:10490" "10180:10580" "10270:10670" "10360:10760"
Or a vectorized option is to replicate the 'trt.start' and then add with sequence of values
v2 <- rep(trt.start, each = 401) + seq_len(401) -1
all.equal(v2, v1)
#[1] TRUE
I want to find out the maximum amount comma had appeared in a row in a single column.
For example,
Cars
1 Bugatti (4)","Ferrari (7)","Audi (10)
2 Toyota (6)
3 Tesla (9)","Mercedes(8)
4 Suzuki (11)","Mitsubishi (19)","Ford (7)","BMW (6)
For the table column above, the maximum number a comma had appeared in a row is 3, and it is on row 4. How do I achieve this on a much more larger data (4000+ rows)?
You can use gregexp() to return a vector of the positions of the comma(s) in each string. Then you can apply the length() function to count up the commas:
sapply(gregexpr(",", df$cars), length)
## 2 1 1 3
To answer the exact question asked, just wrap the above line of code in max() to determine the maximum number of times a comma appeared in one of your strings.
The above actually returns a "1" when a "0" is expected. There is probably a more elegant solution, but here's a function that will handle zeros correctly:
count_commas <- function(x) {
y <- sapply(gregexpr(",", x), as.integer) # get position of commas
y <- lapply(y, function(y) if(y[1] == -1) NULL else y) # replace zeros
return( sapply(y, length) ) # return count of commas
}
count_commas(df$cars)
# 2 0 1 3
My idea is to remove the non-comma characters and calculate the number of chars.
I have no clue which class of object you are using for cars. Assuming your input is
cars <- c(' Bugatti (4)","Ferrari (7)","Audi (10)','Toyota (6)','Tesla (9)","Mercedes(8)','Suzuki (11)","Mitsubishi (19)","Ford (7)","BMW (6)')
then you can use nchar(gsub("[^,]","", cars)) to get the number of commas of each row.
I have an R script where
l2 is an array of all permutations of three letter combinations a-z ie the first values is aaa and the last values is zzz
c <- rpois(n, 5) is a vector of poisson distribution results.
I want to create a data frame that containtns a random number of samples from l2 and concatenates them into a path like this: abc -> vye -> tyb for 3.
df_p <- data.frame('len' = c, 's1' = paste(sample(l2,c), collapse = ' -> '))
I've tried to use the above row to create it, but it returns the same path for every row. The length of this path corresponds to the first value in c.
Figured it out ...
unlist(lapply(c, function(x) paste(sample(l2,x), collapse = ' -> ') ))
... does the trick.
I have an excel file of a list of sequences. How would I go about getting the number of times a letter appears before a letter in square brackets? An example of an entry is below.
GTCCTGGTTGTAGCTGAAGCTCTTCCC[A]CTCCTCCCGATCACTGGGACGTCCTATGT
I'd also like to do this for the letter after the square brackets.
Edit: Apologies for the confusion. Take the example below. Id like to count how many times A, C, G, and T appears immediately before and after the letter in square brackets (for which there is only one per line). So to count the occurences of A[A]A, A[A]C, C[A]A, and so on. The file is in excel, and I'm happy to use any method in excel, R or in Linux.
