This question already has an answer here:
More than one value for "each" argument in "rep" function?
(1 answer)
Closed 2 years ago.
Suppose I have a numeric vector v
v <- 1:5
I want to
rep
v[1] by v[1] times.
v[2] by v[2] times... and so on....
The desired output would be:
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
The following does not work. Got any ideas?
rep(v, each = function(x) v[x])
Many thanks in advance.
We can use rep on itself
rep(v, v)
If we want to specify the argument, use times
rep(v, times = v)
The each would not take anonymous function and it takes only a vector of length 1. According to ?rep
each - non-negative integer. Each element of x is repeated each times. Other inputs will be coerced to an integer or double vector and the first element taken. Treated as 1 if NA or invalid.
Related
this may be a simple question but I'm fairly new to R.
What I want to do is to perform some kind of addition on the indexes of a list, but once I get to a maximum value it goes back to the first value in that list and start over from there.
for example:
x <-2
data <- c(0,1,2,3,4,5,6,7,8,9,10,11)
data[x]
1
data[x+12]
1
data[x+13]
3
or something functionaly equivalent. In the end i want to be able to do something like
v=6
x=8
y=9
z=12
values <- c(v,x,y,z)
data <- c(0,1,2,3,4,5,6,7,8,9,10,11)
set <- c(data[values[1]],data[values[2]], data[values[3]],data[values[4]])
set
5 7 8 11
values <- values + 8
set
1 3 4 7
I've tried some stuff with additon and substraction to the lenght of my list but it does not work well on the lower numbers.
I hope this was a clear enough explanation,
thanks in advance!
We don't need a loop here as vectors can take vectors of length >= 1 as index
data[values]
#[1] 5 7 8 11
NOTE: Both the objects are vectors and not list
If we need to reset the index
values <- values + 8
ifelse(values > length(data), values - length(data) - 1, values)
#[1] 1 3 4 7
This question already has an answer here:
How to add two vectors WITHOUT repeating in R?
(1 answer)
Closed 5 years ago.
Very simple quesiton, very hard in R for a newbie like me.
x <- c(1,2,3,4)
y <- c(1,2,3)
The object lengths are different. So i cant perform, let's say, z <- x + y
What is the best way to approach arithemetic of different object length. Add a 0?
To do this programmatically, you could first put the vectors in a named vector that we can run a grouping function on.
z <- setNames(c(x, y), c(seq_along(x), seq_along(y)))
# 1 2 3 4 1 2 3
# 1 2 3 4 1 2 3
Now we can run a grouping function for the sum:
unname(tapply(z, names(z), sum))
# [1] 2 4 6 4
This question already has answers here:
The difference between bracket [ ] and double bracket [[ ]] for accessing the elements of a list or dataframe
(11 answers)
Closed 7 years ago.
I have the following data frame:
df <- data.frame(a=rep(1:3),b=rep(1:3),c=rep(4:6),d=rep(4:6))
df
a b c d
1 1 1 4 4
2 2 2 5 5
3 3 3 6 6
i would like to have a vector N which determines my window size so for thsi example i will set
N <- 1
I would like to split this dataframe into equal portions of N rows and store the 3 resulting dataframes into a list.
I have the following code:
groupMaker <- function(x, y) 0:(x-1) %/% y
testlist2 <- split(df, groupMaker(nrow(df), N))
The problem is that this code renames my column names by adding an X0. in front
result <- as.data.frame(testlist2[1])
result
X0.a X0.b X0.c X0.d
1 1 1 4 4
>
I would like a code that does the exact same thing but keeps the column names as they are. please keep in mind that my original data has a lot more than 3 rows so i need something that is applicable to a much larger dataframe.
To extract a list element, we can use [[. Also, as each list elements are data.frames, we don't need to explicitly call as.data.frame again.
testlist2[[1]]
We can also use gl to create the grouping variable.
split(df, as.numeric(gl(nrow(df), N, nrow(df))))
Let's say I have a vector a = (1,3,4).
I want to create new vector with integer numbers in range [1,length(a)]. But the i-th number should appear a[i] times.
For the vector a I want to get:
(1,2,2,2,3,3,3,3)
Would you explain me how to implement this operation without several messy concatenations?
You can try rep
rep(seq_along(a), a)
#[1] 1 2 2 2 3 3 3 3
data
a <- c(1,3,4)
This question already has answers here:
Create a Data Frame of Unequal Lengths
(6 answers)
Closed 9 years ago.
I have the following elementary issue in R.
I have a for (k in 1:x){...} cycle which produces numerical vectors whose length depends on k.
For each value of k I produce a single numerical vector.
I would like to collect them as rows of a data frame in R, if possible. In other words, I would like to introduce a data frame data s.t.
for (k in 1:x) {
data[k,] <- ...
}
where the dots represent the command producing the vector with length depending on k.
Unfortunately, as far as I know, the length of the rows of a dataframe in R is constant, as it is a list of vectors of equal length. I have already tried to complete each row with a suitable number of zeroes to arrive at a constant length (in this case equal to x). I would like to work "dynamically", instead.
I do not think that this issue is equivalent to merge vectors of different lengths in a dataframe; due to the if cycle, only 1 vector is known at each step.
Edit
A very easy example of what I mean. For each k, I would like to write the vector whose components are 1,2,...,k and store it as kth row of the dataframe data. In the above setting, I would write
for (k in 1:x) {
data[k,] <- seq(1,k,1)
}
As the length of seq(1,k,1) depends on k the code does not work.
You could consider using ldply from plyr here.
set.seed(123)
#k is the length of each result
k <- sample( 5 , 3 , repl = TRUE )
#[1] 2 4 3
# Make a list of vectors, each a sequence from 1:k
ll <- lapply( k , function(x) seq_len(x) )
#[[1]]
#[1] 1 2
#[[2]]
#[1] 1 2 3 4
#[[3]]
#[1] 1 2 3
# take our list and rbind it into a data.frame, filling in missing values with NA
ldply( ll , rbind)
# 1 2 3 4
#1 1 2 NA NA
#2 1 2 3 4
#3 1 2 3 NA