将函数应用于 data.table 中的每个指定列并通过引用更新 [英] Apply a function to every specified column in a data.table and update by reference

问题描述

我有一个 data.table，我想用它对某些列执行相同的操作.这些列的名称在字符向量中给出.在这个特定示例中，我想将所有这些列乘以 -1.

I have a data.table with which I'd like to perform the same operation on certain columns. The names of these columns are given in a character vector. In this particular example, I'd like to multiply all of these columns by -1.

一些玩具数据和一个指定相关列的向量:

Some toy data and a vector specifying relevant columns:

library(data.table)
dt <- data.table(a = 1:3, b = 1:3, d = 1:3)
cols <- c("a", "b")

现在我正在这样做，循环遍历字符向量:

Right now I'm doing it this way, looping over the character vector:

for (col in 1:length(cols)) {
   dt[ , eval(parse(text = paste0(cols[col], ":=-1*", cols[col])))]
}

有没有办法不使用 for 循环直接执行此操作?

Is there a way to do this directly without the for loop?

推荐答案

这似乎可行:

dt[ , (cols) := lapply(.SD, "*", -1), .SDcols = cols]

结果是

    a  b d
1: -1 -1 1
2: -2 -2 2
3: -3 -3 3

这里有一些技巧:

• 因为 (cols) := 中有括号，所以将结果分配给 cols 中指定的列，而不是分配给一些名为cols"的新变量.
• .SDcols 告诉调用我们只查看那些列，并允许我们使用 .SD，即 S子集与这些列关联的 Data.
• lapply(.SD, ...) 对 .SD 进行操作，这是一个列列表(与所有 data.frames 和 data.tables 一样).lapply 返回一个列表，所以最后 j 看起来像 cols := list(...).

• Because there are parentheses in (cols) :=, the result is assigned to the columns specified in cols, instead of to some new variable named "cols".
• .SDcols tells the call that we're only looking at those columns, and allows us to use .SD, the Subset of the Data associated with those columns.
• lapply(.SD, ...) operates on .SD, which is a list of columns (like all data.frames and data.tables). lapply returns a list, so in the end j looks like cols := list(...).

编辑:正如@Arun 提到的，这是另一种可能更快的方法:

EDIT: Here's another way that is probably faster, as @Arun mentioned:

for (j in cols) set(dt, j = j, value = -dt[[j]])

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