CCCACCCGCCAGGAAGCCGCTATCACTGTCCAAGTTGTCATCGGAACTCC[A]CCAGCCTGTGGACTTGGCCTGGTGCCGCCCATCCCCCTTGCGGTCCTTGC
ACCACTACCCCCTTCCCCACCATCCACCTCAGAAGCAGTCCCAGCCTGCC[A]CCCGCCAGCCCCTGCCCAGCCCTGGCTTTTTGGAAACGGGTCAGGATTGG
TTTGCTTTAAAATACTGCAACCACTCCAGGTAAATCTTCCGCTGCCTATA[A]CCCCGCCAATGAGCCTGCACATCAGGAGAGAAAGGGAAGTAACTCAAGCA
GAAATCTTCTGAAACAGTCTCCAGAAGACTGTCTCCAAATACACAGCAGA[A]CCAGCCAGTCCACAGCACTTTACCTTCTCTATTCTCAGATGGCAATTGAG
GGACTGCCCCAAGGCCCGCAGGGAGGTGGAGCTGCACTGGCGGGCCTCCC[A]GTGCCCGCACATCGTACGGATCGTGGATGTGTACGAGAATCTGTACGCAG
GGCCCAACGCCATCCTGAAACTCACTGACTTTGGCTTTGCCAAGGAAACC[A]CCAGCCACAACTCTTTGACCACTCCTTGTTATACACCGTACTATGTGGGT
TCTGCCTGGTCCGCTGGAGCTGGGCATTGAAGCCCCGCAGCTGCTCAGCC[A]CCTGCCCCGCCATCAAGAAGGCCCCACCGGCCCTGGGAAGGACACCCCTG
TTTGAAGCCCTTATGAACCAAGAAACCTTCGTTCAGGACCTCAAAATCAA[A]CCCCGCCACATGCAGCTCGCAGGCCTGCAGGAGGAAAGACAGGTTAGCAA
CTGCAGCCTACCTGTCCATGTCCCAGGGGGCCGTTGCCAACGCCAACAGC[A]CCCCGCCGCCCTATGAGCGTACCCGCCTCTCCCCACCCCGGGCCAGCTAC
ACTGGCAAACATGTTGAGGACAATGATGGAGGGGATGAGCTTGCATAGGA[A]CCTGCCGTAGGGCCACTGTCCCTGGAGAGCCAAGTGAGCCAGCGAGAAGG
CACCCTCAGAGAAGAAGAAAGGAGCTGAGGAGGAGAAGCCAAAGAGGAGG[A]GGCAGGAGAAGCAGGCAGCCTGCCCCTTCTACAACCACGAGCAGATGGGC
CCAGCCCTGTATGAGGACCCCCCAGATCAGAAAACCTCACCCAGTGGCAA[A]CCTGCCACACTCAAGATCTGCTCTTGGAATGTGGATGGGCTTCGAGCCTG
TTCCTGTGCGCCCCAACAACTCCTTTAGCTGGCCTAAAGTGAAAGGACGG[A]CCTGCCAATGAAAATAGACTTTCAGGGTCTAGCAGAAGGCAAGACCACCA
CTAACACCCGCACGAGCTGCTGGTAGATCTGAATGGCCAAGTCACTCAGC[A]CCTGCCGATACTCAGCCAGGTCAAAATTGGTGAGGCAGTGTTCATTCTGG
AGTTCTGCATCTGGAGCAAATCCTTGGCACTCCCTCATGCTGGCTATCAC[A]CCTGCCACGAATGTGCCATGGCCCAACCCTGCAGTCCATAAAGAAAACAA
CGTGCCCATGCAGCTAGTGCTCTTCCGAGAGGCTATTGAACACAGTGAGC[A]CCTGCCACGCCTATCCCCTTCCCCATCATCTCAGTGATGGGGTATGTCTA
ACAAGGACCTGGCCCTGGGGCAGCCCCTCAGCCCACCTGGTCCCTGCCTT[A]CCCAGCCAGTACTCTCCATCAGCACGGCCGAAGCCCAGCTTGTAGTCATT
You could split the original string into parts. From the start of the string to the first [ and from the first ] to the end of the string.
