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我有一个纬度、经度、开始年份和结束年份的数据框。我想要那个时期每个地点的平均降水量。

现在，我可以一次为一个位置获取此信息，但我想为多个位置自动执行以下操作：

以下是一些先决条件：

#library(xts)
#library(rnoaa)
#options(noaakey = "...") # https://ropensci.org/blog/2014/03/13/rnoaa/ says how to get a API key
#station_data <- ghcnd_stations() # Takes a while to run
statenv <- new.env()
lat_lon_df<-structure(list(lat = c(41.1620277777778, 44.483333, 44.066667
), long = c(-96.4115, -92.533333, -93.5), yrmin = c(2001L, 1983L, 
                                                    1982L), yrmax = c(2010L, 1990L, 1992L), id = c("ithaca", "haycreek", 
                                                                                                   "waseca")), class = "data.frame", row.names = c(1389L, 1395L, 
                                                                                                                                                   1403L))

这是肉。

ll_df<-lat_lon_df[1,]
nearby_station<-meteo_nearby_stations(lat_lon_df = ll_df,
    lat_colname = "lat", lon_colname = "long",
    station_data = station_data, radius = 50, year_min=ll_df[1,"yrmin"],
    year_max=ll_df[1,"yrmax"],limit=1, var="PRCP")


nearby_station<-meteo_nearby_stations(lat_lon_df = ll_df,lat_colname = "lat", lon_colname = "long",
                                          station_data = station_data, radius = 50, year_min=ll_df[1,"yrmin"],
                                          year_max=ll_df[1,"yrmin"],limit=1, var="PRCP")
e <- lapply(nearby_station,function(x)  meteo_pull_monitors(x$id[1])) #get actual data based on monitor id's

ll<-xts(e[[1]]$prcp,order.by=e[[1]]$date)
x<-paste0(ll_df[1,"yrmin"],"/",ll_df[1,"yrmax"]) 
 mean(xts::apply.yearly(na.omit(ll[x]),sum))/10 #divide by 10, put in mm

这将返回 776.23。最终结果应该是一个数据框，现在有一个新列“precip”，如下所示：

     lat      long yrmin yrmax       id    precip
41.16203 -96.41150  2001  2010   ithaca    776.23
44.48333 -92.53333  1983  1990 haycreek    829.65
44.06667 -93.50000  1982  1992   waseca    894.62

必须有一种方法可以简单地按行重复lat_long_df，即 for lat_lon_df[1,]、thenlat_lon_df[2,]和 finally lat_lon_df[3,]。

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1 回答 1

1

一种方法是apply在lat_lon_df.

这是一个例子：

library(xts)
library(rnoaa)

设置 API 密钥

#options(noaakey = "...") # https://ropensci.org/blog/2014/03/13/rnoaa/ says how to get a API key

station_data <- ghcnd_stations() #meta-information about all available GHCND weather stations

apply现在应用您在通话中描述的所有步骤

out <- apply(lat_lon_df, 1, function(x){
  min_year <- x[3] #extract the needed values min_year, max_year and ll_df
  max_year <- x[4] 
  ll_df <- data.frame(lat = as.numeric(x[1]),
                      long = as.numeric(x[2]),
                      id = x[5])
  nearby_station <- meteo_nearby_stations(lat_lon_df = ll_df,
                                          lat_colname = "lat",
                                          lon_colname = "long",
                                          station_data = station_data,
                                          radius = 50,
                                          year_min = min_year,
                                          year_max = max_year,
                                          limit=1,
                                          var="PRCP")
  res <- lapply(nearby_station, function(y) {
    res <- meteo_pull_monitors(y[1]$id)
    }
    )
  ll <- xts(res[[1]]$prcp, order.by=res[[1]]$date)
  x <- paste0(min_year <- x[3],"/",max_year) 
  mean(xts::apply.yearly(na.omit(ll[x]),sum))/10
}
)

data.frame(lat_lon_df, precip = out)
#output
          lat      long yrmin yrmax       id   precip
1389 41.16203 -96.41150  2001  2010   ithaca 776.2300
1395 44.48333 -92.53333  1983  1990 haycreek 829.6500
1403 44.06667 -93.50000  1982  1992   waseca 894.6273

请注意，whenyrmin和yrmaxnot change 可以通过使用meteo_nearby_stationson获得所需的信息lat_lon_df。

您还可以将其定义为命名函数

get_mean_precip <- function(x){
  min_year <- x[3]
  max_year <- x[4]
  ll_df <- data.frame(lat = as.numeric(x[1]),
                      long = as.numeric(x[2]),
                      id = x[5])
  nearby_station <- rnoaa::meteo_nearby_stations(lat_lon_df = ll_df,
                                                 lat_colname = "lat",
                                                 lon_colname = "long",
                                                 station_data = station_data,
                                                 radius = 50,
                                                 year_min = min_year,
                                                 year_max = max_year,
                                                 limit=1,
                                                 var = "PRCP")
  res <- lapply(nearby_station, function(y) {
    res <- rnoaa::meteo_pull_monitors(y[1]$id)
  }
  )
  ll <- xts::xts(res[[1]]$prcp, order.by=res[[1]]$date)
  x <- paste0(min_year <- x[3],"/",max_year) 
  mean(xts::apply.yearly(na.omit(ll[x]),sum))/10
}

并将其用作：

out <- apply(lat_lon_df, 1, get_mean_precip)
于 2018-09-11T18:41:22.350 回答