Avoid for loop when loop has an increment - r

In R, I try systematically to avoid "for" loops and use lapply() family instead.
But how to do so when an iteration contains an increment step ?
For example : is it possible to obtain the same result as below with a lapply approach ?
a <- c()
b <- c()
set.seed(1L) # required for reproducible data
for (i in 1:10){
a <- c(a, sample(c(0,1), 1))
b <- c(b, (paste(a, collapse = "-")))
}
data.frame(a, b)
> data.frame(a, b)
> a b
> 1 0 0
> 2 1 0-1
> 3 0 0-1-0
> 4 0 0-1-0-0
> 5 1 0-1-0-0-1
> 6 0 0-1-0-0-1-0
> 7 0 0-1-0-0-1-0-0
> 8 0 0-1-0-0-1-0-0-0
> 9 1 0-1-0-0-1-0-0-0-1
> 10 1 0-1-0-0-1-0-0-0-1-1
EDIT
My question was very badly redacted. The below new example is much more illustrative : is it anyway to use lapply family if each iteration is calculated from the previous one ?
a <- c()
b <- c()
for (i in 1:10){
a <- c(a, sample(c(0,1), 1))
b <- c(b, (paste(a, collapse = "-")))
}
data.frame(a, b)
> data.frame(a, b)
a b
1 0 0
2 1 0-1
3 0 0-1-0
4 1 0-1-0-1
5 1 0-1-0-1-1
6 1 0-1-0-1-1-1
7 1 0-1-0-1-1-1-1
8 0 0-1-0-1-1-1-1-0
9 1 0-1-0-1-1-1-1-0-1
10 1 0-1-0-1-1-1-1-0-1-1

