find strings in data.frame to fill in new column - r

Question

I used dplyr on my data to create a subset of data like this:
dd <- data.frame(ID = c(700689L, 712607L, 712946L, 735907L, 735908L, 735910L, 735911L, 735912L, 735913L, 746929L, 747540L),
`1` = c("eg", NA, NA, "eg", "eg", NA, NA, NA, NA, "eg", NA),
`2` = c(NA, NA, NA, "sk", "lk", NA, NA, NA, NA, "eg", NA),
`3` = c(NA, NA, NA, "sk", "lk", NA, NA, NA, NA, NA, NA),
`4` = c(NA, NA, NA, "lk", "lk", NA, NA, NA, NA, NA, NA),
`5` = c(NA, NA, NA, "lk", "lk", NA, NA, NA, NA, NA, NA),
`6` = c(NA, NA, NA, "lk", "lk", NA, NA, NA, NA, NA, NA))
I now want to check every column except ID if it contains certain strings. In this example I want to create 1 column with "1" for every ID that contains a column with "eg" and "0" for the rest. Likewise one more column which tells me if there is either a "sk" or "lk" in the other columns. After that the old columns except ID can be removed from the data.frame
The difficult part for me is doing this with a dynamic number of columns, as my dplyr-subset will return different amounts of columns based on the specific case, but I need to check every one that is created in every case. I wanted to use unite first to put all strings together but I will have the same problem then: How can I unite all columns except the first ID one.
If this can be solved within dplyr it would be perfect but any working solution is appreciated.
The result should look like this:
result <- data.frame(ID = c(700689L, 712607L, 712946L, 735907L, 735908L, 735910L, 735911L, 735912L, 735913L, 746929L, 747540L),
with_eg = c(1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0),
with_sk_or_lk = c(0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0))

Answer 1

From your description, you want one column to check for "eg" and another column to check for both "lk" and "sk". If this is the case, then the following base R method will work.
dfNew <- cbind(id=dd[1],
eg=pmin(rowSums(dd[-1] == "eg", na.rm=TRUE), 1),
other=pmin(rowSums(dd[-1] == "sk" | dd[-1] == "lk", na.rm=TRUE), 1))
Here, the presence of "eg" is checked across the entire data.frame (except the id column) and a logical matrix is returned, rowSums adds the TRUE values across the rows, with na.rm removing the NAs, then pmin takes the minimum of the output of rowSums and 1, so that any elements with 2 are replaced by 1 and any values with 0 are preserved.
This same logic is applied to the construction of the "other" variable, except the presence of either "lk" or "sk" are checked in the initial logical matrix. Finally, data.frame returns a 3 column data.frame with the desired values.
This returns
dfNew
ID eg other
1 700689 1 0
2 712607 0 0
3 712946 0 0
4 735907 1 1
5 735908 1 1
6 735910 0 0
7 735911 0 0
8 735912 0 0
9 735913 0 0
10 746929 1 0
11 747540 0 0

Answer 2

Here is an admittedly hacky dplyr/purrr solution. Given that your IDs don't seem like they'll ever equal 'eg', 'sk', or 'lk', I haven't included anything to not search the ID column.
library(dplyr)
library(purrr)
dd %>%
split(.$ID) %>%
map_df(~ data_frame(
ID = .x$ID,
eg = ifelse(any(.x == 'eg', na.rm = TRUE), 1, 0),
other = ifelse(any(.x == 'lk' | .x == 'sk', na.rm = TRUE), 1, 0)
))