【发布时间】：2018-02-24 19:01:55
【问题描述】：
我有包含 6000 个位置的数据框。对于每个地点，我都有 36 年的日降雨量数据。
样本数据：
set.seed(123)
mat <- matrix(round(rnorm(6000*36*365), digits = 2),nrow = 6000*36, ncol = 365)
dat <- data.table(mat)
names(dat) <- rep(paste0("d_",1:365))
dat$loc.id <- rep(1:6000, each = 36)
dat$year <- rep(1980:2015, times = 6000)
我想要为每个位置生成每个月的长期平均降雨量。例如对于 loc.id = 1，1 月、2 月、3 月...12 月的平均降雨量。
假设这个数据被称为df 这是一个数据表
library(dplyr)
这就是我所做的：
loc.list <- unique(dat$loc.id)
my.list <- list() # a list to store results
ptm <- proc.time()
for(i in seq_along(loc.list)){
n <- loc.list[i]
df1 <- dat[dat$loc.id == n,]
df2 <- gather(df1, day, rain, -year) # this melts the data in long format
df3 <- df2 %>% mutate(day = gsub("d_","", day)) %>% # since the day column was in "d_1" format, I converted into integer (1,2,3..365)
mutate(day = as.numeric(as.character(day))) %>% # ensure that day column is numeric. For some reasonson, some NA.s appear.
arrange(year,day) %>% # ensure that they are arranged in order
mutate(month = strptime(paste(year, day), format = "%Y %j")$mon + 1) %>% # assing each day to a month
group_by(year,month) %>% # group by year and month
summarise(month.rain = sum(rain)) %>% # calculate for each location, year and month, total rainfall
group_by(month) %>% # group by month
summarise(month.mean = round(mean(month.rain), digits = 2)) # calculate for each month, the long term mean
my.list[[i]] <- df3
}
proc.time() - ptm
user system elapsed
1036.17 0.20 1040.68
我想问有没有更高效更快的方法来完成这个任务
【问题讨论】：
标签： r dplyr data.table