I have vectors that looks like these variations:
cn1 <- c("Probe","Genes","foo","bar","Probe","Genes","foo","bar")
# 0 1 2 3 4 5 6 7
cn2 <- c("Probe","Genes","foo","bar","qux","Probe","Genes","foo","bar","qux")
# 0 1 2 3 4 5 6 7 8 9
Note that in each vector above consists of two parts. They are separated with "Probe","Genes".
What I want to do is to get the indexes of the first part of the entry in between that separator. Yielding
cn1_id ------> [2,3]
cn2_id ------> [2,3,4]
How can I achieve that in R?
I tried this but it doesn't do what I want:
> split(cn1,c("Probe","Genes"))
$Genes
[1] "Genes" "bar" "Genes" "bar"
$Probe
[1] "Probe" "foo" "Probe" "foo"
Here's a function that you can use. Note that R vectors are 1-based so counting starts at 1 rather than 0.
findidx <- function(x) {
idx <- which(x=="Probe" & c(tail(x,-1),NA)=="Genes")
if (length(idx)>1) {
(idx[1]+2):(idx[2]-1)
} else {
NA # what to return if no match found
}
}
findidx(cn1)
# [1] 3 4
findidx(cn2)
# [1] 3 4 5
You could try between from data.table
indx <- between(cn1, 'Genes', 'Probe')
indx2 <- between(cn2, 'Genes', 'Probe')
which(cumsum(indx)==2)[-1]-1
#[1] 2 3
which(cumsum(indx2)==2)[-1]-1
#[1] 2 3 4
Related
Suppose this is my list of list (I would like to organize the result as my data contains more than 40 results and it is difficult for me to organize them manually).
s <- c(1,2,3)
ss <- c(4,5,6)
S <- list(s,ss)
h <- c(4,8,7)
hh <- c(0,3,4)
H <- list(h,hh)
HH <- list(S,H)
names1 <- c("First","Second")
lapply(setNames(HH, paste0(names1, '_Model')), function(x)
setNames(x, paste0('Res_', seq_along(x))))
#$First_Model
#$First_Model$Res_1
#[1] 1 2 3
#$First_Model$Res_2
#[1] 4 5 6
#$Second_Model
#$Second_Model$Res_1
#[1] 4 8 7
#$Second_Model$Res_2
#[1] 0 3 4
I would like to have the result similar to the following:
#$First_Model
#$First_Model$Res_1
#[1] 1 2 3
#$Second_Model
#$Second_Model$Res_1
#[1] 4 8 7
#$First_Model$Res_2
#[1] 4 5 6
#$Second_Model$Res_2
#[1] 0 3 4
The problem in question is how to rearrange the nested list from "Model No. > Results No." to "Results No. > Model No."
I was going for something similar to Wimpel's answer.
Res_no <- seq_along(HH[[1]]) # results elements
lapply(setNames(Res_no, paste0("Res_", Res_no)), function(x)
lapply(setNames(HH, paste0(names1, '_Model')), `[[`, x)
)
Output
#$Res_1
#$Res_1$First_Model
#[1] 1 2 3
#
#$Res_1$Second_Model
#[1] 4 8 7
#
#
#$Res_2
#$Res_2$First_Model
#[1] 4 5 6
#
#$Res_2$Second_Model
#[1] 0 3 4
The base of this solution is to extract the x-th element of the nested list (seen in the inner lapply() function of the code). You can do this with lapply or purrr:map, as described here.
The outer lapply() function lets you repeat it for all the "Results No."
Something like this perhaps?
# From your code, create a list L
L <- lapply(setNames(HH, paste0(names1, '_Model')), function(x)
setNames(x, paste0('Res_', seq_along(x))))
# get all x-th elements from the list, and add them to new list L2
L2 <- lapply( 1:length(L[[1]]), function(x) {
lapply(L, "[[", x)
})
# set names of L2
names(L2) <- names(L[[1]])
output
# $Res_1
# $Res_1$First_Model
# [1] 1 2 3
#
# $Res_1$Second_Model
# [1] 4 8 7
#
#
# $Res_2
# $Res_2$First_Model
# [1] 4 5 6
#
# $Res_2$Second_Model
# [1] 0 3 4
I want to remove certain vectors from a list. I have for example this:
a<-c(1,2,5)
b<-c(1,1,1)
c<-c(1,2,3,4)
d<-c(1,2,3,4,5)
exampleList<-list(a,b,c,d)
exampleList returns of course:
[[1]]
[1] 1 2 5
[[2]]
[1] 1 1 1
[[3]]
[1] 1 2 3 4
[[4]]
[1] 1 2 3 4 5
Is there a way to remove certain vectors from a list in R. I want to remove all vectors in the list exampleList which contain both 1 and 5(so not only vectors which contain 1 or 5, but both). Thanks in advance!
Use Filter:
filteredList <- Filter(function(v) !(1 %in% v & 5 %in% v), exampleList)
print(filteredList)
#> [[1]]
#> [1] 1 1 1
#>
#> [[2]]
#> [1] 1 2 3 4
Filter uses a functional style. The first argument you pass is a function that returns TRUE for an element you want to keep in the list, and FALSE for an element you want to remove from the list. The second argument is just the list itself.
