If you have a data.frame say, "DF", like this:
DF <- data.frame(x=1:9, y=letters[1:9])
And you've a vector z:
z <- c(0, 1, 0, 0, 0, 0, 0, 0, 0)
Note that the number of rows in your data.frame and the length of the vector has to be the same if you want to add the vector to a data.frame as a new column.
dim(DF) # dimensions of data.frame
# [1] 9 2
length(z) # length of vector
# [1] 9
Now, you can use cbind to get the new column as follows:
cbind(DF, z)
# x y z
# 1 1 a 0
# 2 2 b 1
# 3 3 c 0
# 4 4 d 0
# 5 5 e 0
# 6 6 f 0
# 7 7 g 0
# 8 8 h 0
# 9 9 i 0
If you have a vector whose length is not equal to that of the data.frame rows, then,
z <- c(0, 1, 0, 0, 0, 0, 0) # length is 7
cbind(DF, z)
# Error in data.frame(..., check.names = FALSE) :
# arguments imply differing number of rows: 9, 7
cbind'ing results in error due to unequal lengths. In this case, I could think of a couple ways to store this as a list.
First, you can keep your data.frame DF as such and create a list with its first element as the data.frame and the second as a vector as follows:
my_l <- list(d1 = DF, d2 = z)
# $d1
# x y
# 1 1 a
# 2 2 b
# 3 3 c
# 4 4 d
# 5 5 e
# 6 6 f
# 7 7 g
# 8 8 h
# 9 9 i
#
# $d2
# [1] 0 1 0 0 0 0 0
Alternatively, you can convert your data.frame to a list (a data.frame is internally a list) and create a list whose elements are all vectors as follows:
my_l <- c(as.list(DF), list(z=z))
# $x
# [1] 1 2 3 4 5 6 7 8 9
#
# $y
# [1] a b c d e f g h i
# Levels: a b c d e f g h i
#
# $z
# [1] 0 1 0 0 0 0 0
Note that as.list coerces a data.frame columns to a list with it's names the column names of the data.frame. We then create a new list z and then concatenate using the c operator.
Hope this betters your understanding a bit.
cbind, but I can't be more specific without a reproducible example of what objects you have.tfunctiondataframe$newcolumnname <- string. Your string actually looks a lot like a numeric vector which is entirely different.