CODEWITHSUNDEEP
rsortingdataframe
Rappster
Rappster

Reputation: 13090

Order data frame rows according to vector with specific order

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

Upvotes: 224

Views: 215838

Answers (6)

Francis Barton
Francis Barton

Reputation: 137

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)

Upvotes: 2

eonurk
eonurk

Reputation: 547

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,]

Upvotes: 4

Ronak Shah
Ronak Shah

Reputation: 389235

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)))

Upvotes: 63

Lerong
Lerong

Reputation: 538

I prefer to use ***_join in dplyr whenever I need to match data. One possible try for this

left_join(data.frame(name=target),df,by="name")

Note that the input for ***_join require tbls or data.frame

Upvotes: 30

MattV
MattV

Reputation: 1383

This method is a bit different, it provided me with a bit more flexibility than the previous answer. By making it into an ordered factor, you can use it nicely in arrange and such. I used reorder.factor from the gdata package.

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

require(gdata)
df$name <- reorder.factor(df$name, new.order=target)

Next, use the fact that it is now ordered:

require(dplyr)
df %>%
  arrange(name)
    name value
1    b  TRUE
2    c FALSE
3    a  TRUE
4    d FALSE

If you want to go back to the original (alphabetic) ordering, just use as.character() to get it back to the original state.

Upvotes: 19

Edward
Edward

Reputation: 5537

Try match:

df <- data.frame(name=letters[1:4], value=c(rep(TRUE, 2), rep(FALSE, 2)))
target <- c("b", "c", "a", "d")
df[match(target, df$name),]

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

It will work as long as your target contains exactly the same elements as df$name, and neither contain duplicate values.

From ?match:

match returns a vector of the positions of (first) matches of its first argument 
in its second.

Therefore match finds the row numbers that matches target's elements, and then we return df in that order.

Upvotes: 307

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