Simplify a code in R - r

Question

I have written a code that runs through the columns of a dataframe and returns TRUE if it has a number 1 in any of them, and sends the value to a vector with the same size as a column of the dataframe. I would like to know if there is a way to simplify the code snippet below, since I will have to repeat it for several numbers.
n1 <- (tab[, 2]==1| tab[, 3]==1 | tab[, 4]==1 | tab[, 5]==1 |
tab[, 6]==1 | tab[, 7]==1 | tab[, 8]==1 | tab[, 9]==1 |
tab[, 10]==1 | tab[, 11]==1 | tab[, 12]==1 | tab[, 13]==1 |
tab[, 14]==1 | tab[, 15]==1 | tab[, 16]==1)

Answer 1

One possible solution is the following: you search for == 1 numbers in the dataframe and then Reduce the rows of that with the | operator:
tab <- data.frame(a = 1:10, b = 2:11)
apply(tab == 1, 1, function(x) {
Reduce("|", x)
})
For this example it will give you the ouput of:
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
Or one even simpler solution is:
apply(tab, 1, function(x) {
any(x == 1)
})

Answer 2

The other comments and answers can work, but I suggest they are encouraging bad behavior when dealing with a data.frame. First and foremost is that apply and rowSums expect a matrix as the data, and will happily coerce to such if given a data.frame. If any of the data.frame columns are character, then all columns will be converted to character. Some operations may still work as expected (e.g., == 1 since it will effectively be == "1" ... though some rounding errors may cause undesired effects), but anything mathematic will not work.
As an example,
n <- 20
set.seed(2)
tab <- data.frame(
a = as.character(sample(n, replace = FALSE)),
b1 = sample(5, size = n, replace = TRUE),
b2 = sample(5, size = n, replace = TRUE),
stringsAsFactors = FALSE
)
str(tab)
# 'data.frame': 20 obs. of 3 variables:
# $ a : chr "4" "14" "11" "3" ...
# $ b1: int 4 2 5 1 2 3 1 2 5 1 ...
# $ b2: int 5 2 1 1 5 4 5 2 3 5 ...
apply(tab, 1, function(y) any(y == 1))
# [1] FALSE FALSE TRUE TRUE FALSE FALSE TRUE FALSE FALSE TRUE TRUE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE TRUE
apply(tab, 1, sum)
# Error in FUN(newX[, i], ...) : invalid 'type' (character) of argument
rowSums(tab == 1)
# [1] 0 0 1 2 0 0 1 0 0 1 2 2 0 0 0 0 0 1 0 1
rowSums(tab)
# Error in rowSums(tab) : 'x' must be numeric
There are some easy ways to deal with this. Given your example, it appears that columns 2:16 are numeric and the ones you are concerned about. If that's the case, then you can safely use either one of:
rowSums(tab[,2:16] == 1) # Frank's comment
apply(tab[,2:16], 1, function(y) any(y == 1)) # suggested by You-leee's answer
(the former being fairly specific, the latter can be extended to other functionality). If there's only one non-numeric, once can always do
rowSums(tab[,-1,drop=FALSE] == 1)
apply(tab[,-1,drop=FALSE], 1, function(y) any(y == 1))
A third technique is to determine at run time which columns to choose:
isnum <- sapply(tab, is.numeric)
Reduce(`|`, lapply(tab[isnum], function(y) any(y == 1)))
This was a little more complex, because the return from lapply is a list, but it still works fine. Realize that the use of isnum could be based on column names as well, using something like grepl. This method is fairly robust, too, in that it does not error if none of the columns match.