I am using the Tidyverse package in R. I have a data frame with 20 rows and 500 columns. I want to sort all the columns based on the size of the value in the last row of each column.
Here is an example with just 3 rows and 4 columns:
1 2 3 4,
5 6 7 8,
8 7 9 1
The desired result is:
3 1 2 4,
7 5 6 8,
9 8 7 1
I searched stack overflow but could not find an answer to this type of question.
If we want to use dplyr from tidyverse, we can use slice to get the last row and then use order in decreasing order to subset columns.
library(dplyr)
df[df %>% slice(n()) %>% order(decreasing = TRUE)]
# V3 V1 V2 V4
#1 3 1 2 4
#2 7 5 6 8
#3 9 8 7 1
Whose translation in base R would be
df[order(df[nrow(df), ], decreasing = TRUE)]
data
df <- read.table(text = "1 2 3 4
5 6 7 8
8 7 9 1")
The following reorders the data frame columns by the order of the last-rows values:
df <- data.frame(col1=c(1,5,8),col2=c(2,6,7),col3=c(3,7,9),col4=c(4,8,1))
last_row <- df[nrow(df),]
df <- df[,order(last_row,decreasing = T)]
First, to get the last rows. Then to sort them with the order() function and to return the reordered columns.
>df
col3 col1 col2 col4
1 3 1 2 4
2 7 5 6 8
3 9 8 7 1
In xtfrm.data.frame(x) : cannot xtfrm data frames. See, https://www.r-bloggers.com/2021/02/it-has-always-been-wrong-to-call-order-on-a-data-frame/.
character, causing all columns to be cast as strings, resulting in an alphabetic sort, not a numeric sort.