How to Declare a Vector of Zeros in R

How to declare a vector of zeros in R

You have several options

integer(3)
numeric(3)
rep(0, 3)
rep(0L, 3)

Create a Vector of Length N of the same numbers in R

rep(4.5, 100)

The function rep does the trick

Adding zeros in front of an vector

Something like this?

# create a vector
a <- rnorm(730)
# add the 0
a <- c(rep(0,730), a)

Then you could make a matrix:

m <- cbind(1:length(a), a)

Change zero to ones in vector if surrounded by less than five consecutive zeros

A possible solution with rle which does not change shorts sequences of zero's at the beginning or end of x:

# create the run length encoding
r <- rle(x)

# create an index of which zero's should be changed
i <- r$values == 0 & r$lengths < 5 & 
  c(tail(r$values, -1) == 1, FALSE) & 
  c(FALSE, head(r$values, -1) == 1)

# set the appropriate values to 1
r$values[i] <- 1

# use the inverse of rle to recreate the vector
inverse.rle(r)

which gives:

[1] 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1

Sum elements of a vector beween zeros in R

We can use tapply, creating groups with cumsum and take sum of each group.

new_x <- tapply(x, cumsum(x == 0), sum)
new_x

#  1   2   3   4   5   6   7   8   9  10 
#  0   0   0   0 138   0   0   0 156  14 

Since all numbers are positive we can ignore the ones with 0 and filter the one which has value greater than 0.

new_x[new_x > 0]
#  5   9  10 
#138 156  14 

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We can also follow the same logic with sapply and split

sapply(split(x, cumsum(x==0)), sum)

Create a matrix of zeros and ones from R

We can use model.matrix in base R

model.matrix(~ factor(id) - 1, data_toy)

-output

#   factor(id)1 factor(id)2 factor(id)3
#1           1           0           0
#2           1           0           0
#3           1           0           0
#4           0           1           0
#5           0           1           0
#6           0           0           1

Or use table

with(data_toy, table(seq_along(id), id))

How to define blocks of elements in a vector to zero and looping over the next blocks?

First, I define my vector and matrix.

# Data definitions
a <- c(1:6)
b <- matrix(1:36, nrow = 6, ncol = 6)

Next, I create a function that produces a list of your a values. It defaults to blocks of zeros that are length two.

# Define blocks of zeroes in vector 
define_block <- function(v, n = 2){
  # Check that v is a multiple of n
  if(length(v)%%n)warning("Vector not a multiple of block length")

  # Define blocks of zeroes
  lapply(1:(length(v)/n), function(x)replace(v, ((x-1)*n + 1):(x*n), 0))
}

Here, I apply the function:

# Create lists of blocks
block_list <- define_block(a)

which produces,

# [[1]]
# [1] 0 0 3 4 5 6
# 
# [[2]]
# [1] 1 2 0 0 5 6
# 
# [[3]]
# [1] 1 2 3 4 0 0

Finally, I run through block_list and perform your calculation and bind the results together using do.call with rbind.

# Run through block list and bind result into a matrix
do.call(rbind, lapply(block_list, function(x)x%*%b))


#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]   86  194  302  410  518  626
#[2,]   66  150  234  318  402  486
#[3,]   30   90  150  210  270  330

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