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I have the following data:
library(data.table)
d = data.table(a = c(1:3), b = c(2:4))
and would like to get this result (in a way that would work with arbitrary number of columns):
d[, c := paste0('a_', a, '_b_', b)]
d
# a b c
#1: 1 2 a_1_b_2
#2: 2 3 a_2_b_3
#3: 3 4 a_3_b_4
The following works, but I'm hoping to find something shorter and more legible.
d = data.table(a = c(1:3), b = c(2:4))
d[, c := apply(mapply(paste, names(.SD), .SD, MoreArgs = list(sep = "_")),
1, paste, collapse = "_")]
549
Next, we can use the group_by and mutate functions of the dplyr package to assign a unique ID number to each group of identical values in a column (i.e. x1): By running the previous R programming syntax we have created the tibble shown in the previous RStudio console output.
In this example, I’ll demonstrate how to create a unique ID column by group using the dplyr package. First, we need to install and load the dplyr package: Next, we can use the group_by and mutate functions of the dplyr package to assign a unique ID number to each group of identical values in a column (i.e. x1):
Now we will use the DataFrame.iterrows () function to iterate over each of the row of the given Dataframe and construct a list out of the data of each row. As we can see in the output, we have successfully extracted each row of the given dataframe into a list.
And to begin with your Machine Learning Journey, join the Machine Learning - Basic Level Course Now we will use the DataFrame.iterrows () function to iterate over each of the row of the given Dataframe and construct a list out of the data of each row.
one way, only slightly cleaner:
d[, c := apply(d, 1, function(x) paste(names(d), x, sep="_", collapse="_")) ]
a b c
1: 1 2 a_1_b_2
2: 2 3 a_2_b_3
3: 3 4 a_3_b_4
Here is an approach using do.call('paste'), but requiring only a single call to paste
I will benchmark on a situtation where the columns are integers (as this seems a more sensible test case
N <- 1e4
d <- setnames(as.data.table(replicate(5, sample(N), simplify = FALSE)), letters[seq_len(5)])
f5 <- function(d){
l <- length(d)
o <- c(1L, l + 1L) + rep_len(seq_len(l) -1L, 2L * l)
do.call('paste',c((c(as.list(names(d)),d))[o],sep='_'))}
microbenchmark(f1(d), f2(d),f5(d))
Unit: milliseconds
expr min lq median uq max neval
f1(d) 41.51040 43.88348 44.60718 45.29426 52.83682 100
f2(d) 193.94656 207.20362 210.88062 216.31977 252.11668 100
f5(d) 30.73359 31.80593 32.09787 32.64103 45.68245 100
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