Hi I have the following data structure:
typeof(snp.seq)
[1] "list"
typeof(snp.seq[1])
[1] "list"
snp.seq[[1]]
[1]"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWN DPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKSPMLNLFQEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ"
typeof(snp.seq[[1]])
"character"
> dput(snp.seq)
list("MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKSPMLNLFQEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKGPV",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ")
where snp.seq is of type list, and each element of snp.seq is also a list - as demonstrated in the example.
What I need to do is to specify a number, and return the corresponding letter from the sequence in snp.seq[[1]].
i.e. position 1 is the letter 'M'
I have tried to call it by doing
snp.seq[[1,1]]]
snp.seq[[1],[1]]
of which both do not work as they are calling a position which is out of bounds.
How would I call any letter by giving a number (it's position in the sequence)?
Thanks
With the data in your example -
snp.seq <- list("MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKSPMLNLFQEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKGPV",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ")
##
getLetter <- function(elem,pos){
if(pos > nchar(snp.seq[[elem]])){
stop("Position argument exceeds sequence length")
} else {
substr(snp.seq[[elem]],pos,pos)
}
}
##
So to get the first letter for the first three elements in snp.seq, you would do
> getLetter(1,1)
[1] "M"
> getLetter(2,1)
[1] "M"
> getLetter(3,1)
[1] "M"
> substr(snp.seq[[1]],1,3)
[1] "MYS"
> substr(snp.seq[[2]],1,3)
[1] "MYS"
> substr(snp.seq[[3]],1,3)
[1] "MDY"
(sampling the first few letters of these elements to validate).
You could use letter from Biostrings
library(Biostrings)
letter(snp.seq[[1]], 1)
#[1] "M"
letter(snp.seq[[3]], 3:1)
#[1] "YDM"
letter(snp.seq[[5]], 5:10)
#[1] "NTLRLY"
Try:
snp.seq <- list("MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKSPMLNLFQEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKGPV",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ",
"MYSFNTLRLYLWETIVFFSLAASKEAEAARSAPKPMSPSDFLDKLMGRTSGYDARIRPNFKGPPVNVSCNIFINSFGSIAETTMDYRVNIFLRQQWNDPRLAYNEYPDDSLDLDPSMLDSIWKPDLFFANEKGAHFHEITTDNKLLRISRNGNVLYSIRITLTLACPMDLKNFPMDVQTCIMQLESFGYTMNDLIFEWQEQGAVQVADGLTLPQFILKEEKDLRYCTKHYNTGKFTCIEARFHLERQMGYYLIQMYIPSLLIVILSWISFWINMDAAPARVGLGITTVLTMTTQSSGSRASLPKVSYVKAIDIWMAVCLLFVFSALLEYAAVNFVSRQHKELLRFRRKRRHHKEDEAGEGRFNFSAYGMGPACLQAKDGISVKGANNSNTTNPPPAPSKSPEEMRKLFIQRAKKIDKISRIGFPMAFLIFNMFYWIIYKIVRREDVHNQ")
##
uls = unlist(snp.seq)
substr(uls,1,1)
[1] "M" "M" "M" "M" "M"
substr(uls,1,2)
[1] "MY" "MY" "MD" "MD" "MY"
substr(uls,3,5)
[1] "SFN" "SFN" "YRV" "YRV" "SFN"
Related
I am using R and want to covert following:
"A,B"
to
"A","B" OR 'A','B'
I tried str_replace(), but that's not working out.
Please suggest, thanks.
Update
I tried the suggested answer by d.b. Though it works, but I didn't realize that I should have shared that, I am going to use above solution for vector. I need the values in data with "A,B" to split in order to use it as a vector.
