I'm working with several netCDF files. Each nc file has 33 variables. I need to create a stack for each nc file with these 33 variables to do some calculations, but the only way I know is to convert each variable in a rater and stack them, one by one... Like this:
library(raster)
library(ncdf4)
library(rgdal)
nc_data <- nc_open('./data/GCAM/RAW/93d4aa096b15491b1ba136b46d8063cdca59d253c75d59791b4d4cb6f8a1ae91/Project ID 68344/GCAM-Demeter/GCAM-Harmonized/Mean_Std/GCAM_Demeter_LU_H_ssp1_rcp26_modelmean_2030.nc')
PTF0 <- nc_data$var[[1]]
data1 <- ncvar_get( nc_data, PTF0 )
data1 <- raster(data1)
plot(data1)
Can anyone help to automatize this?? I'm thankful in advance
This is a structure of the NetCDF file, I highlight the files that I need to stack, actually, I need stack just PT1 to PTF8
What you are doing now could be done like tis
f <- 'GCAM_Demeter_LU_H_ssp1_rcp26_modelmean_2030.nc'
library(raster)
r <- raster(f, "PTF0")
(assuming that "PTF0" is a variable name)
But if you want to create a single object for multiple variables in one step, use terra instead
library(terra)
r <- rast(f)
You can specify the variables you want
rr <- rast(f, c("PFT1", "PTF2"))
You can also create a SpatRasterDataSet and then extract the variables you want like this
s <-sds(f)
x <- rast(s[2:8])
Thanks Mr Robert! With these few lines, I solve the issue and I can do all I need.
library(raster)
library(ncdf4)
library(rgdal)
library(terra)
f <- nc_open('./data/GCAM/RAW/93d4aa096b15491b1ba136b46d8063cdca59d253c75d59791b4d4cb6f8a1ae91/Project ID 68344/GCAM-Demeter/GCAM-Harmonized/Mean_Std/GCAM_Demeter_LU_H_ssp1_rcp26_modelmean_2030.nc')
rr <- rast(f$filename)
rrr <- rr[[2:8]]
plot(rrr)
soma <- sum(rrr)
plot(soma)
Now I'll spend more time thinking about how to automate this... because I need to do this operation for each one of NetCDF files...
Related
I have a folder with hundreds rasters with different names. Some names are partially similar, except for the last 5 letters which characterize the year. for example I have
"raster_a_2010.tif"
"raster_a_1990.tif"
"raster_f_2010.tif"
"raster_f_1990.tif"
I need to stack the rasters that share the first part of name, so (raster_a_2010 with raster_a_1990) but I need to do it automatically, without indicating one by one the pattern.
So far I did this but I'm still far from the solution: basically I'm trying to create a list of vectors that recognise each pattern and then I'd like to use this list to create the stack
raster_year <- base::list.files(file.path (dir_years,"raster"))
#list of raster
files_base <- basename(list.files(file.path (dir_years,"raster")))
files_group <- substring(files_base, 1, char(files_base) - 4)
## Group the files by the extracted portion of the base name
files_grouped <- group_by(data.frame(file = raster_year , group = files_group)) files_grouped
V <-as.list(as.data.frame(files_grouped))
pattern <- unique(V$group)
file_vector <- list.files(path = dir_years_raster, pattern =files_grouped$group, full.names = TRUE)
I think you are looking for the pattern argument to list.files.
For example:
ff <- list.files(file.path(dir_years,"raster"), pattern="_a_")
Or if you first read all files you can subset them with grep
ff <- c("raster_a_2010.tif" , "raster_a_1990.tif", "raster_f_2010.tif", "raster_f_1990.tif")
grep("_a_", ff, value=TRUE)
#[1] "raster_a_2010.tif" "raster_a_1990.tif"
I'm trying to teach myself coding skills for spatial data analysis. I've been using Robert Hijmans' document, "Spatial Data in R," and so far, it's been great. To test my skills, I'm messing around with a GPX file I got from my smartwatch during a run, but I'm having issues getting my data into a SpatVector of lines (or a line, more specifically). I haven't been able to find anything online on this topic.
As you can see below with a data sample, the SpatVector "run" has point geometries even though "lines" was specified. From Hijman's example of SpatVectors with lines, I gathered that adding columns with "id" and "part" both equal to 1 does something that enables the data to be converted to a SpatVector with line geometries. Accordingly, in the SpatVector "run2," the geometry is lines.
