Subset based on list of strings using grepl()?

Question

I'm looking to do something seemingly very simple. I would like to subset a data frame in R using the grepl() command -- or something like it -- on several different phrases without constructing a loop.

For example, I'd like to pull out all the rows for anyone named Bob or Mary:

## example data frame:
tmp = structure(list(Name = structure(c(6L, 8L, 9L, 7L, 2L, 3L, 10L, 
1L, 5L, 4L), .Label = c("Alan", "Bob", "bob smith", "Frank", 
"John", "Mary Anne", "mary jane", "Mary Smith", "Potter, Mary", 
"smith, BOB"), class = "factor"), Age = c(31L, 23L, 23L, 55L, 
32L, 36L, 45L, 12L, 43L, 46L), Height = 1:10), .Names = c("Name", 
"Age", "Height"), class = "data.frame", row.names = c(NA, -10L
))

tmp

#           Name Age Height
#1     Mary Anne  31      1
#2    Mary Smith  23      2
#3  Potter, Mary  23      3
#4     mary jane  55      4
#5           Bob  32      5
#6     bob smith  36      6
#7    smith, BOB  45      7
#8          Alan  12      8
#9          John  43      9
#10        Frank  46     10

## this doesn't work
mynames=c('bob','mary')
tmp[grepl(mynames,tmp$Name,ignore.case=T),]

Any ideas would be helpful!

Answer 1

You can combine your mynames vector with the regular expression operator | and use grep.

tmp[grep(paste(mynames, collapse='|'), tmp$Name, ignore.case=TRUE),]

#           Name Age Height
# 1    Mary Anne  31      1
# 2   Mary Smith  23      2
# 3 Potter, Mary  23      3
# 4    mary jane  55      4
# 5          Bob  32      5
# 6    bob smith  36      6
# 7   smith, BOB  45      7