Ok, what i want is find how many numbers after . in a numeric ,and i don't
know if there is already exists a function to do it( i wrote one by myself
which will be showed later).
e.g.
1.234 has 3 numbers after .
1 has 0 number
1.5342 has 4 numbers
And i solved the above format using:
find<-
Michael, thanks, that was it!
I had installed the packages as root and tried them out as root (not a good
idea I know, but I was lazy), while running the the X11 display as user.
Worse is that I have done that many times. One more thing learned.
Thanks again,
Rainer
On Thursday 23 October 2014
I am baffled. I think those were English words but they didn't make any sense
to me. Not was there a reproducible example to turn to. Can you try again?
---
Jeff NewmillerThe . . Go L
I hope someone can explain what the #1 means (and for that matter the
[2] in the debug output below. I can't find anything in the spec that
explains what they mean.
Thanks
> f(1)
debugging in: f(1)
debugging in: f()
debug at #1: {
if (x < 0)
-x
else x
}
Browse[2]>
__
Dear usRers,
Now i want to cal ,e.g.
cal(1.234) will get 3
cal(1) will get 0
cal(1.3045) will get 4
But the difficult part is cal(1.3450) will get 4 not 3.
So, is there anyone happen to know the solution to this problem, or it can't be
solved in R, because 1.340 will always be transformed
Hi Ken,
Many thanks for your advice.
Earlier this morning I stepped my way through the glm.fit function to see where
things were falling over and realised that first and foremost I had my link
function wrong (link and inverse were back to front). I've now fixed this and
can get the model to
A very limited set of attachment types are allowed on this list.. yours was not
one of them. Reading the Posting Guide will inform you about many useful things.
Self-contained R code is the expected mode of communication here.
>From your description, you might consider monotonic spline fitting t
Hello together,
i have a short question. Maybe anyone can help me to create a barplot in R with
the package lattice.
I have the following data as ouput values and the following code:
Data (d):
KOST BudgetIST
1060 -2.18 0
1080 9
In the figure attached "test", I would like to interpolate the empty - "NA"
section to connect the two curves.
1/ I tried using "na.approx" and "na.spline", without success.
na.approx(test,na.rm=FALSE)
na.spline(test,na.rm=FALSE)
In the first case it just draws a straight line between the end a
Hi All,
This isn't a request for help. I just wanted to post that we've recently
completed a new package for R which allows you to do Open Street Map
plotting via Leaflet.
It's not on CRAN (yet), but it is on Github.
http://greentheo.github.io/OpenStreetMapR/
Theodore Van Rooy
royaltyanalytics
Dear list,
I would like to make heatmaps from my data. Acctualy, I have already done
it, but the issue it that is doesn't work well for all files.
All files have the same structure, after importing them I do few
calculations using the same script, to obtain variables for plotting.
You can find t
Perhaps this is a permissions (Xauthority) issue: is the same user running
both the X11 display and the R session?
On Thu, Oct 23, 2014 at 2:40 AM, R wrote:
> I have written some gWidgets scripts before in the past but have a
> different box now (Debian KWheezy) and cannot get gWidgets working
If I understand the problem correctly, then I’d suggest this:
ndays <- nrow(Wb30)
for (iday in 2:ndays) {
Wb30$Water_Balance[iday] <- Wb30$Water_Balance[iday-1] +
Wb30$Rainfall[iday] - Wb30$Evaporation[iday]
Wb30$Water_Balance[iday] <- min(Wb30$Water_Balance[iday], 100)
Wb30$Water_Ba
Hello,
Yes, you can use lapply. Maybe something like the following. Note that
the result is a list with one member per species. (Untested).
ddeg.correlog.list <- lapply(9:11, function(p)
correlog(plant[plant[,p]=="1", 2], plant[plant[,p]=="1", 3],
plant[plant[,p]=="1", 4]))
Hope this help
Hello List,
I have a database which consist of 912 plots. For each plot, I have the
presence/absence information of 260 species of plants and also 5 different
environmental variables (ddeg, mind, srad, slp, topo).