My questions are 1) is adding the "id" and "part" columns necessary? 2) and what do they actually do? I.e. why are these columns necessary? 3) Is there a way to go directly from the original data to a SpatVector of lines? In the process I used to get "run2," I lost all the attributes from the original data, and I don't want to lose them.
Thanks!
library(plotKML)
library(terra)
library(sf)
library(lubridate)
library(XML)
library(raster)
#reproducible example
GPX <- structure(list(lon = c(-83.9626053348184, -83.9625438954681,
-83.962496034801, -83.9624336734414, -83.9623791072518, -83.9622404705733,
-83.9621777739376, -83.9620685577393, -83.9620059449226, -83.9619112294167,
-83.9618398994207, -83.9617654681206, -83.9617583435029, -83.9617464412004,
-83.9617786277086, -83.9617909491062, -83.9618581719697), lat = c(42.4169608857483,
42.416949570179, 42.4169420264661, 42.4169377516955, 42.4169291183352,
42.4169017933309, 42.4168863706291, 42.4168564472347, 42.4168310500681,
42.4167814292014, 42.4167292937636, 42.4166279565543, 42.4166054092348,
42.4164886493236, 42.4163396190852, 42.4162954464555, 42.4161833804101
), ele = c("267.600006103515625", "268.20001220703125", "268.79998779296875",
"268.600006103515625", "268.600006103515625", "268.399993896484375",
"268.600006103515625", "268.79998779296875", "268.79998779296875",
"269", "269", "269.20001220703125", "269.20001220703125", "269.20001220703125",
"268.79998779296875", "268.79998779296875", "269"), time = c("2020-10-25T11:30:32.000Z",
"2020-10-25T11:30:34.000Z", "2020-10-25T11:30:36.000Z", "2020-10-25T11:30:38.000Z",
"2020-10-25T11:30:40.000Z", "2020-10-25T11:30:45.000Z", "2020-10-25T11:30:47.000Z",
"2020-10-25T11:30:51.000Z", "2020-10-25T11:30:53.000Z", "2020-10-25T11:30:57.000Z",
"2020-10-25T11:31:00.000Z", "2020-10-25T11:31:05.000Z", "2020-10-25T11:31:06.000Z",
"2020-10-25T11:31:12.000Z", "2020-10-25T11:31:19.000Z", "2020-10-25T11:31:21.000Z",
"2020-10-25T11:31:27.000Z"), extensions = c("18.011677", "18.011977",
"18.012176", "18.012678", "18.013078", "18.013277", "18.013578",
"18.013877", "17.013977", "17.014278", "17.014478", "17.014677",
"17.014676", "17.014677", "16.014477", "16.014477", "16.014576"
)), row.names = c(NA, 17L), class = "data.frame")
crdref <- "+proj=longlat +datum=WGS84"
run <- vect(GPX, type="lines", crs=crdref)
run
data <- cbind(id=1, part=1, GPX$lon, GPX$lat)
run2 <- vect(data, type="lines", crs=crdref)
run2
There is a vect method for a matrix and one for a data.frame. The data.frame method can only make points (and has no type argument, so that is ignored). I will change that into an informative error and clarify this in the manual.
So to make a line, you could do
library(terra)
g <- as.matrix(GPX[,1:2])
v <- vect(g, "lines")
To add attributes you would first need to determine what they are. You have one line but 17 rows in GPX that need to be reduced to one row. You could just take the first row
att <- GPX[1, -c(1:2)]
But you may prefer to take the average instead
GPX$ele <- as.numeric(GPX$ele)
GPX$extensions <- as.numeric(GPX$extensions)
GPX$time <- as.POSIXct(GPX$time)
att <- as.data.frame(lapply(GPX[, -c(1:2)], mean))
# ele time extensions
#1 268.7412 2020-10-25 17.3078
values(v) <- att
Or in one step
v <- vect(g, "lines", atts=att)
v
#class : SpatVector
#geometry : lines
#dimensions : 1, 3 (geometries, attributes)
#extent : -83.96261, -83.96175, 42.41618, 42.41696 (xmin, xmax, ymin, ymax)
#coord. ref. :
#names : ele time extensions
#type : <num> <chr> <num>
#values : 268.7 2020-10-25 17.31
The id and part columns are not necessary if you make a single line. But you need them when you wish to create multiple lines and or line parts (in a "multi-line").
gg <- cbind(id=rep(1:3, each=6)[-1], part=1, g)
vv <- vect(gg, "lines")
plot(vv, col=rainbow(5), lwd=8)
lines(v)
points(v, cex=2, pch=1)
And with multiple lines you would use id in aggregate to compute attributes for each line.
