基于列名称模式的成对差异 [英] Multiple pairwise differences based on column name patterns

问题描述

我有一个 data.table ，dt：

dt

Id  v1 v2 v3 x1 x2 x3
1   7  1  3  5  6  8
2   1  3  5  6  8  5
3   3  5  6  8  5  1

v1，v2，v3和x1，x2，x3是数字变量

v1, v2, v3 and x1, x2, x3 are numeric variables

我想要从 v列中减去 x列，即计算差异 v1-x1 ， v2-x2 等。在我的真实数据中，我可能有100个这样的变量对。

I want to subtract the 'x' columns from the 'v' columns, i.e. calculate the differences v1 - x1, v2 - x2, etc. In my real data I may have 100s of such pair of variables.

所需的输出：

dt

Id  v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
1   7  1  3  5  6  8   -2     -4    -3
2   1  3  5  6  8  5   -5     -5     0
3   3  5  6  8  5  1   -3      0     5

---

我尝试了以下操作：

---

I've tried out the following:

newnames <- paste0("diff", 1:3)
v <- paste0("v", 1:3)
x <- paste0("x", 1:3)
dt[ , c(newnames) := get(v) - get(x)]

Howeve r，这将导致3个相同的列，都包含差异 v1-x1 。

However, this results in 3 identical columns all containing the difference v1 - x1.

我知道可能的解决方案是

I am aware that a possible solution is something like

dt[ , .(v1 - x1, v2 - x2, v3 - x3)]

但是，如果我必须输入100个不像v1和x1一样简单的名称，这是一个很长的代码，可能会出现很多键入错误。

However this is quite a long code with possible many typing errors if I have to put in 100 names not as simple as v1 and x1.

希望您能为我提供帮助。

I hope you can help me.

推荐答案

您可以按列是否包含 x ，然后取所得数据表的差。

You could split by whether the column contains x and then take the difference of the resulting data tables.

new_cols <- 
  do.call('-', split.default(dt[,-1], grepl('x', names(dt)[-1])))

dt[, paste0('diff', seq_along(new_cols)) := new_cols]

dt
#    Id v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
# 1:  1  7  1  3  5  6  8     2    -5    -5
# 2:  2  1  3  5  6  8  5    -5    -5     0
# 3:  3  3  5  6  8  5  1    -5     0     5

或使用与您可以执行的问题中的代码段相似的逻辑

Or using similar logic to the code snippet in the question you could do

newnames <- paste0("diff",1:3)
v <- paste0("v",1:3)
x <- paste0("x",1:3)

dt[, (newnames) := Map('-', mget(v), mget(x))]

dt
#    Id v1 v2 v3 x1 x2 x3 diff1 diff2 diff3
# 1:  1  7  1  3  5  6  8     2    -5    -5
# 2:  2  1  3  5  6  8  5    -5    -5     0
# 3:  3  3  5  6  8  5  1    -5     0     5

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