我最近从JBGruber获得了this amazing answer,用于排序具有双数值的字符串列,这适用于本文底部的两个数据集:
library(magrittr)
order_cols <- function(dat) {
# look for words to order by
s_ordered <- stringi::stri_extract_all_regex(colnames(dat), "[[:alpha:]]+") %>%
unlist() %>%
unique() %>%
sort()
if (length(s_ordered) > 1) {
# replace words with their alphabetical index
cnames <- stringi::stri_replace_all_fixed(colnames(dat), s_ordered, seq_along(s_ordered), vectorise_all = FALSE)
} else {
cnames <- colnames(dat)
}
cnames %>%
stringi::stri_extract_all_regex("\\d+") %>% # extract all numbers (including the alphabetical index numbers)
lapply(as.numeric) %>%
lapply(sum) %>%
unlist() %>%
order()
}
然而,我注意到,对于以下数据,它并不完全有效,因为它基于这样一个假设,即按顺序排列的数字之和给出了列的写入顺序:
dat_I <- structure(list(`[25,250)`=3L, `[0,25)` = 5L, `[100,250)` = 43L, `[100,500)` = 0L,
`[1000,1000000]` = 20L, `[1000,1500)` = 0L, `[1500,3000)` = 0L,
`[25,100)` = 38L, `[25,50)` = 0L, `[250,500)` = 27L, `[3000,1000000]` = 0L,
`[50,100)` = 0L, `[500,1000)` = 44L, `[500,1000000]` = 0L), row.names = "Type_A", class = "data.frame")
colnames(dat_I )[order_cols(dat_I)]
有没有办法先按第一个元素排序,然后按第二个元素排序?
Old Data
dat_I <- structure(list(`[0,25)` = 5L, `[100,250)` = 43L, `[100,500)` = 0L,
`[1000,1000000]` = 20L, `[1000,1500)` = 0L, `[1500,3000)` = 0L,
`[25,100)` = 38L, `[25,50)` = 0L, `[250,500)` = 27L, `[3000,1000000]` = 0L,
`[50,100)` = 0L, `[500,1000)` = 44L, `[500,1000000]` = 0L), row.names = "Type_A", class = "data.frame")
dat_II <- structure(list(`[0,25) east` = c(1269L, 85L), `[0,25) north` = c(364L,
21L), `[0,25) south` = c(1172L, 97L), `[0,25) west` = c(549L,
49L), `[100,250) east` = c(441L, 149L), `[100,250) north` = c(224L,
45L), `[100,250) south` = c(521L, 247L), `[100,250) west` = c(770L,
124L), `[100,500) east` = c(0L, 0L), `[100,500) north` = c(0L,
0L), `[100,500) south` = c(0L, 0L), `[100,500) west` = c(0L,
0L), `[1000,1000000] east` = c(53L, 0L), `[1000,1000000] north` = c(82L,
0L), `[1000,1000000] south` = c(23L, 0L), `[1000,1000000] west` = c(63L,
0L), `[1000,1500) east` = c(0L, 0L), `[1000,1500) north` = c(0L,
0L), `[1000,1500) south` = c(0L, 0L), `[1000,1500) west` = c(0L,
0L), `[1500,3000) east` = c(0L, 0L), `[1500,3000) north` = c(0L,
0L), `[1500,3000) south` = c(0L, 0L), `[1500,3000) west` = c(0L,
0L), `[25,100) east` = c(579L, 220L), `[25,100) north` = c(406L,
58L), `[25,100) south` = c(1048L, 316L), `[25,100) west` = c(764L,
131L), `[25,50) east` = c(0L, 0L), `[25,50) north` = c(0L, 0L
), `[25,50) south` = c(0L, 0L), `[25,50) west` = c(0L, 0L), `[250,500) east` = c(232L,
172L), `[250,500) north` = c(207L, 40L), `[250,500) south` = c(202L,
148L), `[250,500) west` = c(457L, 153L), `[3000,1000000] east` = c(0L,
0L), `[3000,1000000] north` = c(0L, 0L), `[3000,1000000] south` = c(0L,
0L), `[3000,1000000] west` = c(0L, 0L), `[50,100) east` = c(0L,
0L), `[50,100) north` = c(0L, 0L), `[50,100) south` = c(0L, 0L
), `[50,100) west` = c(0L, 0L), `[500,1000) east` = c(103L, 0L
), `[500,1000) north` = c(185L, 0L), `[500,1000) south` = c(66L,
0L), `[500,1000) west` = c(200L, 0L), `[500,1000000] east` = c(0L,
288L), `[500,1000000] north` = c(0L, 120L), `[500,1000000] south` = c(0L,
229L), `[500,1000000] west` = c(0L, 175L)), row.names = c("A",
"B"), class = "data.frame")