I want to give variables a specific order in an equation in Maxima. This is display purposes only.
For example:
(%i1) E=(h*c)/%lambda;
c h
(%o1) E = -------
%lambda
I want the h and c variables to be in that order when displayed. I looked at ratvars() and ordergreat() but they don't appear to be relevant here.
Thanks for your help.
It appears that declare(<var>, mainvar) was what I was looking for. When mainvar attribute is declared for a variable it "succeeds all other constants and variables".
I was trying this using the STACK plugin for Moodle. I needed to remove the mainvar keyword from the forbidden list in the file casstring.class.php.
Actually, I think ordergreat() is the function you need, maybe you did a sorting before that needed unorder() first ro reset things.
Try
unorder()$ ordergreat (h, c)$ E=(h*c)/%lambda;
and
unorder()$ ordergreat (c, h)$ E=(h*c)/%lambda;
My original intention to do this is to integrate dplyr with shiny
Prior to 0.3 I have used eval(parse(text=....)), do.call() approach.
In 0.3, I saw two more options, for example:
var <- c('disp','hp')
select_(mtcars,.dots = as.lazy_dots(var))
select(mtcars,one_of(var))
but which one is better? I intended to pass the selectInput values from Shiny app to do data transformations through dplyr.
Another question, what will be the right way to join two different dataset with dynamic but different key column? Is there anything I can leverage in 0.3?
for example
col_a, col_b are key variables to join from datasets a & b
left_join(dataset_a,dataset_b, by=c(col_a=col_b))
Thanks.
After a few attempts, here is my solution for the 2nd question, use a function to create a named vector, and then feed to left_join.
joinCol_a = xxx
joinCol_b = xxx
f <- function(a,b){
vec <- c(b)
names(vec) <- a
return(vec)
}
left_join(dataset_a,dataset_b,by=f(joinCol_a,joinCol_b))
I know it's not the best solution but this is what I can think of so far.
I am coding a survey that outputs a .csv file. Within this csv I have some entries that are space delimited, which represent multi-select questions (e.g. questions with more than one response). In the end I want to parse these space delimited entries into their own columns and create headers for them so i know where they came from.
For example I may start with this (note that the multiselect columns have an _M after them):
Q1, Q2_M, Q3, Q4_M
6, 1 2 88, 3, 3 5 99
6, , 3, 1 2
and I want to go to this:
Q1, Q2_M_1, Q2_M_2, Q2_M_88, Q3, Q4_M_1, Q4_M_2, Q4_M_3, Q4_M_5, Q4_M_99
6, 1, 1, 1, 3, 0, 0, 1, 1, 1
6,,,,3,1,1,0,0,0
I imagine this is a relatively common issue to deal with but I have not been able to find it in the R section. Any ideas how to do this in R after importing the .csv ? My general thoughts (which often lead to inefficient programs) are that I can:
(1) pull column numbers that have the special suffix with grep()
(2) loop through (or use an apply) each of the entries in these columns and determine the levels of responses and then create columns accordingly
(3) loop through (or use an apply) and place indicators in appropriate columns to indicate presence of selection
I appreciate any help and please let me know if this is not clear.
I agree with ran2 and aL3Xa that you probably want to change the format of your data to have a different column for each possible reponse. However, if you munging your dataset to a better format proves problematic, it is possible to do what you asked.
process_multichoice <- function(x) lapply(strsplit(x, " "), as.numeric)
q2 <- c("1 2 3 NA 4", "2 5")
processed_q2 <- process_multichoice(q2)
[[1]]
[1] 1 2 3 NA 4
[[2]]
[1] 2 5
The reason different columns for different responses are suggested is because it is still quite unpleasant trying to retrieve any statistics from the data in this form. Although you can do things like
# Number of reponses given
sapply(processed_q2, length)
#Frequency of each response
table(unlist(processed_q2), useNA = "ifany")
EDIT: One more piece of advice. Keep the code that processes your data separate from the code that analyses it. If you create any graphs, keep the code for creating them separate again. I've been down the road of mixing things together, and it isn't pretty. (Especially when you come back to the code six months later.)