For the sake of completeness, there is also the accumulate() function from the purrr package.
So, building on the answers of Sotos and ThomasIsCoding:
df <- data.frame(a = 1:10)
df$b <- purrr::accumulate(df$a, paste, sep = "-")
df
a b
1 1 1
2 2 1-2
3 3 1-2-3
4 4 1-2-3-4
5 5 1-2-3-4-5
6 6 1-2-3-4-5-6
7 7 1-2-3-4-5-6-7
8 8 1-2-3-4-5-6-7-8
9 9 1-2-3-4-5-6-7-8-9
10 10 1-2-3-4-5-6-7-8-9-10
The difference to Reduce() is
that accumulate() is a function verb on its own (no additional parameter accumulate = TRUE required)
and that additional arguments like sep = "-" can be passed on to the mapped function which may help to avoid the creation of an anonymous function.
EDIT
If I understand correctly OP's edit of the question, the OP is asking if a for loop which computes a result iteratively can be replaced by lapply().
This is difficult to answer for me. Here are some thoughts and observations:
First, accumulate() still will work:
set.seed(1L) # required for reproducible data
df <- data.frame(a = sample(0:1, 10L, TRUE))
df$b <- purrr::accumulate(df$a, paste, sep = "-")
df
a b
1 0 0
2 1 0-1
3 0 0-1-0
4 0 0-1-0-0
5 1 0-1-0-0-1
6 0 0-1-0-0-1-0
7 0 0-1-0-0-1-0-0
8 0 0-1-0-0-1-0-0-0
9 1 0-1-0-0-1-0-0-0-1
10 1 0-1-0-0-1-0-0-0-1-1
This is possible because the computation of a can be pulled out off the loop as it does not depend on b.
IMHO, accumulate() and Reduce() do what the OP is looking for but is not called lapply(): They take the result of the previous iteration and combine it with the actual value, for instance
Reduce(`+`, 1:3)
returns the sum of 1, 2, and 3 by iteratively computing (((0 + 1) + 2) + 3). This can be visualised by using the accumulate parameter
Reduce(`+`, 1:3, accumulate = TRUE)
[1] 1 3 6
Second, there is a major difference between a for loop and functions of the lapply() family: lapply(X, FUN, ...) requires a function FUN to be called on each element of X. So, scoping rules for functions apply.
When we transplant the body of the loop into an anonymous function within lapply()
a <- c()
b <- c()
set.seed(1L) # required for reproducible data
lapply(1:10, function(i) {
a <- c(a, sample(c(0,1), 1))
b <- c(b, (paste(a, collapse = "-")))
})
we get
[[1]]
[1] "0"
[[2]]
[1] "1"
[[3]]
[1] "0"
[[4]]
[1] "0"
[[5]]
[1] "1"
[[6]]
[1] "0"
[[7]]
[1] "0"
[[8]]
[1] "0"
[[9]]
[1] "1"
[[10]]
[1] "1"
data.frame(a, b)
data frame with 0 columns and 0 rows data.frame(a, b)
Due to the scoping rules, a and b inside the function are considered as local to the function. No reference is made to a and b defined outside of the function.
This can be fixed by global assignment using the global assignment operator <<-:
a <- c()
b <- c()
set.seed(1L) # required for reproducible data
lapply(1:10, function(i) {
a <<- c(a, sample(c(0,1), 1))
b <<- c(b, (paste(a, collapse = "-")))
})
data.frame(a, b)
a b
1 0 0
2 1 0-1
3 0 0-1-0
4 0 0-1-0-0
5 1 0-1-0-0-1
6 0 0-1-0-0-1-0
7 0 0-1-0-0-1-0-0
8 0 0-1-0-0-1-0-0-0
9 1 0-1-0-0-1-0-0-0-1
10 1 0-1-0-0-1-0-0-0-1-1
However, global assignment is considered bad programming practice and should be avoided, see, e.g., the 6th Circle of Patrick Burns' The R Inferno and many questions on SO.
Third, the way the loop is written grows vectors in the loop. This also is considered bad practice as it requires to copy the data over and over again which may slow down tremendously with increasing size. See, e.g., the 2nd Circle of Patrick Burns' The R Inferno.
However, the original code
a <- c()
b <- c()
set.seed(1L) # required for reproducible data
for (i in 1:10) {
a <- c(a, sample(c(0,1), 1))
b <- c(b, (paste(a, collapse = "-")))
}
data.frame(a, b)
can be re-written as
a <- integer(10)
b <- character(10)
set.seed(1L) # required for reproducible data
for (i in seq_along(a)) {
a[i] <- sample(c(0,1), 1)
b[i] <- if (i == 1L) a[1] else paste(b[i-1], a[i], sep = "-")
}
data.frame(a, b)
Here, vectors are pre-allocated with the required size to hold the result. Elements to update are identified by subscripting.
Calculation of b[i] still depends only the value of the previous iteration b[i-1] and the actual value a[i] as requested by the OP.

Another way is to use Reduce with accumulate = TRUE, i.e.
df$new <- do.call(rbind, Reduce(paste, split(df, seq(nrow(df))), accumulate = TRUE))
which gives,
a new
1 1 1
2 2 1 2
3 3 1 2 3
4 4 1 2 3 4
5 5 1 2 3 4 5
6 6 1 2 3 4 5 6
7 7 1 2 3 4 5 6 7
8 8 1 2 3 4 5 6 7 8
9 9 1 2 3 4 5 6 7 8 9
10 10 1 2 3 4 5 6 7 8 9 10

You can use sapply (lapply would work too but it returns a list) and iterate over every value of a in df and create a sequence and paste the value together.
df <- data.frame(a = 1:10)
df$b <- sapply(df$a, function(x) paste(seq(x), collapse = "-"))
df
# a b
#1 1 1
#2 2 1-2
#3 3 1-2-3
#4 4 1-2-3-4
#5 5 1-2-3-4-5
#6 6 1-2-3-4-5-6
#7 7 1-2-3-4-5-6-7
#8 8 1-2-3-4-5-6-7-8
#9 9 1-2-3-4-5-6-7-8-9
#10 10 1-2-3-4-5-6-7-8-9-10
If there could be non-numerical values in data on which we can not use seq like
df <- data.frame(a =letters[1:10])
In those case, we can use
df$b <- sapply(seq_along(df$a), function(x) paste(df$a[seq_len(x)], collapse = "-"))
df
# a b
#1 a a
#2 b a-b
#3 c a-b-c
#4 d a-b-c-d
#5 e a-b-c-d-e
#6 f a-b-c-d-e-f
#7 g a-b-c-d-e-f-g
#8 h a-b-c-d-e-f-g-h
#9 i a-b-c-d-e-f-g-h-i
#10 j a-b-c-d-e-f-g-h-i-j

