I was trying to calculate mean and SD per month of a variable from an environmental dataset (.nc file of Sea surface temp/day during 2 years) and the loop I used gives me the following error
Error in h(simpleError(msg, call)) :
error in evaluating the argument 'x' in selecting a method for function 'mean': recursive indexing failed at level 2
I have no idea where my error could be but if you are curious I was using the following .nc dataset just for SST for 2018-2019 from copernicus sstdata
Here is the script I used so far and the packages I'm using:
# Load required libraries (install the required libraries using the Packages tab, if necessary)
library(raster)
library(ncdf4)
#Opern the .nc file with the environmental data
ENV = nc_open("SST.nc")
ENV
#create an index of the month for every (daily) capture from 2018 to 2019 (in this dataset)
m_index = c()
for (y in 2018:2019) {
# if bisestile year (do not apply for this data but in case a larger year set is used)
if (y%%4==0) { m_index = c(m_index, rep(1:12 , times = c(31,29,31,30,31,30,31,31,30,31,30,31))) }
# if non-bisestile year
else { m_index = c(m_index, rep(1:12 , times = c(31,28,31,30,31,30,31,31,30,31,30,31))) }
}
length(m_index) # expected length (730)
table(m_index) # expected number of records assigned to each of the twelve months
# computing of monthly mean and standard deviation.
# We first create two empty raster stack...
SST_MM = stack() # this stack will contain the twelve average SST (one per month)
SST_MSD = stack() # this stack will contain the twelve SST st. dev. (one per month)
# We run the following loop (this can take a while)
for (m in 1:12) { # for every month
print(m) # print current month to track the progress of the loop...
sstMean = mean(ENV[[which(m_index==m)]], na.rm=T) # calculate the mean SST for all the records of the current month
sstSd = calc(ENV[[which(m_index==m)]], sd, na.rm=T) # calculate the st. dev. of SST for all the records of the current month
# add the monthly records to the stacks
SST_MM = stack(SST_MM, sstMean)
SST_MSD = stack(SST_MSD, sstSd)
}
And as mentioned, the output of the loop including the error:
SST_MM = stack() # this stack will contain the twelve average SST (one per month)
> SST_MSD = stack() # this stack will contain the twelve SST st. dev. (one per month)
> for (m in 1:12) { # for every month
+
+ print(m) # print current month to track the progress of the loop...
+
+ sstMean = mean(ENV[[which(m_index==m)]], na.rm=T) # calculate the mean SST for all the records of the current month
+ sstSd = calc(ENV[[which(m_index==m)]], sd, na.rm=T) # calculate the st. dev. of SST for all the records of the current month
+
+ # add the monthly records to the stacks
+
+ SST_MM = stack(SST_MM, sstMean)
+ SST_MSD = stack(SST_MSD, sstSd)
+
+ }
[1] 1
**Error in h(simpleError(msg, call)) :
error in evaluating the argument 'x' in selecting a method for function 'mean': recursive indexing failed at level 2**
It seems that you make things too complicated. I think the easiest way to do this is with terra::tapp like this:
library(terra)
x <- rast("SST.nc")
xmn <- tapp(x, "yearmonths", mean)
xsd <- tapp(x, "yearmonths", sd)
or more manually:
library(terra)
x <- rast("SST.nc")
y <- format(time(x),"%Y")
m <- format(time(x),"%m")
ym <- paste0(y, "_", m)
r <- tapp(x, ym, mean)
I have a scRNA dataset with 10 healthy controls and 17 patients. I am doing the comparative analysis. I did the following:
Created 10 seurat objects for 10 healthy controls and merged them to create one (healthy)
Created 17 seurat objects for 17 patients and merged them to create one (patients)
Created a list of the two objects: data <- list (healthy, patients)
Normalize the data:
data <- lapply(data, function(x) {
x <- NormalizeData(x)
x <- FindVariableFeatures(x, selection.method = "vst", nfeatures = 2000)
})
I am getting the following error:
Error in as(object = data, Class = "dgCMatrix") : no method or default for coercing “patchwork” to “dgCMatrix”
Please help
After some trial and error I was able to reproduce your same error running this line of code before your lapply function:
data <- list(p1 + p2 , p2)
Where p1 and p2 are ggplot objects.