I am not entirely sure what you trying to do respectively what your reasons are for coding like this. Thus my advice is more general – so just feel to clarify and I will try to give a more concrete response.
1) I say that you are coding the survey on your own, which is great because it means you have influence on your .csv file. I would NEVER use different kinds of separation in the same .csv file. Just do the naming from the very beginning, just like you suggested in the second block.
Otherwise you might geht into trouble with checkboxes for example. Let's say someone checks 3 out of 5 possible answers, the next only checks 1 (i.e. "don't know") . Now it will be much harder to create a spreadsheet (data.frame) type of results view as opposed to having an empty field (which turns out to be an NA in R) that only needs to be recoded.
2) Another important question is whether you intend to do a panel survey(i.e longitudinal study asking the same participants over and over again) . That (among many others) would be a good reason to think about saving your data to a MySQL database instead of .csv . RMySQL can connect directly to the database and access its tables and more important its VIEWS.
Views really help with survey data since you can rearrange the data in different views, conditional on many different needs.
3) Besides all the personal / opinion and experience, here's some (less biased) literature to get started:
Complex Surveys: A Guide to Analysis Using R (Wiley Series in Survey Methodology
The book is comparatively simple and leaves out panel surveys but gives a lot of R Code and examples which should be a practical start.
To prevent re-inventing the wheel you might want to check LimeSurvey, a pretty decent (not speaking of the templates :) ) tool for survey conductors. Besides I TYPO3 CMS extensions pbsurvey and ke_questionnaire (should) work well too (only tested pbsurvey).
Multiple choice items should always be coded as separate variables. That is, if you have 5 alternatives and multiple choice, you should code them as i1, i2, i3, i4, i5, i.e. each one is a binary variable (0-1). I see that you have values 3 5 99 for Q4_M variable in the first example. Does that mean that you have 99 alternatives in an item? Ouch...
First you should go on and create separate variables for each alternative in a multiple choice item. That is, do:
# note that I follow your example with Q4_M variable
dtf_ins <- as.data.frame(matrix(0, nrow = nrow(<initial dataframe>), ncol = 99))
# name vars appropriately
names(dtf_ins) <- paste("Q4_M_", 1:99, sep = "")
now you have a data.frame with 0s, so what you need to do is to get 1s in an appropriate position (this is a bit cumbersome), a function will do the job...
# first you gotta change spaces to commas and convert character variable to a numeric one
y <- paste("c(", gsub(" ", ", ", x), ")", sep = "")
z <- eval(parse(text = y))
# now you assing 1 according to indexes in z variable
dtf_ins[1, z] <- 1
And that's pretty much it... basically, you would like to reconsider creating a data.frame with _M variables, so you can write a function that does this insertion automatically. Avoid for loops!
Or, even better, create a matrix with logicals, and just do dtf[m] <- 1, where dtf is your multiple-choice data.frame, and m is matrix with logicals.
I would like to help you more on this one, but I'm recuperating after a looong night! =) Hope that I've helped a bit! =)
Thanks for all the responses. I agree with most of you that this format is kind of silly but it is what I have to work with (survey is coded and going into use next week). This is what I came up with from all the responses. I am sure this is not the most elegant or efficient way to do it but I think it should work.
colnums <- grep("_M",colnames(dat))
responses <- nrow(dat)
for (i in colnums) {
vec <- as.vector(dat[,i]) #turn into vector
b <- lapply(strsplit(vec," "),as.numeric) #split up and turn into numeric
c <- sort(unique(unlist(b))) #which values were used
newcolnames <- paste(colnames(dat[i]),"_",c,sep="") #column names
e <- matrix(nrow=responses,ncol=length(c)) #create new matrix for indicators
colnames(e) <- newcolnames
#next loop looks for responses and puts indicators in the correct places
for (i in 1:responses) {
e[i,] <- ifelse(c %in% b[[i]],1,0)
}
dat <- cbind(dat,e)
}
Suggestions for improvement are welcome.