Another way of using Reduce, different to the approach by #Sotos
df$b <- Reduce(function(...) paste(...,sep = "-"), df$a, accumulate = T)
such that
> df
a b
1 1 1
2 2 1-2
3 3 1-2-3
4 4 1-2-3-4
5 5 1-2-3-4-5
6 6 1-2-3-4-5-6
7 7 1-2-3-4-5-6-7
8 8 1-2-3-4-5-6-7-8
9 9 1-2-3-4-5-6-7-8-9
10 10 1-2-3-4-5-6-7-8-9-10

Related

Getting the values in a matrix with row and column names stored in a dataframe [duplicate]

I have a 2D matrix mat with 500 rows × 335 columns, and a data.frame dat with 120425 rows. The data.frame dat has two columns I and J, which are integers to index the row, column from mat. I would like to add the values from mat to the rows of dat.
Here is my conceptual fail:
> dat$matval <- mat[dat$I, dat$J]
Error: cannot allocate vector of length 1617278737
(I am using R 2.13.1 on Win32). Digging a bit deeper, I see that I'm misusing matrix indexing, as it appears that I'm only getting a sub-matrix of mat, and not a single-dimension array of values as I expected, i.e.:
> str(mat[dat$I[1:100], dat$J[1:100]])
int [1:100, 1:100] 20 1 1 1 20 1 1 1 1 1 ...
I was expecting something like int [1:100] 20 1 1 1 20 1 1 1 1 1 .... What is the correct way to index a 2D matrix using indices of row, column to get the values?
Almost. Needs to be offered to "[" as a two column matrix:
dat$matval <- mat[ cbind(dat$I, dat$J) ] # should do it.
There is a caveat: Although this also works for dataframes, they are first coerced to matrix-class and if any are non-numeric, the entire matrix becomes the "lowest denominator" class.
Using a matrix to index as DWin suggests is of course much cleaner, but for some strange reason doing it manually using 1-D indices is actually slightly faster:
# Huge sample data
mat <- matrix(sin(1:1e7), ncol=1000)
dat <- data.frame(I=sample.int(nrow(mat), 1e7, rep=T),
J=sample.int(ncol(mat), 1e7, rep=T))
system.time( x <- mat[cbind(dat$I, dat$J)] ) # 0.51 seconds
system.time( mat[dat$I + (dat$J-1L)*nrow(mat)] ) # 0.44 seconds
The dat$I + (dat$J-1L)*nrow(m) part turns the 2-D indices into 1-D ones. The 1L is the way to specify an integer instead of a double value. This avoids some coercions.
...I also tried gsk3's apply-based solution. It's almost 500x slower though:
system.time( apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat ) ) # 212
Here's a one-liner using apply's row-based operations
> dat <- as.data.frame(matrix(rep(seq(4),4),ncol=2))
> colnames(dat) <- c('I','J')
> dat
I J
1 1 1
2 2 2
3 3 3
4 4 4
5 1 1
6 2 2
7 3 3
8 4 4
> mat <- matrix(seq(16),ncol=4)
> mat
[,1] [,2] [,3] [,4]
[1,] 1 5 9 13
[2,] 2 6 10 14
[3,] 3 7 11 15
[4,] 4 8 12 16
> dat$K <- apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat )
> dat
I J K
1 1 1 1
2 2 2 6
3 3 3 11
4 4 4 16
5 1 1 1
6 2 2 6
7 3 3 11
8 4 4 16
n <- 10
mat <- cor(matrix(rnorm(n*n),n,n))
ix <- matrix(NA,n*(n-1)/2,2)
k<-0
for (i in 1:(n-1)){
for (j in (i+1):n){
k <- k+1
ix[k,1]<-i
ix[k,2]<-j
}
}
o <- rep(NA,nrow(ix))
o <- mat[ix]
out <- cbind(ix,o)

