Dear all,
I am sorry, I did a lot of confusion. I am sorry, I have to relax and stat
all again in order to understand.
If I could I would like to start again, without mixing strategy and waiting
for your advice.
I am really appreciate you help, really really.
Here my new file, a *.csv file (buy th
I am trying to build an R package with Rtools version 3.5.0.4 along with
R-3.5.1
using the following sequence of commands:
File -> Open Project -> Build -> Build Binary Package
I received the following error message:
zip I/O error: No such file or directory
zip error: Temporary file failure (Y
... and the most common source of NA values in time data is wrong timezones.
You really need to make sure the timezone that is assumed when the character
data are converted to POSIXt agrees with the data. In most cases the easiest
way to insure this is to use
Sys.setenv(TZ="US/Pacific")
or wha
Hi Diego,
I think the error is due to NA values in your data file. If I extend
your example and run it, I get no errors:
MyData<-read.table(text="103001930 103001580 103001530
1998-10-01 00:00:00 0.6 0 0
1998-10-01 01:00:00 0.2 0.2 0.2
1998-10-01 02:00:00 0.6 0.2 0.4
1998-10-01 03:00:00 0 0 0.6
19
You can do
Vect[-grep (“foo”, names(vect))]
On Tue, 31 Jul 2018 at 11:12 PM, إبراهيم خطاب Ibrauheem Khat'taub <
barhomopo...@gmail.com> wrote:
> H
> i All,
>
> If I have this vector:
>
> > vect <- c(foo = 11, bar = 2, norf = 45)
>
> I can have a subset that has only "bar and "norf" this way:
> >
Awesome, thanks!
مع خالص الشكر والسلام عليكم.
إبراهيم خطاب
On Tue, 31 Jul 2018 at 13:54, Data Science Classes <
datascienceclas...@gmail.com> wrote:
> You can do
>
> Vect[-grep (“foo”, names(vect))]
>
> On Tue, 31 Jul 2018 at 11:12 PM, إبراهيم خطاب Ibrauheem Khat'taub <
> barhomopo...@gmail.com>
Awesome, Sarah, thanks!
And thanks for the clarification about declaring the version of R.
Best,
Ibrahim
On Tue, 31 Jul 2018 at 13:50, Sarah Goslee wrote:
> Hi,
>
> You need to tell R to look in the names component of your vector. Here
> are three different ways:
>
> vect <- c(foo = 11, bar =
Hi,
You need to tell R to look in the names component of your vector. Here
are three different ways:
vect <- c(foo = 11, bar = 2, norf = 45)
vect[!(names(vect) %in% c("foo"))] # easily generalizable to a longer list
vect[!grepl("foo", names(vect))]
vect[!(names(vect) == "foo")]
There are many
try:
vect[which(names(vect) != "foo")]
On Tue, Jul 31, 2018 at 1:41 PM, إبراهيم خطاب Ibrauheem Khat'taub
wrote:
> H
> i All,
>
> If I have this vector:
>
>> vect <- c(foo = 11, bar = 2, norf = 45)
>
> I can have a subset that has only "bar and "norf" this way:
>> vect[c("bar","norf")]
>
> Now ho
H
i All,
If I have this vector:
> vect <- c(foo = 11, bar = 2, norf = 45)
I can have a subset that has only "bar and "norf" this way:
> vect[c("bar","norf")]
Now how do I achieve the same by asking it for a subset that simply
excludes "foo"? I tried all these, resulting in errors:
vect[-"foo"]
Dear all,
I have still problem with date.
Could you please tel me how to use POSIXct.
Indeed I have found this command:
timeAverage, but I am not able to convert MyDate to properly date.
Thank a lot
I hope to no bother you, at least too much
Diego
On 31 July 2018 at 11:12, Diego Avesani wrot
Dear Jim, Dear all,
thanks a lot.
Unfortunately, I get the following error:
st1_daily<-by(MyData$st1,MyData$date,mean)Error in
tapply(seq_len(0L), list(`MyData$date` = c(913L, 914L, 925L, :
arguments must have same length
This is particularly strange. indeed, if I apply
mean(MyData$str1
Dear all,
I have found the error, my fault. Sorry.
There was an extra come in the headers line.
Thanks again.
If I can I would like to ask you another questions about the imported data.
I would like to compute the daily average of the different date. Basically
I have hourly data, I would like to a
Dear all,
I move to csv file because originally the date where in csv file.
In addition, due to the fact that, as you told me, read.csv is a special
case of read.table, I prefer start to learn from the simplest one.
After that, I will try also the *.txt format.
with read.csv, something strange hap
... or deleting .RData and avoiding creating them in the future. Named save
files (anything.RData) are useful but the autoload feature of .RData files is
not and this particular behavior is one of the reasons why.
On July 31, 2018 3:39:39 AM PDT, PIKAL Petr wrote:
>Hi
>
>However if in .RData is
Hi
However if in .RData is an object created by "yaml" and the package is not
loaded it could cause this message. It was similar with ggplot objects if
ggplot2 is not loaded.
Removing this object and saving session to update .RData can help.
Cheers
Petr
> -Original Message-
> From: R-
.RData does not save any info abut previously loaded packages - so this would
not cause the problem.
Rainer
> On 30 Jul 2018, at 21:27, Rui Barradas wrote:
>
> Hello,
>
> Maybe R is loading a previously saved session.
> Check whether you have a file named .RData in your working directory.
>
Hi Diego,
One way you can get daily means is:
st1_daily<-by(MyData$st1,MyData$date,mean)
st2_daily<-by(MyData$st2,MyData$date,mean)
st3_daily<-by(MyData$st3,MyData$date,mean)
Jim
On Tue, Jul 31, 2018 at 6:51 PM, Diego Avesani wrote:
> Dear all,
> I have found the error, my fault. Sorry.
> There
Quoting Christofer Bogaso :
The data type is defined as bigint
Your query does not specify a number, but a string
(you single-quote the digits). Databases may do type conversion;
for instance, see the MySQL manual:
https://dev.mysql.com/doc/refman/8.0/en/type-conversion.html
Since you se
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