Related
I have the following vector:
X <- c("mama.log", "papa.log", "mimo.png", "mentor.log")
How do I retrieve another vector that only contains elements starting with "m" and ending with ".log"?
you can use grepl with regular expression:
X[grepl("^m.*\\.log", X)]
Try this:
grep("^m.*[.]log$", X, value = TRUE)
## [1] "mama.log" "mentor.log"
A variation of this is to use a glob rather than a regular expression:
grep(glob2rx("m*.log"), X, value = TRUE)
## [1] "mama.log" "mentor.log"
The documentation on the stringr package says:
str_subset() is a wrapper around x[str_detect(x, pattern)], and is equivalent to grep(pattern, x, value = TRUE). str_which() is a wrapper around which(str_detect(x, pattern)), and is equivalent to grep(pattern, x).
So, in your case, the more elegant way to accomplish your task using tidyverse instead of base R is as following.
library(tidyverse)
c("mama.log", "papa.log", "mimo.png", "mentor.log") %>%
str_subset(pattern = "^m.*\\.log")
which produces the output:
[1] "mama.log" "mentor.log"
Using pipes...
library(tidyverse)
c("mama.log", "papa.log", "mimo.png", "mentor.log") %>%
.[grepl("^m.*\\.log$", .)]
[1] "mama.log" "mentor.log"
I'm new to R and am stuck with backreferencing that doesn't seem to work. In:
gsub("\\((\\d+)\\)", f("\\1"), string)
It correctly grabs the number in between parentheses but doesn't apply the (correctly defined, working otherwise) function f to replace the number --> it's actually the string "\1" that passes through to f.
Am I missing something or is it just that R does not handle this? If so, any idea how I could do something similar, i.e. applying a function "on the fly" to the (actually many) numbers that occur in between parentheses in the text I'm parsing?
Thanks a lot for your help.
R does not have the option of applying a function directly to a match via gsub. You'll actually have to extract the match, transform the value, then replace the value. This is relativaly easy with the regmatches function. For example
x<-"(990283)M (31)O (29)M (6360)M"
f<-function(x) {
v<-as.numeric(substr(x,2,nchar(x)-1))
paste0(v+5,".1")
}
m <- gregexpr("\\(\\d+\\)", x)
regmatches(x, m) <- lapply(regmatches(x, m), f)
x
# [1] "990288.1M 36.1O 34.1M 6365.1M"
Of course you can make f do whatever you like just make sure it's vector-friendly. Of course, you could wrap this in your own function
gsubf <- function(pattern, x, f) {
m <- gregexpr(pattern, x)
regmatches(x, m) <- lapply(regmatches(x, m), f)
x
}
gsubf("\\(\\d+\\)", x, f)
Note that in these examples we're not using a capture group, we're just grabbing the entire match. There are ways to extract the capture groups but they are a bit messier. If you wanted to provide an example where such an extraction is required, I might be able to come up with something fancier.
To use a callback within a regex-capable replacement function, you may use either gsubfn or stringr functions.
When choosing between them, note that stringr is based on ICU regex engine and with gsubfn, you may use either the default TCL (if the R installation has tcltk capability, else it is the default TRE) or PCRE (if you pass the perl=TRUE argument).
Also, note that gsubfn allows access to all capturing groups in the match object, while str_replace_all will only allow to manipulate the whole match only. Thus, for str_replace_all, the regex should look like (?<=\()\d+(?=\)), where 1+ digits are matched only when they are enclosed with ( and ) excluding them from the match.
With stringr, you may use str_replace_all:
library(stringr)
string <- "(990283)M (31)O (29)M (6360)M"
## Callback function to increment found number:
f <- function(x) { as.integer(x) + 1 }
str_replace_all(string, "(?<=\\()\\d+(?=\\))", function(m) f(m))
## => [1] "(990284)M (32)O (30)M (6361)M"
With gsubfn, pass perl=TRUE and backref=0 to be able to use lookarounds and just modify the whole match:
gsubfn("(?<=\\()\\d+(?=\\))", ~ f(m), string, perl=TRUE, backref=0)
## => [1] "(990284)M (32)O (30)M (6361)M"
If you have multiple groups in the pattern, remoe backref=0 and enumerate the group value arguments in the callback function declaration:
gsubfn("(\\()(\\d+)(\\))", function(m,n,o) paste0(m,f(n),o), string, perl=TRUE)
^ 1 ^^ 2 ^^ 3 ^ ^^^^^^^ ^^^^
This is for multiple different replacements.
