Reputation: 53
I have a dataframe, say
df = data.frame(x = c("a","a","b","b","b","c","d","t","c","b","t","c","t","a","a","b","d","t","t","c"),
y = c(2,4,5,2,6,2,4,5,2,6,2,4,5,2,6,2,4,5,2,6))
I want to remove only those rows in which one or multiple t
s are directly in between a d
and a c
, in all other cases I want to retain the cases. So for this example, I would like to remove the t
s on row 8, 18 and 19, but keep the others. I have over thousands of cases so doing this manually would be a true horror. Any help is very much appreciated.
Upvotes: 1
Views: 68
Reputation: 12559
Here is another solution with base R:
df = data.frame(x = c("a","a","b","b","b","c","d","t","c","b","t","c","t","a","a","b","d","t","t","c"),
y = c(2,4,5,2,6,2,4,5,2,6,2,4,5,2,6,2,4,5,2,6))
#
s <- paste0(df$x, collapse="")
L <- c(NA, NA)
while (TRUE) {
r <- regexec("dt+c", s)[[1]]
if (r[1]==-1) break
L <- rbind(L, c(pos=r[1]+1, length=attr(r, "match.length")-2))
s <- sub("d(t+)c", "x\\1x", s)
}
L <- L[-1,]
drop <- unlist(apply(L,1, function(x) seq(from=x[1], len=x[2])))
df[-drop, ]
# > drop
# 8 18 19
# > df[-drop, ]
# x y
# 1 a 2
# 2 a 4
# 3 b 5
# 4 b 2
# 5 b 6
# 6 c 2
# 7 d 4
# 9 c 2
# 10 b 6
# 11 t 2
# 12 c 4
# 13 t 5
# 14 a 2
# 15 a 6
# 16 b 2
# 17 d 4
# 20 c 6
With gregexpr()
it is shorter:
s <- paste0(df$x, collapse="")
g <- gregexpr("dt+c", s)[[1]]
L <- data.frame(pos=g+1, length=attr(g, "match.length")-2)
drop <- unlist(apply(L,1, function(x) seq(from=x[1], len=x[2])))
df[-drop, ]
Upvotes: 0
Reputation: 18425
This also works, by collapsing to a string, identifying groups of t's between d and c (or c and d - not sure whether you wanted this option as well), then working out where they are and removing the rows as appropriate.
df = data.frame(x=c("a","a","b","b","b","c","d","t","c","b","t","c","t","a","a","b","d","t","t","c"),
y=c(2,4,5,2,6,2,4,5,2,6,2,4,5,2,6,2,4,5,2,6),stringsAsFactors = FALSE)
dfs <- paste0(df$x,collapse="") #collapse to a string
dfs2 <- do.call(rbind,lapply(list(gregexpr("dt+c",dfs),gregexpr("ct+d",dfs)),
function(L) data.frame(x=L[[1]],y=attr(L[[1]],"match.length"))))
dfs2 <- dfs2[dfs2$x>0,] #remove any -1 values (if string not found)
drop <- unlist(mapply(function(a,b) (a+1):(a+b-2),dfs2$x,dfs2$y))
df2 <- df[-drop,]
Upvotes: 1
Reputation: 14360
One option would be to use rle
to get runs of the same string and then you can use an sapply
to check forward/backward and return all the positions you want to drop:
rle_vals <- rle(as.character(df$x))
drop <- unlist(sapply(2:length(rle_vals$values), #loop over values
function(i, vals, lengths) {
if(vals[i] == "t" & vals[i-1] == "d" & vals[i+1] == "c"){#Check if value is "t", previous is "d" and next is "c"
(sum(lengths[1:i-1]) + 1):sum(lengths[1:i]) #Get row #s
}
},vals = rle_vals$values, lengths = rle_vals$lengths))
drop
#[1] 8 18 19
df[-drop,]
# x y
#1 a 2
#2 a 4
#3 b 5
#4 b 2
#5 b 6
#6 c 2
#7 d 4
#9 c 2
#10 b 6
#11 t 2
#12 c 4
#13 t 5
#14 a 2
#15 a 6
#16 b 2
#17 d 4
#20 c 6
Upvotes: 1