Sorry, that should be
temp.tables <- temp.tables[-(31:32),]
Rui Barradas
Em 28-03-2014 10:58, Rui Barradas escreveu:
Hello,
Maybe the following will help.
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, which = 3)
st
Hello,
Maybe the following will help.
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, which = 3)
str(temp.tables)
temp.tables <- temp.tables[-31,]
temp.tables[,1:10] <- lapply(temp.tables[,1:10], function(x)
as.numeric
Hello,
You are right, the "which" option avoid the selection in my example.
Also, I think ?"[[" might be helpful for you.
Regards.
Pascal
On Fri, Mar 28, 2014 at 3:02 PM, Bill wrote:
> Hi. I found that this works:
> tt=readHTMLTable(url,header=TRUE, as.data.frame=TRUE,which=c(3))
> tt[1,]
> I
Hello,
Your example leads to error. Anyway:
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, header=TRUE)
temp.tables <- temp.tables[[3]]
temp.tables <- temp.tables[1:30,]
HTH,
Pascal
On Fri, Mar 28, 2014 at 2:42 PM, Bill
Hi. I ran some code and I am trying to access the table with the data in
it. I want in particular to delete the 31st row for example
Monthly means and totals:
or remove the names of the columns (or change them). DayT TM Tm
SLPHPP VVV VM VG RA SN TS
Try
my.data.frame[my.data.frame$Buy==1 | my.data.frame$Sell ==1, ]
or
subset(my.data.frame,buy == 1 | sell == 1)
Then take a look at the help page for || i.e., help("||")
to see what you did wrong.
- Phil Spector
I have a data frame with the following columns:
Date Price Buy Sell
The Buy and Sell variables are binary. They are either zero or 1.
Additionally, they are mutually exclusive. Either Buy is 1, Sell is 1 or
they are both 0. But they are never both 1 for a given observation.
I want to see observa
Thanks everyone... the as.character(fileLines[1][1]) solution worked well...
Factors??? the treatment is so far away from what I know.
Cool though...
On Mon, Dec 28, 2009 at 8:55 PM, David Winsemius wrote:
>
> On Dec 28, 2009, at 5:14 PM, Nick Torenvliet wrote:
>
> Consider the following
On Dec 28, 2009, at 5:14 PM, Nick Torenvliet wrote:
Consider the following
fileLines
V1 V2V3V4 V5 V6V7 V8
1 AB 20091224 156.0 156.0 154.00 154.0055 1198
2 AB.C 20091224 156.0 156.0 156.00 156.00 0 0
3 ABF10 20091224 156.0 156.0 156
This is a result of how R treats factors.
There's more than one way to do what I think you're asking for.
I've constructed a smaller version of your data frame to illustrate one
quick way if that's all you need:
> smdat<-
> data.frame(V1=c("AB","AB.C","ABF10"),V2=rep("20091224",3),V3=rep(156.0
?read.csv
On Mon, Dec 28, 2009 at 5:31 PM, Nick Torenvliet
wrote:
> I have the following code
>
> fileList <-list.files(path = ".", pattern = "[^a-z].txt$", all.files =
> FALSE, full.names = FALSE, recursive = FALSE, ignore.case = FALSE)
> for (x in 1:length(fileList)){
>fileLines <- data.fra
Hi Nick,
Your first column is being stored as a factor. You can either take
care when creating the matrix to keep the values in this column as
characters, or you might want to convert the result using
as.character().
HTH,
Tom
On Dec 28, 2009, at 1:14 PM, Nick Torenvliet wrote:
Conside
Consider the following
> fileLines
V1 V2V3V4 V5 V6V7 V8
1 AB 20091224 156.0 156.0 154.00 154.0055 1198
2 AB.C 20091224 156.0 156.0 156.00 156.00 0 0
3 ABF10 20091224 156.0 156.0 156.00 156.0055444
4 ABH10 20091224 156.0 156.0 1
I have the following code
fileList <-list.files(path = ".", pattern = "[^a-z].txt$", all.files =
FALSE, full.names = FALSE, recursive = FALSE, ignore.case = FALSE)
for (x in 1:length(fileList)){
fileLines <- data.frame(read.table(fileList[x]))
print(string)
}
the lines of the file all hav
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