Order data frame rows according to vector with specific order

user_989691 0
Viewed 171500

Is there an easier way to ensure that a data frame's rows are ordered according to a "target" vector as the one I implemented in the short example below?

df <- data.frame(name = letters[1:4], value = c(rep(TRUE, 2), rep(FALSE, 2)))

df
#   name value
# 1    a  TRUE
# 2    b  TRUE
# 3    c FALSE
# 4    d FALSE

target <- c("b", "c", "a", "d")

This somehow seems to be a bit too "complicated" to get the job done:

idx <- sapply(target, function(x) {
    which(df$name == x)
})
df <- df[idx,]
rownames(df) <- NULL

df 
#   name value
# 1    b  TRUE
# 2    c FALSE
# 3    a  TRUE
# 4    d FALSE
dataframe r sorting
6 Answers

We can adjust the factor levels based on target and use it in arrange 

library(dplyr)
df %>% arrange(factor(name, levels = target))

#  name value
#1    b  TRUE
#2    c FALSE
#3    a  TRUE
#4    d FALSE

Or order it and use it in slice

df %>% slice(order(factor(name, levels = target)))
user_39629140

If you don't want to use any libraries and you have reoccurrences in your data, you can use which with sapply as well.

new_order <- sapply(target, function(x,df){which(df$name == x)}, df=df)
df        <- df[new_order,]
user_77312050

Here's a similar system for the situation where you have a variable you want to sort by, initially, but then you want to sort by a secondary variable according to the order that this secondary variable first appears in the initial sort.

In the function below, the initial sort variable is called order_by and the secondary variable is called order_along - as in "order by this variable along its initial order".

library(dplyr, warn.conflicts = FALSE)
df <- structure(
  list(
    msoa11hclnm = c(
      "Bewbush", "Tilgate", "Felpham",
      "Selsey", "Brunswick", "Ratton", "Ore", "Polegate", "Mile Oak",
      "Upperton", "Arundel", "Kemptown"
    ),
    lad20nm = c(
      "Crawley", "Crawley",
      "Arun", "Chichester", "Brighton and Hove", "Eastbourne", "Hastings",
      "Wealden", "Brighton and Hove", "Eastbourne", "Arun", "Brighton and Hove"
    ),
    shape_area = c(
      1328821, 3089180, 3540014, 9738033, 448888, 10152663, 5517102,
      7036428, 5656430, 2653589, 72832514, 826151
    )
  ),
  row.names = c(NA, -12L), class = "data.frame"
)

this does not give me what I need:

df %>% 
  dplyr::arrange(shape_area, lad20nm)
#>    msoa11hclnm           lad20nm shape_area
#> 1    Brunswick Brighton and Hove     448888
#> 2     Kemptown Brighton and Hove     826151
#> 3      Bewbush           Crawley    1328821
#> 4     Upperton        Eastbourne    2653589
#> 5      Tilgate           Crawley    3089180
#> 6      Felpham              Arun    3540014
#> 7          Ore          Hastings    5517102
#> 8     Mile Oak Brighton and Hove    5656430
#> 9     Polegate           Wealden    7036428
#> 10      Selsey        Chichester    9738033
#> 11      Ratton        Eastbourne   10152663
#> 12     Arundel              Arun   72832514

Here’s a function:

order_along <- function(df, order_along, order_by) {
  cols <- colnames(df)
  
  df <- df %>%
    dplyr::arrange({{ order_by }})
  
  df %>% 
    dplyr::select({{ order_along }}) %>% 
    dplyr::distinct() %>% 
    dplyr::full_join(df) %>% 
    dplyr::select(dplyr::all_of(cols))
  
}

order_along(df, lad20nm, shape_area)
#> Joining, by = "lad20nm"
#>    msoa11hclnm           lad20nm shape_area
#> 1    Brunswick Brighton and Hove     448888
#> 2     Kemptown Brighton and Hove     826151
#> 3     Mile Oak Brighton and Hove    5656430
#> 4      Bewbush           Crawley    1328821
#> 5      Tilgate           Crawley    3089180
#> 6     Upperton        Eastbourne    2653589
#> 7       Ratton        Eastbourne   10152663
#> 8      Felpham              Arun    3540014
#> 9      Arundel              Arun   72832514
#> 10         Ore          Hastings    5517102
#> 11    Polegate           Wealden    7036428
#> 12      Selsey        Chichester    9738033

Created on 2021-01-12 by the reprex package (v0.3.0)

user_51689070
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