Removing columns from a data.table in R based on conditions

Question

How do I remove columns based on values in a data.table in R

If I have a data.table row,

dt = data.table("col1" = "a", "col2" = "b", "col3" = "c", 
"col4" = 'd', "col5" = "e", "col6" = 9, "col7" = 0, "col8" = 7,
"col9" = 0, "col10" = 99)

The first 5 columns are categorical, and the 6-10 columns are numerical. The numbers are repeated for all rows for the numerical columns.

I had two doubts

1. How do I remove the columns containing 0s? This column can vary based on inputs - ie sometimes col7 might be 0 sometimes col8 might be 0 etc

2. Once I remove the columns with 0 values, how do I concatenate the rest of the numbers into a single column - in this case, the new column will contain the number 9799

Is there a way to do this without removing the 0 value columns?

For the first part, I tried

cols_chosen = c("col6", "col7","col8","col9","col10")

condition = c(FALSE, dt[, lapply(.SD, function(x) sum(x)< 1), .SDcols = cols_chosen])

dt[, which(condition) := NULL]

While I am getting the correct value for the conditions (a list of 5 logical values), the last command is failing with the error

Error in which(condition) : argument to 'which' is not logical

I had taken the above statements from an earlier answer Remove columns of dataframe based on conditions in R

Answer 1

dt = data.table("col1" = "a", "col2" = "b", "col3" = "c", 
"col4" = 'd', "col5" = "e", "col6" = 9, "col7" = 0, "col8" = 7,
"col9" = 0, "col10" = 99)

not0 = function(x) is.numeric(x) && !anyNA(x) && all(x!=0)
dt[, .(
    ## your categorical columns
    col1, col2, col3, col4, col5,
    ## new column pasted from non-0 numeric columns
    new = as.numeric(paste0(unlist(.SD), collapse=""))
  ),
  ## this filters columns to be provided in .SD column subset
  .SDcols = not0,
  ## we group by each row so it will handle input of multiple rows
  by = .(row=seq_len(nrow(dt)))
  ][, row:=NULL ## this removes extra grouping column
    ][] ## this prints
#   col1 col2 col3 col4 col5  new
#1:    a    b    c    d    e 9799

---

Alternatively if you want to update in place existing table

is0 = function(x) is.numeric(x) && !anyNA(x) && all(x==0)
## remove columns that has 0
dt[, which(sapply(dt, is0)) := NULL]

## add new column
dt[, new := as.numeric(
    paste0(unlist(.SD), collapse="")
  ), .SDcols=is.numeric, by=.(row=seq_len(nrow(dt)))
  ][]
#   col1 col2 col3 col4 col5 col6 col8 col10  new
#1:    a    b    c    d    e    9    7    99 9799