在 R 的嵌套表中，从具有天、月、年的单独列创建日期

Question

I'm working on hydrological data in nested tables. Besides streamflow (Q) I have dates split into 3 columns (one for days, second for months and third for year).

year_2008 <- list("910" = data.frame(Year = c(2008), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
             "950" = data.frame(Year = c(2008), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
             "1012" = data.frame(Year = c(2008), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
              "1087" = data.frame(Year = c(2008), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)))


year_2009 <- list("910" = data.frame(Year = c(2009), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
                  "950" = data.frame(Year = c(2009), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
                  "1012" = data.frame(Year = c(2009), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)),
                  "1087" = data.frame(Year = c(2009), Day=c(1:5), Month= c(1, 1, 1, 1 ,1)))

It's a nested table.

year_list <- list (year_2008, year_2009)
names(year_list) <- c("2008", "2009")

I'm trying to gather the information from those 3 columns into a single one which will show a date (in a new column called "date"). I've created a function to create a date column:

transform_date2 <- function(x) {x$date <- as.Date(with(x, paste(x$Day, x$Month, x$Year, sep="-")), "%d-%m-%Y")}

I tried applying that function to the nested table:

result1 <- lapply(year_list, `[[`, transform_date2)

this results in an error: Error in FUN(X[[i]], ...): invalid subscript type 'closure'

I tried doing a loop:

result2 <- for (i in 1:seq_along(year_list)) {transform_date2}

This results in: Warning message: In 1:seq_along(year_list): numerical expression has 2 elements: only the first used

I feel that I'm having problems accessing the correct "level" of the nested table.

My end goal is to keep the current format (nested table), remove the columns Day, Month, Year in each nested table, create the date column in each and keep the Q column. Please help

Answer 1

Since it's a nested list use nested lapply :

year_list <- lapply(year_list, function(x) lapply(x, transform_date2))

where transform_date2 is:

transform_date2 <- function(x) {x$date <- as.Date(with(x, paste(x$Day, x$Month, x$Year, sep="-")), "%d-%m-%Y");x}