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"
Related
I have the following data
GT-BU7867-09
GT-BU6523-113
GT-BU6452-1
GT-BU8921-12
How do I use R to make the numbers after the hyphen to pad leading zeros so it will have three digits? The resulting format should look like this:
GT-BU7867-009
GT-BU6523-113
GT-BU6452-001
GT-BU8921-012
Base solution:
sapply(strsplit(x,"-"), function(x)
paste(x[1], x[2], sprintf("%03d",as.numeric(x[3])), sep="-")
)
Result:
[1] "GT-BU7867-009" "GT-BU6523-113" "GT-BU6452-001" "GT-BU8921-012"
A solution using stringr and str_pad and strsplit
library(stringr)
x <- readLines(textConnection('GT-BU7867-09
GT-BU6523-113
GT-BU6452-1
GT-BU8921-12'))
unlist(lapply(strsplit(x,'-'),
function(x){
x[3] <- str_pad(x[3], width = 3, side = 'left', pad = '0')
paste0(x, collapse = '-')}))
[1] "GT-BU7867-009" "GT-BU6523-113" "GT-BU6452-001"
[4] "GT-BU8921-012"
Another version using str_pad and str_extract from package stringr
library(stringr)
x <- gsub("[[:digit:]]+$", str_pad(str_extract(x, "[[:digit:]]+$"), 3, pad = "0"), x)
i.e. extract the trailing numbers of x, pad them to 3 with 0s, then substitute these for the original trailing numbers.
I am trying to delete the 5th delimiter in this string:
"Bacteria_Firmicutes_Clostridia_Clostridiales_Rumino_coccaceae_Ruminococcus_Ruminococcus_albus"
so it becomes:
"Bacteria_Firmicutes_Clostridia_Clostridiales_Ruminococcaceae_Ruminococcus_Ruminococcus_albus"
This seems to work, but I feel like there should be a more elegant solution possibly with regex and str_replace
library(stringr)
name <- "Bacteria_Firmicutes_Clostridia_Clostridiales_Rumino_coccaceae_Ruminococcus_Ruminococcus_albus"
index <- str_locate_all(name, "_")[[1]]
str_sub(name, index[5, "start"], index[5, "end"]) <- ""
name
Try gsub:
> gsub("((?:[^_]+_){4}[^_]+)_", "\\1", name)
[1] "Bacteria_Firmicutes_Clostridia_Clostridiales_Ruminococcaceae_Ruminococcus_Ruminococcus_albus"
>
Or a less "pretty" way:
> gsub("([^_]*_[^_]*_[^_]*_[^_]*_[^_]*)_", "\\1", name)
[1] "Bacteria_Firmicutes_Clostridia_Clostridiales_Ruminococcaceae_Ruminococcus_Ruminococcus_albus"
>
Or with the strex library:
> library(strex)
> paste(str_before_nth(name, "_", 5), str_after_nth(name, "_", 5), sep="")
[1] "Bacteria_Firmicutes_Clostridia_Clostridiales_Ruminococcaceae_Ruminococcus_Ruminococcus_albus"
>
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"
Let's say I have the following string:
s <- "ID=MIMAT0027618;Alias=MIMAT0027618;Name=hsa-miR-6859-5p;Derives_from=MI0022705"
I would like to recover the strings between ";" and "=" to get the following output:
[1] "MIMAT0027618" "MIMAT0027618" "hsa-miR-6859-5p" "MI0022705"
Can I use strsplit() with more than one split element?
1) strsplit with matrix Try this:
> matrix(strsplit(s, "[;=]")[[1]], 2)[2,]
[1] "MIMAT0027618" "MIMAT0027618" "hsa-miR-6859-5p" "MI0022705"
2) strsplit with gsub or this use of strsplit with gsub:
> strsplit(gsub("[^=;]+=", "", s), ";")[[1]]
[1] "MIMAT0027618" "MIMAT0027618" "hsa-miR-6859-5p" "MI0022705"
3) strsplit with sub or this use of strsplit with sub:
> sub(".*=", "", strsplit(s, ";")[[1]])
[1] "MIMAT0027618" "MIMAT0027618" "hsa-miR-6859-5p" "MI0022705"
4) strapplyc or this which extracts consecutive non-semicolons after equal signs:
> library(gsubfn)
> strapplyc(s, "=([^;]+)", simplify = unlist)
[1] "MIMAT0027618" "MIMAT0027618" "hsa-miR-6859-5p" "MI0022705"
ADDED additional strplit solutions.