int count = firstPart.Count(f => f == 'a');
count += secondPart.Count(f => f == 'a');
Option Explicit
Sub test()
Dim seq As String
seq = "GTCCTGGTTGTAGCTGAAGCTCTTCCC[A]CTCCTCCCGATCACTGGGACGTCCTATGT"
Debug.Print CountLetter("A", seq)
End Sub
Function CountLetter(letter As String, ByVal sequence As String) As Long
'--- assumes the letter in the brackets is the same as that being counted
Dim allLetters() As String
allLetters = Split("A,C,G,T", ",")
Dim letterToDelete As Variant
For Each letterToDelete In allLetters
If letterToDelete <> letter Then
sequence = Replace(sequence, letterToDelete, "")
End If
Next letterToDelete
CountLetter = Len(sequence) - 1
End Function
x = "GTCCTGGTTGTAGCTGAAGCTCTTCCC[A]CTCCTCCCGATCACTGGGACGTCCTATGT"
#COUNT 'A'
sapply(unlist(strsplit(x,"\\[[A-z]\\]")), function(a) length(unlist(gregexpr("A", a))))
# GTCCTGGTTGTAGCTGAAGCTCTTCCC CTCCTCCCGATCACTGGGACGTCCTATGT
# 3 4
#COUNT 'G'
sapply(unlist(strsplit(x,"\\[[A-z]\\]")), function(a) length(unlist(gregexpr("G", a))))
# GTCCTGGTTGTAGCTGAAGCTCTTCCC CTCCTCCCGATCACTGGGACGTCCTATGT
# 7 6
New R solution (after clarification by OP)
Let's assume the data have been read from Excel into a data.table called los (list of sequences) which has only one column called sequence. Then, the occurences can be counted as follows:
library(data.table)
los[, .N, by = stringr::str_extract(sequence, "[ACGT]\\[[ACGT]\\][ACGT]")]
# stringr N
#1: C[A]C 8
#2: A[A]C 5
#3: C[A]G 1
#4: G[A]G 1
#5: G[A]C 1
#6: T[A]C 1
str_extract() looks for one of the letters A, C, G, T followed by [ followed by one of the letters A, C, G, T followed by ] followed by one of the letters A, C, G, T in column sequence and extracts the matching substrings. Then, los is grouped by the substrings and the number of occurences is counted (.N).
Data
If the Excel file is stored in CSV format then it can be read using data.table's fread() function like this
los <- fread("your_file_name.csv")
(Perhaps, some parameters to fread() might need to be adjusted for the specific file.)
However, some data already are provided in the question. These can be read as character string using fread() as well:
los <- fread("sequence
CCCACCCGCCAGGAAGCCGCTATCACTGTCCAAGTTGTCATCGGAACTCC[A]CCAGCCTGTGGACTTGGCCTGGTGCCGCCCATCCCCCTTGCGGTCCTTGC
ACCACTACCCCCTTCCCCACCATCCACCTCAGAAGCAGTCCCAGCCTGCC[A]CCCGCCAGCCCCTGCCCAGCCCTGGCTTTTTGGAAACGGGTCAGGATTGG
TTTGCTTTAAAATACTGCAACCACTCCAGGTAAATCTTCCGCTGCCTATA[A]CCCCGCCAATGAGCCTGCACATCAGGAGAGAAAGGGAAGTAACTCAAGCA
GAAATCTTCTGAAACAGTCTCCAGAAGACTGTCTCCAAATACACAGCAGA[A]CCAGCCAGTCCACAGCACTTTACCTTCTCTATTCTCAGATGGCAATTGAG
