如何根据特定列的日期值重新排序?

一种选择是转换为 Date 上课然后点菜

# using a pattern, get the column index
i1 <- grep("^\\d{2}", names(df1))
# sort the extracted the column names after converting to 'Date' class
nm1 <-  names(df1)[i1][order(as.Date(names(df1)[i1], '%d/%m/%Y'))]
# get the names of the other columns
nm2 <- setdiff(names(df1), names(df1)[i1])
# concatenate the columns
df2 <- df1[c(nm2, nm1)]
df2
#    retailer department      type 22/03/2015 15/01/2017 25/07/2018 11/01/2019 12/01/2019
#1        1          a completed          4          1          2          4          1
#2        2          b completed          1          1          2          3          4
#3        3          c completed          5          1          2          2          3

数据

df1 <- data.frame(retailer = 1:3, department = letters[1:3], 
 type = 'completed', `11/01/2019` = c(4, 3, 2),
 `12/01/2019` = c(1, 4, 3), `15/01/2017` = 1,
 `25/07/2018` = 2, `22/03/2015` = c(4, 1, 5), check.names = FALSE)