Subsetting rows and columns with given indices [duplicate]

I have a 2D matrix mat with 500 rows × 335 columns, and a data.frame dat with 120425 rows. The data.frame dat has two columns I and J, which are integers to index the row, column from mat. I would like to add the values from mat to the rows of dat.
Here is my conceptual fail:
> dat$matval <- mat[dat$I, dat$J]
Error: cannot allocate vector of length 1617278737
(I am using R 2.13.1 on Win32). Digging a bit deeper, I see that I'm misusing matrix indexing, as it appears that I'm only getting a sub-matrix of mat, and not a single-dimension array of values as I expected, i.e.:
> str(mat[dat$I[1:100], dat$J[1:100]])
int [1:100, 1:100] 20 1 1 1 20 1 1 1 1 1 ...
I was expecting something like int [1:100] 20 1 1 1 20 1 1 1 1 1 .... What is the correct way to index a 2D matrix using indices of row, column to get the values?
Almost. Needs to be offered to "[" as a two column matrix:
dat$matval <- mat[ cbind(dat$I, dat$J) ] # should do it.
There is a caveat: Although this also works for dataframes, they are first coerced to matrix-class and if any are non-numeric, the entire matrix becomes the "lowest denominator" class.
Using a matrix to index as DWin suggests is of course much cleaner, but for some strange reason doing it manually using 1-D indices is actually slightly faster:
# Huge sample data
mat <- matrix(sin(1:1e7), ncol=1000)
dat <- data.frame(I=sample.int(nrow(mat), 1e7, rep=T),
J=sample.int(ncol(mat), 1e7, rep=T))
system.time( x <- mat[cbind(dat$I, dat$J)] ) # 0.51 seconds
system.time( mat[dat$I + (dat$J-1L)*nrow(mat)] ) # 0.44 seconds
The dat$I + (dat$J-1L)*nrow(m) part turns the 2-D indices into 1-D ones. The 1L is the way to specify an integer instead of a double value. This avoids some coercions.
...I also tried gsk3's apply-based solution. It's almost 500x slower though:
system.time( apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat ) ) # 212
Here's a one-liner using apply's row-based operations
> dat <- as.data.frame(matrix(rep(seq(4),4),ncol=2))
> colnames(dat) <- c('I','J')
> dat
I J
1 1 1
2 2 2
3 3 3
4 4 4
5 1 1
6 2 2
7 3 3
8 4 4
> mat <- matrix(seq(16),ncol=4)
> mat
[,1] [,2] [,3] [,4]
[1,] 1 5 9 13
[2,] 2 6 10 14
[3,] 3 7 11 15
[4,] 4 8 12 16
> dat$K <- apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat )
> dat
I J K
1 1 1 1
2 2 2 6
3 3 3 11
4 4 4 16
5 1 1 1
6 2 2 6
7 3 3 11
8 4 4 16
n <- 10
mat <- cor(matrix(rnorm(n*n),n,n))
ix <- matrix(NA,n*(n-1)/2,2)
k<-0
for (i in 1:(n-1)){
for (j in (i+1):n){
k <- k+1
ix[k,1]<-i
ix[k,2]<-j
}
}
o <- rep(NA,nrow(ix))
o <- mat[ix]
out <- cbind(ix,o)

R: How to attribute a specific value to specific elements of a matrix [duplicate]