GGACTGCCCCAAGGCCCGCAGGGAGGTGGAGCTGCACTGGCGGGCCTCCC[A]GTGCCCGCACATCGTACGGATCGTGGATGTGTACGAGAATCTGTACGCAG
GGCCCAACGCCATCCTGAAACTCACTGACTTTGGCTTTGCCAAGGAAACC[A]CCAGCCACAACTCTTTGACCACTCCTTGTTATACACCGTACTATGTGGGT
TCTGCCTGGTCCGCTGGAGCTGGGCATTGAAGCCCCGCAGCTGCTCAGCC[A]CCTGCCCCGCCATCAAGAAGGCCCCACCGGCCCTGGGAAGGACACCCCTG
TTTGAAGCCCTTATGAACCAAGAAACCTTCGTTCAGGACCTCAAAATCAA[A]CCCCGCCACATGCAGCTCGCAGGCCTGCAGGAGGAAAGACAGGTTAGCAA
CTGCAGCCTACCTGTCCATGTCCCAGGGGGCCGTTGCCAACGCCAACAGC[A]CCCCGCCGCCCTATGAGCGTACCCGCCTCTCCCCACCCCGGGCCAGCTAC
ACTGGCAAACATGTTGAGGACAATGATGGAGGGGATGAGCTTGCATAGGA[A]CCTGCCGTAGGGCCACTGTCCCTGGAGAGCCAAGTGAGCCAGCGAGAAGG
CACCCTCAGAGAAGAAGAAAGGAGCTGAGGAGGAGAAGCCAAAGAGGAGG[A]GGCAGGAGAAGCAGGCAGCCTGCCCCTTCTACAACCACGAGCAGATGGGC
CCAGCCCTGTATGAGGACCCCCCAGATCAGAAAACCTCACCCAGTGGCAA[A]CCTGCCACACTCAAGATCTGCTCTTGGAATGTGGATGGGCTTCGAGCCTG
TTCCTGTGCGCCCCAACAACTCCTTTAGCTGGCCTAAAGTGAAAGGACGG[A]CCTGCCAATGAAAATAGACTTTCAGGGTCTAGCAGAAGGCAAGACCACCA
CTAACACCCGCACGAGCTGCTGGTAGATCTGAATGGCCAAGTCACTCAGC[A]CCTGCCGATACTCAGCCAGGTCAAAATTGGTGAGGCAGTGTTCATTCTGG
AGTTCTGCATCTGGAGCAAATCCTTGGCACTCCCTCATGCTGGCTATCAC[A]CCTGCCACGAATGTGCCATGGCCCAACCCTGCAGTCCATAAAGAAAACAA
CGTGCCCATGCAGCTAGTGCTCTTCCGAGAGGCTATTGAACACAGTGAGC[A]CCTGCCACGCCTATCCCCTTCCCCATCATCTCAGTGATGGGGTATGTCTA
ACAAGGACCTGGCCCTGGGGCAGCCCCTCAGCCCACCTGGTCCCTGCCTT[A]CCCAGCCAGTACTCTCCATCAGCACGGCCGAAGCCCAGCTTGTAGTCATT")
Old solution (before clarification by OP) - left here for reference
This is a solution in base R with help of the stringr package which will work with a "list" of sequences (a data.frame), any single letter enclosed in square brackets, and arbitrary lengths of the sequences. It assumes that the data already have been read from file into a data.frame which is named los here.
# create data: data frame with two sequences
los <- data.frame(
sequence = c("GTCCTGGTTGTAGCTGAAGCTCTTCCC[A]CTCCTCCCGATCACTGGGACGTCCTATGT",
"GTCCTGGTTGTAGCTGAAGCTCTTCCCACT[C]CTCCCGATCACTGGGACGTCCTATGT"))
# split sequences in three parts
mat <- stringr::str_split_fixed(los$sequence, "[\\[\\]]", n = 3)
los$letter <- mat[, 2]
los$n_before <- stringr::str_count(mat[, 1], mat[, 2])
los$n_after <- stringr::str_count(mat[, 3], mat[, 2])
print(los)
# sequence letter n_before n_after
#1 GTCCTGGTTGTAGCTGAAGCTCTTCCC[A]CTCCTCCCGATCACTGGGACGTCCTATGT A 3 4
#2 GTCCTGGTTGTAGCTGAAGCTCTTCCCACT[C]CTCCCGATCACTGGGACGTCCTATGT C 9 9
Note this code works best if there is exactly one pair of square brackets in each sequence. Any additional brackets will be ignored.
It will also work if there is more than just one letter enclosed in brackets, e.g., [GT].
