Hi,
Try:
indx <- grep("Test",test_df[,1]) ##assuming that there is some pattern
res <- within(test_df[-indx,],titel <- rep(test_df$titel[indx],
diff(c(indx,nrow(test_df)+1))-1))
## If you need to change the class
res[] <- lapply(res,function(x) if(any(grepl("[[:alpha:]]",x))) as.character(x)
else as.numeric(as.character(x)))
##Using data.frame(cbind()), etc. creates
A.K.
On Saturday, November 16, 2013 11:14 AM, Stageexp <ron...@gmx.net> wrote:
Hi guys, I am a total newbie to R, so I hope this isn't a totally dumb
question. I have a dataframe with a title in one row and the corresponding
values in the next rows. Let's take this example:
test_df <- data.frame(cbind(titel = "", x = 4:5, y = 1:2))
test_df = rbind(cbind(titel="1.Test", x="", y=""), test_df,
cbind(titel="2.Test", x="", y=""), test_df, cbind(titel="3.Test", x="",
y=""), test_df)
test_df
titel x y
1 1.Test
2 4 1
3 5 2
4 2.Test
5 4 1
6 5 2
7 3.Test
8 4 1
9 5 2
What I want to have is:
titel x y
2 1.Test 4 1
3 1.Test 5 2
5 2.Test 4 1
6 2.Test 5 2
8 3.Test 4 1
9 3.Test 5 2
In my example, the title is in every third line, but in my real data there
is no pattern. Each title has at least one line but can have x lines.
I was able to solve my problem in a for loop with the following code:
test_df$titel <- as.character(test_df$titel)
for (i in 1:nrow(test_df))
{
if (nchar(test_df$titel[i])==0){
test_df$titel[i]=test_df$titel[i-1]
}
}
test_df <- subset(test_df,test_df$x!="")
The problem is, I have a lot of data and the for loop is obviously very
slow. Is there a more elegant way to achieve the same? I think I have to use
the apply function, but I don't know how to use it with just one column.
--
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Sent from the R help mailing list archive at Nabble.com.
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