I have a 2D matrix mat with 500 rows × 335 columns, and a data.frame dat with 120425 rows. The data.frame dat has two columns I and J, which are integers to index the row, column from mat. I would like to add the values from mat to the rows of dat.
Here is my conceptual fail:
> dat$matval <- mat[dat$I, dat$J]
Error: cannot allocate vector of length 1617278737
(I am using R 2.13.1 on Win32). Digging a bit deeper, I see that I'm misusing matrix indexing, as it appears that I'm only getting a sub-matrix of mat, and not a single-dimension array of values as I expected, i.e.:
> str(mat[dat$I[1:100], dat$J[1:100]])
int [1:100, 1:100] 20 1 1 1 20 1 1 1 1 1 ...
I was expecting something like int [1:100] 20 1 1 1 20 1 1 1 1 1 .... What is the correct way to index a 2D matrix using indices of row, column to get the values?
Almost. Needs to be offered to "[" as a two column matrix:
dat$matval <- mat[ cbind(dat$I, dat$J) ] # should do it.
There is a caveat: Although this also works for dataframes, they are first coerced to matrix-class and if any are non-numeric, the entire matrix becomes the "lowest denominator" class.
Using a matrix to index as DWin suggests is of course much cleaner, but for some strange reason doing it manually using 1-D indices is actually slightly faster:
# Huge sample data
mat <- matrix(sin(1:1e7), ncol=1000)
dat <- data.frame(I=sample.int(nrow(mat), 1e7, rep=T),
J=sample.int(ncol(mat), 1e7, rep=T))
system.time( x <- mat[cbind(dat$I, dat$J)] ) # 0.51 seconds
system.time( mat[dat$I + (dat$J-1L)*nrow(mat)] ) # 0.44 seconds
The dat$I + (dat$J-1L)*nrow(m) part turns the 2-D indices into 1-D ones. The 1L is the way to specify an integer instead of a double value. This avoids some coercions.
...I also tried gsk3's apply-based solution. It's almost 500x slower though:
system.time( apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat ) ) # 212
Here's a one-liner using apply's row-based operations
> dat <- as.data.frame(matrix(rep(seq(4),4),ncol=2))
> colnames(dat) <- c('I','J')
> dat
I J
1 1 1
2 2 2
3 3 3
4 4 4
5 1 1
6 2 2
7 3 3
8 4 4
> mat <- matrix(seq(16),ncol=4)
> mat
[,1] [,2] [,3] [,4]
[1,] 1 5 9 13
[2,] 2 6 10 14
[3,] 3 7 11 15
[4,] 4 8 12 16
> dat$K <- apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat )
> dat
I J K
1 1 1 1
2 2 2 6
3 3 3 11
4 4 4 16
5 1 1 1
6 2 2 6
7 3 3 11
8 4 4 16
n <- 10
mat <- cor(matrix(rnorm(n*n),n,n))
ix <- matrix(NA,n*(n-1)/2,2)
k<-0
for (i in 1:(n-1)){
for (j in (i+1):n){
k <- k+1
ix[k,1]<-i
ix[k,2]<-j
}
}
o <- rep(NA,nrow(ix))
o <- mat[ix]
out <- cbind(ix,o)