I know this is an old question, but I found the usage of lookaround regular expressions quite elegant for this problem:
library(stringr)
your_string <- '/this/file/name.txt'
result <- str_extract(string = your_string, pattern = "(?<=/)[^/]*(?=\\.)")
result
In words,
The (?<=...) part looks before the desired string for a... (in this case a forward slash).
The [^/]* then looks for as many characters in a row that are not a forward slash (in this case name.txt).
The (?=...) then looks after the desired string for a ... (in this case the special period character, which needs to be escaped as \\.).
This also works on dataframes:
library(dplyr)
strings <- c('/this/file/name1.txt', 'tis/other/file/name2.csv')
df <- as.data.frame(strings) %>%
mutate(name = str_extract(string = strings, pattern = "(?<=/)[^/]*(?=\\.)"))
# Optional
names <- df %>% pull(name)
Or, in your case:
your_string <- "ID=MIMAT0027618;Alias=MIMAT0027618;Name=hsa-miR-6859-5p;Derives_from=MI0022705"
result <- str_extract(string = your_string, pattern = "(?<=;Alias=)[^;]*(?=;)")
result # Outputs 'MIMAT0027618'
How do I remove part of a string? For example in ATGAS_1121 I want to remove everything before _.
Use regular expressions. In this case, you can use gsub:
gsub("^.*?_","_","ATGAS_1121")
[1] "_1121"
This regular expression matches the beginning of the string (^), any character (.) repeated zero or more times (*), and underscore (_). The ? makes the match "lazy" so that it only matches are far as the first underscore. That match is replaced with just an underscore. See ?regex for more details and references
You can use a built-in for this, strsplit:
> s = "TGAS_1121"
> s1 = unlist(strsplit(s, split='_', fixed=TRUE))[2]
> s1
[1] "1121"
strsplit returns both pieces of the string parsed on the split parameter as a list. That's probably not what you want, so wrap the call in unlist, then index that array so that only the second of the two elements in the vector are returned.
Finally, the fixed parameter should be set to TRUE to indicate that the split parameter is not a regular expression, but a literal matching character.
If you're a Tidyverse kind of person, here's the stringr solution:
R> library(stringr)
R> strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
R> strings %>% str_replace(".*_", "_")
[1] "_1121" "_1432" "_1121"
# Or:
R> strings %>% str_replace("^[A-Z]*", "")
[1] "_1121" "_1432" "_1121"
Here's the strsplit solution if s is a vector:
> s <- c("TGAS_1121", "MGAS_1432")
> s1 <- sapply(strsplit(s, split='_', fixed=TRUE), function(x) (x[2]))
> s1
[1] "1121" "1432"
Maybe the most intuitive solution is probably to use the stringr function str_remove which is even easier than str_replace as it has only 1 argument instead of 2.
The only tricky part in your example is that you want to keep the underscore but its possible: You must match the regular expression until it finds the specified string pattern (?=pattern).
See example:
strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
strings %>% stringr::str_remove(".+?(?=_)")
[1] "_1121" "_1432" "_1121"
Here the strsplit solution for a dataframe using dplyr package
col1 = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
col2 = c("T", "M", "A")
df = data.frame(col1, col2)
df
col1 col2
1 TGAS_1121 T
2 MGAS_1432 M
3 ATGAS_1121 A
df<-mutate(df,col1=as.character(col1))
df2<-mutate(df,col1=sapply(strsplit(df$col1, split='_', fixed=TRUE),function(x) (x[2])))
df2
col1 col2
1 1121 T
2 1432 M
3 1121 A