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Access items in a list using R, merge after splitting, and apply

I am just starting out in the process of clearing strings with R and still recognized a ton. The part of the current project that I'm stuck (I think) is pretty much related to referring to the values ​​in the list. I am trying to use the split-apply-comb method and am having problems returning to the form of data that I want at the end.

I tried various permutations which(), unlist()and any()but just can't get what I'm looking for, so I'm sitting asking for your help. Thank.

I have a long list of lines ...

> dput(x) 

list(c("ACETATE", "RINGERS"), "ACETYLCYSTEINE", c("ACETYLSALICYLIC", 
"ACID"), "ADIRAMYCIN", c("ADRENALINE", "00003901"), "ALEMTUZUMAB")

> x

[[1]]
[1] "ACETATE" "RINGERS"

[[2]]
[1] "ACETYLCYSTEINE"

[[3]]
[1] "ACETYLSALICYLIC" "ACID"           

[[4]]
[1] "ADIRAMYCIN"

[[5]]
[1] "ADRENALINE" "00003901"  

[[6]]
[1] "ALEMTUZUMAB"

I wrote simple code to remove certain words from the list ...

bad.words <- c("ACID", "BASE", "HYDROCHOLIRDE", "STUFF")
a <- lapply(x, function(x) x %in% bad.words) 

> a

[[1]]
[1] FALSE FALSE

[[2]]
[1] FALSE

[[3]]
[1] FALSE  TRUE

[[4]]
[1] FALSE

[[5]]
[1] FALSE FALSE

[[6]]
[1] FALSE

I would like to be able to create zso that it is xwithout bad.words.

> z

[[1]]
[1] "ACETATE" "RINGERS"

[[2]]
[1] "ACETYLCYSTEINE"

[[3]]
[1] "ACETYLSALICYLIC"           

[[4]]
[1] "ADIRAMYCIN"

[[5]]
[1] "ADRENALINE" "00003901"  

[[6]]
[1] "ALEMTUZUMAB"
+4
1

:

lapply(x, function(z) z[! (z %in% bad.words)])


lapply(x, function(z) setdiff(z,bad.words))

.

+4

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