I have a column of strings that I would like to remove everything after the last '.' like so:
ENST00000338167.9
ABCDE.42927.6
ENST00000265393.10
ABCDE.43577.3
ENST00000370826.3
I would like to replace remove the '.' and everything after for the 'ENST' entries only
eg:
ENST00000338167
ABCDE.42927.6
ENST00000265393
ABCDE.43577.3
ENST00000370826
I can do
function(x) sub("\\.[^.]*$", "", x)
if I try
function(x) sub("ENST*\\.[^.]*$", "", x)
this isn't quite working and I don't fully understand the regex commands.
We can use combination of ifelse, grepl and sub. We first check if the string consists of "ENST" string and if it does then remove everything after "." using sub.
ifelse(grepl("^ENST", x), sub("\\..*", "", x), x)
#[1] "ENST00000338167" "ABCDE.42927.6" "ENST00000265393" "ABCDE.43577.3"
#[5] "ENST00000370826"
data
x <- c("ENST00000338167.9","ABCDE.42927.6","ENST00000265393.10",
"ABCDE.43577.3","ENST00000370826.3")
We can use a capture group inside a single gsub call
gsub("(^ENST\\d+)\\.\\d+", "\\1", df[, 1])
#[1] "ENST00000338167" "ABCDE.42927.6" "ENST00000265393" "ABCDE.43577.3"
#[5] "ENST00000370826"
Sample data
df <- read.table(text =
"ENST00000338167.9
ABCDE.42927.6
ENST00000265393.10
ABCDE.43577.3
ENST00000370826.3", header = F)
We can use data.table to specify the logical condition in i while updating the j
library(data.table)
setDT(df)[grepl("^ENST", Col1), Col1 := sub("\\.[^.]+$", "", Col1)]
df
# Col1
#1: ENST00000338167
#2: ABCDE.42927.6
#3: ENST00000265393
#4: ABCDE.43577.3
#5: ENST00000370826
data
df <- structure(list(Col1 = c("ENST00000338167.9", "ABCDE.42927.6",
"ENST00000265393.10", "ABCDE.43577.3", "ENST00000370826.3")), row.names = c(NA,
-5L), class = "data.frame")
We can use startsWith and sub combination:
Data:
df=read.table(text="ENST00000338167.9
ABCDE.42927.6
ENST00000265393.10
ABCDE.43577.3
ENST00000370826.3",header=F)
# if string starts with ENST then remove everything after . (dot) in the
# string else print the string as it is.
ifelse(startsWith(as.character(df[,1]),"ENST"),sub("*\\..*", "", df$V1),
as.character(df[,1]))
Output:
[1] "ENST00000338167" "ABCDE.42927.6" "ENST00000265393" "ABCDE.43577.3" "ENST00000370826"
Related
How can I show which special character was a match in each row of the single column dataframe?
Sample dataframe:
a <- data.frame(name=c("foo","bar'","ip_sum","four","%23","2_planet!","#abc!!"))
determining if the string has a special character:
a$name_cleansed <- gsub("([-./&,])|[[:punct:]]","\\1",a$name) #\\1 puts back the exception we define (dash and slash)
a <- a %>% mutate(has_special_char=if_else(name==name_cleansed,FALSE,TRUE))
You can use str_extract if we want only first special character.
library(stringr)
str_extract(a$name,'[[:punct:]]')
#[1] NA "'" "_" NA "%" "_" "#"
If we need all of the special characters we can use str_extract_all.
sapply(str_extract_all(a$name,'[[:punct:]]'), function(x) toString(unique(x)))
#[1] "" "'" "_" "" "%" "_, !" "#, !"
To exclude certain symbols, we can use
exclude_symbol <- c('-', '.', '/', '&', ',')
sapply(str_extract_all(a$name,'[[:punct:]]'), function(x)
toString(setdiff(unique(x), exclude_symbol)))
We can use grepl here for a base R option:
a$has_special_char <- grepl("(?![-./&,])[[:punct:]]", a$name, perl=TRUE)
a$special_char <- ifelse(a$has_special_char, sub("^.*([[:punct:]]).*$", "\\1", a$name), NA)
a
name has_special_char special_char
1 foo FALSE <NA>
2 bar' TRUE '
3 ip_sum TRUE _
4 four FALSE <NA>
5 %23 TRUE %
6 2_planet! TRUE !
7 #abc!! TRUE !
