Às 22:54 de 11/06/2023, javad bayat escreveu:
Dear Rui;
Many thanks for your email. I used one of your codes,
"data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it works
correctly for me.
Actually I need to expand the codes so as to consider all "Levels" in the
"Layer" column. There are more than hundred levels in the Layer column.
If I use your provided code, I have to write it hundred of time as below:
data2$LU[which(data2$Layer == "Level 1")] <- "Park";
data2$LU[which(data2$Layer == "Level 2")] <- "Agri";
...
...
...
.
Is there any other way to expand the code in order to consider all of the
levels simultaneously? Like the below code:
data2$LU[which(data2$Layer == c("Level 1","Level 2", "Level 3", ...))] <-
c("Park", "Agri", "GS", ...)
Sincerely
On Sun, Jun 11, 2023 at 1:43 PM Rui Barradas <ruipbarra...@sapo.pt> wrote:
Às 21:05 de 11/06/2023, javad bayat escreveu:
Dear R users;
I am trying to fill a column based on a specific value in another column
of
a dataframe, but it seems there is a problem with the codes!
The "Layer" and the "LU" are two different columns of the dataframe.
How can I fix this?
Sincerely
for (i in 1:nrow(data2$Layer)){
if (data2$Layer == "Level 12") {
data2$LU == "Park"
}
}
Hello,
There are two bugs in your code,
1) the index i is not used in the loop
2) the assignment operator is `<-`, not `==`
Here is the loop corrected.
for (i in 1:nrow(data2$Layer)){
if (data2$Layer[i] == "Level 12") {
data2$LU[i] <- "Park"
}
}
But R is a vectorized language, the following two ways are the idiomac
ways of doing what you want to do.
i <- data2$Layer == "Level 12"
data2$LU[i] <- "Park"
# equivalent one-liner
data2$LU[data2$Layer == "Level 12"] <- "Park"
If there are NA's in data2$Layer it's probably safer to use ?which() in
the logical index, to have a numeric one.
i <- which(data2$Layer == "Level 12")
data2$LU[i] <- "Park"
# equivalent one-liner
data2$LU[which(data2$Layer == "Level 12")] <- "Park"
Hope this helps,
Rui Barradas
Hello,
You don't need to repeat the same instruction 100+ times, there is a way
of assigning all new LU values at the same time with match().
This assumes that you have the new values in a vector.
Values <- sort(unique(data2$Layer))
Names <- c("Park", "Agri", "GS")
i <- match(data2$Layer, Values)
data2$LU <- Names[i]
Hope this helps,
Rui Barradas
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