Calculate all the absolute differences between 6 columns of a table using mutate? [duplicate]

问题

I have a table with 6 columns Z1 to Z6, and I want to calculate the absolute value of the difference between each of these columns.

So far, I enumerate all the differences in a mutate command:

FactArray <- FactArray %>% mutate(diff12 = abs(Z1-Z2), 
                                  diff13 = abs(Z1-Z3),
                                  diff14 = abs(Z1-Z4),
                                  diff15 = abs(Z1-Z5),
                                  diff16 = abs(Z1-Z6),
                                  diff23 = abs(Z2-Z3),
                                  diff24 = abs(Z2-Z4),
                                  diff25 = abs(Z2-Z5),
                                  diff26 = abs(Z2-Z6),
                                  diff34 = abs(Z3-Z4),
                                  diff35 = abs(Z3-Z5),
                                  diff36 = abs(Z3-Z6),
                                  diff46 = abs(Z4-Z6),
                                  diff56 = abs(Z5-Z6))

But I realise this is error prone and will have to be rewritten if I have a different number of columns.

Is there any way to do this "automatically"? I mean in a way such as it would adjust itself if I am considering any number of columns?

Best,

Damien

回答1:

You can generate all possible combination of the columns using combn and subtract them.

cols <- paste0('Z', 1:6)
combn(cols, 2, function(x) abs(df[[x[1]]]  - df[[x[2]]]))

Here's using a small reproducible example also adding appropriate column names.

set.seed(123)
df <- data.frame(Z1 = sample(10, 4), Z2 = sample(10, 4), Z3 = sample(10,4))
cols <- paste0('Z', 1:3)
new_cols <- combn(cols, 2, paste0, collapse = "_")
df[new_cols] <- combn(cols, 2, function(x) abs(df[[x[1]]]  - df[[x[2]]]))
df

#  Z1 Z2 Z3 Z1_Z2 Z1_Z3 Z2_Z3
#1  3  6  6     3     3     0
#2 10  5  9     5     1     4
#3  2  4  2     2     0     2
#4  8 10  3     2     5     7

回答2:

Here is a function that can be used in a magrittr pipe.

fun <- function(X, cols, pref = "diff", sep = "", collapse = ""){
  f <- function(x) abs(x[[1]] - x[[2]])
  cmb <- combn(X[cols], 2, f)
  out <- as.data.frame(cmb)
  nms <- combn(cols, 2, paste, collapse = collapse)
  nms <- paste(pref, nms, sep = sep)
  names(out) <- nms
  out
}

library(dplyr)

df1 %>% fun(1:6)
df1 %>% bind_cols(df1 %>% fun(1:6))

Data

set.seed(2020)
df1 <- replicate(6, sample(10))
df1 <- as.data.frame(df1)

回答3:

It's easiest if you first make your data tidy. The approach below is easily generalised and robust.

library(tidyverse)

# Generate test data
df <- tibble(id=1:5, z1=rnorm(5), z2=rnorm(5), z3=rnorm(5), z4=rnorm(5), z5=rnorm(5), z6=rnorm(5))
# Tidy data
tidyDF1 <- df %>% pivot_longer(values_to="Value1", names_pattern="z(\\d)", names_to="Index1", cols=starts_with("z")) 
# Copy the tidy data
tidyDF2 <- tidyDF1 %>% 
            rename(Value2=Value1, Index2=Index1) %>% 
            mutate(Index2=as.integer(Index2))
# Create a copies of each value, one for each column in the original data 
tidyDF1 <- tidyDF1 %>% expand(nesting(id, Index1, Value1), Index2=1:6)
joinDF <- tidyDF1 %>% 
            left_join(tidyDF2, by=c("id", "Index2")) %>% # Create all pairwise combinations
            mutate(Diff=Value1 - Value2) %>%             # Calculate differences
            filter(Index1 != Index2)                     # Remove comparisons with self
# Present the results in "wide format"
joinDF %>% 
  pivot_wider(
    id_cols="id",
    values_from="Diff",
    names_from=c("Index1", "Index2"),
    names_prefix="Diff",
    names_sep=""
)

来源：https://stackoverflow.com/questions/62572690/calculate-all-the-absolute-differences-between-6-columns-of-a-table-using-mutate

标签

dplyr

mutate