How to squeeze in missing values into a vector

Let me try to make this question as general as possible.
Let's say I have two variables a and b.
a <- as.integer(runif(20, min = 0, max = 10))
a <- as.data.frame(a)
b <- as.data.frame(a[c(-7, -11, -15),])
So b has 17 observations and is a subset of a which has 20 observations.
My question is the following: how I would use these two variables to generate a third variable c which like a has 20 observations but for which observations 7, 11 and 15 are missing, and for which the other observations are identical to b but in the order of a?
Or to put it somewhat differently: how could I squeeze in these missing observations into variable b at locations 7, 11 and 15?
It seems pretty straightforward (and it probably is) but I have been not getting this to work for a bit too long now.
1) loop Try this loop:
# test data
set.seed(123) # for reproducibility
a <- as.integer(runif(20, min = 0, max = 10))
a <- as.data.frame(a)
b <- as.data.frame(a[c(-7, -11, -15),])
# lets work with vectors
A <- a[[1]]
B <- b[[1]]
j <- 1
C <- A
for(i in seq_along(A)) if (A[i] == B[j]) j <- j+1 else C[i] <- NA
which gives:
> C
[1] 2 7 4 8 9 0 NA 8 5 4 NA 4 6 5 NA 8 2 0 3 9
2) Reduce Here is a loop-free version:
f <- function(j, a) j + (a == B[j])
r <- Reduce(f, A, acc = TRUE)
ifelse(duplicated(r), NA, A)
giving:
[1] 2 7 4 8 9 0 NA 8 5 4 NA 4 6 5 NA 8 2 0 3 9
3) dtw. Using dtw in the package of the same name we can get a compact loop-free one-liner:
library(dtw)
ifelse(duplicated(dtw(A, B)$index2), NA, A)
giving:
[1] 2 7 4 8 9 0 NA 8 5 4 NA 4 6 5 NA 8 2 0 3 9
REVISED Added additional solutions.
Here's a more complicated way of doing it, using the Levenshtein distance algorithm, that does a better job on more complicated examples (it also seemed faster in a couple of larger tests I tried):
# using same data as G. Grothendieck:
set.seed(123) # for reproducibility
a <- as.integer(runif(20, min = 0, max = 10))
a <- as.data.frame(a)
b <- as.data.frame(a[c(-7, -11, -15),])
A = a[[1]]
B = b[[1]]
# compute the transformation between the two, assigning infinite weight to
# insertion and substitution
# using +1 here because the integers fed to intToUtf8 have to be larger than 0
# could also adjust the range more dynamically based on A and B
transf = attr(adist(intToUtf8(A+1), intToUtf8(B+1),
costs = c(Inf,1,Inf), counts = TRUE), 'trafos')
C = A
C[substring(transf, 1:nchar(transf), 1:nchar(transf)) == "D"] <- NA
#[1] 2 7 4 8 9 0 NA 8 5 4 NA 4 6 5 NA 8 2 0 3 9
More complex matching example (where the greedy algorithm would perform poorly):
A = c(1,1,2,2,1,1,1,2,2,2)
B = c(1,1,1,2,2,2)
transf = attr(adist(intToUtf8(A), intToUtf8(B),
costs = c(Inf,1,Inf), counts = TRUE), 'trafos')
C = A
C[substring(transf, 1:nchar(transf), 1:nchar(transf)) == "D"] <- NA
#[1] NA NA NA NA 1 1 1 2 2 2
# the greedy algorithm would return this instead:
#[1] 1 1 NA NA 1 NA NA 2 2 2
The data frame version, which isn't terribly different from G.'s above.
(Assumes a,b setup as above).
j <- 1
c <- a
for (i in (seq_along(a[,1]))) {
if (a[i,1]==b[j,1]) {
j <- j+1
} else
{
c[i,1] <- NA
}
}