It looks to me that in your data list you don't have Seurat objects.
You should check for any mistakes in the code that you have used to generate your list of seurat objects.
I hope this helps :)
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 am having a problem using augPred after I have created an nlme object. My data is 529 rows long with no NA's, and is grouped with this function:
RSize.gD.100 <- groupedData(area ~ size | ID, data = RSize)
Under the grouping factor “ID”, this is a character column with 6 levels.
I create my nlsList object with the selfStarter NLS.expoDecay():
fm1R.list <- nlsList(area ~ NLS.expoDecay(size, a, k) | ID, data = RSize.gD.100)
and then my nlme with:
fm1R.nlme <- nlme(fm1R.list)
Following the methods of Pinheiro and Bates (2000), I try to plot my predicted values against my actual values using:
plot(augPred(fm1R.nlme, level=0:1))
But it returns this error:
Error in [[<-.data.frame(*tmp*, nm, value = c(2L, 2L, 2L, 2L, 2L, :
replacement has 529 rows, data has 6
I cannot find any suggestions on other threads. Any help would be much appreciated!!
Thank you!
I have used codeproject to get share data from yahoo
( http://www.codeproject.com/Articles/37550/Stock-quote-and-chart-from-Yahoo-in-C ).
In yahoo finance, there are 'Key Statistics' which i would like to use, but are not available by this means (e.g. data at http://uk.finance.yahoo.com/q/ks?s=BNZL.L ). Is there any way to get this information directly? I would really rather not screen scrape if possible.
I am using C#/.NET4.
You can use my lib for .NET Yahoo! Managed. There you have the MaasOne.Finance.YahooFinance.CompanyStatisticsDownload class to do exactly what you want.
p/s: You need to use the latest version (0.10.1). v0.10.0.2 is obsolete with Key Statistics Download.
I landed on this question while searching for an answer couple of days ago, thought of providing an answer I created in R (and shared it on R-Bloggers). I know that the answer I am providing is not in C# but XPath and XML are supported in every language so you can use this approach there. The URL to the blog is - http://www.r-bloggers.com/pull-yahoo-finance-key-statistics-instantaneously-using-xml-and-xpath-in-r/
#######################################################################
##Alternate method to download all key stats using XML and x_path - PREFERRED WAY
#######################################################################
setwd("C:/Users/i827456/Pictures/Blog/Oct-25")
require(XML)
require(plyr)
getKeyStats_xpath <- function(symbol) {
yahoo.URL <- "http://finance.yahoo.com/q/ks?s="
html_text <- htmlParse(paste(yahoo.URL, symbol, sep = ""), encoding="UTF-8")
#search for <td> nodes anywhere that have class 'yfnc_tablehead1'
nodes <- getNodeSet(html_text, "/*//td[#class='yfnc_tablehead1']")
if(length(nodes) > 0 ) {
measures <- sapply(nodes, xmlValue)
#Clean up the column name
measures <- gsub(" *[0-9]*:", "", gsub(" \\(.*?\\)[0-9]*:","", measures))
#Remove dups
dups <- which(duplicated(measures))
#print(dups)
for(i in 1:length(dups))
measures[dups[i]] = paste(measures[dups[i]], i, sep=" ")
#use siblings function to get value
values <- sapply(nodes, function(x) xmlValue(getSibling(x)))
df <- data.frame(t(values))
colnames(df) <- measures
return(df)
} else {
break
}
}
tickers <- c("AAPL")
stats <- ldply(tickers, getKeyStats_xpath)
rownames(stats) <- tickers
write.csv(t(stats), "FinancialStats_updated.csv",row.names=TRUE)
#######################################################################
If you don't mind using the key statistics from BarChart.com, here is a simple function script:
library(XML)
getKeyStats <- function(symbol) {
barchart.URL <- "http://www.barchart.com/profile.php?sym="
barchart.URL.Suffix <- "&view=key_statistics"
html_table <- readHTMLTable(paste(barchart.URL, symbol, barchart.URL.Suffix, sep = ""))
df_keystats = html_table[[5]]
print(df_keystats)
}