通过引用将所有列合并到data.table中 [英] Merging all column by reference in a data.table

问题描述

我想通过引用将两个 data.table 合并在一起，而不必写下我要合并的所有变量。这是一个了解我的需求的简单示例：

I would like to merge two data.table together by reference without having to write down all variables I want to merge. Here is a simple example to understand my needs :

set.seed(20170711)
(a <- data.table(v_key=seq(1, 5), key="v_key"))
#   v_key
#1:     1
#2:     2
#3:     3
#4:     4
#5:     5

a_backup <- copy(a)

(b <- data.table(v_key=seq(1, 5), v1=runif(5), v2=runif(5), v3=runif(5), key="v_key"))
#   v_key          v1        v2          v3
#1:     1 0.141804303 0.1311052 0.354798849
#2:     2 0.425955903 0.3635612 0.950234261
#3:     3 0.001070379 0.4615936 0.359660693
#4:     4 0.453054854 0.5768500 0.008470552
#5:     5 0.951767837 0.1649903 0.565894298

我想将 b 的每一列复制到通过引用 a 而不指定列名。

I want to copy every columns of b into a by reference without specifying the column names.

我可以执行以下操作，但是可以复制该对象无缘无故，降低了我的程序的性能并增加了所需的RAM：

I could do the following, but that would make a copy of the object for no reason, reducing the performance of my program and increasing the RAM needed :

(a  <- a[b])
#   v_key          v1        v2          v3
#1:     1 0.141804303 0.1311052 0.354798849
#2:     2 0.425955903 0.3635612 0.950234261
#3:     3 0.001070379 0.4615936 0.359660693
#4:     4 0.453054854 0.5768500 0.008470552
#5:     5 0.951767837 0.1649903 0.565894298

另一个选择（没有无用的副本）是指定 b ，结果如下：

Another option (without useless copy) would be to specify the name of every column of b, resulting in the following :

a <- copy(a_backup)
a[b, `:=`(v1=v1, v2=v2, v3=v3)][]
#   v_key          v1        v2          v3
#1:     1 0.141804303 0.1311052 0.354798849
#2:     2 0.425955903 0.3635612 0.950234261
#3:     3 0.001070379 0.4615936 0.359660693
#4:     4 0.453054854 0.5768500 0.008470552
#5:     5 0.951767837 0.1649903 0.565894298

简而言之，我想提高第二个示例（无用副本）的效率我不得不在 b 中指定每个列的名称。

In brief, I would like to have the efficiency of my second example (no useless copy) without having to specify every column names in b.

我想我可以找到一种使用 colnames（）和 get（）函数的组合，但是我想知道是否有更清洁的方法

I guess I could find a way of doing it using a combination of the colnames() and get() functions, but I am wondering if there is a cleaner way to do it, syntax is so important for me.

推荐答案

如您所写，名称的组合和 mget 可以带您到达那里。

As you wrote, a combination of colnames and mget could get you there.

考虑一下：

# retrieve the column names from b - without the key ('v_key')
thecols = setdiff(colnames(b), key(b))

# assign them to a
a[b, (thecols) := mget(thecols)]

这看起来还不错，不是吗？

This is not too bad-looking, is it?

此外，我认为目前 data.table 。但是我很高兴被证明是错误的：）

Besides, I don't think another syntax is currently implemented with data.table. But I would be happy to be proven wrong :)

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