insert elements in a vector in R

I have a vector in R,
a = c(2,3,4,9,10,2,4,19)
let us say I want to efficiently insert the following vectors, b, and c,
b = c(2,1)
d = c(0,1)
right after the 3rd and 7th positions (the "4" entries), resulting in,
e = c(2,3,4,2,1,9,10,2,4,0,1,19)
How would I do this efficiently in R, without recursively using cbind or so.
I found a package R.basic but its not part of CRAN packages so I thought about using a supported version.
Try this:
result <- vector("list",5)
result[c(TRUE,FALSE)] <- split(a, cumsum(seq_along(a) %in% (c(3,7)+1)))
result[c(FALSE,TRUE)] <- list(b,d)
f <- unlist(result)
identical(f, e)
#[1] TRUE
EDIT: generalization to arbitrary number of insertions is straightforward:
insert.at <- function(a, pos, ...){
dots <- list(...)
stopifnot(length(dots)==length(pos))
result <- vector("list",2*length(pos)+1)
result[c(TRUE,FALSE)] <- split(a, cumsum(seq_along(a) %in% (pos+1)))
result[c(FALSE,TRUE)] <- dots
unlist(result)
}
> insert.at(a, c(3,7), b, d)
[1] 2 3 4 2 1 9 10 2 4 0 1 19
> insert.at(1:10, c(4,7,9), 11, 12, 13)
[1] 1 2 3 4 11 5 6 7 12 8 9 13 10
> insert.at(1:10, c(4,7,9), 11, 12)
Error: length(dots) == length(pos) is not TRUE
Note the bonus error checking if the number of positions and insertions do not match.
You can use the following function,
ins(a, list(b, d), pos=c(3, 7))
# [1] 2 3 4 2 1 9 10 2 4 0 1 4 19
where:
ins <- function(a, to.insert=list(), pos=c()) {
c(a[seq(pos[1])],
to.insert[[1]],
a[seq(pos[1]+1, pos[2])],
to.insert[[2]],
a[seq(pos[2], length(a))]
)
}
Here's another function, using Ricardo's syntax, Ferdinand's split and #Arun's interleaving trick from another question:
ins2 <- function(a,bs,pos){
as <- split(a,cumsum(seq(a)%in%(pos+1)))
idx <- order(c(seq_along(as),seq_along(bs)))
unlist(c(as,bs)[idx])
}
The advantage is that this should extend to more insertions. However, it may produce weird output when passed invalid arguments, e.g., with any(pos > length(a)) or length(bs)!=length(pos).
You can change the last line to unname(unlist(... if you don't want a's items named.
The straightforward approach:
b.pos <- 3
d.pos <- 7
c(a[1:b.pos],b,a[(b.pos+1):d.pos],d,a[(d.pos+1):length(a)])
[1] 2 3 4 2 1 9 10 2 4 0 1 19
Note the importance of parenthesis for the boundaries of the : operator.
After using Ferdinand's function, I tried to write my own and surprisingly it is far more efficient.
Here's mine :
insertElems = function(vect, pos, elems) {
l = length(vect)
j = 0
for (i in 1:length(pos)){
if (pos[i]==1)
vect = c(elems[j+1], vect)
else if (pos[i] == length(vect)+1)
vect = c(vect, elems[j+1])
else
vect = c(vect[1:(pos[i]-1+j)], elems[j+1], vect[(pos[i]+j):(l+j)])
j = j+1
}
return(vect)
}
tmp = c(seq(1:5))
insertElems(tmp, c(2,4,5), c(NA,NA,NA))
# [1] 1 NA 2 3 NA 4 NA 5
insert.at(tmp, c(2,4,5), c(NA,NA,NA))
# [1] 1 NA 2 3 NA 4 NA 5
And there's the benchmark result :
> microbenchmark(insertElems(tmp, c(2,4,5), c(NA,NA,NA)), insert.at(tmp, c(2,4,5), c(NA,NA,NA)), times = 10000)
Unit: microseconds
expr min lq mean median uq max neval
insertElems(tmp, c(2, 4, 5), c(NA, NA, NA)) 9.660 11.472 13.44247 12.68 13.585 1630.421 10000
insert.at(tmp, c(2, 4, 5), c(NA, NA, NA)) 58.866 62.791 70.36281 64.30 67.923 2475.366 10000
my code works even better for some cases :
> insert.at(tmp, c(1,4,5), c(NA,NA,NA))
# [1] 1 2 3 NA 4 NA 5 NA 1 2 3
# Warning message:
# In result[c(TRUE, FALSE)] <- split(a, cumsum(seq_along(a) %in% (pos))) :
# number of items to replace is not a multiple of replacement length
> insertElems(tmp, c(1,4,5), c(NA,NA,NA))
# [1] NA 1 2 3 NA 4 NA 5
Here's an alternative that uses append. It's fine for small vectors, but I can't imagine it being efficient for large vectors since a new vector is created upon each iteration of the loop (which is, obviously, bad). The trick is to reverse the vector of things that need to be inserted to get append to insert them in the correct place relative to the original vector.
a = c(2,3,4,9,10,2,4,19)
b = c(2,1)
d = c(0,1)
pos <- c(3, 7)
z <- setNames(list(b, d), pos)
z <- z[order(names(z), decreasing=TRUE)]
for (i in seq_along(z)) {
a <- append(a, z[[i]], after = as.numeric(names(z)[[i]]))
}
a
# [1] 2 3 4 2 1 9 10 2 4 0 1 19

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