You can add a column to your data using various techniques. The quotes below come from the “Details” section of the relevant help text, [[.data.frame
.
Data frames can be indexed in several modes. When
[
and[[
are used with a single vector index (x[i]
orx[[i]]
), they index the data frame as if it were a list.
my.dataframe["new.col"] <- a.vector
my.dataframe[["new.col"]] <- a.vector
The data.frame method for
$
, treatsx
as a list
my.dataframe$new.col <- a.vector
When
[
and[[
are used with two indices (x[i, j]
andx[[i, j]]
) they act like indexing a matrix
my.dataframe[ , "new.col"] <- a.vector
Since the method for data.frame
assumes that if you don’t specify if you’re working with columns or rows, it will assume you mean columns.
For your example, this should work:
# make some fake data
your.df <- data.frame(no = c(1:4, 1:7, 1:5), h_freq = runif(16), h_freqsq = runif(16))
# find where one appears and
from <- which(your.df$no == 1)
to <- c((from-1)[-1], nrow(your.df)) # up to which point the sequence runs
# generate a sequence (len) and based on its length, repeat a consecutive number len times
get.seq <- mapply(from, to, 1:length(from), FUN = function(x, y, z) {
len <- length(seq(from = x[1], to = y[1]))
return(rep(z, times = len))
})
# when we unlist, we get a vector
your.df$group <- unlist(get.seq)
# and append it to your original data.frame. since this is
# designating a group, it makes sense to make it a factor
your.df$group <- as.factor(your.df$group)
no h_freq h_freqsq group
1 1 0.40998238 0.06463876 1
2 2 0.98086928 0.33093795 1
3 3 0.28908651 0.74077119 1
4 4 0.10476768 0.56784786 1
5 1 0.75478995 0.60479945 2
6 2 0.26974011 0.95231761 2
7 3 0.53676266 0.74370154 2
8 4 0.99784066 0.37499294 2
9 5 0.89771767 0.83467805 2
10 6 0.05363139 0.32066178 2
11 7 0.71741529 0.84572717 2
12 1 0.10654430 0.32917711 3
13 2 0.41971959 0.87155514 3
14 3 0.32432646 0.65789294 3
15 4 0.77896780 0.27599187 3
16 5 0.06100008 0.55399326 3