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I have a list like L (comes from a vector splitting).
L <- strsplit(c("1 5 9", "", "3 7 11", ""), " ")
# [[1]]
# [1] "1" "5" "9"
#
# [[2]]
# character(0)
#
# [[3]]
# [1] "3" "7" "11"
#
# [[4]]
# character(0)
When I do an ordinary rbind as follows, I'm losing all the character(0) rows.
do.call(rbind, L)
# [,1] [,2] [,3]
# [1,] "1" "5" "9"
# [2,] "3" "7" "11"
Do I always have to do a lapply like the following or have I missed something?
do.call(rbind, lapply(L, function(x)
if (length(x) == 0) rep("", 3) else x))
# [,1] [,2] [,3]
# [1,] "1" "5" "9"
# [2,] "" "" ""
# [3,] "3" "7" "11"
# [4,] "" "" ""
Base R answers are preferred.
794
Row bind using Rbind() & bind_rows() in R. Rbind() function in R row binds the data frames which is a simple joining or concatenation of two or more dataframes (tables) by row wise. In other words, Rbind in R appends or combines vector, matrix or data frame by rows.
To join two data frames (datasets) vertically, use the rbind function. The two data frames must have the same variables, but they do not have to be in the same order. If data frameA has variables that data frameB does not, then either: Delete the extra variables in data frameA or.
So, the column names in both the data frames must be the same if you want to use rbind().
do. call(rbind, ls) gives you the same output as Reduce(rbind, ls) . The later is less efficient, but it serves to show how you are iterating over the objects in ls rather than manipulating ls (which is a concatenated list of 2 lists) directly.
If you use lapply you don't have to worry about length so you can skip the rep part it will automatically be recycled across columns.
do.call(rbind, lapply(L, function(x) if (length(x) == 0) "" else x))
# [,1] [,2] [,3]
#[1,] "1" "5" "9"
#[2,] "" "" ""
#[3,] "3" "7" "11"
#[4,] "" "" ""
Another option using same logic as @NelsonGon we can replace the empty lists with blank and then rbind.
L[lengths(L) == 0] <- ""
do.call(rbind, L)
# [,1] [,2] [,3]
#[1,] "1" "5" "9"
#[2,] "" "" ""
#[3,] "3" "7" "11"
#[4,] "" "" ""
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