Data:
a <- data.frame(name=c("foo","bar'","ip_sum","four","%23","2_planet!","#abc!!"))
The above logic returns, arbitrarily, the first symbol character, if present, in each name, otherwise returning NA. It reuses the has_special_char column to determine if a symbol occurs in the name already.
Edit:
If you want a column which shows all special characters, then use:
a$all_special_char <- ifelse(a$has_special_char, gsub("[^[:punct:]]+", "", a$name), NA)
Base R regex solution using (caret) not "^" operator:
gsub("(^[-./&,])|[^[:punct:]]", "", a$name)
Also if you want a data.frame returned:
within(a, {
special_char <- gsub("(^[-./&,])|[^[:punct:]]", "", name);
has_special_char <- special_char != ""})
If you only want unique special characters per name as in #Ronak Shah's answer:
within(a, {
special_char <- sapply(gsub("(^[-./&,])|[^[:punct:]]", "", a$name),
function(x){toString(unique(unlist(strsplit(x, ""))))});
has_special_char <- special_char != ""
}
I have the following strings:
strings <- c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG")
I want to cut off the string, as soon as the number of occurances of A, G and N reach a certain value, say 3. In that case, the result should be:
some_function(strings)
c("ABBSDGN", "AABSDG", "AGN", "GGG")
I tried to use the stringi, stringr and regex expressions but I can't figure it out.
You can accomplish your task with a simple call to str_extract from the stringr package:
library(stringr)
strings <- c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG")
str_extract(strings, '([^AGN]*[AGN]){3}')
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"
The [^AGN]*[AGN] portion of the regex pattern says to look for zero or more consecutive characters that are not A, G, or N, followed by one instance of A, G, or N. The additional wrapping with parenthesis and braces, like this ([^AGN]*[AGN]){3}, means look for that pattern three times consecutively. You can change the number of occurrences of A, G, N, that you are looking for by changing the integer in the curly braces:
str_extract(strings, '([^AGN]*[AGN]){4}')
# [1] "ABBSDGNHN" NA "AGNA" "GGGDSRTYHG"
There are a couple ways to accomplish your task using base R functions. One is to use regexpr followed by regmatches:
m <- regexpr('([^AGN]*[AGN]){3}', strings)
regmatches(strings, m)
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"
Alternatively, you can use sub:
sub('(([^AGN]*[AGN]){3}).*', '\\1', strings)
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"
Here is a base R option using strsplit
sapply(strsplit(strings, ""), function(x)
paste(x[1:which.max(cumsum(x %in% c("A", "G", "N")) == 3)], collapse = ""))
#[1] "ABBSDGN" "AABSDG" "AGN" "GGG"
Or in the tidyverse
library(tidyverse)
map_chr(str_split(strings, ""),
~str_c(.x[1:which.max(cumsum(.x %in% c("A", "G", "N")) == 3)], collapse = ""))
Identify positions of pattern using gregexpr then extract n-th position (3) and substring everything from 1 to this n-th position using subset.
nChars <- 3
pattern <- "A|G|N"
# Using sapply to iterate over strings vector
sapply(strings, function(x) substr(x, 1, gregexpr(pattern, x)[[1]][nChars]))
PS:
If there's a string that doesn't have 3 matches it will generate NA, so you just need to use na.omit on the final result.
This is just a version without strsplit to Maurits Evers neat solution.
sapply(strings,
function(x) {
raw <- rawToChar(charToRaw(x), multiple = TRUE)
idx <- which.max(cumsum(raw %in% c("A", "G", "N")) == 3)
paste(raw[1:idx], collapse = "")
})
## ABBSDGNHNGA AABSDGDRY AGNAFG GGGDSRTYHG
## "ABBSDGN" "AABSDG" "AGN" "GGG"
Or, slightly different, without strsplit and paste:
test <- charToRaw("AGN")
sapply(strings,
function(x) {
raw <- charToRaw(x)
idx <- which.max(cumsum(raw %in% test) == 3)
rawToChar(raw[1:idx])
})
Interesting problem. I created a function (see below) that solves your problem. It's assumed that there are just letters and no special characters in any of your strings.
reduce_strings = function(str, chars, cnt){
# Replacing chars in str with "!"
chars = paste0(chars, collapse = "")
replacement = paste0(rep("!", nchar(chars)), collapse = "")
str_alias = chartr(chars, replacement, str)
# Obtain indices with ! for each string
idx = stringr::str_locate_all(pattern = '!', str_alias)
# Reduce each string in str
reduce = function(i) substr(str[i], start = 1, stop = idx[[i]][cnt, 1])
result = vapply(seq_along(str), reduce, "character")
return(result)
}
# Example call
str = c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG")
chars = c("A", "G", "N") # Characters that are counted
cnt = 3 # Count of the characters, at which the strings are cut off
reduce_strings(str, chars, cnt) # "ABBSDGN" "AABSDG" "AGN" "GGG"
I need to replace the 6,7,8th position to "_". In substring, I mentioned the start and stop position. It didn't work.
