R - adding a value to a new row if column name matches a string

Question

I have a very large data.table with the structure below:

variant ID1 ID2 ID3 ID4 .... ID80000
123     0    1   2   1         0
321     1    2   1   1         1
543     1    1   2   1         1
6542    1    0   0   1         0  
243     1    0   2   1         1
654     0    1   1   2         1 
342     1    2   1   2         1

I would like to add a row with a 0 or 1 dependant on whether the ID in a string is present in the above table.

Say I have the a list: a <- c("ID42", "ID4")

The desired output would be:

variant ID1 ID2 ID3 ID4 .... ID80000
123     0    1   2   1         0
321     1    2   1   1         1
543     1    1   2   1         1
6542    1    0   0   1         0  
243     1    0   2   1         1
654     0    1   1   2         1 
342     1    2   1   2         1
present 0    1   0   1         0

A bit like this question but in rows: Create new conditional column if string contains elements from list

I have tried:

df$present <- colnames(df) %in% id_list 

But I get an error as it saws there are too many rows.

Many thanks for your help

Would an ifelse be helpful here?

Answer 1

I created a similar structure as your df, and append a row to the dataframe.

df <- data.frame(matrix(ncol=10, nrow=0)) 
colnames(df) <- c("variant", rep(paste0("ID",2:ncol(df))))
a <- c("ID42", "ID4")
df[nrow(df) + 1,] <- c("present", ifelse(colnames(df) %in% a, 1, 0)[-1])
df

variant ID2 ID3 ID4 ID5 ID6 ID7 ID8 ID9 ID10
present   0   0   1   0   0   0   0   0    0

Answer 2

Thanks to @sashahafner above:

I got df['present',] <-colnames(df) %in% id_list  

to work, not sure it's the best solution but hey ho!