The dataframe looks like this:
Plot_NumberX Y ddeg mi
csiro.au> writes:
> I'm trying to fit a binomial GLM with user defined
link
function (negative exponential), however I seem to
> be unable to find the correct starting values to
initialise such a model. I've tried taking starting
> values from a logistic and log models fit to the
same data an
I am using R and quantmod to get stock and option quotes. However, it
has stopped working. I expect the following
function call to produce a list of options:
getOptionChain( "XOM", Exp = "2015-01-20" )
However, I get the following error messages:
Error in lapply(strsplit(opt, ""), f
Sorry... That last expression was backward...
Wb30$ValidWB <- with( Wb30, 0 == cumsum( Water_Balance < 0 | 100 <
Water_Balance ) )
---
Jeff NewmillerThe . . Go Live...
DCN:B
Counting chickens after they have left the coop is not going to work. If your
inputs push w outside the limits of physics then your input data are invalid.
Arbitrarily forcing w to fit in that case partially ignores the inputs
anyway... and since there are many ways for the data to be invalid yo
Dear Duncan,
Those condition should be there and also look at Rainfall and evaporation
columns.
If i change it to be like the following loop, it can't do it.
The problem is how to include those conditions and also respect the formula?
wb=c()
for (w in 1:length(Wb30$Water_Balance)){
if(w<0){
On 23/10/2014, 8:56 AM, Robert Sherry wrote:
> I am trying to get the current price of gold for my application. I am
> using the library quantmod. The
> R commands I use are:
> getMetals(c('XAU'), from=Sys.Date(), autoassign = FALSE )
> XAUUSD$XAU.USD[1,1]
>
> I would expect the value
I am trying to get the current price of gold for my application. I am
using the library quantmod. The
R commands I use are:
getMetals(c('XAU'), from=Sys.Date(), autoassign = FALSE )
XAUUSD$XAU.USD[1,1]
I would expect the value in XAUUSD$XAU.USD[1,1] to be a scalar but it
comes back
On 23/10/2014, 8:33 AM, Frederic Ntirenganya wrote:
> Dear All,
>
> I want to calculate water balance using the following formula:
> Water balance today = Water balance yesterday + Rainfall − Evaporation
>
> This is a sample of data I am using:
>
> head(Wb30)
> May Rainfall Evaporation Water_B
Dear All,
I want to calculate water balance using the following formula:
Water balance today = Water balance yesterday + Rainfall − Evaporation
This is a sample of data I am using:
head(Wb30)
May Rainfall Evaporation Water_Balance
1 70 5 0
2 8 10
> I am not taking this course for a grade, just self interest. I am stock with
> making the function. is there any where I can see how is done.
Go to the HTML help for R and read "An Introduction to R", section 10, "Writing
your own functions" (the clue's in the title). Don't miss out section 10
Hello,
Try
aggregate(Rain ~ Year + Month, data = dat, FUN = sum)
Hope this helps,
Rui Barradas
Em 23-10-2014 01:29, Hafizuddin Arshad escreveu:
Dear R users,
Can someone help me on this? I would like to find the sum of the Rain if
the Month appears more than once. For example in row 3 and
I have written some gWidgets scripts before in the past but have a different
box now (Debian KWheezy) and cannot get gWidgets working. It may be an obvious
mistake but auntie Google (who has helped me a lot to get as far as I am now)
leaves me in the dark now.
Here is where I am stuck:
- - - - -
PST8PDT is the valid timezone in my machine and I have verified it.
I am followed your suggestion. there is no difference in the output.
*> Sys.setenv(TZ="America/Los_Angeles");*
*> Sys.getenv("TZ");*
* TZ *
*"America/Los_Angeles" *
*> as.POSIXlt(Sys.Date());*
*[1] "2014-10-22
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