> a=c("UHI786KJRH2V", "TYR324FHASJKDG","DHA927NFSYFN34")
> substring(a, 6,8) <- "_"
> a
[1] "UHI78_KJRH2V" "TYR32_FHASJKDG" "DHA92_NFSYFN34"
I need UHI78_RH2V TYR32_ASJKDG DHA92_SYFN34
Using sub, we can match on the pattern (?<=^.{5}).{3}, and then replace it by a single underscore:
a <- c("UHI786KJRH2V", "TYR324FHASJKDG","DHA927NFSYFN34")
out <- sub("(?<=^.{5}).{3}", "_", a, perl=TRUE)
out
[1] "UHI78_RH2V" "TYR32_ASJKDG" "DHA92_SYFN34"
Demo
We could also try doing substring operations here, but we would have to do some splicing:
out <- paste0(substr(a, 1, 5), "_", substr(a, 9, nchar(a)))
1) str_sub<- The str_sub<- replacement function in the stringr package can do that.
library(stringr)
str_sub(a, 6, 8) <- "_"
a
## [1] "UHI78_RH2V" "TYR32_ASJKDG" "DHA92_SYFN34"
2 Base R With only base R you could do this. It replaces the entire string with the match to the first capture group, an underscore and the match to the second capture group.
sub("(.....)...(.*)", "\\1_\\2", a)
## [1] "UHI78_RH2V" "TYR32_ASJKDG" "DHA92_SYFN34"
That regex could also be written as "(.{5}).{3}(.*)" .
3) separate/unite If a is a column in a data frame then we could use dplyr and tidyr to do this:
library(dplyr)
library(tidyr)
DF <- data.frame(a)
DF %>%
separate(a, into = c("pre", "junk", "post"), sep = c(5, 8)) %>%
select(-junk) %>%
unite(a)
giving:
a
1 UHI78_RH2V
2 TYR32_ASJKDG
3 DHA92_SYFN34
From the documentation:
If the portion to be replaced is longer than the replacement string, then only the portion the length of the string is replaced.
So we could do something like this:
substring(a, 6,8) <- "_##"
sub("#+", "", a)
[1] "UHI78_RH2V" "TYR32_ASJKDG" "DHA92_SYFN34"
I would like to create new colnames for my dataframe MirAligner consisting of the part before the first _ in the original colnames. This is what I tried:
unlist(strsplit(as.character(colnames(MirAligner)),'_',fixed=TRUE))
Column names
head(colnames(MirAligner))
[1] "na-008_S52_L003_R1_001.mir.fa.gz" "na-014_S99_L005_R1_001.mir.fa.gz" "na015_S114_L005_R1_001.mir.fa.gz" [4] "na-015_S50_L003_R1_001.mir.fa.gz" "na-018_S147_L007_R1_001.mir.fa.gz" "na020_S162_L007_R1_001.mir.fa.gz"
Expected output:
na-008 na-014 na015
We can use sub
sub('_.*', '', str1)
#[1] "na-014" "na015" "na-015" "na-018" "na020"
data
str1 <- c("na-014_S99_L005_R1_001.mir.fa.gz",
"na015_S114_L005_R1_001.mir.fa.gz",
"na-015_S50_L003_R1_001.mir.fa.gz",
"na-018_S147_L007_R1_001.mir.fa.gz",
"na020_S162_L007_R1_001.mir.fa.gz")
gsub("^(.*?)_.*", "\\1", try5)
#[1] "na-008" "na-014" "na015"
Using strsplit within sapply:
#myColNames <- colnames(MirAligner)
myColNames <- c("na-008_S52_L003_R1_001.mir.fa.gz", "na-014_S99_L005_R1_001.mir.fa.gz")
sapply(strsplit(myColNames, "_", fixed = TRUE), "[[", 1)
#output
# [1] "na-008" "na-014"
Or using read.table:
read.table(text = myColNames, sep = "_", stringsAsFactors = FALSE)[, "V1"]
This string is a ticker for a bond: OAT 3 25/32 7/17/17. I want to extract the coupon rate which is 3 25/32 and is read as 3 + 25/32 or 3.78125. Now I've been trying to delete the date and the name OAT with gsub, however I've encountered some problems.