text="foo(200) (300)bar (400)foo (500)bar (600)foo (700)bar"
f=function(x)
{
return(as.numeric(x[[1]])+5)
}
a=strsplit(text,"\\(\\K\\d+",perl=T)[[1]]
b=f(str_extract_all(text,perl("\\(\\K\\d+")))
paste0(paste0(a[-length(a)],b,collapse=""),a[length(a)]) #final output
#[1] "foo(205) (305)bar (405)foo (505)bar (605)foo (705)bar"
Here's a way by tweaking a bit stringr::str_replace(), in the replace argument, just use a lambda formula as the replace argument, and reference the captured group not by ""\\1" but by ..1, so your gsub("\\((\\d+)\\)", f("\\1"), string) will become str_replace2(string, "\\((\\d+)\\)", ~f(..1)), or just str_replace2(string, "\\((\\d+)\\)", f) in this simple case :
str_replace2 <- function(string, pattern, replacement, type.convert = TRUE){
if(inherits(replacement, "formula"))
replacement <- rlang::as_function(replacement)
if(is.function(replacement)){
grps_mat <- stringr::str_match(string, pattern)[,-1, drop = FALSE]
grps_list <- lapply(seq_len(ncol(grps_mat)), function(i) grps_mat[,i])
if(type.convert) {
grps_list <- type.convert(grps_list, as.is = TRUE)
replacement <- rlang::exec(replacement, !!! grps_list)
replacement <- as.character(replacement)
} else {
replacement <- rlang::exec(replacement, !!! grps_list)
}
}
stringr::str_replace(string, pattern, replacement)
}
str_replace2(
"foo (4)",
"\\((\\d+)\\)",
sqrt)
#> [1] "foo 2"
str_replace2(
"foo (4) (5)",
"\\((\\d+)\\) \\((\\d+)\\)",
~ sprintf("(%s)", ..1 * ..2))
#> [1] "foo (20)"
Created on 2020-01-24 by the reprex package (v0.3.0)
I'm new to R and am stuck with backreferencing that doesn't seem to work. In:
gsub("\\((\\d+)\\)", f("\\1"), string)
It correctly grabs the number in between parentheses but doesn't apply the (correctly defined, working otherwise) function f to replace the number --> it's actually the string "\1" that passes through to f.
Am I missing something or is it just that R does not handle this? If so, any idea how I could do something similar, i.e. applying a function "on the fly" to the (actually many) numbers that occur in between parentheses in the text I'm parsing?
Thanks a lot for your help.
R does not have the option of applying a function directly to a match via gsub. You'll actually have to extract the match, transform the value, then replace the value. This is relativaly easy with the regmatches function. For example
x<-"(990283)M (31)O (29)M (6360)M"
f<-function(x) {
v<-as.numeric(substr(x,2,nchar(x)-1))
paste0(v+5,".1")
}
m <- gregexpr("\\(\\d+\\)", x)
regmatches(x, m) <- lapply(regmatches(x, m), f)
x
# [1] "990288.1M 36.1O 34.1M 6365.1M"
Of course you can make f do whatever you like just make sure it's vector-friendly. Of course, you could wrap this in your own function
gsubf <- function(pattern, x, f) {
m <- gregexpr(pattern, x)
regmatches(x, m) <- lapply(regmatches(x, m), f)
x
}
gsubf("\\(\\d+\\)", x, f)
Note that in these examples we're not using a capture group, we're just grabbing the entire match. There are ways to extract the capture groups but they are a bit messier. If you wanted to provide an example where such an extraction is required, I might be able to come up with something fancier.
To use a callback within a regex-capable replacement function, you may use either gsubfn or stringr functions.
When choosing between them, note that stringr is based on ICU regex engine and with gsubfn, you may use either the default TCL (if the R installation has tcltk capability, else it is the default TRE) or PCRE (if you pass the perl=TRUE argument).
Also, note that gsubfn allows access to all capturing groups in the match object, while str_replace_all will only allow to manipulate the whole match only. Thus, for str_replace_all, the regex should look like (?<=\()\d+(?=\)), where 1+ digits are matched only when they are enclosed with ( and ) excluding them from the match.
With stringr, you may use str_replace_all:
library(stringr)
string <- "(990283)M (31)O (29)M (6360)M"
## Callback function to increment found number:
f <- function(x) { as.integer(x) + 1 }
str_replace_all(string, "(?<=\\()\\d+(?=\\))", function(m) f(m))
## => [1] "(990284)M (32)O (30)M (6361)M"
With gsubfn, pass perl=TRUE and backref=0 to be able to use lookarounds and just modify the whole match:
gsubfn("(?<=\\()\\d+(?=\\))", ~ f(m), string, perl=TRUE, backref=0)
## => [1] "(990284)M (32)O (30)M (6361)M"
If you have multiple groups in the pattern, remoe backref=0 and enumerate the group value arguments in the callback function declaration:
gsubfn("(\\()(\\d+)(\\))", function(m,n,o) paste0(m,f(n),o), string, perl=TRUE)
^ 1 ^^ 2 ^^ 3 ^ ^^^^^^^ ^^^^
This is for multiple different replacements.
