使用列表列对data.table进行分组

I have a really big problem and looping through the data.table to do what I want is too slow, so I am trying to get around looping. Let assume I have a data.table as follows:

我有一个非常大的问题,并循环data.table来做我想要的太慢,所以我试图绕过循环。假设我有一个data.table如下:

a <- data.table(i = c(1,2,3), j = c(2,2,6), k = list(c("a","b"),c("a","c"),c("b")))

> a
  i j   k
1: 1 2 a,b
2: 2 2 a,c
3: 3 6   b

And I want to group based on the values in k. So something like this:

我想根据k中的值进行分组。所以这样的事情:

a[, sum(j), by = k]

right now I am getting the following error:

现在我收到以下错误:

 Error in `[.data.table`(a, , sum(i), by = k) : 
 The items in the 'by' or 'keyby' list are length (2,2,1). Each must be same length as rows in x or number of rows returned by i (3).

The answer I am looking for is to group first all the rows having "a" in column k and calculate sum(j) and then all rows having "b" and so on. So the desired answer would be:

我正在寻找的答案是首先将列k中具有“a”的所有行分组并计算sum(j),然后计算所有具有“b”的行,依此类推。所以答案是:

k V1 
a 4
b 8
c 2

Any hint how to do it efficiently? I cant melt the column K by repeating the rows since the size of the data.table would be too big for my case.

任何提示如何有效地做到这一点?我不能通过重复行来融化列K,因为data.table的大小对我来说太大了。

3 个解决方案

#1

I think this might work:

我认为这可能有效:

a[, .(k = unlist(k)), by=.(i,j)][,sum(j),by=k]

   k V1
1: a  4
2: b  8
3: c  2

#2

If we are using tidyr, a compact option would be

如果我们使用tidyr,那么紧凑的选择就是

library(tidyr)
unnest(a, k)[, sum(j) ,k]
#   k V1
#1: a  4
#2: b  8
#3: c  2

Or using the dplyr/tidyr pipes

或者使用dplyr / tidyr管道

unnest(a, k) %>%
       group_by(k) %>%
       summarise(V1 = sum(j))
#     k    V1
#   <chr> <dbl>
#1     a     4
#2     b     8
#3     c     2

#3

Since by-group operations can be slow, I'd consider...

由于分组操作可能很慢,我会考虑......

dat = a[rep(1:.N, lengths(k)), c(.SD, .(k = unlist(a$k))), .SDcols=setdiff(names(a), "k")]

   i j k
1: 1 2 a
2: 1 2 b
3: 2 2 a
4: 2 2 c
5: 3 6 b

We're repeating rows of cols i:j to match the unlisted k. The data should be kept in this format instead of using a list column, probably. From there, as in @MikeyMike's answer, we can dat[, sum(j), by=k].

我们重复cols i:j的行以匹配未列出的k。数据应该以这种格式保存,而不是使用列表列。从那里开始,就像在@ MikeyMike的回答中一样,我们可以使用dat [,sum(j),by = k]。

In data.table 1.9.7+, we can similarly do

在data.table 1.9.7+中,我们也可以这样做

dat = a[, c(.SD[rep(.I, lengths(k))], .(k = unlist(k))), .SDcols=i:j]

#1