similar to: Converting .grib to excel file

Richard, Many thanks for your email. I had attached the grib file to the original email to R help team but it seems you did not receive it. Unfortunately, I do not know how to reduce the volume or extract some of the grib file data to send it for you. The file has the volume of 6 Megabyte. I can send it by email. The file has 6 met parameters and Date (day/month/year hour:minute). I want the

Dear Bert and Sara; I have searched on the internet and found some way to do this like python. But python is so complicated to me as it needs many steps to be done for reading and converting it. I will try terra package to convert it. On Tue, 25 Jun 2024, 15:15 javad bayat, <j.bayat194 at gmail.com> wrote: > Richard, > Many thanks for your email. > I had attached the grib file to

I'm now inclined to go with 'search for "convert GRIB to CSV". https://confluence.ecmwf.int/display/CKB/How+to+convert+GRIB+to+CSV is the first line. I know that's not an R solution, but using software specifically developed for encoding, decoding, extracting, &c GRIB file by the European Centre for Medium-Range Weather Forecasts and actively maintained, with an example

Whoops, sorry, you *did* answer "what went wrong". > param_names <- c("param1", "param2", "param3", "param4", "param5", "param6") > extracted_data <- extract(raster_data, param_names, df = TRUE) #Error in (function (classes, fdef, mtable) : #unable to find an inherited method for function ?extract? for signature

For years I used Deducer (Package Deducer) developed by Ian Fellows.? This is still available and partially remains mostly functional.? He was working on a new GUI years ago but appears no longer to be in the pipeline I think replaced by R-Studio. > Send R-help mailing list submissions to > r-help at r-project.org > > To subscribe or unsubscribe via the World Wide Web, visit >

The volume of a polygon = 0. Polyhedra have volumes. This may be irrelevant, but if the lake is cylindrical == constant cross sectional area at all depths, then height doubles when the volume does and vice versa. Otherwise you have to know how area varies with height or use more sensible approximations thereto. Cheers, Bert On Tue, Dec 5, 2023, 20:13 javad bayat <j.bayat194 at

Dear all; I am trying to calculate the volume of a polygon shapefile according to a DEM raster. I have provided some codes at the end of this email.I dont know if the codes are correct or not. Following this, I have another question too. I want to know if the volume of the reservoir rises or doubles, what would be the elevation? I would be more than happy if anyone could help me. Sincerely "

hi all, how can i saving R output to docx or Jpeg format? [[alternative HTML version deleted]]

Hi, As already mentioned, this is a great topic for R-sig-geo, where you'd probably get specialist answers like the lake morphology package https://cran.r-project.org/web/packages/lakemorpho/index.html which is explicitly designed for this kind of question. Sarah On Tue, Dec 5, 2023 at 11:13?PM javad bayat <j.bayat194 at gmail.com> wrote: > > Dear all; > I am trying to

Dear R users I am trying to find a formula to calculate the angle of a polyline. Is there a way to do this? Many thanks. -- Best Regards Javad Bayat M.Sc. Environment Engineering Alternative Mail: bayat194 at yahoo.com [[alternative HTML version deleted]]

Dear all; Thank you for your reply. David has explained an interesting method. David I have DEM file of the region and I have extracted the xyz data from that. Also I can extract bathymetry data as xyz file. I have calculated the storage (volume) of reservoir at the current elevation. But the method I have used to calculate the volume is different from your method. I have crop DEM by the reservoir

Homework? -- Sent from my Android device with K-9 Mail. Please excuse my brevity. On April 7, 2024 8:27:18 AM EDT, javad bayat <j.bayat194 at gmail.com> wrote: >Dear all; >I have a question about the water level of a reservoir, when the volume >changed or doubled. >There is a DEM file with the highest elevation 1267 m. The lowest elevation >is 1230 m. The current volume of

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"

Dear all; I have a question about the water level of a reservoir, when the volume changed or doubled. There is a DEM file with the highest elevation 1267 m. The lowest elevation is 1230 m. The current volume of the reservoir is 7,000,000 m3 at 1240 m. Now I want to know what would be the water level if the volume rises to 1250 m? or what would be the water level if the volume doubled (14,000,000

Water engineer here. The standard approach is to 1) get the storage vs. elevation data from the designers of the reservoir or, barring that, 2) get the bathymetry data from USBR or state DWR, or, if available, get the DEM data from USGS if the survey was done before the reservoir was built or 3) get a boat+sonar with GPS? +lots of time and survey the bottom elevation yourself. Put the xyz

John, Your reaction was what my original reaction was until I realized I had to find out what a DEM file was and that contains enough of the kind of depth-dimension data you describe albeit what may be a very irregular cross section to calculate for areas and thence volumes. If I read it correctly, this can be a very real-world problem worthy of a solution, such as in places like California

Assuming your polyline is defined by two vectors, one for the x coordinates, one for the y coordinates, you can try the following library(NISTunits) polyangles <- function(xV,yV) { stopifnot( (length(xV)==length(yV)) && (length(xV) >= 3)) v <- function(i) { c( xV[i]-xV[i-1], yV[i]-yV[i-1])} vlen <- function(v) { sqrt(sum(v*v)) } lV <- rep(NA_real_,length(xV)) for

?s 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)){ >

A polyline by definition has many angles, so your question is ill-formed. And this is a question about math, not R, so is off topic here. I suggest reading Wikipedia. -- Sent from my phone. Please excuse my brevity. On January 29, 2018 11:10:02 PM PST, javad bayat <j.bayat194 at gmail.com> wrote: >Dear R users >I am trying to find a formula to calculate the angle of a polyline. Is

Aside from the fact that the original question might well be a class exercise (or homework), the question is unanswerable given the data given by the original poster. One needs to know the dimensions of the reservoir, above and below the current waterline. Are the sides, above and below the waterline smooth? Is the region currently above the waterline that can store water a mirror image of the