retrieving names(table) following subset where only 1 observation - r

Question

My issue is that when I try to retrieve names(myresults) after subsetting a table I get null when the returned subset has only 1 result. Rather than returning a character vector of row names r returns an integer (in this case of 1).
Here is a table
head(tbl)
1 2 3 4 5 6
afford 0 1 0 0 0 0
app 0 0 0 1 0 0
back 0 1 0 0 0 0
cancel 0 0 0 0 1 0
charg 0 0 0 0 0 1
download 0 0 0 0 0 1
I have been subsetting the table within a loop to return a table for each group. If a term belongs to a group it has a value of 1:
for (i in 1:ncol(tbl)) {
t <- tbl[which(tbl[,i]==1),i]
nam <- names(t)
df <- as.data.frame(nam)
names(df) <- paste0("Cluster ",i)
print(kable(df))
}
This loop seems to work OK when there are more than one instance of a term returned by which(). But the group 4, which has only 1 term "app" gives me issues. Here's an example on group 3, which works as expected then on group 4, which does not:
> t <- tbl[which(tbl[,4]==1),4] # only 1 observation meets this criteria
> t
[1] 1
> t <- tbl[which(tbl[,3]==1),3] # 3 observations meet this criteria
> t
aword cat dog
1 1 1
So I can get names(t) for tbl[,3] where it has 3 returned instances but not for tbl[,4] which only has 1.
> t <- fintab[which(fintab[,4]==1),4]
> names(t)
NULL # expected "app"
> t <- fintab[which(fintab[,4]==1),4]
> names(t)
[1] "aword" "cat" "dog"
How can I get names(t) when I have only 1 instance returned like in the example?
Some further context following comment below:
> str(tbl)
'table' int [1:33, 1:6] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:33] "aword" "app" "cat" "dog" ...
..$ : chr [1:6] "1" "2" "3" "4" ...
>
and
> dput(tbl)
structure(c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
1L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L,
1L, 0L, 1L, 1L, 0L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 0L,
1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L,
1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L,
0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L,
0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L, 0L,
0L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 1L), .Dim = c(33L, 6L), .Dimnames = structure(list(
c("aword", "app", "back", "cancel", "charg", "download",
"enough", "expens", "get", "great", "just", "like", "love",
"cat", "dog", "bla", "month", "much", "need",
"never", "phone", "pleas", "blabla", "realli", "term", "sign",
"thank", "time", "triangle", "use", "want", "will", "work"), c("1",
"2", "3", "4", "5", "6")), .Names = c("", "")), class = "table")

Answer 1

As we are subsetting a single column, we get the logical index (tbl[,4] ==1 - no need to wrap with which unless there are NAs. In that case, the which remove those NAs) and use that to subset the column vector.
tbl[,4][tbl[,4]==1]
# app
# 1
tbl[,3][tbl[,3]==1]
# cat blabla time
# 1 1 1