I am trying to expand on this answer, by creating a solution that works both on the new_dat and the old_dat.
New Data
new_dat <- 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("Andere akkerbouwbedrijven",
"Andere combinatiebedrijven"), class = "data.frame")
Old data and original Solution
old_dat <- 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")
The solution makes uses of the fact that the sum of the two numbers in each column name added provide the correct order.
ord <- gsub("\\[|\\]|\\)", "", colnames(new_dat)) %>%
strsplit(",") %>%
lapply(as.numeric) %>%
lapply(sum) %>%
unlist %>%
order()
colnames(dat)[ord]
New approach
The new data not only has to numerical values but also a string value (east, north, south, west). I realised that I could use the same solution if I give east a value of 1, north of 2 and so on. The sum of the three numbers than still provides the correct order.
I have been having some trouble adapting the code though.
ord <- gsub("\\[|\\]|\\)", "", colnames(new_dat)) %>%
# provides "0,25 east", "0,25 north" etc
strsplit(",") %>%
# provides "0" and "25 east", "0" and "25 north" etc
lapply(as.numeric) %>%
lapply(sum) %>%
# SHOULD provide 0 25 1 (east), 0 25 2 (north) etc
unlist %>%
order()
The issue lies in splitting the string in 3 parts, and convert the directions to a number, IF, and ONLY IF, there are three parts. Otherwise it should just use the two. How should I do this?
CodePudding user response：
To build on your solution you can do,
ord <- gsub("\\D ", ",", stri_replace_all_regex(names(new_dat), '[A-Za-z]', 1:4)) %>%
strsplit(",") %>%
lapply(as.numeric) %>%
lapply(sum, na.rm = TRUE) %>%
unlist() %>%
order()
> names(new_dat)[ord]
[1] "[0,25) east" "[0,25) south" "[0,25) north" "[0,25) west" "[25,50) east" "[25,50) south" "[25,50) north" "[25,50) west" "[25,100) east" "[25,100) south"
[11] "[25,100) north" "[25,100) west" "[50,100) east" "[50,100) south" "[50,100) north" "[50,100) west" "[100,250) east" "[100,250) south" "[100,250) north" "[100,250) west"
[21] "[100,500) east" "[100,500) south" "[100,500) north" "[100,500) west" "[250,500) east" "[250,500) south" "[250,500) north" "[250,500) west" "[500,1000) east" "[500,1000) south"
[31] "[500,1000) north" "[500,1000) west" "[1000,1500) east" "[1000,1500) south" "[1000,1500) north" "[1000,1500) west" "[1500,3000) east" "[1500,3000) south" "[1500,3000) north" "[1500,3000) west"
[41] "[500,1000000] east" "[500,1000000] south" "[500,1000000] north" "[500,1000000] west" "[1000,1000000] east" "[1000,1000000] south" "[1000,1000000] north" "[1000,1000000] west" "[3000,1000000] east" "[3000,1000000] south"
[51] "[3000,1000000] north" "[3000,1000000] west"
CodePudding user response：
Maybe a little overkill but with this one you don't need to find the patterns "east", "south" etc.
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()
}
In the first part of the function, I extract strings from the colnames and order them. Their order is then used to replace the words in the colnames with their indexes. Afterwards, I extract numeric values and pretty much follow your initial approach. I put this in a function to make it easier to use:
colnames(new_dat)[order_cols(new_dat)]
#> [1] "[0,25) east" "[0,25) north" "[0,25) south"
#> [4] "[0,25) west" "[25,50) east" "[25,50) north"
#> [7] "[25,50) south" "[25,50) west" "[25,100) east"
#> [10] "[25,100) north" "[25,100) south" "[25,100) west"
#> [13] "[50,100) east" "[50,100) north" "[50,100) south"
#> [16] "[50,100) west" "[100,250) east" "[100,250) north"
#> [19] "[100,250) south" "[100,250) west" "[100,500) east"
#> [22] "[100,500) north" "[100,500) south" "[100,500) west"
#> [25] "[250,500) east" "[250,500) north" "[250,500) south"
#> [28] "[250,500) west" "[500,1000) east" "[500,1000) north"
#> [31] "[500,1000) south" "[500,1000) west" "[1000,1500) east"
#> [34] "[1000,1500) north" "[1000,1500) south" "[1000,1500) west"
#> [37] "[1500,3000) east" "[1500,3000) north" "[1500,3000) south"
#> [40] "[1500,3000) west" "[500,1000000] east" "[500,1000000] north"
#> [43] "[500,1000000] south" "[500,1000000] west" "[1000,1000000] east"
#> [46] "[1000,1000000] north" "[1000,1000000] south" "[1000,1000000] west"
#> [49] "[3000,1000000] east" "[3000,1000000] north" "[3000,1000000] south"
#> [52] "[3000,1000000] west"
colnames(dat)[order_cols(dat)]
#> [1] "[0,25)" "[25,50)" "[25,100)" "[50,100)"
#> [5] "[100,250)" "[100,500)" "[250,500)" "[500,1000)"
#> [9] "[1000,1500)" "[1500,3000)" "[500,1000000]" "[1000,1000000]"
#> [13] "[3000,1000000]"
Created on 2022-05-06 by the reprex package (v2.0.1)
P.S.: If you are using a newer version of R (>= 4.10), you can use the native pipe (|>) instead of magrittr's %>%.