Using strsplit
> data
[1] "A,B"
> test <- strsplit(x = data, split = ",")
> test
[[1]]
[1] "A" "B"
above test won't be useful because I can't use it for following:
> output_1 <- c(test)
> outputFinalData <- outputFinal[outputFinal$Column %in% output_1,]
outputFinalData is empty with above process. But is not empty when I do:
> output_2 <- c("A", "B")
> outputFinalData <- outputFinal[outputFinal$Column %in% output_2,]
Also, output_1 and output_2 are not same:
> output_1
[[1]]
[1] "Bin_14" "Bin_15"
> output_2
[1] "Bin_14" "Bin_15"
> output_1 == output_2
[1] FALSE FALSE
Use strsplit:
> data = "A,B"
> strsplit(x=data,split=",")
[[1]]
[1] "A" "B"
Note that it returns a list with a vector. The list is length one because you asked it to split one string. If you ask it to split two strings you get a list of length 2:
> data = c("A,B","Foo,bar")
> strsplit(x=data,split=",")
[[1]]
[1] "A" "B"
[[2]]
[1] "Foo" "bar"
So if you know you are only going to have one thing to split you can get a vector of the parts by taking the first element:
> data = "A,B"
> strsplit(x=data,split=",")[[1]]
[1] "A" "B"
However it might be more efficient to do a load of splits in one go and put the bits in a matrix. As long as you can be sure everything splits into the same number of parts, then something like:
> data = c("A,B","Foo,bar","p1,p2")
> do.call(rbind,(strsplit(x=data,split=",")))
[,1] [,2]
[1,] "A" "B"
[2,] "Foo" "bar"
[3,] "p1" "p2"
>
Gets you the two parts in columns of a matrix that you can then add to a data frame if that's what you need.
I am currently trying to subset a list in R from a dataframe. My current attempt looks like:
list.level <- unique(buckets$group)
bucket.group <- vector("list",length(list.level))
for(i in list.level){
bucket.group[[i]] <- subset(buckets$group,buckets$group == i)
}
However, instead of filling the list it seems to create a duplicate list of the same amount of rows, returning:
[[1]]
NULL
[[2]]
NULL
...
NULL
[[22]]
NULL
[[23]]
NULL
$A
[1] "A"
$C
[1] "C" "C" "C"
$D
[1] "D" "D" "D"
...
$AJ
[1] "AJ" "AJ" "AJ" "AJ" "AJ"
$AK
[1] "AK" "AK"
A should be filling into 1, C into 2, etc. etc. How do I get these to fill in the original rows rather than creating extra rows at the bottom of the list?
Here is what is going on. Suppose your buckets$group is c("a","a","b","b").
list.level <- unique(buckets$group)
Now list.level is c("a","b")
bucket.group <- vector("list",length(list.level))
Since length(list.level) is 2, now your bucket.group is a list of 2 NULL elements, their names are 1 and 2.
for(i in list.level){
Recalling the value of list.level, it is the same as for i in c("a","b").
bucket.group[[i]] <- subset(buckets$group,buckets$group == i)
Since i loops over "a" and "b", you now fill bucket.group[["a"]] and bucket.group[["b"]], while bucket.group[[1]] and bucket.group[[2]] remain intact.
To fix this, you should write instead
list.level <- unique(buckets$group) # ok, this was correct
bucket.group <- list() # just empty list
for(i in 1:length(list.level)){
bucket.group[[i]] <- buckets$group[buckets$group == list.level[[i]] ]
}
I think the issue is with your for statement.
Your code is like this:
list.level<-letters[1:10]
> for(i in list.level) print(i)
[1] "a"
[1] "b"
[1] "c"
[1] "d"
[1] "e"
[1] "f"
[1] "g"
[1] "h"
[1] "i"
[1] "j"
It assigns each element in list.level to i, so i is a letter. When you do
bucket.group[[i]] <- subset(buckets$group,buckets$group == i)
in the first iteration, i is a letter. So it looks for a list element called bucket.group[["a"]] and does not find it, so it creates it and stores the data there. If instead you use seq_along
for(i in seq_along(list.level)) print(i)
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 6
[1] 7
[1] 8
[1] 9
[1] 10
now i will alway be a number and the code will do what you want.
So use seq_along instead.
this should work:
list.level <- unique(buckets$group)
bucket.group <- vector("list",length(list.level))
for(i in 1:length(list.level)){
bucket.group[[i]] <- subset(buckets$group,buckets$group == list.level[i])
}
I'd like to find the midpoint of any word after the following is done to the word:
>x = 'hello'
>y = strsplit(x, '')
>y
[[1]]
[1] "h" "e" "l" "l" "o"
>z = unlist(y)
>z
[1] "h" "e" "l" "l" "o"
Doing this then allows for :
> z[1]
[1] "h"
> z[4]
[1] "l"
The difference being that before z=unlist(y) when you try z[index] you get back NA, example:
> x = 'hello'
> strsplit(x, '')
[[1]]
[1] "h" "e" "l" "l" "o"
> x[1]
[1] "hello"
> x[2]
[1] NA
Anyways, what I want to do is find the mid point of words that are in this format so that the output would be something like:
"l"
in the case of the word "hello". Also, in this example we have a word with 5 letters allowing to easily designate a single character as the midpoint but for a word like "bake" I would like to designate both "a" and "k" together as the midpoint.