I'm confessing that I'm addicted to Hadley Wickham's stringr package because I have difficulties to remember the inconsistently named base R functions for string maninpulation like strsplit, grepl, sub, match, gregexpr, etc. To understand what I mean please have a look at the Usage and See Also sections of ?grep and compare to stringr.
I would think that R packages for bioinformatics, such as seqinr or Biostrings, would be a good starting point. However, here's a "roll your own" solution.
First step: get your data from Excel into R. I will assume that file mydata.xlsx contains one sheet with a column of sequence and no header. You need to adapt this for your file and sheet format.
library(readxl)
sequences <- read_excel("mydata.xlsx", col_names = FALSE)
colnames(sequences) <- "sequence"
Now you need a function to extract the base in square brackets and the bases at -1 and +1. This function uses the stringr package to extract bases using regular expressions.
get_bases <- function(seq) {
require(stringr)
require(magrittr)
subseqs <- str_match(seq, "^([ACGT]+)\\[([ACGT])\\]([ACGT]+)$")
bases <- list(
before = subseqs[, 2] %>% str_sub(-1, -1),
base = subseqs[, 3],
after = subseqs[, 4] %>% str_sub(1, 1)
)
return(bases)
}
Now you can pass the column of sequences to the function to generate a list of lists, which can be converted to a data frame.
library(purrr)
sequences_df <- lapply(sequences, get_bases) %>%
map_df(as.data.frame, stringsAsFactors = FALSE)
head(sequences_df, 3)
before base after
1 C A C
2 C A C
3 A A C
The last step is to use functions from dplyr and tidyr to count up the bases.
library(tidyr)
sequences_df %>%
gather(position, letter, -base) %>%
group_by(base, position, letter) %>%
tally() %>%
spread(position, n) %>%
select(base, letter, before, after)
Result using your 17 example sequences. I would use better names than I did if I were you: base = the base in square brackets, letter = the base being counted, before = count at -1, after = count at +1.
base letter before after
* <chr> <chr> <int> <int>
1 A A 5 NA
2 A C 9 15
3 A G 2 2
4 A T 1 NA
I have the following code for finding out a pattern (consecutively repeated substring) in a string, say 0110110110000. The output patterns are 011 and 110, since they are both repeated within the string. What changes can be done to the following code?
I'd like to identify substrings that start from any position in a given string, and which repeat for at least a threshold number of times. In the above mentioned string, the threshold is three (th = 3). The repeated string should be the maximal repeated string. In the above string, 110 and 011 both satisfy these conditions.
Here's my attempt at doing this:
reps <- function(s, n) paste(rep(s, n), collapse = "") # repeat s n times
find.string <- function(string, th = 3, len = floor(nchar(string)/th)) {
for(k in len:1) {
pat <- paste0("(.{", k, "})", reps("\\1", th-1))
r <- regexpr(pat, string, perl = TRUE)
if (attr(r, "capture.length") > 0) break
}
if (r > 0) substring(string, r, r + attr(r, "capture.length")-1) else ""
}
You can do this with regex:
s <- '0110110110000'
thr <- 3
m <- gregexpr(sprintf('(?=(.+)(?:\\1){%s,})', thr-1), s, perl=TRUE)
unique(mapply(function(x, y) substr(s, x, x+y-1),
attr(m[[1]], 'capture.start'),
attr(m[[1]], 'capture.length')))
# [1] "011" "110" "0"
The pattern in the gregexpr uses a positive lookahead to prevent characters from being consumed by the match (and so allowing overlapping matches, such as with the 011 and 110). We use a repeated (at least thr - 1 times) backreference to the captured group to look for repeated substrings.
Then we can extract the matched substrings by taking start positions and lengths from the attributes of the result of gregexpr, i.e. the object m.
You didn't specify a minimum string length, so this returns 0 as one of the repeated substrings. If you have a minimum and/or maximum substring length in mind, you can modify the first subexpression of the regex. For example, the following would match only substrings with at least 2 characters.
sprintf('(?=(.{2,})(?:\\1){%s,})', thr-1)