Às 17:18 de 13/06/2023, javad bayat escreveu:
Dear Rui;
Hi. I used your codes, but it seems it didn't work for me.
pat <- c("_esmdes|_Des Section|0")
dim(data2)
[1] 281549 9
grep(pat, data2$Layer)
dim(data2)
[1] 281549 9
What does grep function do? I expected the function to remove 3 rows of the
dataframe.
I do not know the reason.
On Mon, Jun 12, 2023 at 5:16 PM Rui Barradas <ruipbarra...@sapo.pt> wrote:
Às 23:13 de 12/06/2023, javad bayat escreveu:
Dear Rui;
Many thanks for the email. I tried your codes and found that the length
of
the "Values" and "Names" vectors must be equal, otherwise the results
will
not be useful.
For some of the characters in the Layer column that I do not need to be
filled in the LU column, I used "NA".
But I need to delete some of the rows from the table as they are useless
for me. I tried this code to delete entire rows of the dataframe which
contained these three value in the Layer column: It gave me the following
error.
data3 = data2[-grep(c("_esmdes","_Des Section","0"), data2$Layer),]
Warning message:
In grep(c("_esmdes", "_Des Section", "0"), data2$Layer) :
argument 'pattern' has length > 1 and only the first element will
be
used
data3 = data2[!grepl(c("_esmdes","_Des Section","0"), data2$Layer),]
Warning message:
In grepl(c("_esmdes", "_Des Section", "0"), data2$Layer) :
argument 'pattern' has length > 1 and only the first element will be
used
How can I do this?
Sincerely
On Sun, Jun 11, 2023 at 5:03 PM Rui Barradas <ruipbarra...@sapo.pt>
wrote:
Às 13:18 de 11/06/2023, Rui Barradas escreveu:
Às 22:54 de 11/06/2023, javad bayat escreveu:
Dear Rui;
Many thanks for your email. I used one of your codes,
"data2$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.
Sorry, this is not clear. I mean
This assumes that you have the new values in a vector, the vector Names
below. The vector of values to be matched is created from the data.
Rui Barradas
Values <- sort(unique(data2$Layer))
Names <- c("Park", "Agri", "GS")
i <- match(data2$Layer, Values)
data2$LU <- Names[i]
Hope this helps,
Rui Barradas
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Hello,
Please cc the r-help list, R-Help is threaded and this can in the future
be helpful to others.
You can combine several patters like this:
pat <- c("_esmdes|_Des Section|0")
grep(pat, data2$Layer)
or, programatically,
pat <- paste(c("_esmdes","_Des Section","0"), collapse = "|")
Hope this helps,
Rui Barradas
Hello,
I only posted a corrected grep statement, the complete code should be
pat <- c("_esmdes|_Des Section|0")
data3 <- data2[-grep(pat, data2$Layer),]
Sorry for the confusion.
Hope this helps,
Rui Barradas
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
