I have a vector where I want to replace one element with multiple element, I am able to replace with one but not multuiple, can anyone help?
For example I have
data <- c('a', 'x', 'd')
> data
[1] "a" "x" "d"
I want to replace "x" with "b", "c" to get
[1] "a" "b" "c" "d"
However
gsub('x', c('b', 'c'), data)
gives me
[1] "a" "b" "d"
Warning message:
In gsub("x", c("b", "c"), data) :
argument 'replacement' has length > 1 and only the first element will
be used
Here's how I would tackle it:
data <- c('a', 'x', 'd')
lst <- as.list(data)
unlist(lapply(lst, function(x) if(x == "x") c("b", "c") else x))
# [1] "a" "b" "c" "d"
We're making use of the fact that list-structures are more flexible than atomic vectors. We can replace a length-1 list-element with a length>1 element and then unlist the result to go back to an atomic vector.
Since you want to replace exact matches of "x" I prefer not to use sub/gsub in this case.
You may try this , although I believe the accepted answer is great:
unlist(strsplit(gsub("x", "b c", data), split = " "))
Logic: Replacing "x" with "b c" with space and then doing the strsplit, once its splitted we can convert is again back to vector using unlist.
This is a bit tricky of a problem because in your replacement you also want to grow your vector. That being said, I believe this should work:
replacement <- c("b","c")
new_data <- rep(data, times = ifelse(data=="x", length(replacement), 1))
new_data[new_data=="x"] <- replacement
new_data
#[1] "a" "b" "c" "d"
This will also work if you have multiple "x"s in your vector like:
data <- c("a","x","d","x")
Another approach:
data <- c('a', 'x', 'd')
pattern <- "x"
replacement <- c("b", "c")
sub_one_for_many <- function(pattern, replacement, x) {
removal_index <- which(x == pattern)
if (removal_index > 1) {
result <- c(x[1:(removal_index-1)], replacement, x[(removal_index+1):length(x)])
} else if (removal_index == 1) {
result <- c(replacement, x[2:length(x)])
}
return(result)
}
answer <- sub_one_for_many(pattern, replacement, data)
Output:
> answer
[1] "a" "b" "c" "d"
Related
I have a vector v. I would like to replace everything that doesn't equal S or D. It can be replaced by x.
v <- c("S","D","hxbasxhas","S","cbsdc")
result
r <- c("S","D","x","S","x")
A stringr approach is possible with a negative look-around.
Using str_replace:
library(stringr)
str_replace(v, "^((?!S|D).)*$", "x")
Result:
[1] "S" "D" "x" "S" "x"
You can negate %in% :
v <- c("S","D","hxbasxhas","S","cbsdc")
v[!v %in% c('S', 'D')] <- 'x'
v
#[1] "S" "D" "x" "S" "x"
Or use forcats::fct_other :
forcats::fct_other(v, c('S', 'D'), other_level = 'x')
I believe that the which statement is what you are looking for:
v[which(v!="S" & v!="D")]<-"x"
I am trying to use combn to divide a group of n = 20 different units into 3 groups of unequal size -- 4, 6 and 10. Then I am trying to validate for values that must be together within a group -- if one element from the pair exists in the group then the other should also be in the group. If one is not in the group then neither should be in the group. In this fashion, I'd like to evaluate the groups in order to find all possible valid solutions where the rules are true.
x <- letters[1:20]
same_group <- list(
c("a", "c"),
c("d", "f"),
c("b", "k", "r")
)
combinations_list <- combn(x, 4, simplify = F)
validate_combinations <- function(x) all(c("a", "c") %in% x) | !any(c("a", "c") %in% x)
valid_combinations <- keep(combinations_list, validate_combinations)
In this way I'd like to combine -> reduce each group until I have a list of all valid combinations. I'm not sure how to combine combinations_list, validate_combinations, and the same_group to check all same_group "rules" against the combinations in the table. The furthest I can get is to check against one combination c("a", "c"), which when run against keep(combinations_list, validate_combinations) is indeed giving me the output I want.
I think once I can do this, I can then use the unpicked values in another combn function for the group of 6 and the group of 10.
We can change the function to accept variable group
validate_combinations <- function(x, group) all(group %in% x) | !any(group %in% x)
then for each group subset the combinations_list which satisfy validate_combinations
lapply(same_group, function(x) combinations_list[
sapply(combinations_list, function(y) validate_combinations(y, x))])
#[[1]]
#[[1]][[1]]
#[1] "a" "b" "c" "d"
#[[1]][[2]]
#[1] "a" "b" "c" "e"
#[[1]][[3]]
#[1] "a" "b" "c" "f"
#[[1]][[4]]
#[1] "a" "b" "c" "g"
#[[1]][[5]]
#[1] "a" "b" "c" "h"
#[[1]][[6]]
#[1] "a" "b" "c" "i"
#[[1]][[7]]
#[1] "a" "b" "c" "j"
#[[1]][[8]]
#[1] "a" "b" "c" "k"
#......
I have a list of elements (letter here in the example)
(l <- list(letters[1:2], letters[2:3]))
# [[1]]
# [1] "a" "b"
# [[2]]
# [1] "b" "c"
And another elements
(r <- letters[2])
# [1] "b"
I create a function that delete that elements appearing before one element and itself (here "b").
out = lapply(l, function(x) x[-c(1,which(x == "b"))])
Filter(length, out)
#[[1]]
#[1] "c"
Now my question is in case I have a list of elements r not only one "b", how can I loop all the list:
for example :
r
[1] a
[2] b
I would like to have a result like this
[1] c
Thank you
Cheers
We can use %in%
Filter(length, lapply(l, function(x)
x[-seq(tail(which(x %in% r),1))]))
#[[1]]
#[1] "c"
data
r <- c('a', 'b')
I started with a matrix
Xray Stay Leave
[1,] "H" "H" "H"
[2,] "A" "L" "O"
And I have the following vector:
[1] "H" "L"
I want to get the output
"Stay".
I tried this:
which(vec %in% matrix )
but that gives me the following output:
[1] 1 2
Seems to be just telling me the rows that it finds the H and L in. I need the column name of the one that is an exact match.
Another approach:
vec <- c("H", "L")
colnames(mat)[colMeans(mat == vec) == 1]
# [1] "Stay"
where mat is the name of your matrix.
Assuming m is your matrix, you could do
> vec <- c("H", "A")
> colnames(m)[apply(m, 2, identical, vec)]
NOTE: identical used here because the original post says "I need the column name of the one that is an exact match"
This should return a logical vector:
logv <- apply(mat, 2, function(x) identical(vec,x))
Then this will select the correct column name:
dimnames(mat)[[2]][logv]
[1] "Stay"
Test case:
mat <- matrix( c( "H","H", "H", "A","L","O") ,2, byrow=TRUE,
dimnames=list(NULL, c('Xray', 'Stay', 'Leave') ) )
You could try:
If mat and vec are the matrix and vector
colnames(mat)[table(mat %in% vec, (seq_along(mat)-1)%/%nrow(mat) +1)[2,] >1]
#[1] "Stay"
I tried using the paste command but it returns the same vector?
x = c("a","b","c")
y = paste(x)
y
[1] "a" "b" "c"
length(y)
[1] 3
I want a single character of "abc"
The collapse="" options is your friend:
> x <- c("a", "b", "c")
> paste(x, collapse="")
[1] "abc"
>
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