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'
Related
I have string, which should be split into parts from "random" locations. Split occurs always from next comma after colon.
My idea was to find colons with
stringr::str_locate_all(test, ":") %>%
unlist()
then find commas
stringr::str_locate_all(test, ",") %>%
unlist()
and from there to figure out position where it should be split up, but could not find suitable way to do it. Feels like there is always 6 characters after colon before the comma, but I can't be sure about that for whole data.
Here is example string:
dput(test)
"AA,KK,QQ,JJ,TT,99,88:0.5083,66,55:0.8303,AK,AQ,AJs,AJo:0.9037,ATs:0.0024,ATo:0.5678"
Here is what result should be
dput(result)
c("AA,KK,QQ,JJ,TT,99,88:0.5083", "66,55:0.8303", "AK,AQ,AJs,AJo:0.9037",
"ATs:0.0024", "ATo:0.5678")
Perehaps we can use regmatches like below
> regmatches(test, gregexpr("(\\w+,?)+:[0-9.]+", test))[[1]]
[1] "AA,KK,QQ,JJ,TT,99,88:0.5083" "66,55:0.8303"
[3] "AK,AQ,AJs,AJo:0.9037" "ATs:0.0024"
[5] "ATo:0.5678"
here is one option with strsplit - replace the , after the digit followed by the . and one or more digits (\\d+) with a new delimiter using gsub and then split with strsplit in base R
result1 <- strsplit(gsub("([0-9]\\.[0-9]+),", "\\1;", test), ";")[[1]]
-checking
> identical(result, result1)
[1] TRUE
If the number of characters are fixed, use a regex lookaround
result1 <- strsplit(test, "(?<=:.{6}),", perl = TRUE)[[1]]
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 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 have an string called PATTERN:
PATTERN <- "MODEL_Name.model-OUTCOME_any.outcome-IMP_number"
and I would like to parse the string using a pattern matching function, like grep, sub, ... to obtain a string variable MODEL equal to "Name.model", a string variable OUTCOME equal to "any.outcome" and an integer variable IMP equal to number.
If MODEL, OUTCOME and IMP were all integers, I could get the values using function sub:
PATTERN <- "MODEL_002-OUTCOME_007-IMP_001"
pattern_build <- "MODEL_([0-9]+)-OUTCOME_([0-9]+)-IMP_([0-9]+)"
MODEL <- as.integer(sub(pattern_build, "\\1", PATTERN))
OUTCOME <- as.integer(sub(pattern_build, "\\2", PATTERN))
IMP <- as.integer(sub(pattern_build, "\\3", PATTERN))
Do you have any idea of how to match the string contained in variable PATTERN?
Possible tricky patterns are:
PATTERN <- "MODEL_PS2-OUTCOME_stroke_i-IMP_001"
PATTERN <- "MODEL_linear-model-OUTCOME_stroke_i-IMP_001"
A solution which is also able to deal with the 'tricky' patterns:
PATTERN <- "MODEL_linear-model-OUTCOME_stroke_i-IMP_001"
lst <- strsplit(PATTERN, '([A-Z]+_)')[[1]][2:4]
lst <- sub('-$','',lst)
which gives:
> lst
[1] "linear-model" "stroke_i" "001"
And if you want that in a dataframe:
df <- as.data.frame.list(lst)
names(df) <- c('MODEL','OUTCOME','IMP')
which gives:
> df
MODEL OUTCOME IMP
1 linear-model stroke_i 001
A minimal-regex approach,
sapply(strsplit(PATTERN, '-'), function(i) sub('(.*?_){1}', '', i))
# [,1]
#[1,] "PS2"
#[2,] "stroke_i"
#[3,] "001"
You may use a pattern with capturing groups matching any chars, as few as possible between known delimiting substrings:
MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)
See the regex demo. Note that the last .* is greedy since you get all the rest of the string into this capture.
You may precise this pattern to only allow matching expected characters (say, to match digits into the last capturing group, use ([0-9]+) rather than (.*).
Use it with, say, str_match from stringr:
> library(stringr)
> x <- "MODEL_Name.model-OUTCOME_any.outcome-IMP_number"
> res <- str_match(x, "MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)")
> res[,2]
[1] "Name.model"
> res[,3]
[1] "any.outcome"
> res[,4]
[1] "number"
>
A base R solution using the same regex will involve a regmatches / regexec:
> res <- regmatches(x, regexec("MODEL_(.*?)-OUTCOME_(.*?)-IMP_(.*)", x))[[1]]
> res[2]
[1] "Name.model"
> res[3]
[1] "any.outcome"
> res[4]
[1] "number"
>
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