我有这个data.frame:
df <- structure(list(att_number = structure(1:3, .Label = c("0", "1",
"2"), class = "factor"), `1` = structure(c(2L, 3L, 1L), .Label = c("1026891",
"412419", "424869"), class = "factor"), `10` = structure(c(2L,
1L, 3L), .Label = c("235067", "546686", "92324"), class = "factor"),
`2` = structure(c(3L, 1L, 2L), .Label = c("12729", "7569",
"9149"), class = "factor")), .Names = c("att_number", "1",
"10", "2"), row.names = c(NA, -3L), class = "data.frame")
看起来这个数字作为列名。
att_number 1 10 2
0 412419 546686 9149
1 424869 235067 12729
2 1026891 92324 7569
在dplyr链中,我想按升序排列列,如下所示:
att_number 1 2 10
0 412419 9149 546686
1 424869 12729 235067
2 1026891 7569 7569
我尝试使用select_
,但它不想按计划工作。有关如何做到这一点的任何想法?这是我的微弱尝试:
names_order <- names(df)[-1] %>%
as.numeric %>%
.[order(.)] %>%
as.character %>%
c('att_number', .)
df %>%
select_(.dots = names_order)
Error: Position must be between 0 and n
答案 0 :(得分:2)
您需要在数字列名称周围返回刻度以停止选择尝试将它们解释为列位置:
library(tidyverse)
sort_names <- function(data) {
name <- names(data)
chars <- keep(name, grepl, pattern = "[^0-9]") %>% sort()
nums <- discard(name, grepl, pattern = "[^0-9]") %>%
as.numeric() %>%
sort() %>%
sprintf("`%s`", .)
select_(data, .dots = c(chars, nums))
}
sort_names(df)