We can use sapply on every list element and remove those elements where both the values 1 and 5 are present.
exampleList[!sapply(exampleList, function(x) any(x == 1) & any(x == 5))]
#[[1]]
#[1] 1 1 1
#[[2]]
#[1] 1 2 3 4
Here a solution with two steps:
exampleList<-list(a=c(1,2,5), b=c(1,1,1), c=c(1,2,3,4), d=c(1,2,3,4,5))
L <- lapply(exampleList, function(x) if (!all(c(1,5) %in% x)) x)
L[!sapply(L, is.null)]
# $b
# [1] 1 1 1
#
# $c
# [1] 1 2 3 4
Here is a one-step variant without any definition of a new function
exampleList[!apply(sapply(exampleList, '%in%', x=c(1,5)), 2, all)]
(... but it has two calls to apply-functions)
I want to remove part of the list where it is a complete set of the other part of the list. For example, B intersect A and E intersect C, therefore B and E should be removed.
MyList <- list(A=c(1,2,3,4,5), B=c(3,4,5), C=c(6,7,8,9), E=c(7,8))
MyList
$A
[1] 1 2 3 4 5
$B
[1] 3 4 5
$C
[1] 6 7 8 9
$E
[1] 7 8
MyListUnique <- RemoveSubElements(MyList)
MyListUnique
$A
[1] 1 2 3 4 5
$C
[1] 6 7 8 9
Any ideas ? Any know function to do it ?
As long as your data is not too huge, you can use an approach like the following:
# preparation
MyList <- MyList[order(lengths(MyList))]
idx <- vector("list", length(MyList))
# loop through list and compare with other (longer) list elements
for(i in seq_along(MyList)) {
idx[[i]] <- any(sapply(MyList[-seq_len(i)], function(x) all(MyList[[i]] %in% x)))
}
# subset the list
MyList[!unlist(idx)]
#$C
#[1] 6 7 8 9
#
#$A
#[1] 1 2 3 4 5
Similar to the other answer, but hopefully clearer, using a helper function and 2 sapplys.
#helper function to determine a proper subset - shortcuts to avoid setdiff calculation if they are equal
is.proper.subset <- function(x,y) !setequal(x,y) && length(setdiff(x,y))==0
#double loop over the list to find elements which are proper subsets of other elements
idx <- sapply(MyList, function(x) any(sapply(MyList, function(y) is.proper.subset(x,y))))
#filter out those that are proper subsets
MyList[!idx]
$A
[1] 1 2 3 4 5
$C
[1] 6 7 8 9
I like to report a specific column from every dataframe from a list of dataframes. Any ideas? This is my code:
# Create dissimilarity matrix
df.diss<-dist(t(df[,6:11]))
mds.table<-list() # empty list of values
for(i in 1:6){ # For Loop to iterate over a command in a function
a<-mds(pk.diss,ndim=i, type="ratio", verbose=TRUE,itmax=1000)
mds.table[[i]]<-a # Store output in empty list
}
Now here is where I'm having trouble. After storing the values, I'm unable to call a specific column from every dataframe from the list.
# This function should call every $stress column from each data frame.
lapply(mds.table, function(x){
mds.table[[x]]$stress
})
Thanks again!
you are very close:
set.seed(1)
l_df <- lapply(1:5, function(x){
data.frame(a = sample(1:5,5), b = sample(1:5,5))
})
lapply(l_df, function(x){
x[['a']]
})
[[1]]
[1] 2 5 4 3 1
[[2]]
[1] 2 1 3 4 5
[[3]]
[1] 5 1 2 4 3
[[4]]
[1] 3 5 2 1 4
[[5]]
[1] 5 3 4 2 1
I would like to convert nested list like this:
l <- list(A=list(a=list(1),b=list(2)),
B=list(cd=list(c=list(3,4,5),d=list(6,7,8)),e=list(c(9,10))))
into list
o <- list(A=c(1,2),A.a=1,A.b=2,B=c(3:10),
B.cd=c(3:8),B.cd.c=c(3:5),B.cd.d=c(6:8),B.e=c(9:10))
At each list level values from nested lists should be concatenated.
Clearly a case for a recursive function, but getting the return values to unlist properly is tricky. Here's a function that will do it; it doesn't get the names quite right but that's easily fixed afterwards.
unnest <- function(x) {
if(is.null(names(x))) {
list(unname(unlist(x)))
}
else {
c(list(all=unname(unlist(x))), do.call(c, lapply(x, unnest)))
}
}
Output from unnest(l) is
$all
[1] 1 2 3 4 5 6 7 8 9 10
$A.all
[1] 1 2
$A.a
[1] 1
$A.b
[1] 2
$B.all
[1] 3 4 5 6 7 8 9 10
$B.cd.all
[1] 3 4 5 6 7 8
$B.cd.c
[1] 3 4 5
$B.cd.d
[1] 6 7 8
$B.e
[1] 9 10
and can be massaged into your desired output with
out <- unnest(l)
names(out) <- sub("\\.*all", "", names(out))
out[-1]
To not recurse when there's only one element, try
unnest2 <- function(x) {
if(is.null(names(x)) | length(x)==1) {
list(unname(unlist(x)))
} else {
c(list(all=unname(unlist(x))), do.call(c, lapply(x, unnest2)))
}
}