英文:
Merge hourly ERA5 netcdf files into daily files with R
问题
Sure, here are the translated parts of your text:
"I have hourly data for maximum temperature from 1970-2022 for all month in NetCDF format (NetCDF of each month contains data of 24 hours from 1970-2022). Can anyone help me out in aggregating hourly data to daily data with R? I have already tried it by directly converting the hourly data to daily and downloading it but it is unable to download such a large dataset."
"I am trying to aggregate the hourly data to daily of that month in R."
英文:
I have hourly data for maximum temperature from 1970-2022 for all month in NetCDF format (NetCDF of each month contains data of 24 hours from 1970-2022). Can anyone help me out in aggregating hourly data to daily data with R? I have already tried it by directly converting the hourly data to daily and downloading it but it is unable to download such large dataset. (here's the code that I had tried).
wf_set_key(service = "cds")
data=c.retrieve(
'reanalysis-era5-single-levels',
{
'product_type': 'reanalysis',
'variable': 'maximum_2m_temperature_since_previous_post_processing',
'year': [
'1970', '1971', '1972',
'1973', '1974', '1975',
'1976', '1977', '1978',
'1979', '1980',
],
'month': [
'03','04',
'05', '06',
],
'day': [
'01', '02', '03',
'04', '05', '06',
'07', '08', '09',
'10', '11', '12',
'13', '14', '15',
'16', '17', '18',
'19', '20', '21',
'22', '23', '24',
'25', '26', '27',
'28', '29', '30','31',
],
'time': [
'00:00', '01:00', '02:00',
'03:00', '04:00', '05:00',
'06:00', '07:00', '08:00',
'09:00', '10:00', '11:00',
'12:00', '13:00', '14:00',
'15:00', '16:00', '17:00',
'18:00', '19:00', '20:00',
'21:00', '22:00', '23:00',
],
'area': [
38, 67, 6,
99
],
'format': 'netcdf',
},
'day_mean'=ct.climate.daily_mean(data,keep_attrs=True)
if count == 1:
day_mean_all=day_mean
else:
day_mean_all=ct.cube.concat([day_mean_all, day_mean], dim='time')
count = count + 1
return day_mean_all
'download.nc')
I am trying to aggregate the hourly data to daily of that month in R.
library(ncdf4)
ncpath <- "D:/MAX_TEMP/"
ncname <- "adaptor.mars.internal-1681202164.1038315-25242-15-2a718a58-dcd5-4470-9fd2-ddbdede30875_march"
ncfname <- paste(ncpath, ncname, ".nc", sep="")
ncin <- nc_open(ncfname)
print(ncin)
library(dplyr)
a1<-ncname %>%
group_by(time) %>%
summarize(Mean_Max_Temp = mean(expver))
#Error in UseMethod("group_by")
答案1
得分: 2
请参考Robert Hijmans在这里给出的详细解释。
我假设你熟悉使用raster brick函数来读取你的netcdf文件（也可以使用terra包，但现在我们先使用raster::brick）。然后，你需要按照每天的日月来分组图层，以便在时间维度上进行聚合，而你的netcdf文件中的时间是每小时的。
最后，使用如上链接中所示的stackApply()函数。希望这对你有所帮助！
英文:
Please see the nice explanation given by Robert Hijmans here
I suppose you are familiar with raster brick function to read your netcdf file (Could use package terra as well, but stick to raster::brick for now). You will then need to group the layers by day-mon in other to aggregate over the time dimension, which is hourly in your netcdf.
Finally, use the stackApply () function as shown in the above link. Hope this helps!
答案2
得分: 0
你可以使用 terra::tapp 来实现这个。
英文:
You can use terra::tapp for that.
library(terra)
r <- rast("adaptor.mars.internal-1681202164.1038315-25242-15-2a718a58-dcd5-4470-9fd2-ddbdede30875_march.nc")
x <- tapp(r, "days", mean)
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