text="foo(200) (300)bar (400)foo (500)bar (600)foo (700)bar"
f=function(x)
{
return(as.numeric(x[[1]])+5)
}
a=strsplit(text,"\\(\\K\\d+",perl=T)[[1]]
b=f(str_extract_all(text,perl("\\(\\K\\d+")))
paste0(paste0(a[-length(a)],b,collapse=""),a[length(a)]) #final output
#[1] "foo(205) (305)bar (405)foo (505)bar (605)foo (705)bar"
Here's a way by tweaking a bit stringr::str_replace(), in the replace argument, just use a lambda formula as the replace argument, and reference the captured group not by ""\\1" but by ..1, so your gsub("\\((\\d+)\\)", f("\\1"), string) will become str_replace2(string, "\\((\\d+)\\)", ~f(..1)), or just str_replace2(string, "\\((\\d+)\\)", f) in this simple case :
str_replace2 <- function(string, pattern, replacement, type.convert = TRUE){
if(inherits(replacement, "formula"))
replacement <- rlang::as_function(replacement)
if(is.function(replacement)){
grps_mat <- stringr::str_match(string, pattern)[,-1, drop = FALSE]
grps_list <- lapply(seq_len(ncol(grps_mat)), function(i) grps_mat[,i])
if(type.convert) {
grps_list <- type.convert(grps_list, as.is = TRUE)
replacement <- rlang::exec(replacement, !!! grps_list)
replacement <- as.character(replacement)
} else {
replacement <- rlang::exec(replacement, !!! grps_list)
}
}
stringr::str_replace(string, pattern, replacement)
}
str_replace2(
"foo (4)",
"\\((\\d+)\\)",
sqrt)
#> [1] "foo 2"
str_replace2(
"foo (4) (5)",
"\\((\\d+)\\) \\((\\d+)\\)",
~ sprintf("(%s)", ..1 * ..2))
#> [1] "foo (20)"
Created on 2020-01-24 by the reprex package (v0.3.0)
I have a vector of a binary string:
a<-c(0,0,0,1,0,1)
I would like to convert this vector into decimal.
I tried using the compositions package and the unbinary() function, however, this solution and also most others that I have found on this site require g-adic string as input argument.
My question is how can I convert a vector rather than a string to decimal?
to illustrate the problem:
library(compositions)
unbinary("000101")
[1] 5
This gives the correct solution, but:
unbinary(a)
unbinary("a")
unbinary(toString(a))
produces NA.
You could try this function
bitsToInt<-function(x) {
packBits(rev(c(rep(FALSE, 32-length(x)%%32), as.logical(x))), "integer")
}
a <- c(0,0,0,1,0,1)
bitsToInt(a)
# [1] 5
here we skip the character conversion. This only uses base functions.
It is likely that
unbinary(paste(a, collapse=""))
would have worked should you still want to use that function.
There is a one-liner solution:
Reduce(function(x,y) x*2+y, a)
Explanation:
Expanding the application of Reduce results in something like:
Reduce(function(x,y) x*2+y, c(0,1,0,1,0)) = (((0*2 + 1)*2 + 0)*2 + 1)*2 + 0 = 10
With each new bit coming next, we double the so far accumulated value and add afterwards the next bit to it.
Please also see the description of Reduce() function.
If you'd like to stick to using compositions, just convert your vector to a string:
library(compositions)
a <- c(0,0,0,1,0,1)
achar <- paste(a,collapse="")
unbinary(achar)
[1] 5
This function will do the trick.
bintodec <- function(y) {
# find the decimal number corresponding to binary sequence 'y'
if (! (all(y %in% c(0,1)))) stop("not a binary sequence")
res <- sum(y*2^((length(y):1) - 1))
return(res)
}
I want to replace non-ascii characters (for now, only spanish), by their ascii equivalent. If I have "á", I want to replace it with "a" and so on.
I built this function (works fine), but I don't want to use a loop (including internal loops like sapply).
latin2ascii<-function(x) {
if(!is.character(x)) stop ("input must be a character object")
require(stringr)
mapL<-c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA<-c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")
for(y in 1:length(mapL)) {
x<-str_replace_all(x,mapL[y],mapA[y])
}
x
}
Is there an elegante way to solve it? Any help, suggestion or modification is appreciated
gsubfn() in the package of the same name is really nice for this sort of thing:
library(gsubfn)
# Create a named list, in which:
# - the names are the strings to be looked up
# - the values are the replacement strings
mapL <- c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA <- c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")
# ll <- setNames(as.list(mapA), mapL) # An alternative to the 2 lines below
ll <- as.list(mapA)
names(ll) <- mapL
# Try it out
string <- "ÍÓáÚ"
gsubfn("[áéíóúÁÉÍÓÚñÑüÜ]", ll, string)
# [1] "IOaU"
Edit:
G. Grothendieck points out that base R also has a function for this:
A <- paste(mapA, collapse="")
L <- paste(mapL, collapse="")
chartr(L, A, "ÍÓáÚ")
# [1] "IOaU"
I like the version by Josh, but I thought I might add another 'vectorized' solution. It returns a vector of unaccented strings. It also only relies on the base functions.
x=c('íÁuÚ','uíÚÁ')
mapL<-c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA<-c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")
split=strsplit(x,split='')
m=lapply(split,match,mapL)
mapply(function(split,m) paste(ifelse(is.na(m),split,mapA[m]),collapse='') , split, m)
# "iAuU" "uiUA"