使用mutate计算表格的6列之间的所有绝对差值? [英] Calculate all the absolute differences between 6 columns of a table using mutate?

问题描述

我有一个包含6列Z1到Z6列的表，我想计算这些列之间的差的绝对值.

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.

到目前为止，我列举了mutate命令中的所有差异:

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))

但是我意识到这很容易出错，如果我的列数不同，则必须重写.

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

有什么方法可以自动"执行此操作?我的意思是，如果我考虑任意数量的列，它会自行调整?

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?

最好

达明(Damien)

推荐答案

您可以使用 combn 生成所有可能的列组合，然后减去它们.

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]]]))

---

这里使用的是一个小的可复制示例，还添加了适当的列名.

---

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

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