This is the code to delete the date:
tkr.bond <- 'OAT 3 25/32 7/17/17'
tkr.ptrn <- '[0-9][[:punct:]][0-9][[:punct:]][0-9]'
gsub(tkr.ptrn, "", tkr.bond)
However it gets me the same string. When I use [0-9][[:punct:]][0-9] in the pattern I manage to delete part of the date, however it also deletes the fraction part of the coupon rate for the bond.
The tricky thing is to find a solution that doesn't involve the pattern of the coupon because the tickers have this form: Name Coupon Date, so, using a specific pattern for the coupon may limit the scope of the solution. For example, if the ticker is this way OAT 0 7/17/17, the coupon is zero.
Just replace first and last word with an empty string.
> tkr.bond <- 'OAT 3 25/32 7/17/17'
> gsub("^\\S+\\s*|\\s*\\S+$", "", tkr.bond)
[1] "3 25/32"
OR
Use gsubfn function in-order to use a function in the replacement part.
> gsubfn("^\\S+\\s+(\\d+)\\s+(\\d+)/(\\d+).*", ~ as.numeric(x) + as.numeric(y)/as.numeric(z), tkr.bond)
[1] "3.78125"
Update:
> tkr.bond1 <- c(tkr.bond, 'OAT 0 7/17/17')
> m <- gsub("^\\S+\\s*|\\s*\\S+$", "", tkr.bond1)
> gsubfn(".+", ~ eval(parse(text=x)), gsub("\\s+", "+", m))
[1] "3.78125" "0"
Try
eval(parse(text=sub('[A-Z]+ ([0-9]+ )([0-9/]+) .*', '\\1 + \\2', tkr.bond)))
#[1] 3.78125
Or you may need
sub('^[A-Z]+ ([^A-Z]+) [^ ]+$', '\\1', tkr.bond)
#[1] "3 25/32"
Update
tkr.bond1 <- c(tkr.bond, 'OAT 0 7/17/17')
v1 <- sub('^[A-Z]+ ([^A-Z]+) [^ ]+$', '\\1', tkr.bond1)
unname(sapply(sub(' ', '+', v1), function(x) eval(parse(text=x))))
#[1] 3.78125 0.00000
Or
vapply(strsplit(tkr.bond1, ' '), function(x)
eval(parse(text= paste(x[-c(1, length(x))], collapse="+"))), 0)
#[1] 3.78125 0.00000
Or without the eval(parse
vapply(strsplit(gsub('^[^ ]+ | [^ ]+$', '', tkr.bond1), '[ /]'), function(x) {
x1 <- as.numeric(x)
sum(x1[1], x1[2]/x1[3], na.rm=TRUE)}, 0)
#[1] 3.78125 0.00000
Similar to akrun's answer, using sub with a replacement. How it works: you put your "desired" pattern inside parentheses and leave the rest out (while still putting regex characters to match what's there and that you don't wish to keep). Then when you say replacement = "\\1" you indicate that the whole string must be substituted by only what's inside the parentheses.
sub(pattern = ".*\\s(\\d\\s\\d+\\/\\d+)\\s.*", replacement = "\\1", x = tkr.bond, perl = TRUE)
# [1] "3 25/32"
Then you can change it to numerical:
temp <- sub(pattern = ".*\\s(\\d\\s\\d+\\/\\d+)\\s.*", replacement = "\\1", x = tkr.bond, perl = TRUE)
eval(parse(text=sub(" ","+",x = temp)))
# [1] 3.78125
You can also use strsplit here. Then evaluate components excluding the first and the last. Like this
> tickers <- c('OAT 3 25/32 7/17/17', 'OAT 0 7/17/17')
>
> unlist(lapply(lapply(strsplit(tickers, " "),
+ function(x) {x[-length(x)][-1]}),
+ function(y) {sum(
+ sapply(y, function (z) {eval(parse(text = z))}) )} ) )
[1] 3.78125 0.00000