Try
f1 <- function(str1){
N <- nchar(str1)
if(!N%%2){
res <- substr(str1, N/2, (N/2)+1)
}
else{
N1 <- median(sequence(N))
res <- substr(str1, N1, N1)
}
res
}
f1('bake')
#[1] "ak"
f1('hello')
#[1] "l"
Another option. get_middle assumes the word has already been split into characters, as per your description:
get_middle <- function(x) {
mid <- (length(x) + 1) / 2
x[unique(c(ceiling(mid), floor(mid)))]
}
Then:
words <- c("bake", "hello")
lapply(strsplit(words, ""), get_middle)
Produces:
[[1]]
[1] "k" "a"
[[2]]
[1] "l"
You could try this:
midpoint <- function(word) {
# Split the word into a vector of letters
split <- strsplit(word, "")[[1]]
# Get the number of letters in the word
n <- nchar(word)
# Get the two middle letters for words of even length,
# otherwise get the single middle letter
if (n %% 2 == 0) {
c(split[n/2], split[n/2+1])
} else {
split[ceiling(n/2)]
}
}
In the case of a word of even length, the middle two characters are returned as a vector.
midpoint("hello")
#[1] "l"
midpoint("bake")
#[1] "a" "k"
How about:
mid<-function(str)substr(str,(nchar(str)+1)%/%2,(nchar(str)+2)%/%2)
Or slightly more legibly:
mid2<-function(str){
n1<-nchar(str)+1
substr(str,n1%/%2,(n1+1)%/%2)
}
> mid("bake")
[1] "ak"
> mid("hello")
[1] "l"
This has the advantage that it immediately vectorizes:
> mid(c("bake","hello"))
[1] "ak" "l"
It is slower than #akrun's solution for long words, but my second version is faster; apparently counting characters can be costly for longer strings.
If you want the final product in a list, you can just strsplit the result:
mid3<-function(str)strsplit(mid2(str),"")
word = c("bake","hello")
print(nchar(word))
q = ifelse (nchar(word)%%2==0, substr(word,nchar(word)/2,nchar(word)/2+1),substr(word,nchar(word)/2+1,nchar(word)/2+1))
print(q)
[1] 4 5
[1] "ak" "l"
Let's say I have two vectors:
a1=c("a","b")
a2=c("x","y")
Now in a 'for' loop, I want to access the first element of each vector:
for(i in c(a1,a2)) {
print(i[1])
}
If I run the above code, I get:
[1] "a"
[1] "b"
[1] "x"
[1] "y"
But I just want:
[1] "a"
[1] "x"
More surprisingly, if I want to access the second element:
for(i in c(a1,a2)) {
print(i[2])
}
I get:
[1] "NA"
[1] "NA"
[1] "NA"
[1] "NA"
Any help will be highly appreciated.
Because c(a1, a2) = c("a","b","x","y") -- passing multiple atomic vectors to c causes them to get collapsed. Use list(a1, a2) in the loop instead.
I would like to use the characters in a vector as the names of character objects
aiming to get
first as say "d","e","a","t" etc.
tried this approach but am clearly missing some function to apply to x[i]
x <- c("first","second","third"..)
for (i in 1:length(x)) {
x[i] <- sample(letters,4)
}
TIA
The function you are looking for is assign():
> x <- c("first","second","third")
> for (i in 1:length(x)) {
+ assign(x[i], sample(letters,4))
+ }
>
> ls()
[1] "first" "i" "second" "third" "x"
> first
[1] "t" "d" "u" "j"
> second
[1] "o" "i" "p" "l"
> third
[1] "w" "v" "r" "n"
As an alternative, you could build these vectors as different elements of a list:
> mylist <- list()
> for (i in 1:length(x)) {
+ mylist[[x[i]]] <- sample(letters,4)
+ }
> mylist
$first
[1] "e" "l" "y" "d"
$second
[1] "t" "o" "k" "h"
$third
[1] "g" "x" "p" "b"
You don't say what you will be doing with this object. You may get the simplest structure by using a named vector:
names(x) <- x
x[] <- sample(letters, 4)
If you do not use the paired bracket on the LHS, the whole vector gets replaced and the names will be lost. You can now access the values with quoted names:
> x
first second third fourth
"w" "c" "r" "x"
> x["second"]
second
"c"